Distribution Estimation of a Sum Random Variable from Noisy Samples

Abstract

Let X, Y be independent continuous univariate random variables with unknown distributions. Suppose we observe two independent random samples $X'_1, \ldots , X'_n$ and $Y'_1, \ldots , Y'_m$ from the distributions of $X' = X+\zeta$ and $Y'=Y+\eta$, respectively. Here $\zeta$, $\eta$ are random noises and have known distributions. This paper is devoted to an estimation for unknown cumulative distribution function (cdf) $F_{X+Y}$ of the sum $X+Y$ on the basis of the samples. We suggest a nonparametric estimator of $F_{X+Y}$ and demonstrate its consistency with respect to the root mean squared error. Some upper and minimax lower bounds on convergence rate are derived when the cdf’s of X, Y belong to Sobolev classes and when the noises are Fourier-oscillating, supersmooth and ordinary smooth, respectively. Particularly, if the cdf’s of X, Y have the same smoothness degrees and $n=m$, our estimator is minimax optimal in order when the noises are Fourier-oscillating as well as supersmooth. A numerical example is also given to illustrate our method.

1 Introduction

In some fields of applications, many quantities of interest can be modeled as sum random variables. For example, in wireless communications, sum of current fluctuations in tunnel junctions can be represented as the sum of lognormal distributed random variables, see, e.g., [14]. In insurance, the total amount of claims on a portfolio of insurance contracts is described by the sum of all claims on the individual policies, e.g., see [7]. In view of applications, one is often interested in distributions of sum random variables; however, the distributions are often not available in a closed form, and this arises statistical problems to estimate the distributions. Historically, this problem type was first studied by Zijaeva [18], who considered nonparametric density estimation for sum of two independent random variables based on two random samples taken from underlying distributions. In a particular case where component random variables have the same distributions, several estimation methods based only on one random sample have also been developed, see, e.g., [1, 7, 15] and the references therein.

In practice, direct observations of component random variables can be not available due to the presence of measurement errors. Hence, it is necessary to consider the problem type in a certain noisy data situation. Up to present, however, we have not yet found any research regarding directly such situations, and hence, this gap must be filled. In the present paper, we fill partially the gap by considering the problem of estimating cumulative distribution function (cdf for short) of a sum of two independent random variables from noisy data containing additional random noises. Precisely, the problem under our consideration here is as follows. Let X, Y defined on the same probability space $(\Omega , \mathcal {F}, \mathbb {P})$ be independent real-valued continuous random variables with unknown distributions. Suppose we only observe the random variables $X'$ and $Y'$ generated by the model

$$\begin{aligned} X' = X+ \zeta , \quad Y'=Y+\eta . \end{aligned}$$

(1.1)

Here $\zeta$ and $\eta$ are continuous random noises distributed, respectively, with known densities $f_{\zeta }$ and $f_{\eta }$, called noise densities. Assume also that the variables X, Y, $\zeta$ and $\eta$ are mutually independent. Let $X_1', \ldots , X_n'$ and $Y_1', \ldots , Y_m'$ be two samples of independent and identically distributed (i.i.d.) observations drawn from the distributions of $X'$ and $Y'$, respectively. Our aim is to estimate nonparametrically the cdf $F_{X+Y}$ of the sum variable $X+Y$ on the basis of the observations.

The independence assumption of X and Y implies that the target cdf $F_{X+Y}$ can be represented in the form

$$\begin{aligned} F_{X+Y}(x) = \int _{-\infty }^{\infty } F_X(x-u) \mathrm{d}F_Y(u), \end{aligned}$$

where $F_X$ and $F_Y$ stand, respectively, for the cdf’s of X and Y. On the other hand, nonparametric estimations of $F_{X}$, $F_{Y}$ from the samples $(X'_1, \ldots , X'_n)$, $(Y'_1, \ldots , Y'_m)$, respectively, are known as cdf deconvolution problems in one sample measurement error model, which have been studied in Gaffey [8], Fan [6], Meister [13], Dattner et al. [5], Dattner and Reiser [4], Trong and Phuong [2]. Therefore, our problem can be viewed as a cdf deconvolution problem in two samples measurement error model. It should be noted that our problem is connected with a few previous studies. Concretely, Dattner [3] and Trong et al. [16] considered the problem of estimating the probability $\mathbb {P}(X<Y)$ which has arisen from stress–strength models, see, e.g., [12]. The latter problem is a special case of our problem because $\mathbb {P}(X<Y)$ is in fact the value of the cdf of the variable $X+(-Y)$ at zero.

An important point for deconvolution problems in one or two sample(s) measurement error model is that the convergence of deconvolved estimators depends strongly on the decay of noise characteristic functions. Roughly speaking, the convergence will be slower when the characteristic functions decay faster. We recall here that the characteristic function of a random variable U is defined as $\phi _U(t):= \mathbb {E}(\mathrm {e}^{\mathrm {i}tU})$, for $t\in \mathbb {R}$ and $\mathrm {i}= \sqrt{-1}$. Two types of noises are often considered in the deconvolution literature: supersmooth noises and ordinary smooth noises; see [6]. A noise is said to be supersmooth (ordinary smooth, respectively) if its characteristic function decays with an exponential rate (a polynomial rate, respectively). For example, the Gaussian and Cauchy noises are supersmooth, whereas the Laplace, Gamma, symmetric Gamma, Chi-square and exponential noises are ordinary smooth. Notably, both supersmooth and ordinary smooth noises have non-vanishing and monotonically decaying characteristic functions, and hence, they are referred to as standard noises. In contrast with the mentioned noises, there exist some noises whose characteristic functions may have periodic zeros on the real line and show oscillatory behaviors. Such noises are said to be Fourier-oscillating. Among the latter noises, we can cite uniform noises or sum of a Gaussian noise with a uniform noise.

In the current work, our goal is to develop an estimation procedure for the cdf $F_{X+Y}$ that works well whether the noises $\zeta$, $\eta$ are Fourier-oscillating or standard. Our idea is as follows. First, we write the cdf $F_{X+Y}$ with respect to the characteristic functions $\phi _{X'}$, $\phi _{Y'}$, $\phi _{\zeta }$, $\phi _{\eta }$ and estimate $\phi _{X'}$, $\phi _{Y'}$ by using the availability of the samples $X_1', \ldots , X_n'$ and $Y_1', \ldots , Y_m'$. Afterwards, by applying the ridge-parameter regularization method (see [10]), we propose a nonparametric estimator of $F_{X+Y}$ that depends on two parameters. Under certain conditions imposed on the parameters, our estimator is shown to be consistent respect to root mean squared error. We then derive some upper and minimax lower bounds on convergence rate of our estimator when the cdf’s $F_X$, $F_Y$ are assumed to be in Sobolev classes and when $\zeta$, $\eta$ are the Fourier-oscillating noises as well as the standard noises.

The rest of the present paper is structured as follows. In Sect. 2, we discuss methodology for estimating $F_{X+Y}$. In Sect. 3, we establish some theoretical convergence results of proposed estimator, such as mean consistency, upper and lower bounds on convergence rate. In Sect. 4, we present a short simulation study to illustrate the convergence of our estimator. Finally, all proofs are gathered in Sect. 5.

2 The Estimator

We first introduce some notions and notations. The Fourier transform of a function $f \in L^1(\mathbb {R})$ is defined as $\phi _f (t):= \int _{-\infty }^{\infty } f(x) \mathrm {e}^{\mathrm {i}tx} \mathrm{d}x$, $t\in \mathbb {R}$. Hence, if f is density of a random variable U, then $\phi _{f} = \phi _U$. Note that if $f\in L^1(\mathbb {R}) \cap L^2(\mathbb {R})$, then $\int _{-\infty }^{\infty }|\phi _{f}(t)|^2 \mathrm{d}t = 2\pi \int _{-\infty }^{\infty } |f(x)|^2\mathrm{d}x$, called Parseval identity. The convolution of two measurable functions f, g on $\mathbb {R}$ is the function $(f *g)(x):= \int _{-\infty }^{\infty } f(x-u)g(u) \mathrm{d}u$. For a function $\varphi : \mathbb {R} \rightarrow \mathbb {R}$, the notation $\mathrm {supp}(\varphi )$ stands for the support of $\varphi$, i.e., the closure in $\mathbb {R}$ of the set $\{x\in \mathbb {R}: \varphi (x) \ne 0\}$. The Lebesgue measure of a measurable set $A \subset \mathbb {R}$ is denoted by $\lambda (A)$. The notations $\mathfrak {R}\{z\}$, $\mathfrak {I}\{z\}$ and $\bar{z}$ denote the real part, imaginary part and conjugate of a complex number z, respectively. For positive parameters $a_{n,m}$ and $b_{n,m}$ depending on the sample sizes n, m, the notation $a_{n,m} \le \mathcal {O}(b_{n,m})$ means that there exists a constant $c>0$ independent of n, m such that $a_{n,m} \le c\, b_{n,m}$ for large n, m.

We now discuss our estimation method. Since X, Y are continuous type random variables, the sum variable $X+Y$ is also continuous type. Hence, the cdf $F_{X+Y}$ can be represented in the form (see, e.g., [9])

$$\begin{aligned} F_{X+Y}(x) = \frac{1}{2} - \frac{1}{\pi }\int _{0}^{\infty } \frac{1}{t}\mathfrak {I}\{\phi _{X+Y}(t) \mathrm {e}^{-\mathrm {i}tx}\} \mathrm{d}t, \quad x\in \mathbb {R}. \end{aligned}$$

(2.1)

In view of (2.1), replacing the unknown quantity $\phi _{X+Y}(t)$ by a suitable empirical counterpart on the basis of the available data $X'_1, \ldots , X'_n$, $Y'_1, \ldots , Y'_m$ yields a desired estimator for $F_{X+Y}(x)$. For estimating $\phi _{X+Y}(t)$, we first apply the independence assumption of the variables X, Y, $\zeta$ and $\eta$ in (1.1) to obtain that

$$\begin{aligned}\phi _{X+Y}(t) = \phi _{X}(t)\phi _{Y}(t), \quad \phi _{X'}(t) = \phi _{X}(t) \phi _{\zeta }(t), \quad \phi _{Y'}(t) = \phi _{Y}(t) \phi _{\eta }(t), \,\,\, \text {for all }t\in \mathbb {R},\end{aligned}$$

so

$$\begin{aligned} \phi _{X+Y}(t) = \frac{\phi _{X'}(t)\phi _{Y'}(t)}{\phi _{\zeta }(t) \phi _{\eta }(t)}, \,\,\, \text {for all }t \notin Z(\phi _{\zeta }) \cup Z(\phi _{\eta }), \end{aligned}$$

(2.2)

where

$$\begin{aligned} Z(\phi _{\zeta }):= \{t\in \mathbb {R}: \phi _{\zeta }(t) =0 \}, \quad Z(\phi _{\eta }):= \{t\in \mathbb {R}: \phi _{\eta }(t) =0 \}. \end{aligned}$$

(2.3)

For every $t\in \mathbb {R}$, we define

$$\begin{aligned}\widehat{\phi }_{X'}(t) := \frac{1}{n}\sum _{j=1}^{n}\mathrm {e}^{\mathrm {i}tX'_j}, \quad \widehat{\phi }_{Y'}(t) := \frac{1}{m}\sum _{k=1}^{m}\mathrm {e}^{\mathrm {i}tY'_k}.\end{aligned}$$

Obviously, $\mathbb {E} \widehat{\phi }_{X'}(t) = \phi _{X'}(t)$ and $\mathbb {E}\widehat{\phi }_{Y'}(t) = \phi _{Y'}(t)$, for all $t\in \mathbb {R}$. Moreover, for every $t\in \mathbb {R}$, it follows from the strong law of large numbers that $\widehat{\phi }_{X'}(t) \overset{a.s.}{\rightarrow } \phi _{X'}(t)$ as $n\rightarrow \infty$, and $\widehat{\phi }_{Y'}(t) \overset{a.s.}{\rightarrow } \phi _{Y'}(t)$ as $m\rightarrow \infty$. Here $\overset{a.s.}{\rightarrow }$ denotes the almost sure convergence. Hence, $\widehat{\phi }_{X'}(t)$ and $\widehat{\phi }_{Y'}(t)$ are considered as the empirical counterparts of $\phi _{X'}(t)$ and $\phi _{Y'}(t)$, respectively. Thus, in (2.2), by plugging $\widehat{\phi }_{X'}(t)$ in $\phi _{X'}(t)$, $\widehat{\phi }_{Y'}(t)$ in $\phi _{Y'}(t)$, we obtain a naive estimator for $\phi _{X+Y}(t)$ in the form

$$\begin{aligned}\widetilde{\phi }_{X+Y}(t) := \frac{\widehat{\phi }_{X'}(t) \widehat{\phi }_{Y'}(t)}{\phi _{\zeta }(t) \phi _{\eta }(t)}, \quad t\notin Z(\phi _{\zeta }) \cup Z(\phi _{\eta }).\end{aligned}$$

Although $\widetilde{\phi }_{X+Y}(t)$ is an unbiased estimator of $\phi _{X+Y}(t)$ at every $t\notin Z(\phi _{\zeta }) \cup Z(\phi _{\eta })$, it is unstable when |t| is large enough as well as when t varies in a small neighborhood around possible zeros of $\phi _{\zeta }$ or $\phi _{\eta }$. It is because that $\lim _{|t|\rightarrow \infty } \phi _{\zeta }(t) \phi _{\eta }(t) = 0$, and $\lim _{t\rightarrow s} \phi _{\zeta }(t) \phi _{\eta }(t) = 0$ if $s\in Z(\phi _{\zeta }) \cup Z(\phi _{\eta })$. Therefore, replacing $\phi _{X+Y}(t)$ in (2.1) by $\widetilde{\phi }_{X+Y}(t)$ without any regularization may lead to a non-definiteness estimator, and there is some necessity to regularize $\widetilde{\phi }_{X+Y}(t)$ before the plug-in principle is applied. For this purpose, inspired from the ridge-parameter regularization method in Hall and Meister [10], we propose to replace $\widetilde{\phi }_{X+Y}(t)$ for $t\in \mathbb {R}$ by

$$\begin{aligned}\widehat{\phi }_{X+Y; \delta ,a}(t) := \frac{\phi _{\zeta }(-t) \phi _{\eta }(-t)\widehat{\phi }_{X'}(t) \widehat{\phi }_{Y'}(t)}{\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta |t|^a\}},\end{aligned}$$

where $\delta ,a>0$ are parameters. The function $\rho _{\delta ,a}(t):= \delta |t|^a$ in the denominator is called a ridge function and used to avoid divisions by zero or by numbers very close to zero. We have coincidence of $\widehat{\phi }_{X+Y; \delta ,a}(t)$ and $\widetilde{\phi }_{X+Y}(t)$ whenever $|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 \ge \rho _{\delta ,a}(t)$. Otherwise, the ridge regularization becomes efficient if $|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 < \rho _{\delta ,a}(t)$. Observe that $\widehat{\phi }_{X+Y; \delta ,a}$ is continuous on $\mathbb {R}$, $\widehat{\phi }_{X+Y; \delta ,a}(0)=1$, $\widehat{\phi }_{X+Y; \delta ,a}(t) = 0$ if $t\in Z(\phi _{\zeta }) \cup Z(\phi _{\eta })$, $\lim _{t\rightarrow s} \widehat{\phi }_{X+Y; \delta ,a}(t) = 0$ if $s\in \{\infty \} \cup Z(\phi _{\zeta }) \cup Z(\phi _{\eta })$; in addition, $\lim _{\delta \rightarrow 0^+} \mathbb {E} \widehat{\phi }_{X+Y; \delta ,a}(t) = \phi _{X+Y}(t)$, for all $t\notin Z(\phi _{\zeta }) \cup Z(\phi _{\eta })$. Hence, it is reasonable for using $\widehat{\phi }_{X+Y; \delta ,a}(t)$ to estimate $\phi _{X+Y}(t)$. Now, in (2.1), plugging $\widehat{\phi }_{X+Y; \delta ,a}(t)$ in $\phi _{X+Y}(t)$ yields the final estimator for $F_{X+Y}(x)$ as follows:

$$\begin{aligned} \widehat{F}_{X+Y; \delta ,a}(x) := \frac{1}{2} - \frac{1}{nm}\sum _{j=1}^{n} \sum _{k=1}^{m} U_{j,k; \delta ,a}(x) \end{aligned}$$

(2.4)

with

$$\begin{aligned} U_{j,k; \delta ,a}(x):= \frac{1}{\pi }\int _{0}^{\infty } \frac{1}{t}\mathfrak {I}\left\{ \frac{\phi _{\zeta }(-t) \phi _{\eta }(-t)\mathrm {e}^{\mathrm {i}t(X'_j + Y'_k - x)}}{\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}}\right\} \mathrm{d}t. \end{aligned}$$

(2.5)

Remark 1

The problem of estimating $F_X$ from the sample $X_1', \ldots , X_n'$ was studied in Dattner et al. [5] and Dattner and Reiser [4] with ordinary smooth and supersmooth noises. The authors estimated $\phi _{X}(t)$ by the truncated estimator $\widehat{\phi }_{X;T}(t):= \frac{\widehat{\phi }_{X'}(t)}{\phi _{\zeta }(t)} \mathbb {I}_{(0,T)}(t)$, with $T>0$. Here $\mathbb {I}_{(0,T)}$ denotes the indicator function of (0, T). The idea of using a cut-off parameter can be applicable to our problem in a specific case. More concretely, if $\phi _{\zeta }$, $\phi _{\eta }$ do not vanish on $\mathbb {R}$, then we can also estimate $\phi _{X+Y}(t)$ for $t>0$ by

$$\begin{aligned}\widehat{\phi }_{X+Y;T}(t):= \frac{\widehat{\phi }_{X'}(t) \widehat{\phi }_{Y'}(t)}{\phi _{\zeta }(t) \phi _{\eta }(t)} \mathbb {I}_{(0,T)}(t),\end{aligned}$$

with a suitably chosen parameter $T > 0$. Observe that if $|\phi _{\zeta }|$, $|\phi _{\eta }|$ are monotonically decreasing functions on $[0,\infty )$ and $T>0$ is chosen such that $|\phi _{\zeta }(T)\phi _{\eta }(T)|^2 = \delta T^a$, then $\widehat{\phi }_{X+Y;T}(t) = \widehat{\phi }_{X+Y;\delta ,a}(t)$ for any $t\in (0,T)$.

3 Convergence Results

3.1 Mean Consistency

In this part, we study consistency of the estimator $\widehat{F}_{X+Y; \delta ,a}$ with respect to root mean squared error (RMSE) defined by $\sqrt{\mathbb {E}|\widehat{F}_{X+Y; \delta ,a}(x) - F_{X+Y}(x)|^2}$. Here the notation $\mathbb {E}$ denotes the expectation with respect to the joint distribution of $X'_1, \ldots , X'_n$, $Y'_1, \ldots , Y'_m$. First of all, we need to establish general upper bounds for the bias and variance of $\widehat{F}_{X+Y; \delta ,a}$. To this end, we make the following assumptions:

• (A.1) There exist $c_0, \tau , t_0>0$ such that $|\phi _{\zeta }(t)\phi _{\eta }(t)|^2 \ge 1- c_0 |t|^{\tau }$, for all $|t|\le t_0$.

• (A.2) $C_{\zeta }:= \int _{-\infty }^{\infty } |\phi _{\zeta }(t)|\mathrm{d}t < \infty$ and $C_{\eta }:= \int _{-\infty }^{\infty } |\phi _{\eta }(t)|\mathrm{d}t < \infty$.

The assumption (A.1) describes behavior of the function $\phi _{\zeta }\phi _{\eta }$ in a neighborhood of the point $t=0$. It is satisfied if the function is smooth at the point. For instance, when $\zeta \sim \mathrm {N}(0, \upsilon _{\zeta }^2)$ (i.e., the Gaussian distribution with zero mean and $\upsilon _{\zeta }^2$ variance) and $\eta \sim \mathrm {N}(0, \upsilon _{\eta }^2)$, one has $1- |\phi _{\zeta }(t)\phi _{\eta }(t)|^2 = 1 - \mathrm {e}^{-(\upsilon _{\zeta }^2 + \upsilon _{\eta }^2)t^2} \le (\upsilon _{\zeta }^2 + \upsilon _{\eta }^2)t^2$ for all $t \in \mathbb {R}$, so we have verified (A.1) with $c_0 \equiv \upsilon _{\zeta }^2 + \upsilon _{\eta }^2$ and $\tau \equiv 2$.

Another example is the case $\zeta , \eta \sim \mathrm {U}(-1,1)$, the uniform distribution on $(-1,1)$. In that case, for any $t \ne 0$, one has

$$\begin{aligned}\phi _{\zeta }(t) = \frac{\sin (t)}{t} = \frac{1}{t} \sum _{\ell =0}^{\infty } \frac{(-1)^{\ell } t^{2\ell +1}}{(2\ell +1)!} = 1+ t\sum _{\ell =1}^{\infty } \frac{(-1)^{\ell } t^{2\ell -1}}{(2\ell +1)!} =: 1+ t\varphi (t).\end{aligned}$$

Similarly, $\phi _{\eta }(t) = 1+ t\varphi (t)$. Therefore, for $0<|t| \le 1$, applying the binomial theorem gives

$$\begin{aligned}1- |\phi _{\zeta }(t)\phi _{\eta }(t)|^2 = 1 - [1+t\varphi (t)]^4 = -\sum _{k=1}^{4} \left( {\begin{array}{c}4\\ k\end{array}}\right) [t\varphi (t)]^k \le |t| \sum _{k=1}^{4} \left( {\begin{array}{c}4\\ k\end{array}}\right) |\varphi (t)|^k.\end{aligned}$$

In view of the estimate $\sup _{|t| \le 1} |\varphi (t)| \le \sum _{\ell =1}^{\infty } [(2\ell +1)!]^{-1} = \mathrm {sinh}(1)$, we conclude that (A.1) is satisfied with $c_0 \equiv \sum _{k=1}^{4} \left( {\begin{array}{c}4\\ k\end{array}}\right) \left( \mathrm {e}-5/2\right) ^k$, $\tau \equiv 1$ and $t_0 \equiv 1$.

In what follows, we denote

$$\begin{aligned}t_1 := \min \{t_0; (2c_0)^{-1/\tau }; 1\},\end{aligned}$$

where $t_0$, $c_0$ and $\tau$ are as in (A.1).

Proposition 3.1

Consider the estimator $\widehat{F}_{X+Y; \delta ,a}$ as in (2.4), (2.5) with $0<\delta <1$, $a>0$. Then,

$$\begin{aligned} \sup _{x\in \mathbb {R}}|\mathbb {E}\widehat{F}_{X+Y; \delta ,a}(x) - F_{X+Y}(x)| \le \frac{1}{\pi } B_{\delta ,a}, \end{aligned}$$

(3.1)

with

$$\begin{aligned}B_{\delta ,a}:= \int _{0}^{\infty } \left| 1- \frac{|\phi _{\zeta }(t)|^2 |\phi _{\eta }(t)|^2}{\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}} \right| \frac{|\phi _X(t)| |\phi _{Y}(t)|}{t} \mathrm{d}t.\end{aligned}$$

Moreover, if the assumptions  (A.1) and (A.2) are satisfied, then there exists a constant $C\equiv C(C_{\zeta }, C_{\eta }, \tau )>0$ such that

$$\begin{aligned} \sup _{x\in \mathbb {R}}\mathrm {Var}\widehat{F}_{X+Y; \delta ,a}(x) \le C\left( \frac{1}{n} + \frac{1}{m} + \frac{V_{1;\delta ,a}}{nm} + \frac{V_{2;\delta ,a}}{n} + \frac{ V_{3;\delta ,a}}{n} + \frac{V_{4;\delta ,a}}{m} + \frac{V_{5;\delta ,a}}{m} \right) , \end{aligned}$$

(3.2)

where

$$\begin{aligned}&V_{1;\delta ,a}:= \int _{t_1}^{\infty } \frac{|\phi _{\zeta }(t)|^2 |\phi _{\eta }(t)|^2}{t^2 (\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\})^2} \mathrm{d}t,\\&V_{2;\delta ,a}:= \int _{t_1}^{\infty } \frac{|\phi _{\zeta }(t)|^2 |\phi _{\eta }(t)|^4 |\phi _{Y}(t)|^2}{t^2 (\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\})^2} \mathrm{d}t, \\&V_{3;\delta ,a}:= \int _{t_1}^{\infty } \frac{|\phi _{\zeta }(t)| |\phi _{\eta }(t)|^2|\phi _{Y}(t)|}{t\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}} \mathrm{d}t,\\&V_{4;\delta ,a}:= \int _{t_1}^{\infty } \frac{|\phi _{\zeta }(t)|^4 |\phi _{\eta }(t)|^2 |\phi _{X}(t)|^2}{t^2 (\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\})^2} \mathrm{d}t , \\&V_{5;\delta ,a}:= \int _{t_1}^{\infty } \frac{|\phi _{\zeta }(t)|^2 |\phi _{\eta }(t)||\phi _{X}(t)|}{t\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}} \mathrm{d}t. \end{aligned}$$

To give mean consistency of the estimator $\widehat{F}_{X+Y; \delta ,a}$, we add the following assumptions:

• (A.3) $\lambda (Z(\phi _{\zeta })) = \lambda (Z(\phi _{\eta })) =0$, where $Z(\phi _{\zeta })$, $Z(\phi _{\eta })$ are as in (2.3).

• (A.4) $\int _{t_1}^{\infty } t^{-1}|\phi _X(t)||\phi _{Y}(t)|\mathrm{d}t < \infty$.

There are many distributions satisfying (A.3), such as the Gaussian, Cauchy, Laplace, Gamma, uniform, triangular distributions and many others. We emphasize that the assumption (A.3) ensures identifiability of $F_{X+Y}$, i.e., the unique reconstruction of the target cdf $F_{X+Y}$ from $F_{X'}$ and $F_{Y'}$, the cdf’s of $X'$ and $Y'$, respectively.

Now we are ready to state a general upper bound for the RMSE as well as a consistency result of the estimator $\widehat{F}_{X+Y; \delta , a}$.

Theorem 3.2

Let the assumptions  (A.1) and  (A.2) hold. Moreover, suppose $0<a\le a_{*}$, where $a_{*}$ is a positive constant independent of n, m. The,n there exists a constant $C_{*}>0$ depending only on $t_1$, $a_{*}$ such that

$$\begin{aligned} \sup _{x\in \mathbb {R}}\sqrt{\mathbb {E}|\widehat{F}_{X+Y; \delta ,a}(x) - F_{X+Y}(x)|^2} \le C_{*}\left( B_{\delta ,a} + \frac{1}{\min \{1; \sqrt{a}\} \sqrt{\delta \min \{n;m\}}} \right) , \end{aligned}$$

(3.3)

where $B_{\delta ,a}$ is as in Proposition 3.1.

In addition, let the assumptions  (A.3) and (A.4) hold. If the parameter $\delta$ depends on n, m in such a way that $\lim _{n,m\rightarrow \infty }\delta = 0$ and $\lim _{n,m\rightarrow \infty } \delta \min \{n;m\} =\infty$, then

$$\begin{aligned}\lim _{n,m\rightarrow \infty }\sup _{x\in \mathbb {R}}\sqrt{\mathbb {E}|\widehat{F}_{X+Y; \delta ,a}(x) - F_{X+Y}(x)|^2} = 0.\end{aligned}$$

Remark 2

The parameter a of the estimator $\widehat{F}_{X+Y; \delta ,a}$ is given and fixed. On the other hand, in view of the estimate

$$\begin{aligned}B_{\delta ,a} > \int _{1}^{\infty } \left| 1- \frac{|\phi _{\zeta }(t)|^2 |\phi _{\eta }(t)|^2/(\delta t^a)}{\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 /(\delta t^a) ; 1\}} \right| \frac{|\phi _X(t)| |\phi _{Y}(t)|}{t} \mathrm{d}t,\end{aligned}$$

we infer that

$$\begin{aligned}\mathop {\liminf }\limits _{a \rightarrow \infty } B_{\delta ,a} \ge \int _{1}^{\infty } \frac{|\phi _X(t)| |\phi _{Y}(t)|}{t} \mathrm{d}t,\end{aligned}$$

which implies that the right-hand side in (3.3) would not converge to zero if we let $a\rightarrow \infty$. Hence, the parameter a cannot be arbitrarily large. This explains why we require the condition $0<a\le a_{*}$ in the latter theorem.

3.2 Rates of Convergence

Under some certain conditions on the parameters $\delta$ and a, Theorem 3.2 gives the convergence of $\sup _{x\in \mathbb {R}}\sqrt{\mathbb {E}|\widehat{F}_{X+Y; \delta ,a}(x) - F_{X+Y}(x)|^2}$ when the samples sizes n, m are increased. A further question is that how does the convergence depend on n, m? To answer this question, we have to add some further assumptions on the functions $F_{X}$, $F_Y$, $\phi _{\zeta }$ and $\phi _{\eta }$.

First, concerning $F_{X}$ and $F_Y$, we define $\mathcal {S}(\alpha ,L)$ with $\alpha >-1/2$, $L>0$ to be the class of all cdf’s F whose its corresponding density f satisfies $\int _{-\infty }^{\infty } |\phi _{f}(t)|^2 (1+t^2)^{\alpha }\mathrm{d}t \le L$. The class $\mathcal {S}(\alpha ,L)$ contains the cdf’s of the Gaussian, Cauchy, Laplace and Gamma distributions, among others. This class contains absolutely continuous cdf’s if $\alpha >-1/2$, whereas it contains cdf’s with bounded continuous densities if $\alpha >1/2$. Hereafter, we refer to $\mathcal {S}(\alpha ,L)$ as the Sobolev class of univariate cdf’s. This class was mentioned in several pieces of research, such as Dattner et al. [5], Dattner and Reiser [4], Trong and Phuong [2]. Note that, by Lemma 5.3, the assumption (A.4) becomes evident if $F_X \in \mathcal {S}(\alpha _X, L_X)$ and $F_Y \in \mathcal {S}(\alpha _Y, L_Y)$ for $\alpha _X, \alpha _Y > -1/2$ and $L_X, L_Y>0$. Now we set

$$\begin{aligned}&\mathcal {R}[\widehat{F}_{X+Y; \delta ,a}; \mathcal {S}(\alpha _X, L_X), \mathcal {S}(\alpha _Y, L_Y)] \\&\quad := \sup _{F_X \in \mathcal {S}(\alpha _X, L_X), \, F_Y \in \mathcal {S}(\alpha _Y, L_Y)} \sup _{x\in \mathbb {R}} \sqrt{\mathbb {E}|\widehat{F}_{X+Y; \delta ,a}(x) - F_{X+Y}(x)|^2},\end{aligned}$$

called maximal risk of the estimator $\widehat{F}_{X+Y; \delta ,a}$ on the classes $\mathcal {S}(\alpha _X, L_X)$, $\mathcal {S}(\alpha _Y, L_Y)$ of $F_X$, $F_Y$, respectively.

With respect to the noises $\zeta$, $\eta$, we first present the following assumption:

• (A.5) There exist positive constants $c_{1,\zeta }$, $c_{1,\eta }$, $a_{\zeta }$, $a_{\eta }$, $\mu _{\zeta }$, $\mu _{\eta }$, $\nu _{\zeta }$, $\nu _{\eta }$, $\vartheta _{\zeta }$, $\vartheta _{\eta }$, $\kappa _{\zeta }$ and $\kappa _{\eta }$ such that

  $$\begin{aligned}&|\phi _{\zeta }(t)| \ge c_{1,\zeta }|\sin (a_{\zeta } t)|^{\mu _{\zeta }} (1+|t|)^{-\nu _{\zeta }}\, \mathrm {e}^{-\kappa _{\zeta }|t|^{\vartheta _{\zeta }}}, \,\,\, \text {for all }t\in \mathbb {R},\\&|\phi _{\eta }(t)| \ge c_{1,\eta }|\sin (a_{\eta } t)|^{\mu _{\eta }} (1+|t|)^{-\nu _{\eta }}\, \mathrm {e}^{-\kappa _{\eta }|t|^{\vartheta _{\eta }}}, \,\,\, \text {for all }t\in \mathbb {R}.\end{aligned}$$

To give an example for the assumption, we mention the case $f_{\zeta } = f_{\eta } = f_{\mathrm {U}(-1,1)} *f_{\mathrm {N}(0,\sigma ^2)}$. Then, $\phi _{\zeta }(t) = \phi _{\eta }(t) = t^{-1}\sin (t)\mathrm {e}^{-\sigma ^2 t^2/2}$ for any $t\ne 0$; thus, $\phi _{\zeta }$ and $\phi _{\eta }$ satisfy (A.5) with $c_{1,\zeta }=c_{1,\eta }=1$, $a_{\zeta }=a_{\eta }=1$, $\mu _{\zeta }=\mu _{\eta } =1$, $\nu _{\zeta }=\nu _{\eta } =1$, $\vartheta _{\zeta }=\vartheta _{\eta } =2$ and $\kappa _{\zeta } = \kappa _{\eta }= \sigma ^2/2$.

It should be noted that the assumption (A.5) implies that $Z(\phi _{\zeta }) \subset \{k\pi /a_{\zeta }: k \in \mathbb {Z}, k\ne 0\}$ and $Z(\phi _{\eta }) \subset \{l\pi /a_{\eta }: l \in \mathbb {Z}, l\ne 0\}$. So possible zeros of $\phi _{\zeta }$ and $\phi _{\eta }$ are all isolated, and hence (A.3) is also satisfied. In (A.5), the parameters $\nu _{\zeta }$, $\nu _{\eta }$, $\vartheta _{\zeta }$ and $\vartheta _{\eta }$ control behavior tails of $\phi _{\zeta }$, $\phi _{\eta }$, whereas the parameters $\mu _{\zeta }$, $\mu _{\eta }$ describe orders of possible isolated zeros of the functions. In the current work, the noises $\zeta$, $\eta$ from (A.5) are said to be Fourier-oscillating.

The following theorem establishes an upper bound on the convergence rate of the maximal risk when the noises $\zeta$, $\eta$ satisfy (A.1), (A.2) and (A.5).

Theorem 3.3

Let $\alpha _X, \alpha _Y>-1/2$ and $L_X, L_Y>0$. Suppose that (A.1),  (A.2) and  (A.5) are satisfied. Let the estimator $\widehat{F}_{X+Y; \delta ,a}$ be given by (2.4), (2.5) and associated with the parameters $\delta = (\min \{n;m\})^{-\varrho }$ and $a \in (0, a_{*}]$, where $0<\varrho <1$ and $a_{*}$ is as in Theorem 3.2. Then, we obtain

$$\begin{aligned}&\mathcal {R}[\widehat{F}_{X+Y; \delta ,a}; \mathcal {S}(\alpha _X, L_X), \mathcal {S}(\alpha _Y, L_Y)] \\&\quad \le \mathcal {O}\left( \frac{1}{\min \{1; \sqrt{a}\}} \left( \ln (\min \{n;m\})\right) ^{-(\alpha _X + \alpha _Y +1)/\max \{\vartheta _{\zeta }; \vartheta _{\eta }\}} \right) .\end{aligned}$$

Although the logarithm rate in Theorem 3.3 is improved when n, m are increased, it indicates that the convergence of our estimator is very slow under the assumptions (A.1), (A.2) and (A.5). This rate depends on the parameters $\alpha _X$, $\alpha _Y$, $\vartheta _{\zeta }$ and $\vartheta _{\eta }$. It becomes better when $\alpha _X$ or $\alpha _Y$ are larger; however, the increasing of $\vartheta _{\zeta }$ or $\vartheta _{\eta }$ leads to a worse rate. Also, we emphasize that this rate is independent of the parameters $\mu _{\zeta }$, $\mu _{\eta }$, $a_{\zeta }$ and $a_{\eta }$ in (A.5), so the convergence rate is not affected by possible zeros of the functions $\phi _{\zeta }$, $\phi _{\eta }$. In particular, when $\alpha _X = \alpha _Y = \alpha >-1/2$, $\vartheta _{\zeta } = \vartheta _{\eta } = \vartheta >0$, $L_X = L_Y = L>0$ and $n=m$, the rate becomes $\mathcal {O}((\ln n)^{-(2\alpha +1)/\vartheta })$, which is faster than the minimax optimal rate $\mathcal {O}((\ln n)^{-(\alpha +1/2)/\vartheta })$ established for the cdf deconvolution problem in one sample measurement error model with supersmooth noise of order $\vartheta$, see, e.g., [4].

Remark 3

Our estimator depends on two parameters $\delta$ and a, where a is given and fixed in the interval $(0, a_{*}]$, while $\delta$ must be selected according to n, m. In Theorem 3.3, the selection of $\delta$ does not require any knowledge of the class parameters $\alpha _X$, $\alpha _Y$, $L_X$ and $L_Y$, so the resulting estimator is fully data-driven.

In the sequel, we consider the cases where $\zeta$, $\eta$ are ordinary smooth or supersmooth, the standard cases of $\zeta$, $\eta$. These cases have been studied in a large number of works in the deconvolution literature. For this purpose, we introduce the following assumptions:

• (A.6) There exist positive constants $k_{1,\zeta }$, $k_{1,\eta }$, $d_{\zeta }$, $d_{\eta }$, $\gamma _{\zeta }$ and $\gamma _{\eta }$ such that

  $$\begin{aligned}&|\phi _{\zeta }(t)| \ge k_{1,\zeta }\,\mathrm {e}^{-d_{\zeta }|t|^{\gamma _{\zeta }}}, \,\,\, \text {for all }t\in \mathbb {R},\\&|\phi _{\eta }(t)| \ge k_{1,\eta }\, \mathrm {e}^{-d_{\eta }|t|^{\gamma _{\eta }}}, \,\,\, \text {for all }t\in \mathbb {R}.\end{aligned}$$

• (A.7) There exist constants $\ell _{1,\zeta }, \ell _{1,\eta } > 0$ and $\beta _{\zeta }, \beta _{\eta }>1$ such that

  $$\begin{aligned}&|\phi _{\zeta }(t)| \ge \ell _{1,\zeta } (1+|t|)^{-\beta _{\zeta }}, \,\,\, \text {for all }t\in \mathbb {R},\\&|\phi _{\eta }(t)| \ge \ell _{1,\eta } (1+|t|)^{-\beta _{\eta }}, \,\,\, \text {for all }t\in \mathbb {R}.\end{aligned}$$

The assumptions (A.6) and (A.7) correspond to supersmooth and ordinary smooth noises, respectively. For example, if $\zeta , \eta \sim \mathrm {N}(0, \sigma ^2)$, then (A.6) is satisfied with $k_{1,\zeta } = k_{1,\eta } = 1$, $d_{\zeta } = d_{\eta } = \sigma ^2/2$ and $\gamma _{\zeta } = \gamma _{\eta } = 2$. When $\zeta$, $\eta$ have the Laplace distribution with location parameter 0 and scale parameter $s>0$, (A.7) is satisfied for $\ell _{1,\zeta } = \ell _{1,\eta } = 1/\max \{1;s^2\}$ and $\beta _{\zeta } = \beta _{\eta }=2$. Note that the constraint $\beta _{\zeta }, \beta _{\eta }>1$ in (A.7) is a necessary condition for the integrability of $\phi _{\zeta }$, $\phi _{\eta }$, as mentioned in (A.2).

Theorem 3.4

Let $\alpha _X, \alpha _Y>-1/2$ and $L_X, L_Y>0$. Consider the estimator $\widehat{F}_{X+Y; \delta ,a}$ with $a \in (0, a_{*}]$, where $a_{*}$ is as in Theorem 3.2.

• (a) Suppose that  (A.1),  (A.2) and  (A.6) are satisfied. Choosing $\delta = (\min \{n;m\})^{-\varrho }$ $(0<\varrho <1)$, we obtain

  $$\begin{aligned}&\mathcal {R}[\widehat{F}_{X+Y; \delta ,a}; \mathcal {S}(\alpha _X, L_X), \mathcal {S}(\alpha _Y, L_Y)] \\&\quad \le \mathcal {O}\left( \frac{1}{\min \{1; \sqrt{a}\}} \left( \ln (\min \{n;m\})\right) ^{-(\alpha _X + \alpha _Y +1)/\max \{\gamma _{\zeta }; \gamma _{\eta }\}} \right) .\end{aligned}$$

• (b) Suppose that  (A.1),  (A.2) and  (A.7) are satisfied. Choosing

  $$\begin{aligned}\delta = (\min \{n;m\})^{-(a + 2\beta _{\zeta } + 2\beta _{\eta })/(a + 2\beta _{\zeta } + 2\beta _{\eta } + 2\alpha _X + 2\alpha _Y + 2)},\end{aligned}$$

  we obtain

  $$\begin{aligned}&\mathcal {R}[\widehat{F}_{X+Y; \delta ,a}; \mathcal {S}(\alpha _X, L_X), \mathcal {S}(\alpha _Y, L_Y)] \\&\le \mathcal {O}\left( \frac{1}{\min \{1; \sqrt{a}\}} (\min \{n;m\})^{-(\alpha _X + \alpha _Y + 1)/(a + 2\beta _{\zeta } + 2\beta _{\eta } + 2\alpha _X + 2\alpha _Y + 2)} \right) . \end{aligned}$$

• (c) Suppose that  (A.1),  (A.2), the first condition of  (A.6) and the second condition of  (A.7) are satisfied. Choosing $\delta = (\min \{n;m\})^{-\omega }$ $(0<\omega <1)$, we obtain

  $$\begin{aligned}&\mathcal {R}[\widehat{F}_{X+Y; \delta ,a}; \mathcal {S}(\alpha _X, L_X), \mathcal {S}(\alpha _Y, L_Y)] \\&\quad \le \mathcal {O}\left( \frac{1}{\min \{1; \sqrt{a}\}} \left( \ln (\min \{n;m\})\right) ^{-(\alpha _X + \alpha _Y +1)/\gamma _{\zeta }} \right) .\end{aligned}$$

Remark 4

We have some comments on the results of Theorem 3.4 as follows.

• (a) As in Theorem 3.3, all the convergence rates derived in Theorem 3.4 are improved when n, m are increased. Also, they become better when $\alpha _X$ or $\alpha _Y$ are larger. In Part (b) where both $\zeta$ and $\eta$ are ordinary smooth, we derive an algebraic rate which improves the logarithm rates obtained in Part (a) and Part (c).

• (b) The upper bounds of Part (a) and Part (c) in the case $1\le a \le a_{*}$ are faster than those in the case $0<a<1$.

With Part (b), we rewrite the upper bound in the form

$$\begin{aligned}\mathcal {R}[\widehat{F}_{X+Y; \delta ,a}; \mathcal {S}(\alpha _X, L_X), \mathcal {S}(\alpha _Y, L_Y)] \le \mathcal {O}\left( \frac{1}{\min \{1; \sqrt{a}\}} (\min \{n;m\})^{-k_0/(a + \ell _0)} \right) ,\end{aligned}$$

where $k_0:= \alpha _X + \alpha _Y + 1$ and $\ell _0 := 2\beta _{\zeta } + 2\beta _{\eta } + 2\alpha _X + 2\alpha _Y + 2$. Observe that the upper bound in the case $1\le a \le a_{*}$ is the best as possible when a is the smallest as possible, namely $a = 1$, and

$$\begin{aligned}\mathcal {R}[\widehat{F}_{X+Y; \delta _1, a}; \mathcal {S}(\alpha _X, L_X), \mathcal {S}(\alpha _Y, L_Y)] \le \mathcal {O}\left( (\min \{n;m\})^{-k_0/(1 + \ell _0)} \right) ,\end{aligned}$$

with $\delta _1 := (\min \{n;m\})^{-(1 + 2\beta _{\zeta } + 2\beta _{\eta })/(2\beta _{\zeta } + 2\beta _{\eta } + 2\alpha _X + 2\alpha _Y + 3)}$. In the case $0< a <1$, the upper bound in Part (b) is the best as $a = a_{n,m}$, where

$$\begin{aligned}a_{n,m}:=\frac{\ell _0^2}{k_0 \ln (\min \{n;m\}) - \ell _0 + \sqrt{(k_0 \ln (\min \{n;m\}))^2 - 2k_0 \ell _0 \ln (\min \{n;m\})}},\end{aligned}$$

and

$$\begin{aligned}&\inf _{0<a < 1} \mathcal {R}[\widehat{F}_{X+Y; \delta _2,a}; \mathcal {S}(\alpha _X, L_X), \mathcal {S}(\alpha _Y, L_Y)] \\&\quad \le \mathcal {O}\left( \frac{1}{\sqrt{a_{n,m}}} (\min \{n;m\})^{-k_0/(a_{n,m} + \ell _0)} \right) \\&\quad \le \mathcal {O}\left( (\min \{n;m\})^{-k_0/\ell _0} (\ln (\min \{n;m\}))^{1/2} \right) , \end{aligned}$$

with $\delta _2 := (\min \{n;m\})^{-(a_{n,m} + 2\beta _{\zeta } + 2\beta _{\eta })/(a_{n,m} + 2\beta _{\zeta } + 2\beta _{\eta } + 2\alpha _X + 2\alpha _Y + 2)}$.

Since

$$\begin{aligned}\lim _{n,m\rightarrow \infty } \frac{(\min \{n;m\})^{-k_0/\ell _0} (\ln (\min \{n;m\}))^{1/2}}{(\min \{n;m\})^{-k_0/(1 + \ell _0)}} =0,\end{aligned}$$

we conclude that the convergence rate of the upper bound in Part (b) is fastest as $a = a_{n,m}$ and

$$\begin{aligned}&\inf _{0<a \le a_{*}} \mathcal {R}[\widehat{F}_{X+Y; \delta _2,a}; \mathcal {S}(\alpha _X, L_X), \mathcal {S}(\alpha _Y, L_Y)] \nonumber \\&\quad \le \mathcal {O}\left( (\min \{n;m\})^{-k_0/\ell _0} (\ln (\min \{n;m\}))^{1/2} \right) . \end{aligned}$$

(3.4)

We ask that whether the convergence rates derived in Theorems 3.3 and 3.4 are optimal or not. We will answer this question by establishing some lower bounds on convergence rate for so-called minimax risk over the classes $\mathcal {S}(\alpha _X, L_X)$ of $F_X$ and $\mathcal {S}(\alpha _Y, L_Y)$ of $F_Y$, defined as $\inf _{\widehat{F}_{X+Y}} \mathcal {R}[\widehat{F}_{X+Y}; \mathcal {S}(\alpha _{X}, L_{X}), \mathcal {S}(\alpha _{Y}, L_{Y})]$. Here the infimum is taken over all possible estimators $\widehat{F}_{X+Y}$, based on the data $X_1', \ldots X_n'$, $Y_1', \ldots , Y_m'$, of $F_{X+Y}$. Note that if there exists a positive sequence $\{\psi _{n,m}\}_{n,m}$ with $\lim _{n,m\rightarrow \infty } \psi _{n,m} = 0$ such that

$$\begin{aligned}c_1 \psi _{n,m} \le \inf _{\widehat{F}_{X+Y}} \mathcal {R}[\widehat{F}_{X+Y}; \mathcal {S}(\alpha _{X}, L_{X}), \mathcal {S}(\alpha _{Y}, L_{Y})] \le c_2 \psi _{n,m}\end{aligned}$$

for large n, m and for some constants $c_1,c_2>0$ independent of n, m, then the sequence is called an optimal rate of convergence. In particular, an estimator $\widetilde{F}_{X+Y}$ satisfying

$$\begin{aligned}\mathcal {R}[\widetilde{F}_{X+Y}; \mathcal {S}(\alpha _{X}, L_{X}), \mathcal {S}(\alpha _{Y}, L_{Y})] \le c_{*} \psi _{n,m},\end{aligned}$$

where $\{\psi _{n,m}\}_{n,m}$ is the optimal rate of convergence and $c_{*}>0$ is a constant, is called a minimax optimal estimator in order.

For the last-mentioned purpose, we need the following assumptions:

• (A.8) Suppose that $\phi _{\zeta }$, $\phi _{\eta }$ are continuously differentiable on $\mathbb {R}$ and that there are positive constants $\widetilde{c}_{\zeta }$, $\widetilde{c}_{\eta }$, $k_{\zeta }$, $k_{\eta }$, $\upsilon _{\zeta }$ and $\upsilon _{\eta }$ such that

  $$\begin{aligned}&\max \{|\phi _{\zeta }(t)|; \, |\phi _{\zeta }'(t)| \} \le \widetilde{c}_{\zeta }\mathrm {e}^{-k_{\zeta }|t|^{\upsilon _{\zeta }}} , \,\,\, \text {for all }t\in \mathbb {R},\\&\max \{|\phi _{\eta }(t)|; \, |\phi _{\eta }'(t)| \} \le \widetilde{c}_{\eta }\mathrm {e}^{-k_{\eta }|t|^{\upsilon _{\eta }}} , \,\,\, \text {for all }t\in \mathbb {R}.\end{aligned}$$

  Here the notations $\phi _{\zeta }'$ and $\phi _{\eta }'$ stand, respectively, for the first derivatives of $\phi _{\zeta }$ and $\phi _{\eta }$.

• (A.9) Suppose that $\phi _{\zeta }$, $\phi _{\eta }$ are continuously differentiable on $\mathbb {R}$ and that there exist constants $\widetilde{\ell }_{\zeta }, \widetilde{\ell }_{\eta }> 0$, $\widetilde{\beta }_{\zeta }, \widetilde{\beta }_{\eta } >0$ such that

  $$\begin{aligned}&\max \{|\phi _{\zeta }(t)|; \, |\phi _{\zeta }'(t)| \} \le \widetilde{\ell }_{\zeta } (1+|t|)^{-\widetilde{\beta }_{\zeta }} , \,\,\, \text {for all }t\in \mathbb {R},\\&\max \{|\phi _{\eta }(t)|; \, |\phi _{\eta }'(t)| \} \le \widetilde{\ell }_{\eta } (1+|t|)^{-\widetilde{\beta }_{\eta }} , \,\,\, \text {for all }t\in \mathbb {R}.\end{aligned}$$

Theorem 3.5

Let $L_X = L_Y = L>0$ be sufficiently large and $\alpha _X = \alpha _Y = \alpha >1/2$.

• (a) Under the assumption  (A.8), there exists a constant $C_{*,1}>0$ independent of n, m such that

  $$\begin{aligned}\inf _{\widehat{F}_{X+Y}} \mathcal {R}[\widehat{F}_{X+Y}; \mathcal {S}(\alpha , L), \mathcal {S}(\alpha , L)] \ge C_{*,1} (\ln (\max \{n;m\}))^{-(2\alpha + 1)/\min \{\upsilon _{\zeta }; \upsilon _{\eta }\}}\end{aligned}$$

  for large n, m.

• (b) Under the assumption  (A.9), there exists a constant $C_{*,2}>0$ independent of n, m such that

  $$\begin{aligned}\inf _{\widehat{F}_{X+Y}} \mathcal {R}[\widehat{F}_{X+Y}; \mathcal {S}(\alpha , L), \mathcal {S}(\alpha , L)] \ge C_{*,2} (\max \{n;m\})^{-(2\alpha +1)/(2\alpha + 2\min \{\widetilde{\beta }_{\zeta }; \widetilde{\beta }_{\eta }\})}\end{aligned}$$

  for large n, m.

From Theorems 3.3, 3.4 and 3.5, we conclude that our estimator $\widehat{F}_{X+Y; \delta ,a}$ is minimax optimal in order in the particular cases as follows: (i) $\alpha _X = \alpha _Y = \alpha >1/2$, $n=m$, $\phi _{\zeta }$ and $\phi _{\eta }$ satisfy (A.1), (A.5) and (A.8) with $\vartheta _{\zeta } = \vartheta _{\eta } = \upsilon _{\zeta } = \upsilon _{\eta } = \vartheta >0$ (the minimax risk is of order $\mathcal {O}((\ln n)^{-(2\alpha +1)/\vartheta })$); (ii) $\alpha _X = \alpha _Y = \alpha >1/2$, $n=m$, $\phi _{\zeta }$ and $\phi _{\eta }$ satisfy (A.1), (A.6) and (A.8) with $\gamma _{\zeta } = \gamma _{\eta } = \upsilon _{\zeta } = \upsilon _{\eta } = \gamma >0$ (the minimax risk is of order $\mathcal {O}((\ln n)^{-(2\alpha +1)/\gamma })$). Note also that our estimator is adaptive in the two mentioned cases. Nevertheless, in the case where $\phi _{\zeta }$ and $\phi _{\eta }$ satisfy (A.1), (A.7) and (A.9), i.e., both $\zeta$ and $\eta$ are ordinary smooth, if $\beta _{\zeta } = \beta _{\eta } = \widetilde{\beta }_{\zeta } = \widetilde{\beta }_{\eta } = \beta > 1$, $\alpha _X = \alpha _Y = \alpha _Y>1/2$ and $n=m$, then the combination of (3.4) and Theorem 3.5(b) gives

$$\begin{aligned}&C_{*,2} n^{-(2\alpha +1)/(2\alpha + 2\beta )} \le \inf _{\widehat{F}_{X+Y}} \mathcal {R}[\widehat{F}_{X+Y}; \mathcal {S}(\alpha , L), \mathcal {S}(\alpha , L)] \\&\quad \le \inf _{0<a \le a_{*}} \mathcal {R}[\widehat{F}_{X+Y; \delta _2,a}; \mathcal {S}(\alpha , L), \mathcal {S}(\alpha , L)] \\&\quad \le \mathcal {O}\left( n^{-(2\alpha +1)/(4\alpha + 4\beta +2)} (\ln n)^{1/2} \right) , \end{aligned}$$

and this shows that our estimator is not minimax optimal in this case.

Remark 5

In Theorem 3.5, the lower bounds are derived with $\alpha _X = \alpha _Y = \alpha >1/2$ and large $L>0$. If $1/2<\alpha _X < \alpha _Y$, then $\mathcal {S}(\alpha _Y, L) \subset \mathcal {S}(\alpha _X, L)$ and hence,

$$\begin{aligned}\mathcal {R}:= \inf _{\widehat{F}_{X+Y}} \mathcal {R}[\widehat{F}_{X+Y}; \mathcal {S}(\alpha _X, L), \mathcal {S}(\alpha _Y, L)] \ge \inf _{\widehat{F}_{X+Y}} \mathcal {R}[\widehat{F}_{X+Y}; \mathcal {S}(\alpha _Y, L), \mathcal {S}(\alpha _Y, L)] .\end{aligned}$$

Consequently,

$$\begin{aligned} \mathcal {R} \ge {\left\{ \begin{array}{ll} {c} \, (\ln (\max \{n;m\}))^{-(2\alpha _Y + 1)/\min \{\upsilon _{\zeta }; \upsilon _{\eta }\}}&{} \,\, \text {under (A.8)},\\ {c} \, (\max \{n;m\})^{-(2\alpha _Y +1)/(2\alpha _Y + 2\min \{\widetilde{\beta }_{\zeta }; \widetilde{\beta }_{\eta }\})} &{} \,\, \text {under (A.9)}, \end{array}\right. } \end{aligned}$$

for some constant $c>0$. Similarly, if $1/2<\alpha _Y < \alpha _X$, then

$$\begin{aligned} \mathcal {R} \ge {\left\{ \begin{array}{ll} {c} \, (\ln (\max \{n;m\}))^{-(2\alpha _X + 1)/\min \{\upsilon _{\zeta }; \upsilon _{\eta }\}}&{} \,\, \text {under (A.8)},\\ {c} \, (\max \{n;m\})^{-(2\alpha _X +1)/(2\alpha _X + 2\min \{\widetilde{\beta }_{\zeta }; \widetilde{\beta }_{\eta }\})} &{} \,\,\text {under (A.9)}, \end{array}\right. } \end{aligned}$$

for some constant $c>0$. Therefore, with $\alpha _X \ne \alpha _Y$, $\alpha _X, \alpha _Y >1/2$ and large $L>0$, we have

$$\begin{aligned} \mathcal {R} \ge {\left\{ \begin{array}{ll} {c} \, (\ln (\max \{n;m\}))^{-(2\max \{\alpha _X; \alpha _Y\} + 1)/\min \{\upsilon _{\zeta }; \upsilon _{\eta }\}}&{} \,\,\text {under (A.8)},\\ {c} \, (\max \{n;m\})^{-(2\max \{\alpha _X; \alpha _Y\} +1)/(2\max \{\alpha _X; \alpha _Y\} + 2\min \{\widetilde{\beta }_{\zeta }; \widetilde{\beta }_{\eta }\})} &{} \,\,\text {under (A.9)}, \end{array}\right. } \end{aligned}$$

for some constant $c>0$.

4 Numerical Experiment

We now conduct a numerical experiment to illustrate the convergence of the estimator $\widehat{F}_{X+Y; \delta ,a}$ given by (2.4) when increasing the samples sizes n, m. We assume that $X \sim \mathrm {Gamma}(1,1)$ and $Y \sim \mathrm {Gamma}(3,1)$ are independent random variables. Then, $X+Y \sim \mathrm {Gamma}(4,1)$. Here $\mathrm {Gamma}(k,\theta )$ denotes the Gamma distribution with shape parameter $k>0$ and scale parameter $\theta >0$. Denote by $\sigma _X$, $\sigma _Y$, $\sigma _{\zeta }$ and $\sigma _{\eta }$ the standard deviations of X, Y, $\zeta$ and $\eta$, respectively. We will choose the random noises $\zeta$ and $\eta$ such that the signal ratios $\sigma _{\zeta } / \sigma _X$ and $\sigma _{\eta }/\sigma _Y$ are equal to 0.5 and 1.0, corresponding to $50\%$ and $100\%$ noise contamination, respectively. We consider the two following cases of $\zeta$, $\eta$:

• (1) Case 1 $\zeta \sim \mathrm {N}(0, \sigma _{\zeta }^2)$ and $\eta \sim \mathrm {N}(0, \sigma _{\eta }^2)$. In that case, $\phi _{\zeta }(t)= \mathrm {e}^{-\sigma _{\zeta }^2 t^2/2}$, $\phi _{\eta }(t)= \mathrm {e}^{-\sigma _{\eta }^2 t^2/2}$. Hence, $\phi _{\zeta }$, $\phi _{\eta }$ satisfy the assumption (A.6). In the setup, we take:

  • $\sigma _{\zeta } = 1/2$, $\sigma _{\eta } = \sqrt{3}/2$ ($50\%$ contamination).

  • $\sigma _{\zeta } = 1$, $\sigma _{\eta } = \sqrt{3}$ ($100\%$ contamination).

• (2) Case 2 $\zeta = \zeta _1 + \zeta _2$ where $\zeta _1 \sim \mathrm {U}(-1/2, 1/2)$ and $\zeta _2 \sim \mathrm {N}(0, \sigma _{\zeta _2}^2)$ are independent; $\eta = \eta _1 + \eta _2$ where $\eta _1 \sim \mathrm {U}(-1/2,1/2)$ and $\eta _2 \sim \mathrm {N}(0, \sigma _{\eta _2}^2)$ are independent. In that case,

  $$\begin{aligned}\phi _{\zeta }(t) = \frac{2\sin (t/2)}{t}\mathrm {e}^{-\sigma _{\zeta _2}^2 t^2/2}, \quad \phi _{\eta }(t)= \frac{2\sin (t/2)}{t}\mathrm {e}^{-\sigma _{\eta _2}^2 t^2/2}.\end{aligned}$$

  Hence, $\phi _{\zeta }$, $\phi _{\eta }$ satisfy the assumption (A.5). In the setup, we take:

  • $\sigma _{\zeta _2} = \sqrt{1/6}$, $\sigma _{\eta _2} = \sqrt{2/3}$ ($50\%$ contamination).

  • $\sigma _{\zeta _2} = \sqrt{11/12}$, $\sigma _{\eta _2} = \sqrt{35/12}$ ($100\%$ contamination).

In each case of the noises, we generate randomly 100 independent samples $(X_{p,1}, \ldots , X_{p,n})$, $p=1,\ldots , 100$, of size n from the distribution of X and generate randomly 100 independent samples $(Y_{p,1}, \ldots , Y_{p,m})$, $p=1,\ldots , 100$, of size m from the distribution of Y. For each $p = 1,\ldots , 100$, the sample $(X_{p,1}, \ldots , X_{p,n})$ is added by a sample $(\zeta _{p,1}, \ldots , \zeta _{p,n})$ generated from the distribution of $\zeta$ to obtain the sample $(X'_{p,1}, \ldots , X'_{p,n})$. We also obtain the samples $(Y'_{p,1}, \ldots , Y'_{p,m})$, $p=1,\ldots ,100$, by the same way. Then, we have the approximation

$$\begin{aligned}\sqrt{\mathbb {E}|\widehat{F}_{X+Y; \delta ,a}(x) - F_{X+Y}(x)|^2} \approx \sqrt{\frac{1}{100} \sum _{p=1}^{100} |\widehat{F}_{X+Y; \delta ,a; p}(x) - F_{X+Y}(x)|^2}=: \epsilon (x),\end{aligned}$$

where

$$\begin{aligned}\widehat{F}_{X+Y; \delta ,a; p}(x) := \frac{1}{2} - \frac{1}{nm}\sum _{j=1}^{n} \sum _{k=1}^{m} \frac{1}{\pi }\int _{0}^{\infty } \frac{1}{t}\mathfrak {I}\left\{ \frac{\phi _{\zeta }(-t) \phi _{\eta }(-t)\mathrm {e}^{\mathrm {i}t(X'_{p,j} + Y'_{p,k} - x)}}{\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}}\right\} \mathrm{d}t.\end{aligned}$$

The quantity $\epsilon (x)$ is called empirical RMSE of $\widehat{F}_{X+Y; \delta ,a}$ at x. In our setup, we take $a=1$. In addition, as mentioned in Theorem 3.3, we choose $\delta = (\min \{n;m\})^{-1/2}$. Note that we only compute $\epsilon (x)$ at $q_{0.1}$, $q_{0.25}$, $q_{0.5}$, $q_{0.75}$ and $q_{0.9}$, which correspond to the percentiles 0.1, 0.25, 0.5, 0.75 and 0.9 of the distribution of $X+Y$.

Tables 1 and 2 present $\epsilon (x)$ for Case 1 and Case 2 of $\zeta$, $\eta$. In these tables, the values of $\epsilon (x)$ corresponding to $50\%$ and $100\%$ noise contamination are evaluated with respect to some different values of n and m, including $n = m=100$, $n=m=250$ and $n=m=500$. We can see that at every $x\in \{q_{0.1}, q_{0.25}, q_{0.5}, q_{0.75}, q_{0.9}\}$ and $n,m \in \{100, 250, 500\}$, the values of $\epsilon (x)$ for the case $50\%$ contamination are smaller than those for the case $100\%$ contamination. Moreover, the values of $\epsilon (x)$ decrease to zero when increasing n, m. The results are compatible with the mean consistency of the estimator $\widehat{F}_{X+Y; \delta ,a}$.

Table 1 $\epsilon (x) \times 100$ ($50\%$ contamination) – $\epsilon (x) \times 100$ ($100\%$ contamination) for Case 1.

Table 2 $\epsilon (x) \times 100$ ($50\%$ contamination) – $\epsilon (x) \times 100$ ($100\%$ contamination) for Case 2.

5 Proofs

To prove Proposition 3.1, we need the following lemmas.

Lemma 5.1

Let $0<\delta <1$, $a>0$ and $u \in \mathbb {R}$. Under the assumption  (A.1), we have

$$\begin{aligned}\left| \int _{0}^{t_1} \frac{\sin [s(X'_j + Y'_k - u)]}{s\max \{|\phi _{\zeta }(s) \phi _{\eta }(s)|^2 ; \delta s^a\}} \mathrm{d}s \right| \le 3+\frac{\ln 2}{\tau },\end{aligned}$$

for any $j \in \{1,\ldots ,n\}$ and $k \in \{1, \ldots , m\}$.

Proof

By (A.1), we have for any $t\in (0, t_1)$ that

$$\begin{aligned}|1- \max \{|\phi _{\zeta }(t)\phi _{\eta }(t)|^2 ; \delta t^a\} | \le 1-|\phi _{\zeta }(t)\phi _{\eta }(t)|^2 \le c_0 t^{\tau } <c_0 t_1^{\tau } \le \frac{1}{2},\end{aligned}$$

so we have the following expansion:

$$\begin{aligned}\frac{1}{\max \{|\phi _{\zeta }(t)\phi _{\eta }(t)|^2 ; \delta t^a\}} =1 + \sum _{k=1}^{\infty } (1-\max \{|\phi _{\zeta }(t)\phi _{\eta }(t)|^2 ; \delta t^a\})^{k} =: 1 + r_{\zeta ,\eta }(t).\end{aligned}$$

Thus,

$$\begin{aligned} \left| \int _{0}^{t_1} \frac{\sin [t(X'_j + Y'_k - u)]}{t\max \{|\phi _{\zeta }(t)\phi _{\eta }(t)|^2 ; \delta t^a\}} \mathrm{d}t \right|&= \left| \int _{0}^{t_1} \frac{\sin [t(X'_j + Y'_k - u)]}{t}[1 + r_{\zeta ,\eta }(t)] \mathrm{d}t \right| \nonumber \\&\le Q_{1} + Q_{2}, \end{aligned}$$

(5.1)

where

$$\begin{aligned}Q_{1}:= \left| \int _{0}^{t_1} \frac{\sin [t(X'_j + Y'_k - u)]}{t}\mathrm{d}t \right| , \quad Q_{2}:= \int _{0}^{t_1} \frac{|r_{\zeta ,\eta }(t)|}{t}\mathrm{d}t.\end{aligned}$$

As known $\sup _{s>0} \left| \int _{0}^{s} \frac{\sin (w)}{w}\mathrm{d}w \right| \le 3$ (see, e.g., [11]), so

$$\begin{aligned} Q_{1} = \left| \int _{0}^{t_1(X'_j + Y'_k - u)} \frac{\sin (w)}{w}\mathrm{d}w \right| \le 3. \end{aligned}$$

(5.2)

By (A.1) and the equality $\sum _{k=1}^{\infty }\frac{x^k}{k} = \ln [(1-x)^{-1}]$ for all $0<x<1$, we obtain

$$\begin{aligned} Q_{2}&\le \int _{0}^{t_1} \frac{1}{t} \sum _{k=1}^{\infty } (1-|\phi _{\zeta }(t)\phi _{\eta }(t)|^2)^{k} \mathrm{d}t \le \int _{0}^{t_1} \frac{1}{t} \sum _{k=1}^{\infty } (c_0 t^{\tau })^{k} \mathrm{d}t = \sum _{k=1}^{\infty } c_0^k \int _{0}^{t_1} t^{k\tau - 1} \mathrm{d}t \nonumber \\&= \frac{1}{\tau } \sum _{k=1}^{\infty } \frac{(c_0 t_1^{\tau })^{k}}{k} = \frac{1}{\tau } \ln \left( \frac{1}{1- c_0 t_1^{\tau }}\right) \le \frac{\ln 2}{\tau }. \end{aligned}$$

(5.3)

Combining (5.1) with (5.2) and (5.3) gives the desired result. $\square$

Lemma 5.2

Let $U_{j,k; \delta ,a}(x)$ be as in (2.5). Under the assumptions  (A.1) and  (A.2), we have

$$\begin{aligned}&\mathbb {E}(U_{1,1;\delta ,a}(x))^2 \le \frac{2}{\pi ^2}\left( 3+\frac{\ln 2}{\tau }\right) ^2 + \frac{2\min \{C_{\zeta }; C_{\eta }\}}{\pi ^2} V_{1;\delta ,a},\\&|\mathbb {E}(U_{1,1;\delta ,a}(x) U_{1,2;\delta ,a}(x)) | \le \frac{1}{\pi ^2} \left( 3+\frac{\ln 2}{\tau }\right) ^2 + \frac{C_{\zeta }}{\pi ^2} V_{2;\delta ,a} + \frac{2}{\pi ^2} \left( 3+\frac{\ln 2}{\tau }\right) V_{3;\delta ,a},\\&|\mathbb {E}(U_{1,1;\delta ,a}(x) U_{2,1;\delta ,a}(x)) | \le \frac{1}{\pi ^2} \left( 3+\frac{\ln 2}{\tau } \right) ^2 + \frac{C_{\eta }}{\pi ^2} V_{4;\delta ,a} + \frac{2}{\pi ^2} \left( 3+\frac{\ln 2}{\tau } \right) V_{5;\delta ,a},\end{aligned}$$

where $V_{j;\delta ,a}$’s are as in Proposition 3.1.

Proof

Denote by $f_{X'_j}$ and $f_{Y'_k}$ the densities of $X'_j$ and $Y'_k$, respectively.

Estimate $\mathbb {E}(U_{1,1;\delta ,a}(x))^2$: By the inequality $(u+v)^2 \le 2(u^2 + v^2)$ for all $u,v\in \mathbb {R}$, we have

$$\begin{aligned} \mathbb {E}(U_{1,1;\delta ,a}(x))^2&= \frac{1}{\pi ^2} \mathbb {E}\left( \int _{0}^{t_1} \frac{1}{t}\mathfrak {I}\left\{ \frac{\phi _{\zeta }(-t) \phi _{\eta }(-t)\mathrm {e}^{\mathrm {i}t(X'_1 + Y'_1 - x)}}{\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}}\right\} \mathrm{d}t\right. \nonumber \\&\left. \qquad \quad + \int _{t_1}^{\infty } \frac{1}{t}\mathfrak {I}\left\{ \frac{\phi _{\zeta }(-t) \phi _{\eta }(-t)\mathrm {e}^{\mathrm {i}t(X'_1 + Y'_1 - x)}}{\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}}\right\} \mathrm{d}t \right) ^2 \nonumber \\&\le \frac{2}{\pi ^2} (I_{0}^{t_1} + I_{t_1}^{\infty }), \end{aligned}$$

(5.4)

where

$$\begin{aligned}I_{u}^{v}:= \mathbb {E}\left( \int _{u}^{v} \frac{\mathfrak {I}\left\{ \phi _{\zeta }(-t) \phi _{\eta }(-t)\mathrm {e}^{\mathrm {i}t(X'_1 + Y'_1 - x)}\right\} }{t\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}} \mathrm{d}t \right) ^2 , \quad 0<u\le v \le \infty .\end{aligned}$$

(i) Estimate $I_{0}^{t_1}$. We have

$$\begin{aligned}&\mathfrak {I}\left\{ \phi _{\zeta }(-t) \phi _{\eta }(-t) \mathrm {e}^{\mathrm {i}t(X'_1 + Y'_1 - x)} \right\} \\&\quad = \mathfrak {I}\left\{ \left( \int _{-\infty }^{\infty }f_{\zeta }(u)\mathrm {e}^{-\mathrm {i}tu}\mathrm{d}u\right) \left( \int _{-\infty }^{\infty }f_{\eta }(v)\mathrm {e}^{-\mathrm {i}tv}\mathrm{d}v\right) \mathrm {e}^{\mathrm {i}t(X'_1 + Y'_1 - x)} \right\} \\&\quad = \int _{-\infty }^{\infty }\int _{-\infty }^{\infty } f_{\zeta }(u) f_{\eta }(v) \sin [t(X'_1 + Y'_1 - x -u -v)] \mathrm{d}u \mathrm{d}v, \end{aligned}$$

which together with the Fubini theorem to give

$$\begin{aligned} I_{0}^{t_1}&= \mathbb {E}\left( \int _{0}^{t_1} \frac{1}{t\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty } \right. \\&\quad \left. f_{\zeta }(u) f_{\eta }(v) \sin [t(X'_1 + Y'_1 - x -u -v)] \mathrm{d}u \mathrm{d}v \mathrm{d}t \right) ^2 \\&= \mathbb {E}\left( \int _{-\infty }^{\infty }\int _{-\infty }^{\infty } f_{\zeta }(u) f_{\eta }(v) \int _{0}^{t_1} \frac{\sin [t(X'_1 + Y'_1 - x -u -v)]}{t\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}} \mathrm{d}t \mathrm{d}u \mathrm{d}v\right) ^2 \\&\le \mathbb {E}\left( \int _{-\infty }^{\infty }\int _{-\infty }^{\infty } f_{\zeta }(u) f_{\eta }(v) \left| \int _{0}^{t_1} \frac{\sin [t(X'_1 + Y'_1 - x -u -v)]}{t\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}} \mathrm{d}t \right| \mathrm{d}u \mathrm{d}v\right) ^2. \end{aligned}$$

After that applying Lemma 5.1 we obtain

$$\begin{aligned} I_{0}^{t_1} \le \mathbb {E}\left( \int _{-\infty }^{\infty }\int _{-\infty }^{\infty } f_{\zeta }(u) f_{\eta }(v) \left( 3+\frac{\ln 2}{\tau } \right) \mathrm{d}u \mathrm{d}v\right) ^2 = \left( 3+\frac{\ln 2}{\tau }\right) ^2. \end{aligned}$$

(5.5)

(ii) Estimate $I_{t_1}^{\infty }$. By the Fubini theorem,

$$\begin{aligned}&I_{t_1}^{\infty } = \mathbb {E}\int _{t_1}^{\infty }\int _{t_1}^{\infty } \frac{1}{t}\mathfrak {I}\left\{ \frac{\phi _{\zeta }(-t) \phi _{\eta }(-t)\mathrm {e}^{\mathrm {i}t(X'_1 + Y'_1 - x)}}{\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}}\right\} \\&\quad \frac{1}{s}\mathfrak {I}\left\{ \frac{\phi _{\zeta }(-s) \phi _{\eta }(-s)\mathrm {e}^{\mathrm {i}s(X'_1 + Y'_1 - x)}}{\max \{|\phi _{\zeta }(s) \phi _{\eta }(s)|^2 ; \delta s^a\}}\right\} \mathrm{d}t \mathrm{d}s \\&\quad = \int _{-\infty }^{\infty } \int _{-\infty }^{\infty } \biggl (\int _{t_1}^{\infty }\int _{t_1}^{\infty } \frac{1}{t}\mathfrak {I}\left\{ \frac{\phi _{\zeta }(-t) \phi _{\eta }(-t)\mathrm {e}^{\mathrm {i}t(u + v - x)}}{\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}}\right\} \\&\qquad \times \frac{1}{s}\mathfrak {I}\left\{ \frac{\phi _{\zeta }(-s) \phi _{\eta }(-s)\mathrm {e}^{\mathrm {i}s(u + v - x)}}{\max \{|\phi _{\zeta }(s) \phi _{\eta }(s)|^2 ; \delta s^a\}}\right\} \mathrm{d}t \mathrm{d}s\biggr ) f_{X'_1}(u) f_{Y'_1}(v) \mathrm{d}u \mathrm{d}v \\&\quad = \int _{-\infty }^{\infty } \int _{-\infty }^{\infty } \left( \int _{t_1}^{\infty }\int _{t_1}^{\infty } \frac{A(u,v,t,s)}{ts\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\} \max \{|\phi _{\zeta }(s) \phi _{\eta }(s)|^2 ; \delta s^a\}} \mathrm{d}t \mathrm{d}s \right) \\&\qquad f_{X'_1}(u) f_{Y'_1}(v) \mathrm{d}u \mathrm{d}v\\&\quad = \int _{t_1}^{\infty }\int _{t_1}^{\infty } \frac{ \int _{-\infty }^{\infty } \int _{-\infty }^{\infty } A(u,v,t,s) f_{X'_1}(u) f_{Y'_1}(v) \mathrm{d}u \mathrm{d}v }{ts\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\} \max \{|\phi _{\zeta }(s) \phi _{\eta }(s)|^2 ; \delta s^a\}} \mathrm{d}t \mathrm{d}s \end{aligned}$$

with

$$\begin{aligned}A(u,v,t,s) := \mathfrak {I}\left\{ \phi _{\zeta }(-t) \phi _{\eta }(-t) \mathrm {e}^{\mathrm {i}t(u+v - x)}\right\} \mathfrak {I}\left\{ \phi _{\zeta }(-s) \phi _{\eta }(-s) \mathrm {e}^{\mathrm {i}s(u+v - x)}\right\} .\end{aligned}$$

We have

$$\begin{aligned} A(u,v,t,s)&= \int _{-\infty }^{\infty } \int _{-\infty }^{\infty } \int _{-\infty }^{\infty } \int _{-\infty }^{\infty } f_{\zeta }(p)f_{\eta }(q) f_{\zeta }(w)f_{\eta }(z)\\&\quad \sin [t(u+v-p-q-x)] \sin [s(u+v-w-z-x)]\mathrm{d}p \mathrm{d}q \mathrm{d}w \mathrm{d}z. \end{aligned}$$

Using the formula $\sin {u}\sin {v} = [\cos (u-v) - \cos (u+v)]/2$ and the Fubini theorem, we obtain

$$\begin{aligned}&\int _{-\infty }^{\infty } \int _{-\infty }^{\infty } A(u,v,t,s) f_{X'_1}(u) f_{Y'_1}(v) \mathrm{d}u \mathrm{d}v \\&\quad = \frac{1}{2}\mathfrak {R}\left\{ \phi _{X_1'}(t-s)\phi _{Y_1'}(t-s)\phi _{\zeta }(-t) \phi _{\eta }(-t) \phi _{\zeta }(s) \phi _{\eta }(s) \mathrm {e}^{\mathrm {i}(s-t)x}\right\} \\&\quad - \frac{1}{2}\mathfrak {R}\left\{ \phi _{X_1'}(t+s)\phi _{Y_1'}(t+s)\phi _{\zeta }(-t) \phi _{\eta }(-t) \phi _{\zeta }(-s) \phi _{\eta }(-s) \mathrm {e}^{-\mathrm {i}(s+t)x}\right\} , \end{aligned}$$

so

$$\begin{aligned} I_{t_1}^{\infty }&= \frac{1}{2} \int _{t_1}^{\infty }\int _{t_1}^{\infty } \frac{\mathfrak {R}\left\{ \phi _{X_1'}(t-s)\phi _{Y_1'}(t-s)\phi _{\zeta }(-t) \phi _{\eta }(-t) \phi _{\zeta }(s) \phi _{\eta }(s) \mathrm {e}^{\mathrm {i}(s-t)x}\right\} }{ ts\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\} \max \{|\phi _{\zeta }(s) \phi _{\eta }(s)|^2 ; \delta s^a\} } \mathrm{d}t \mathrm{d}s \\&\quad - \frac{1}{2} \int _{t_1}^{\infty }\int _{t_1}^{\infty } \frac{\mathfrak {R}\left\{ \phi _{X_1'}(t+s)\phi _{Y_1'}(t+s)\phi _{\zeta }(-t) \phi _{\eta }(-t) \phi _{\zeta }(-s) \phi _{\eta }(-s) \mathrm {e}^{-\mathrm {i}(s+t)x}\right\} }{ ts\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\} \max \{|\phi _{\zeta }(s) \phi _{\eta }(s)|^2 ; \delta s^a\} } \mathrm{d}t \mathrm{d}s. \end{aligned}$$

This implies

$$\begin{aligned} |I_{t_1}^{\infty }|&\le \frac{1}{2} \int _{t_1}^{\infty }\int _{t_1}^{\infty } \frac{|\phi _{X_1'}(t-s)| |\phi _{Y_1'}(t-s)| |\phi _{\zeta }(t)| |\phi _{\eta }(t)| |\phi _{\zeta }(s)| |\phi _{\eta }(s)|}{ ts\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\} \max \{|\phi _{\zeta }(s) \phi _{\eta }(s)|^2 ; \delta s^a\} } \mathrm{d}t \mathrm{d}s \\&\quad + \frac{1}{2} \int _{t_1}^{\infty }\int _{t_1}^{\infty } \frac{|\phi _{X_1'}(t+s)| |\phi _{Y_1'}(t+s)| |\phi _{\zeta }(t)| |\phi _{\eta }(t)| |\phi _{\zeta }(s)| |\phi _{\eta }(s)|}{ ts\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\} \max \{|\phi _{\zeta }(s) \phi _{\eta }(s)|^2 ; \delta s^a\} } \mathrm{d}t \mathrm{d}s. \end{aligned}$$

Using the Cauchy-Schwarz inequality yields

$$\begin{aligned} |I_{t_1}^{\infty }|&\le \frac{1}{2} \int _{t_1}^{\infty }\int _{t_1}^{\infty } \frac{|\phi _{X_1'}(t-s)| |\phi _{Y_1'}(t-s)| |\phi _{\zeta }(t)|^2 |\phi _{\eta }(t)|^2}{ t^2 (\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\})^2 } \mathrm{d}t \mathrm{d}s \\&\quad + \frac{1}{2} \int _{t_1}^{\infty }\int _{t_1}^{\infty } \frac{|\phi _{X_1'}(t+s)| |\phi _{Y_1'}(t+s)| |\phi _{\zeta }(t)|^2 |\phi _{\eta }(t)|^2}{ t^2 (\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\})^2 } \mathrm{d}t \mathrm{d}s. \end{aligned}$$

Since

$$\begin{aligned}&\int _{t_1}^{\infty } |\phi _{X_1'}(t-s)| |\phi _{Y_1'}(t-s)| \mathrm{d}s \le \int _{-\infty }^{\infty } |\phi _{\zeta }(t)| |\phi _{\eta }(t)| \mathrm{d}t \le \min \{C_{\zeta }; C_{\eta }\}<\infty ,\\&\int _{t_1}^{\infty } |\phi _{X_1'}(t+s)| |\phi _{Y_1'}(t+s)| \mathrm{d}s \le \int _{-\infty }^{\infty } |\phi _{\zeta }(t)| |\phi _{\eta }(t)| \mathrm{d}t \le \min \{C_{\zeta }; C_{\eta }\} <\infty ,\end{aligned}$$

we deduce

$$\begin{aligned} |I_{t_1}^{\infty }| \le \min \{C_{\zeta }; C_{\eta }\} V_{1;\delta ,a}, \end{aligned}$$

(5.6)

where $V_{1;\delta ,a}$ is as in Proposition 3.1. From (5.4), (5.5) and (5.6), we obtain that

$$\begin{aligned}\mathbb {E}(U_{1,1;\delta ,a}(x))^2 \le \frac{2}{\pi ^2}\left( 3+ \frac{\ln 2}{\tau } \right) ^2 + \frac{2\min \{C_{\zeta }; C_{\eta }\}}{\pi ^2} V_{1;\delta ,a}.\end{aligned}$$

Estimate $|\mathbb {E}(U_{1,1;\delta ,a}(x) U_{1,2;\delta ,a}(x))|$: We have

$$\begin{aligned}&|\mathbb {E}(U_{1,1;\delta ,a}(x) U_{1,2;\delta ,a}(x)) |\\&\quad =\left| \frac{1}{\pi ^2} \mathbb {E}\int _{0}^{\infty }\int _{0}^{\infty } \frac{ \mathfrak {I}\left\{ \phi _{\zeta }(-t) \phi _{\eta }(-t)\mathrm {e}^{\mathrm {i}t(X'_1 + Y'_1 - x)}\right\} \mathfrak {I}\left\{ \phi _{\zeta }(-s) \phi _{\eta }(-s)\mathrm {e}^{\mathrm {i}s(X'_1 + Y'_2 - x)}\right\} }{ts\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\} \max \{|\phi _{\zeta }(s) \phi _{\eta }(s)|^2 ; \delta s^a\}} \mathrm{d}t \mathrm{d}s \right| \\&\quad \le \frac{1}{\pi ^2}(|I_{0,0}^{t_1, t_1}| + |I_{t_1,t_1}^{\infty , \infty }| + |I_{0,t_1}^{t_1, \infty }| + |I_{t_1,0}^{\infty , t_1}|), \end{aligned}$$

where

$$\begin{aligned}I_{u,w}^{v,z}:= \mathbb {E}\int _{u}^{v}\int _{w}^{z} \frac{ \mathfrak {I}\left\{ \phi _{\zeta }(-t) \phi _{\eta }(-t)\mathrm {e}^{\mathrm {i}t(X'_1 + Y'_1 - x)}\right\} \mathfrak {I}\left\{ \phi _{\zeta }(-s) \phi _{\eta }(-s)\mathrm {e}^{\mathrm {i}s(X'_1 + Y'_2 - x)}\right\} }{ts\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\} \max \{|\phi _{\zeta }(s) \phi _{\eta }(s)|^2 ; \delta s^a\}} \mathrm{d}t \mathrm{d}s .\end{aligned}$$

(iii) Estimate $|I_{0,0}^{t_1, t_1}|$: We have

$$\begin{aligned}&\mathfrak {I}\left\{ \phi _{\zeta }(-t) \phi _{\eta }(-t) \mathrm {e}^{\mathrm {i}t(X'_1 + Y'_1 - x)} \right\} \\&\quad = \int _{-\infty }^{\infty }\int _{-\infty }^{\infty } f_{\zeta }(u) f_{\eta }(v) \sin [t(X'_1 + Y'_1 - x -u -v)] \mathrm{d}u \mathrm{d}v, \\&\mathfrak {I}\left\{ \phi _{\zeta }(-s) \phi _{\eta }(-s) \mathrm {e}^{\mathrm {i}s(X'_1 + Y'_2 - x)} \right\} \\&\quad = \int _{-\infty }^{\infty }\int _{-\infty }^{\infty } f_{\zeta }(w) f_{\eta }(z) \sin [s(X'_1 + Y'_2 - x -w -z)] \mathrm{d}w \mathrm{d}z. \end{aligned}$$

By the equalities and the Fubini theorem, we deduce

$$\begin{aligned}&|I_{0,0}^{t_1, t_1}| \le \mathbb {E} \int _{-\infty }^{\infty }\int _{-\infty }^{\infty } \int _{-\infty }^{\infty }\int _{-\infty }^{\infty } f_{\zeta }(u) f_{\eta }(v) f_{\zeta }(w) f_{\eta }(z) \\&\qquad \times \left| \int _{0}^{t_1} \frac{\sin [t(X'_1 + Y'_1 - x -u -v)]}{t\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}} \mathrm{d}t \right| \left| \int _{0}^{t_1} \frac{\sin [s(X'_1 + Y'_2 - x -w -z)]}{s\max \{|\phi _{\zeta }(s) \phi _{\eta }(s)|^2 ; \delta s^a\}} \mathrm{d}s \right| \\&\qquad \mathrm{d}u \mathrm{d}v \mathrm{d}w \mathrm{d}z. \end{aligned}$$

Using Lemma 5.1 yields

$$\begin{aligned}&|I_{0,0}^{t_1, t_1}| \le \int _{-\infty }^{\infty }\int _{-\infty }^{\infty } \int _{-\infty }^{\infty }\int _{-\infty }^{\infty } f_{\zeta }(u) f_{\eta }(v) f_{\zeta }(w) f_{\eta }(z) \left( 3+\frac{\ln 2}{\tau } \right) ^2 \\&\mathrm{d}u \mathrm{d}v \mathrm{d}w \mathrm{d}z = \left( 3+\frac{\ln 2}{\tau } \right) ^2.\end{aligned}$$

(iv) Estimate $|I_{t_1,t_1}^{\infty , \infty }|$: We have

$$\begin{aligned}&I_{t_1,t_1}^{\infty , \infty } = \mathbb {E}\int _{t_1}^{\infty }\int _{t_1}^{\infty } \frac{1}{t}\mathfrak {I}\left\{ \frac{\phi _{\zeta }(-t) \phi _{\eta }(-t)\mathrm {e}^{\mathrm {i}t(X'_1 + Y'_1 - x)}}{\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}}\right\} \\&\quad \frac{1}{s}\mathfrak {I}\left\{ \frac{\phi _{\zeta }(-s) \phi _{\eta }(-s)\mathrm {e}^{\mathrm {i}s(X'_1 + Y'_2 - x)}}{\max \{|\phi _{\zeta }(s) \phi _{\eta }(s)|^2 ; \delta s^a\}}\right\} \mathrm{d}t \mathrm{d}s \\&\quad = \int _{-\infty }^{\infty } \int _{-\infty }^{\infty } \int _{-\infty }^{\infty } \biggl (\int _{t_1}^{\infty }\int _{t_1}^{\infty } \frac{1}{t}\mathfrak {I}\left\{ \frac{\phi _{\zeta }(-t) \phi _{\eta }(-t)\mathrm {e}^{\mathrm {i}t(u + v - x)}}{\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}}\right\} \\&\qquad \times \frac{1}{s}\mathfrak {I}\left\{ \frac{\phi _{\zeta }(-s) \phi _{\eta }(-s)\mathrm {e}^{\mathrm {i}s(u + w - x)}}{\max \{|\phi _{\zeta }(s) \phi _{\eta }(s)|^2 ; \delta s^a\}}\right\} \mathrm{d}t \mathrm{d}s\biggr ) f_{X'_1}(u) f_{Y'_1}(v) f_{Y'_2}(w) \mathrm{d}u \mathrm{d}v \mathrm{d}w \\&\quad = \int _{-\infty }^{\infty } \int _{-\infty }^{\infty } \int _{-\infty }^{\infty } \left( \int _{t_1}^{\infty }\int _{t_1}^{\infty } \frac{B(u,v,w,t,s)}{ts\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\} \max \{|\phi _{\zeta }(s) \phi _{\eta }(s)|^2 ; \delta t^a\}} \mathrm{d}t \mathrm{d}s \right) \\&f_{X'_1}(u) f_{Y'_1}(v) f_{Y'_2}(w) \mathrm{d}u \mathrm{d}v \mathrm{d}w\\&\quad = \int _{t_1}^{\infty }\int _{t_1}^{\infty } \frac{ \int _{-\infty }^{\infty } \int _{-\infty }^{\infty } \int _{-\infty }^{\infty } B(u,v,w,t,s) f_{X'_1}(u) f_{Y'_1}(v) f_{Y'_2}(w) \mathrm{d}u \mathrm{d}v \mathrm{d}w }{ts\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\} \max \{|\phi _{\zeta }(s) \phi _{\eta }(s)|^2 ; \delta t^a\}} \mathrm{d}t \mathrm{d}s \end{aligned}$$

with

$$\begin{aligned}B(u,v,w,t,s) := \mathfrak {I}\left\{ \phi _{\zeta }(-t) \phi _{\eta }(-t) \mathrm {e}^{\mathrm {i}t(u+v - x)}\right\} \mathfrak {I}\left\{ \phi _{\zeta }(-s) \phi _{\eta }(-s) \mathrm {e}^{\mathrm {i}s(u+w - x)}\right\} .\end{aligned}$$

Using the formula $\sin (u)\sin (v) = [\cos (u-v) - \cos (u+v)]/2$ and the Fubini theorem, we obtain

$$\begin{aligned}&\int _{-\infty }^{\infty } \int _{-\infty }^{\infty } \int _{-\infty }^{\infty } B(u,v,w,t,s) f_{X'_1}(u) f_{Y'_1}(v) f_{Y'_2}(w) \mathrm{d}u \mathrm{d}v \mathrm{d}w\\&= \frac{1}{2}\mathfrak {R}\left\{ \phi _{X_1'}(t-s)\phi _{Y_1'}(t) \phi _{Y_2'}(-s)\phi _{\zeta }(-t) \phi _{\eta }(-t) \phi _{\zeta }(s) \phi _{\eta }(s) \mathrm {e}^{\mathrm {i}(s-t)x}\right\} \\&\quad - \frac{1}{2}\mathfrak {R}\left\{ \phi _{X_1'}(t+s)\phi _{Y_1'}(t) \phi _{Y_2'}(s)\phi _{\zeta }(-t) \phi _{\eta }(-t) \phi _{\zeta }(-s) \phi _{\eta }(-s) \mathrm {e}^{-\mathrm {i}(s+t)x}\right\} , \end{aligned}$$

so

$$\begin{aligned} I_{t_1,t_1}^{\infty , \infty }&= \frac{1}{2} \int _{t_1}^{\infty }\int _{t_1}^{\infty } \frac{\mathfrak {R}\left\{ \phi _{X_1'}(t-s)\phi _{Y_1'}(t) \phi _{Y_2'}(-s)\phi _{\zeta }(-t) \phi _{\eta }(-t) \phi _{\zeta }(s) \phi _{\eta }(s) \mathrm {e}^{\mathrm {i}(s-t)x}\right\} }{ ts\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\} \max \{|\phi _{\zeta }(s) \phi _{\eta }(s)|^2 ; \delta t^a\} } \mathrm{d}t \mathrm{d}s \\&\quad - \frac{1}{2} \int _{t_1}^{\infty }\int _{t_1}^{\infty } \frac{\mathfrak {R}\left\{ \phi _{X_1'}(t+s)\phi _{Y_1'}(t) \phi _{Y_2'}(s)\phi _{\zeta }(-t) \phi _{\eta }(-t) \phi _{\zeta }(-s) \phi _{\eta }(-s) \mathrm {e}^{-\mathrm {i}(s+t)x}\right\} }{ ts\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\} \max \{|\phi _{\zeta }(s) \phi _{\eta }(s)|^2 ; \delta t^a\} } \mathrm{d}t \mathrm{d}s. \end{aligned}$$

This leads to

$$\begin{aligned} |I_{t_1,t_1}^{\infty , \infty }|&\le \frac{1}{2} \int _{t_1}^{\infty }\int _{t_1}^{\infty } \frac{|\phi _{X_1'}(t-s)| |\phi _{Y_1'}(t)| |\phi _{Y_2'}(s)| |\phi _{\zeta }(t)| |\phi _{\eta }(t)| |\phi _{\zeta }(s)| |\phi _{\eta }(s)|}{ ts\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\} \max \{|\phi _{\zeta }(s) \phi _{\eta }(s)|^2 ; \delta t^a\} } \mathrm{d}t \mathrm{d}s \\&\quad + \frac{1}{2} \int _{t_1}^{\infty }\int _{t_1}^{\infty } \frac{|\phi _{X_1'}(t+s)| |\phi _{Y_1'}(t)| |\phi _{Y_2'}(s)| |\phi _{\zeta }(t)| |\phi _{\eta }(t)| |\phi _{\zeta }(s)| |\phi _{\eta }(s)|}{ ts\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\} \max \{|\phi _{\zeta }(s) \phi _{\eta }(s)|^2 ; \delta t^a\} } \mathrm{d}t \mathrm{d}s. \end{aligned}$$

Using the Cauchy-Schwarz inequality yields

$$\begin{aligned} |I_{t_1,t_1}^{\infty , \infty }|&\le \frac{1}{2} \int _{t_1}^{\infty }\int _{t_1}^{\infty } \frac{|\phi _{X_1'}(t-s)| |\phi _{Y_1'}(t)|^2 |\phi _{\zeta }(t)|^2 |\phi _{\eta }(t)|^2}{ t^2 (\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\})^2 } \mathrm{d}t \mathrm{d}s \\&\quad + \frac{1}{2} \int _{t_1}^{\infty }\int _{t_1}^{\infty } \frac{|\phi _{X_1'}(t+s)| |\phi _{Y_1'}(t)|^2 |\phi _{\zeta }(t)|^2 |\phi _{\eta }(t)|^2}{ t^2 (\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\})^2 } \mathrm{d}t \mathrm{d}s. \end{aligned}$$

Since

$$\begin{aligned}&\int _{t_1}^{\infty } |\phi _{X_1'}(t-s)| \mathrm{d}s \le \int _{-\infty }^{\infty } |\phi _{\zeta }(t)| \mathrm{d}t = C_{\zeta },\\&\int _{t_1}^{\infty } |\phi _{X_1'}(t+s)| \mathrm{d}s \le \int _{-\infty }^{\infty } |\phi _{\zeta }(t)| \mathrm{d}t = C_{\zeta },\end{aligned}$$

we deduce

$$\begin{aligned}|I_{t_1,t_1}^{\infty , \infty }| \le C_{\zeta } V_{2;\delta ,a},\end{aligned}$$

where $V_{2;\delta ,a}$ is as in Proposition 3.1.

(v) Estimate $|I_{0,t_1}^{t_1,\infty }|$: We have

$$\begin{aligned}&I_{0,t_1}^{t_1,\infty } = \mathbb {E}\int _{0}^{t_1}\int _{t_1}^{\infty } \frac{1}{t}\mathfrak {I}\left\{ \frac{\phi _{\zeta }(-t) \phi _{\eta }(-t)\mathrm {e}^{\mathrm {i}t(X'_1 + Y'_1 - x)}}{\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}}\right\} \\&\quad \frac{1}{s}\mathfrak {I}\left\{ \frac{\phi _{\zeta }(-s) \phi _{\eta }(-s)\mathrm {e}^{\mathrm {i}s(X'_1 + Y'_2 - x)}}{\max \{|\phi _{\zeta }(s) \phi _{\eta }(s)|^2 ; \delta s^a\}}\right\} \mathrm{d}t \mathrm{d}s \\&\quad = \int _{-\infty }^{\infty } \int _{-\infty }^{\infty } \int _{-\infty }^{\infty } \biggl (\int _{0}^{t_1}\int _{t_1}^{\infty } \frac{1}{t}\mathfrak {I}\left\{ \frac{\phi _{\zeta }(-t) \phi _{\eta }(-t)\mathrm {e}^{\mathrm {i}t(u + v - x)}}{\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}}\right\} \\&\qquad \times \frac{1}{s}\mathfrak {I}\left\{ \frac{\phi _{\zeta }(-s) \phi _{\eta }(-s)\mathrm {e}^{\mathrm {i}s(u + w - x)}}{\max \{|\phi _{\zeta }(s) \phi _{\eta }(s)|^2 ; \delta s^a\}}\right\} \mathrm{d}t \mathrm{d}s\biggr ) f_{X'_1}(u) f_{Y'_1}(v) f_{Y'_2}(w) \mathrm{d}u \mathrm{d}v \mathrm{d}w \\&\quad = \int _{-\infty }^{\infty } \int _{-\infty }^{\infty } \biggl (\int _{0}^{t_1} \frac{1}{t}\mathfrak {I}\left\{ \frac{\phi _{\zeta }(-t) \phi _{\eta }(-t)\mathrm {e}^{\mathrm {i}t(u + v - x)}}{\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}}\right\} \mathrm{d}t \\&\qquad \times \int _{t_1}^{\infty }\frac{1}{s}\mathfrak {I}\left\{ \frac{\phi _{\zeta }(-s) \phi _{\eta }(-s)\mathrm {e}^{\mathrm {i}s(u - x)}\phi _{Y'}(s)}{\max \{|\phi _{\zeta }(s) \phi _{\eta }(s)|^2 ; \delta s^a\}}\right\} \mathrm{d}s\biggr ) f_{X'_1}(u) f_{Y'_1}(v) \mathrm{d}u \mathrm{d}v. \end{aligned}$$

Therefore,

$$\begin{aligned} |I_{0,t_1}^{t_1,\infty }|&\le \int _{-\infty }^{\infty } \int _{-\infty }^{\infty } \left| \int _{0}^{t_1} \frac{1}{t}\mathfrak {I}\left\{ \frac{\phi _{\zeta }(-t) \phi _{\eta }(-t)\mathrm {e}^{\mathrm {i}t(u + v - x)}}{\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}}\right\} \mathrm{d}t \right| f_{X'_1}(u) f_{Y'_1}(v) \mathrm{d}u \mathrm{d}v \\&\times \int _{t_1}^{\infty } \frac{|\phi _{\zeta }(s)| |\phi _{\eta }(s)|^2|\phi _{Y}(s)|}{s\max \{|\phi _{\zeta }(s) \phi _{\eta }(s)|^2 ; \delta s^a\}} \mathrm{d}s. \end{aligned}$$

Because

$$\begin{aligned}&\left| \int _{0}^{t_1} \frac{1}{t}\mathfrak {I}\left\{ \frac{\phi _{\zeta }(-t) \phi _{\eta }(-t)\mathrm {e}^{\mathrm {i}t(u + v - x)}}{\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}}\right\} \mathrm{d}t \right| \\&\quad \le \int _{-\infty }^{\infty }\int _{-\infty }^{\infty } f_{\zeta }(p) f_{\eta }(q)\left| \int _{0}^{t_1} \frac{\sin [t(u+v-x-p-q)]}{t\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}} \mathrm{d}t \right| \mathrm{d}p \mathrm{d}q \le 3+\frac{\ln 2}{\tau } , \end{aligned}$$

we infer that

$$\begin{aligned}|I_{0,t_1}^{t_1,\infty }| \le \left( 3+\frac{\ln 2}{\tau } \right) V_{3;\delta ,a},\end{aligned}$$

where $V_{3;\delta ,a}$ is as in Proposition 3.1.

(vi) Estimate $|I_{t_1,0}^{\infty ,t_1}|$: The same steps as in the estimate of $|I_{0,t_1}^{t_1,\infty }|$ may be repeated to obtain

$$\begin{aligned}|I_{t_1,0}^{\infty ,t_1}| \le \left( 3+\frac{\ln 2}{\tau } \right) V_{3;\delta ,a}.\end{aligned}$$

From the estimates of $|I_{0,0}^{t_1,t_1}|$, $|I_{t_1,t_1}^{\infty ,\infty }|$, $|I_{0,t_1}^{t_1,\infty }|$ and $|I_{t_1,0}^{\infty ,t_1}|$, we derive

$$\begin{aligned}|\mathbb {E}(U_{1,1;\delta ,a}(x) U_{1,2;\delta ,a}(x)) | \le \frac{1}{\pi ^2}\left( 3+\frac{\ln 2}{\tau } \right) ^2 + \frac{C_{\zeta }}{\pi ^2} V_{2;\delta ,a} + \frac{2}{\pi ^2} \left( 3+\frac{\ln 2}{\tau } \right) V_{3;\delta ,a}.\end{aligned}$$

Estimate $|\mathbb {E}(U_{1,1;\delta ,a}(x) U_{2,1;\delta ,a}(x))|$: The estimate of $|\mathbb {E}(U_{1,1;\delta ,a}(x) U_{2,1;\delta ,a}(x))|$ is similar to the estimate of $|\mathbb {E}(U_{1,1;\delta ,a}(x) U_{1,2;\delta ,a}(x))|$. $\square$

5.1 Proof of Proposition 3.1

By the Fubini theorem, we have

$$\begin{aligned} \mathbb {E}\widehat{F}_{X+Y; \delta ,a}(x)&= \frac{1}{2} - \frac{1}{nm} \sum _{j=1}^{n}\sum _{k=1}^{m} \mathbb {E}U_{j,k;\delta ,a}(x) = \frac{1}{2} - \mathbb {E}U_{1,1;\delta ,a}(x) \\&= \frac{1}{2} - \frac{1}{\pi }\int _{0}^{\infty } \frac{1}{t}\mathfrak {I}\left\{ \frac{|\phi _{\zeta }(t)|^2 |\phi _{\eta }(t)|^2 \phi _X(t) \phi _{Y}(t)\mathrm {e}^{-\mathrm {i}tx}}{\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}}\right\} \mathrm{d}t. \end{aligned}$$

Combining this equality with (2.1) gives

$$\begin{aligned}&\sup _{x\in \mathbb {R}}|\mathbb {E}\widehat{F}_{X+Y; \delta ,a}(x) - F_{X+Y}(x)| \\&\quad = \sup _{x\in \mathbb {R}}\left| \frac{1}{\pi }\int _{0}^{\infty } \frac{1}{t}\mathfrak {I}\left\{ \left( 1- \frac{|\phi _{\zeta }(t)|^2 |\phi _{\eta }(t)|^2}{\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}}\right) \phi _X(t) \phi _{Y}(t) \mathrm {e}^{-\mathrm {i}tx}\right\} \mathrm{d}t\right| \\&\quad \le \frac{1}{\pi }\int _{0}^{\infty } \left| 1- \frac{|\phi _{\zeta }(t)|^2 |\phi _{\eta }(t)|^2}{\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}} \right| \frac{|\phi _X(t)| |\phi _{Y}(t)|}{t} \mathrm{d}t. \end{aligned}$$

So (3.1) is verified.

Next we bound $\sup _{x\in \mathbb {R}}\mathrm {Var}\widehat{F}_{X+Y; \delta ,a}(x)$. Indeed, we have

$$\begin{aligned} \mathrm {Var}\widehat{F}_{X+Y; \delta ,a}(x)&= \frac{1}{(nm)^2} \mathrm {Cov}\left( \sum _{j=1}^{n}\sum _{k=1}^{m} U_{j,k;\delta ,a}(x), \sum _{p=1}^{n}\sum _{q=1}^{m} U_{p,q;\delta ,a}(x)\right) \\&= \frac{1}{(nm)^2} \sum _{j=1}^{n}\sum _{k=1}^{m} \sum _{p=1}^{n}\sum _{q=1}^{m} \mathrm {Cov}(U_{j,k;\delta ,a}(x), U_{p,q;\delta ,a}(x)) \\&= \frac{1}{(nm)^2} (A_1 + A_2 + A_3 + A_4), \end{aligned}$$

with

$$\begin{aligned}&A_1 := \sum _{j,k,p,q; j = p, k = q} \mathrm {Cov}(U_{j,k;\delta ,a}(x), U_{p,q;\delta ,a}(x)), \\&A_2 := \sum _{j,k,p,q; j = p, k \ne q} \mathrm {Cov}(U_{j,k;\delta ,a}(x), U_{p,q;\delta ,a}(x)), \\&A_3 := \sum _{j,k,p,q; j \ne p, k = q} \mathrm {Cov}(U_{j,k;\delta ,a}(x), U_{p,q;\delta ,a}(x)), \\&A_4 := \sum _{j,k,p,q; j \ne p, k \ne q} \mathrm {Cov}(U_{j,k;\delta ,a}(x), U_{p,q;\delta ,a}(x)). \end{aligned}$$

Here $\mathrm {Cov}$ denotes the covariance. We see that

$$\begin{aligned}&A_1 = nm \mathrm {Cov}(U_{1,1;\delta ,a}(x), U_{1,1;\delta ,a}(x)) = nm \mathrm {Var}U_{1,1;\delta ,a}(x), \\&A_2 = nm(m-1) \mathrm {Cov}(U_{1,1;\delta ,a}(x), U_{1,2;\delta ,a}(x)),\\&A_3 = nm(n-1) \mathrm {Cov}(U_{1,1;\delta ,a}(x), U_{2,1;\delta ,a}(x)),\\&A_4 = 0. \end{aligned}$$

Therefore,

$$\begin{aligned} \mathrm {Var}\widehat{F}_{X+Y; \delta ,a}(x)&= \frac{\mathrm {Var}U_{1,1;\delta ,a}(x) + (m-1) \mathrm {Cov}(U_{1,1;\delta ,a}(x), U_{1,2;\delta ,a}(x))}{nm} \\&\quad + \frac{(n-1) \mathrm {Cov}(U_{1,1;\delta ,a}(x), U_{2,1;\delta ,a}(x))}{nm} \\&\le \frac{1}{nm}\mathbb {E}(U_{1,1;\delta ,a}(x))^2 + \frac{1}{n}|\mathbb {E}(U_{1,1;\delta ,a}(x) U_{1,2;\delta ,a}(x))| \\&\quad + \frac{1}{m}|\mathbb {E}(U_{1,1;\delta ,a}(x) U_{2,1;\delta ,a}(x))|. \end{aligned}$$

Combining the latter estimate with Lemma 5.2, we obtain

$$\begin{aligned} \sup _{x\in \mathbb {R}}\mathrm {Var}\widehat{F}_{X+Y; \delta ,a}(x)&\le \frac{1}{nm}\left[ \frac{2}{\pi ^2}\left( 3+\frac{\ln 2}{\tau } \right) ^2 + \frac{2\min \{C_{\zeta }; C_{\eta }\}}{\pi ^2} V_{1;\delta ,a} \right] \\&\quad + \frac{1}{n} \bigg [\frac{1}{\pi ^2}\left( 3+\frac{\ln 2}{\tau } \right) ^2 + \frac{C_{\zeta }}{\pi ^2} V_{2;\delta ,a} + \frac{2}{\pi ^2} \left( 3+\frac{\ln 2}{\tau } \right) V_{3;\delta ,a} \biggr ] \\&\quad + \frac{1}{m} \bigg [\frac{1}{\pi ^2}\left( 3+\frac{\ln 2}{\tau } \right) ^2 + \frac{C_{\eta }}{\pi ^2} V_{4;\delta ,a} + \frac{2}{\pi ^2} \left( 3+\frac{\ln 2}{\tau } \right) V_{5;\delta ,a} \biggr ]. \end{aligned}$$

From this bound, we find a constant $C>0$ depending on $C_{\zeta }$, $C_{\eta }$, $\tau$ so that the estimate (3.2) is satisfied. The proposition is proved. $\square$

5.2 Proof of Theorem 3.2

First, we verify the estimate (3.3). Using the standard bias-variance decomposition and the inequality $\sqrt{u+v} \le \sqrt{u} + \sqrt{v}$ for $u,v\ge 0$, we obtain

$$\begin{aligned} \sup _{x\in \mathbb {R}}\sqrt{\mathbb {E}|\widehat{F}_{X+Y; \delta ,a}(x) - F_{X+Y}(x)|^2}&\le \sup _{x\in \mathbb {R}}|\mathbb {E}\widehat{F}_{X+Y; \delta ,a}(x) - F_{X+Y}(x)| \nonumber \\&+ \sup _{x\in \mathbb {R}} \sqrt{\mathrm {Var}\widehat{F}_{X+Y; \delta ,a}(x)}. \end{aligned}$$

(5.7)

We need to bound $\sup _{x\in \mathbb {R}}\sqrt{\mathrm {Var}\widehat{F}_{X+Y; \delta ,a}(x)}$. Indeed, we have

$$\begin{aligned}V_{1;\delta ,a} \le \frac{1}{\delta } \int _{t_1}^{\infty } t^{-(a+2)} \mathrm{d}t = \frac{t_1^{-(a+1)}}{(a+1)\delta } \le \frac{t_1^{-(a_{*}+1)}}{\delta }.\end{aligned}$$

Similarly, we also derive that $V_{2;\delta ,a}, V_{4;\delta ,a} \le t_1^{-(a_{*}+1)}/ \delta$ and $V_{3;\delta ,a}, V_{5;\delta ,a} \le t_1^{-a_{*}}/(a\delta )$. From the estimates of $V_{j;\delta ,a}$’s and the estimate (3.2), we deduce that

$$\begin{aligned} \sup _{x\in \mathbb {R}}\mathrm {Var}\widehat{F}_{X+Y; \delta ,a}(x)&\le C\left( \frac{1}{n} + \frac{1}{m} + \frac{t_1^{-(a_{*}+1)}}{nm\delta } + \frac{t_1^{-(a_{*}+1)}}{n\delta } + \frac{t_1^{-a_{*}}}{an\delta } + \frac{t_1^{-(a_{*}+1)}}{m\delta } + \frac{t_1^{-a_{*}}}{am\delta } \right) \\&\le C\left( \frac{1}{n} + \frac{1}{m} + \frac{3}{2} t_1^{-(a_{*}+1)}\left( \frac{1}{n} + \frac{1}{m}\right) \frac{1}{\delta } + \frac{t_1^{-a_{*}}}{a}\left( \frac{1}{n} + \frac{1}{m}\right) \frac{1}{\delta } \right) \\&\le \frac{2C}{\min \{1;a\}}\left( 1 + \frac{3}{2}t_1^{-(a_{*}+1)} + t_1^{-a_{*}}\right) \frac{1}{\delta \min \{n;m\}}. \end{aligned}$$

From the latter estimate, there exists a constant $\widetilde{C} \equiv \widetilde{C}(a_{*}, t_1)>0$ such that

$$\begin{aligned} \sup _{x\in \mathbb {R}}\sqrt{\mathrm {Var}\widehat{F}_{X+Y; \delta ,a}(x)} \le \frac{\widetilde{C}}{ \min \{1; \sqrt{a}\} \sqrt{\delta \min \{n;m\}}}. \end{aligned}$$

(5.8)

From (5.7), (3.1) and (5.8), we derive (3.3).

Now we prove the remaining statement of the present theorem. Using the inequality $0\le \max \{u;v\} - u \le v$ (for $u,v\ge 0$), we have

$$\begin{aligned} B_{\delta ,a}&\le \int _{0}^{\infty } \frac{\delta t^{a-1} |\phi _X(t)| |\phi _{Y}(t)|}{\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}} \mathrm{d}t \\&\le \int _{0}^{t_1} \frac{\delta t^{a-1} }{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2} \mathrm{d}t + \int _{t_1}^{\infty } \frac{\delta t^{a-1} |\phi _X(t)| |\phi _{Y}(t)|}{\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}} \mathrm{d}t. \end{aligned}$$

By (A.1), for any $t\in (0,t_1)$, we have $|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 \ge 1 - c_0 t_1^{\tau } \ge 1/2$, so

$$\begin{aligned}\int _{0}^{t_1} \frac{\delta t^{a-1}}{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2}\mathrm{d}t \le 2\delta \int _{0}^{t_1} t^{a-1} \mathrm{d}t = \frac{2t_1^{a} \delta }{a} \le \frac{2\delta }{a}.\end{aligned}$$

Moreover, it follows from (A.3) that $\lambda (Z(\phi _{\zeta }) \cup Z(\phi _{\eta })) =0$ and hence

$$\begin{aligned}&\int _{t_1}^{\infty } \frac{\delta t^{a-1}|\phi _X(t)| |\phi _{Y}(t)|}{\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}}\mathrm{d}t \\&\quad = \int _{(t_1, \infty ) \setminus [Z(\phi _{\zeta }) \cup Z(\phi _{\eta })]} \frac{\delta t^{a}}{\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}}\cdot \frac{|\phi _X(t)| |\phi _{Y}(t)|}{t}\mathrm{d}t .\end{aligned}$$

Thus,

$$\begin{aligned}B_{\delta ,a} \le \frac{2\delta }{a} + \int _{(t_1, \infty ) \setminus [Z(\phi _{\zeta }) \cup Z(\phi _{\eta })]} \frac{\delta t^{a}}{\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}}\cdot \frac{|\phi _X(t)| |\phi _{Y}(t)|}{t}\mathrm{d}t .\end{aligned}$$

From the latter estimate, the assumptions (A.4) and $\lim _{n,m\rightarrow \infty }\delta = 0$, we apply the Lebesgue dominated convergence theorem to give $\lim _{n,m\rightarrow \infty } B_{\delta ,a} =0$. Combining this with (3.3) and the assumption $\lim _{n,m\rightarrow \infty }\delta \min \{n;m\} = \infty$, the desired result follows. $\square$

Lemma 5.3

Suppose that $F_{X}\in \mathcal {S}(\alpha _X, L_X)$ and $F_{Y}\in \mathcal {S}(\alpha _Y, L_Y)$ with $\alpha _X, \alpha _Y>-1/2$ and $L_X, L_Y>0$. Then, for any $M>1$ we have

$$\begin{aligned}\int _{M}^{\infty }\frac{|\phi _{X}(t)| |\phi _{Y}(t)|}{t}\mathrm{d}t \le D M^{-(\alpha _X + \alpha _Y +1)},\end{aligned}$$

where

$$\begin{aligned}D := {\left\{ \begin{array}{ll} \frac{1}{2}\sqrt{L_X L_Y} &{} \text {for }\,\,\, \alpha _X + \alpha _Y \ge 0,\\ 2^{-(\alpha _X + \alpha _Y +2)/2}\sqrt{L_X L_Y} &{} \text {for }\,\,\, -1<\alpha _X + \alpha _Y <0. \end{array}\right. }\end{aligned}$$

Proof

For brevity, we denote $I_{M}:= \int _{M}^{\infty } t^{-1} |\phi _{X}(t)| |\phi _{Y}(t)|\mathrm{d}t$. We write

$$\begin{aligned}I_{M} = \int _{M}^{\infty }|\phi _{X}(t)|(1+t^2)^{\alpha _X /2} |\phi _{Y}(t)| (1+t^2)^{\alpha _Y /2} \frac{(1+t^2)^{-(\alpha _X + \alpha _Y)/2}}{t} \mathrm{d}t.\end{aligned}$$

If $\alpha _X + \alpha _Y \ge 0$, we have

$$\begin{aligned} I_{M}&\le M^{-(\alpha _X + \alpha _Y +1)} \int _{M}^{\infty }|\phi _{X}(t)|(1+t^2)^{\alpha _X /2} |\phi _{Y}(t)| (1+t^2)^{\alpha _Y /2} \mathrm{d}t \\&\le M^{-(\alpha _X + \alpha _Y +1)} \sqrt{\int _{M}^{\infty }|\phi _{X}(t)|^2(1+t^2)^{\alpha _X }\mathrm{d}t \int _{M}^{\infty } |\phi _{Y}(t)|^2 (1+t^2)^{\alpha _Y} \mathrm{d}t }\\&\le \frac{1}{2}\sqrt{L_X L_Y} \, M^{-(\alpha _X + \alpha _Y +1)}, \end{aligned}$$

where we have used the Cauchy-Schwarz inequality and the assumptions $F_{X}\in \mathcal {S}(\alpha _X, L_X)$, $F_{Y}\in \mathcal {S}(\alpha _Y, L_Y)$.

Likewise, if $-1< \alpha _X + \alpha _Y <0$, then

$$\begin{aligned} I_{M}&\le 2^{-(\alpha _X + \alpha _Y)/2} \int _{M}^{\infty }|\phi _{X}(t)|(1+t^2)^{\alpha _X /2} |\phi _{Y}(t)| (1+t^2)^{\alpha _Y /2} t^{-(\alpha _X + \alpha _Y +1)}\mathrm{d}t \\&\le 2^{-(\alpha _X + \alpha _Y)/2} M^{-(\alpha _X + \alpha _Y +1)} \int _{M}^{\infty }|\phi _{X}(t)|(1+t^2)^{\alpha _X /2} |\phi _{Y}(t)| (1+t^2)^{\alpha _Y /2}\mathrm{d}t \\&\le 2^{-(\alpha _X + \alpha _Y +2)/2}\sqrt{L_X L_Y} \, M^{-(\alpha _X + \alpha _Y +1)} . \end{aligned}$$

From the two cases, we derive the result of the lemma. $\square$

5.3 Proof of Theorem 3.3

Suppose $F_{X}\in \mathcal {S}(\alpha _X, L_X)$ and $F_{Y}\in \mathcal {S}(\alpha _Y, L_Y)$. Put $G_{\delta ,a}:= \{t> 0: |\phi _{\zeta }(t)|^2 |\phi _{\eta }(t)|^2 < \delta t^a\}$. We then have

$$\begin{aligned} B_{\delta ,a}= & {} \int _{G_{\delta ,a}} \left| 1- \frac{|\phi _{\zeta }(t)|^2 |\phi _{\eta }(t)|^2}{\max \{|\phi _{\zeta }(t) \phi _{\eta }(t)|^2 ; \delta t^a\}} \right| \frac{|\phi _X(t)| |\phi _{Y}(t)|}{t} \mathrm{d}t\\\le & {} \int _{G_{\delta ,a}} \frac{|\phi _X(t)| |\phi _{Y}(t)|}{t} \mathrm{d}t. \end{aligned}$$

Since the functions $\phi _{\zeta }$, $\phi _{\eta }$ are continuous on $\mathbb {R}$ and $\phi _{\zeta }(0) = \phi _{\eta }(0) = 1$, there exists a positive constant $t_{\zeta ,\eta }$ depending on $\phi _{\zeta }$, $\phi _{\eta }$ such that $|\phi _{\zeta }(t)| \ge 1/2$ and $|\phi _{\eta }(t)| \ge 1/2$, for all $|t| \le t_{\zeta ,\eta }$. Let $R> t_{\zeta ,\eta }$, $0<\rho <1/16$ and $0<\delta < R^{-a}\rho$. We see that

$$\begin{aligned} G_{\delta ,a}&= \{t \in (0, R]: |\phi _{\zeta }(t)|^2 |\phi _{\eta }(t)|^2< \delta t^a\} \cup \{t>R: |\phi _{\zeta }(t)|^2 |\phi _{\eta }(t)|^2 < \delta t^a\} \\&\subset \{t \in (0, R]: |\phi _{\zeta }(t)|^2 |\phi _{\eta }(t)|^2 \le \rho \} \cup (R,\infty ) \\&= \{t \in (t_{\zeta ,\eta }, R]: |\phi _{\zeta }(t)|^2 |\phi _{\eta }(t)|^2 \le \rho \} \cup (R,\infty ) \\&\subset L_{\phi _{\zeta }}(R, \rho ^{1/4}) \cup L_{\phi _{\eta }}(R, \rho ^{1/4}) \cup (R,\infty ), \end{aligned}$$

where $L_{\phi _{\zeta }}(R, \rho ^{1/4}):= \{t\in (t_{\zeta ,\eta }, R]: |\phi _{\zeta }(t)| \le \rho ^{1/4}\}$. Hence,

$$\begin{aligned}&B_{\delta ,a} \le \int _{L_{\phi _{\zeta }}(R, \rho ^{1/4})} \frac{|\phi _X(t)| |\phi _{Y}(t)|}{t} \mathrm{d}t + \int _{L_{\phi _{\eta }}(R, \rho ^{1/4})} \frac{|\phi _X(t)| |\phi _{Y}(t)|}{t} \mathrm{d}t \\&\quad + \int _{R}^{\infty } \frac{|\phi _X(t)| |\phi _{Y}(t)|}{t} \mathrm{d}t .\end{aligned}$$

It follows from the estimate $|\phi _X(t)||\phi _Y(t)|/t \le 1/t_{\zeta ,\eta }$ for all $t\in L_{\phi _{\zeta }}(R, \rho ^{1/4}) \cup L_{\phi _{\eta }}(R, \rho ^{1/4})$ and Lemma 5.3 that

$$\begin{aligned}B_{\delta ,a} \le \mathcal {O}\left( \lambda (L_{\phi _{\zeta }}(R, \rho ^{1/4})) + \lambda (L_{\phi _{\eta }}(R,\rho ^{1/4})) + R^{-(\alpha _X + \alpha _Y +1)} \right) .\end{aligned}$$

Under the assumption (A.5), by the same arguments as in the proof of Lemma 4 in Trong and Phuong [2], we derive

$$\begin{aligned}&\lambda (L_{\phi _{\zeta }}(R, \rho ^{1/4}) \le \mathcal {O}\left( R^{\nu _{\zeta }/\mu _{\zeta } +1} \mathrm {e}^{\kappa _{\zeta } R^{\vartheta _{\zeta }}/\mu _{\zeta }} \rho ^{1/(4\mu _{\zeta })}\right) ,\\&\lambda (L_{\phi _{\eta }}(R, \rho ^{1/4})) \le \mathcal {O}\left( R^{\nu _{\eta }/\mu _{\eta } +1} \mathrm {e}^{\kappa _{\eta } R^{\vartheta _{\eta }}/\mu _{\eta }} \rho ^{1/(4\mu _{\eta })}\right) .\end{aligned}$$

Therefore,

$$\begin{aligned}B_{\delta ,a} \le \mathcal {O}\left( R^{-(\alpha _X + \alpha _Y +1)} + R^{\nu ^{*}/\mu _{*} +1} \mathrm {e}^{\kappa ^{*} R^{\vartheta ^{*}/\mu _{*}}} \rho ^{1/(4\mu ^{*})} \right) ,\end{aligned}$$

where $\nu ^{*} := \max \{\nu _{\zeta }; \nu _{\eta }\}$, $\mu _{*} := \min \{\mu _{\zeta }; \mu _{\eta }\}$, $\kappa ^{*} := \max \{\kappa _{\zeta }; \kappa _{\eta }\}$, $\vartheta ^{*} := \max \{\vartheta _{\zeta }; \vartheta _{\eta }\}$ and $\mu ^{*} := \max \{\mu _{\zeta }; \mu _{\eta }\}$.

From the latter estimate of $B_{\delta ,a}$ and the estimate (3.3), we conclude for $R> t_{\zeta ,\eta }$, $0<\rho <1/16$, $0<\delta < R^{-a}\rho$ and $a\in (0, a_{*}]$ that

$$\begin{aligned}&\mathcal {R}[\widehat{F}_{X+Y; \delta ,a}; \mathcal {S}(\alpha _X, L_X), \mathcal {S}(\alpha _Y, L_Y)] \\&\quad \le \mathcal {O}\left( \frac{1}{\min \{1; \sqrt{a}\}} \left( R^{-(\alpha _X + \alpha _Y +1)} + R^{\nu ^{*}/\mu _{*} +1} \mathrm {e}^{\kappa ^{*} R^{\vartheta ^{*}/\mu _{*}}} \rho ^{1/(4\mu ^{*})} + \frac{1}{\sqrt{\delta \min \{n;m\}}} \right) \right) . \end{aligned}$$

Replacing $\delta = (\min \{n;m\})^{-\varrho }, \quad \rho = (\min \{n;m\})^{-\varrho _1}, \quad R = (\varrho _2\ln (\min \{n;m\}))^{1/\vartheta ^{*}}$ with $0<\varrho <1$, $0<\varrho _1 < \varrho$, $0< \varrho _2 < \varrho _1 \mu _{*}/ (4\mu ^{*} \kappa ^{*})$ to the latter estimate, the desired result follows. $\square$

5.4 Proof of Theorem 3.4

(a) When $\phi _{\zeta }$, $\phi _{\eta }$ satisfy (A.6), they also satisfy (A.5) with $c_{1,\zeta } \equiv k_{1,\zeta }$, $c_{1,\eta } \equiv k_{1,\eta }$, $\kappa _{\zeta } \equiv d_{\zeta }$, $\kappa _{\eta } \equiv d_{\eta }$, $\vartheta _{\zeta } \equiv \gamma _{\zeta }$, $\vartheta _{\eta } \equiv \gamma _{\eta }$ and with arbitrary positive numbers $a_{\zeta }$, $a_{\eta }$, $\mu _{\zeta }$, $\mu _{\eta }$, $\nu _{\zeta }$ and $\nu _{\eta }$. Hence, applying Theorem 3.3 yields

$$\begin{aligned}&\mathcal {R}[\widehat{F}_{X+Y; \delta ,a}; \mathcal {S}(\alpha _X, L_X), \mathcal {S}(\alpha _Y, L_Y)] \\&\quad \le \mathcal {O}\left( \frac{1}{\min \{1; \sqrt{a}\}} \left( \ln (\min \{n;m\})\right) ^{-(\alpha _X + \alpha _Y +1)/\max \{\gamma _{\zeta }; \gamma _{\eta }\}} \right) \end{aligned}$$

under the selection $\delta = (\min \{n;m\})^{-\varrho }$, with $0<\varrho <1$.

(b) Suppose $F_{X}\in \mathcal {S}(\alpha _X, L_X)$ and $F_{Y}\in \mathcal {S}(\alpha _Y, L_Y)$. As in the proof of Theorem 3.3, we have $B_{\delta ,a} \le \int _{G_{\delta ,a}} t^{-1}|\phi _X(t)| |\phi _{Y}(t)| \mathrm{d}t$, in which $G_{\delta ,a} := \{t>0: |\phi _{\zeta }(t) \phi _{\eta }(t)|^2 < \delta t^a\}$.

Let $0<\delta < \ell _{1,\zeta }^{2} \ell _{1,\eta }^{2}/ 4^{\beta _{\zeta } + \beta _{\eta }}$. For any $t\in G_{\delta ,a}$, we have

$$\begin{aligned}\ell _{1,\zeta }^{2} \ell _{1,\eta }^{2}/\delta< t^a (1+t)^{2\beta _{\zeta } + 2\beta _{\eta }} < 4^{\beta _{\zeta } + \beta _{\eta }} t^{a + 2\beta _{\zeta } + 2\beta _{\eta }},\end{aligned}$$

so $t> [\ell _{1,\zeta }^{2} \ell _{1,\eta }^{2}/(4^{\beta _{\zeta } + \beta _{\eta }} \delta )]^{1/(a + 2\beta _{\zeta } + 2\beta _{\eta })} =: T$. Hence, $G_{\delta ,a} \subset (T, \infty )$. This yields

$$\begin{aligned} B_{\delta ,a} \le \int _{T}^{\infty } \frac{|\phi _X(t)| |\phi _{Y}(t)|}{t} \mathrm{d}t \le \mathcal {O}\left( T^{-(\alpha _X + \alpha _Y + 1)}\right) = \mathcal {O}\left( \delta ^{(\alpha _X + \alpha _Y + 1)/(a + 2\beta _{\zeta } + 2\beta _{\eta })}\right) , \end{aligned}$$

where we have also applied Lemma 5.3.

From the latter estimate of $B_{\delta ,a}$ and the estimate (3.3), we conclude for $0<\delta < \ell _{1,\zeta }^{2} \ell _{1,\eta }^{2}/ 4^{\beta _{\zeta } + \beta _{\eta }}$ and $a\in (0, a_{*}]$ that

$$\begin{aligned}&\mathcal {R}[\widehat{F}_{X+Y; \delta ,a}; \mathcal {S}(\alpha _X, L_X), \mathcal {S}(\alpha _Y, L_Y)] \\&\quad \le \mathcal {O}\left( \frac{1}{\min \{1; \sqrt{a}\}} \left( \delta ^{(\alpha _X + \alpha _Y + 1)/(a + 2\beta _{\zeta } + 2\beta _{\eta })} + \frac{1}{\sqrt{\delta \min \{n;m\}}} \right) \right) . \end{aligned}$$

Finally, by replacing $\delta = (\min \{n;m\})^{-(a + 2\beta _{\zeta } + 2\beta _{\eta })/(2\alpha _X + 2\alpha _Y + 2 + a + 2\beta _{\zeta } + 2\beta _{\eta })}$ to the latter estimate, we derive the desired result of this part.

(c) Suppose $F_{X}\in \mathcal {S}(\alpha _X, L_X)$ and $F_{Y}\in \mathcal {S}(\alpha _Y, L_Y)$. As in the proof of the part (b), we have $B_{\delta ,a} \le \int _{G_{\delta ,a}} t^{-1} |\phi _X(t)| |\phi _{Y}(t)| \mathrm{d}t$, where $G_{\delta ,a} := \{t>0: |\phi _{\zeta }(t) \phi _{\eta }(t)|^2 < \delta t^a\}$.

Let $0<\delta < k_{1,\zeta }^2 \ell _{1,\eta }^2 /(4^{\beta _{\eta }}\mathrm {e}^{2d_{\zeta }})$. For any $t\in G_{\delta ,a}$, we have

$$\begin{aligned}k_{1,\zeta }^2 \ell _{1,\eta }^2 /\delta< t^a (1+t)^{2\beta _{\eta }} \mathrm {e}^{2d_{\zeta } t^{\gamma _{\zeta }}}< 4^{\beta _{\eta }} t^{a+2\beta _{\eta }} \mathrm {e}^{2d_{\zeta } t^{\gamma _{\zeta }}},\end{aligned}$$

which gives

$$\begin{aligned}\ln (k_{1,\zeta }^2 \ell _{1,\eta }^2 /(4^{\beta _{\eta }}\delta )) < (a+ 2\beta _{\eta }) \ln t + 2d_{\zeta } t^{\gamma _{\zeta }} \le (a_{*} + 2\beta _{\eta }) \ln t + 2d_{\zeta } t^{\gamma _{\zeta }}.\end{aligned}$$

Hence, $2d_{\zeta } t^{\gamma _{\zeta }} > 2^{-1} \ln (k_{1,\zeta }^2 \ell _{1,\eta }^2 /(4^{\beta _{\eta }}\delta ))$ or $(a_{*} + 2\beta _{\eta }) \ln t > 2^{-1} \ln (k_{1,\zeta }^2 \ell _{1,\eta }^2 /(4^{\beta _{\eta }}\delta ))$. If $2d_{\zeta } t^{\gamma _{\zeta }} > 2^{-1} \ln (k_{1,\zeta }^2 \ell _{1,\eta }^2 /(4^{\beta _{\eta }}\delta ))$, then $t> M$, with $M:= [(4d_{\zeta })^{-1} \ln (k_{1,\zeta }^2 \ell _{1,\eta }^2 /(4^{\beta _{\eta }}\delta ))]^{1/\gamma _{\zeta }}$. If $(a_{*} + 2\beta _{\eta }) \ln t > 2^{-1} \ln (k_{1,\zeta }^2 \ell _{1,\eta }^2 /(4^{\beta _{\eta }}\delta ))$, then $t>(k_{1,\zeta }^2 \ell _{1,\eta }^2 /(4^{\beta _{\eta }}\delta ))^{1/(2a_{*} + 4\beta _{\eta })} > M$, for all $\delta >0$ small enough. Thus, $G_{\delta ,a} \subset (M, \infty )$ when $\delta >0$ is small enough. This gives

$$\begin{aligned} B_{\delta ,a} \le \int _{M}^{\infty } \frac{|\phi _X(t)| |\phi _{Y}(t)|}{t} \mathrm{d}t \le \mathcal {O}\left( M^{-(\alpha _X + \alpha _Y + 1)}\right) = \mathcal {O}\left( (\ln (1/\delta ))^{-(\alpha _X + \alpha _Y + 1)/\beta _{\zeta }}\right) , \end{aligned}$$

where we have also applied Lemma 5.3. From the latter estimate of $B_{\delta ,a}$ and the estimate (3.3), we conclude for small $\delta >0$ and $a\in (0, a_{*}]$ that

$$\begin{aligned}&\mathcal {R}[\widehat{F}_{X+Y; \delta ,a}; \mathcal {S}(\alpha _X, L_X), \mathcal {S}(\alpha _Y, L_Y)] \\&\quad \le \mathcal {O}\left( \frac{1}{\min \{1; \sqrt{a}\}} \left( (\ln (1/\delta ))^{-(\alpha _X + \alpha _Y + 1)/\beta _{\zeta }} + \frac{1}{\sqrt{\delta \min \{n;m\}}} \right) \right) . \end{aligned}$$

Finally, by replacing $\delta = (\min \{n;m\})^{-\omega }$ ($0<\omega <1$) to the latter estimate, we derive the desired result of this part. $\square$

5.5 Proof of Theorem 3.5

Dattner [3] considered the problem of estimating $F_{X-Y}(0) \equiv \mathbb {P}(X<Y)$ and also gave a minimax lower bound for the quantity $\inf _{\widehat{F}_{X-Y}} \sup _{F_X \in \mathcal {S}(\alpha , L)} \sup _{F_Y \in \mathcal {S}(\alpha , L)} \sqrt{\mathbb {E}|\widehat{F}_{X-Y}(0) - F_{X-Y}(0)|^2}$ when $\phi _{\zeta }$, $\phi _{\eta }$ satisfy (A.8) with $\upsilon _{\zeta } = \upsilon _{\eta } = \upsilon >0$. The proof for the result of Dattner [3] can be modified with a few changes to apply in our case. For the readers’ convenience we reproduce the proof here.

Let $H_0 : \mathbb {R} \rightarrow \mathbb {R}$ be a function satisfying the following properties: $H_0$ is twice continuously differentiable on $\mathbb {R}$, $H_0(t) \ge 0$ for all $t\in \mathbb {R}$, $\mathrm {supp}(H_0) \subset [1,2]$, and $H_0^{(k)}(1) = H_0^{(k)}(2) = 0$ for $k\in \{0,1\}$. Put $H(t) := H_0(t)$ with $t\ge 0$, and $H(t) := - H_0(-t)$ with $t<0$. Then, H is twice continuously differentiable on $\mathbb {R}$, $\mathrm {supp}(H) \subset [-2,-1] \cup [1,2]$, $H^{(k)}(1) = H^{(k)}(2) = 0$ for $k\in \{0,1\}$, and $H(-t) = -H(t)$ for all $t\in \mathbb {R}$. Let $J(x) := \frac{1}{\pi } \int _{1}^{2} H(t) \sin (tx) \mathrm{d}t$, $x\in \mathbb {R}$. Then, J is a real-valued function, $J(-x) = -J(x)$, and $\phi _{J}(t) = \mathrm {i} H(t)$. Next, put

$$\begin{aligned} \widetilde{J}(x) := \int _{-\infty }^{x} J(w) \mathrm{d}w, \quad x\in \mathbb {R}.\end{aligned}$$

Observe that the function $\widetilde{J}$ is symmetric around zero. Indeed, for any $x\in \mathbb {R}$,

$$\begin{aligned} \widetilde{J}(-x)= & {} \int _{-\infty }^{-x} J(w) \mathrm{d}w = -\int _{x}^{\infty } J(v) \mathrm{d}v = -\left( \int _{-\infty }^{\infty } J(v) \mathrm{d}v - \int _{-\infty }^{x} J(v) \mathrm{d}v\right) \\= & {} \widetilde{J}(x).\end{aligned}$$

We set $J_{*}(x):= \int _{-\infty }^{\infty } \widetilde{J}(u) J(u+x) \mathrm{d}u$, $x\in \mathbb {R}$. Since $J_{*}(-x) = -J_{*}(x)$ for all $x\in \mathbb {R}$ and $J_{*}$ is continuous on $\mathbb {R}$, we can choose a constant $c>0$ so that $|J_{*}(c)|>0$.

Suppose that b, T are positive real numbers depending on n, m such that $\lim _{n,m\rightarrow \infty } b = 0$, $\lim _{n,m\rightarrow \infty } T = \infty$ and $b^2 T^{2\alpha -1} \le \mathcal {O}(1)$. We will choose the parameters later. Define

$$\begin{aligned} \psi (x):= & {} \frac{1}{\pi (1+x^2)}, \quad g_X(x) := b J(Tx), \quad g_Y(x) := b J(-Tx + c),\\ f_{X,1}(x):= & {} \psi (x), \quad f_{X,2}(x) := f_{X,1}(x) + g_{X}(x),\\ f_{Y,1}(x):= & {} \psi \left( -x+ \frac{c}{T}\right) , \quad f_{Y,2}(x) := f_{Y,1}(x) + g_{Y}(x),\\ F_{X+Y,j}(x):= & {} \int _{-\infty }^{x} (f_{X,j} *f_{Y,j})(u) \mathrm{d}u, \quad j \in \{1,2\}.\end{aligned}$$

Clearly, $f_{X,1}$ is a density with $\phi _{f_{X,1}}(t) = \phi _{\psi }(t) = \mathrm {e}^{-|t|}$, $t\in \mathbb {R}$. Since J is an odd function around zero, we have $\int _{-\infty }^{\infty } J(u)\mathrm{d}u =0$, so $\int _{-\infty }^{\infty } f_{X,2}(x)\mathrm{d}x = \int _{-\infty }^{\infty } f_{X,1}(x)\mathrm{d}x =1$. For any $|x| \le 1$, we have $f_{X,1}(x) \ge 1/(2\pi )$ and

$$\begin{aligned}|g_X(x)| = b|J(Tx)| = \frac{b}{\pi }\left| \int _{1}^{2} H(t) \sin (Ttx) \mathrm{d}t\right| \le \frac{b}{\pi } \int _{1}^{2} |H(t)| \mathrm{d}t,\end{aligned}$$

so, for large n, m,

$$\begin{aligned}f_{X,2}(x) = f_{X,1}(x) + g_X(x) \ge \frac{1}{2\pi } - \frac{b}{\pi } \int _{1}^{2} |H(t)| \mathrm{d}t >0.\end{aligned}$$

For $|x|>1$, using integration by parts, the property $H^{(k)}(1) = H^{(k)}(2) = 0$ for $k\in \{0,1\}$ we have

$$\begin{aligned}|g_X(x)| = \frac{b}{\pi T^2 x^2} \left| \int _{1}^{2} H''(t) \sin (Ttx) \mathrm{d}t\right| \le \frac{2b}{\pi T^2 (1+x^2)} \int _{1}^{2} |H''(t)| \mathrm{d}t,\end{aligned}$$

thus, for large n, m,

$$\begin{aligned}f_{X,2}(x) = f_{X,1}(x) + g_X(x) \ge \frac{1}{\pi (1+x^2)}\left( 1 - \frac{2b}{T^2} \int _{1}^{2} |H''(t)| \mathrm{d}t \right) > 0.\end{aligned}$$

Hence, we have verified the non-negativity of $f_{X,2}$. So $f_{X,2}$ is also a density.

A direct computation gives $\phi _{g_X}(t) = \frac{\mathrm {i}b}{T}H\left( \frac{t}{T}\right)$. From the relation and the inequality $|z_1 + z_2| \le 2|z_1|^2 + 2|z_2|^2$ for all $z_1, z_2 \in \mathbb {C}$, we have

$$\begin{aligned}&\int _{-\infty }^{\infty } |\phi _{f_{X,2}}(t)|^2 (1+t^2)^{\alpha } \mathrm{d}t\\&\quad \le 2\int _{-\infty }^{\infty } |\phi _{f_{X,1}}(t)|^2 (1+t^2)^{\alpha } \mathrm{d}t + 2\int _{-\infty }^{\infty } |\phi _{g_X}(t)|^2 (1+t^2)^{\alpha } \mathrm{d}t \\&\quad = 2\int _{-\infty }^{\infty } \mathrm {e}^{-2|t|} (1+t^2)^{\alpha } \mathrm{d}t + \frac{4b^2}{T^2}\int _{0}^{\infty } \left| H\left( \frac{t}{T}\right) \right| ^2 (1+ t^2)^{\alpha } \mathrm{d}t \\&\quad \le 2\int _{-\infty }^{\infty } \mathrm {e}^{-2|t|} (1+t^2)^{\alpha } \mathrm{d}t + 4b^2 T^{2\alpha -1}\int _{1}^{2} |H(s)|^2 (1+s^2)^{\alpha }\mathrm{d}s. \end{aligned}$$

From this estimate and the condition $b^2 T^{2\alpha -1} \le \mathcal {O}(1)$, there exists a constant $L>0$ large enough independent of n, m so that $\int _{-\infty }^{\infty } |\phi _{f_{X,2}}(t)|^2 (1+t^2)^{\alpha } \mathrm{d}t \le L$. Of course, we also have $\int _{-\infty }^{\infty } |\phi _{f_{X,1}}(t)|^2 (1+t^2)^{\alpha } \mathrm{d}t \le L$. Therefore, we conclude that $F_{X,j} \in \mathcal {S}(\alpha ,L)$ (for $j \in \{1,2\}$) for large n, m. By the same arguments, $F_{Y,j} \in \mathcal {S}(\alpha ,L)$ (for $j \in \{1,2\}$) for large n, m.

Denote $\mathbf {X}':= (X'_1, \ldots , X'_n)$ and $\mathbf {Y}':= (Y'_1, \ldots , Y'_m)$. Suppose that $\widehat{F}_{X+Y}$ is an arbitrary estimator of $F_{X+Y}$ based on the samples $\mathbf {X}'$ and $\mathbf {Y}'$. For each $i \in \{1,2\}$, let $\mathbf {P}_{\mathbf {X}',i}$, $\mathbf {P}_{\mathbf {Y}',i}$ be, respectively, the probability distributions of the samples $\mathbf {X}'$ and $\mathbf {Y}'$ with respect to $F_{X,i}$ and $F_{Y,i}$. By Theorems 2.1 and 2.2 of Tsybako [17], there exists a constant $C>0$ independent of n, m such that

$$\begin{aligned} \mathcal {R}[\widehat{F}_{X+Y}; \mathcal {S}(\alpha , L), \mathcal {S}(\alpha , L)] \ge C|F_{X+Y,1}(0) - F_{X+Y,2}(0)| \end{aligned}$$

(5.9)

if the chi-squared distance $\chi ^2(\mathbf {P}_{\mathbf {X}',1} \times \mathbf {P}_{\mathbf {Y}',1}, \mathbf {P}_{\mathbf {X}',2} \times \mathbf {P}_{\mathbf {Y}',2})$ is bounded above.

For the difference $|F_{X+Y,1}(0) - F_{X+Y,2}(0)|$, we have by the Fubini theorem that

$$\begin{aligned}&|F_{X+Y,1}(0) - F_{X+Y,2}(0)| \\&\quad = \left| \int _{-\infty }^{0} [(f_{X,1}*f_{Y,1})(x) - (f_{X,2}*f_{Y,2})(x)] \mathrm{d}x\right| \\&\quad = \left| \int _{-\infty }^{0} \left[ \int _{-\infty }^{\infty } f_{X,1}(x-u) f_{Y,1}(u) \mathrm{d}u - \int _{-\infty }^{\infty } f_{X,2}(x-u) f_{Y,2}(u) \mathrm{d}u \right] \mathrm{d}x\right| \\&\quad = \left| \int _{-\infty }^{\infty } \left( \int _{-\infty }^{0} f_{X,1}(x-u) \mathrm{d}x \right) f_{Y,1}(u)\mathrm{d}u - \int _{-\infty }^{\infty } \left( \int _{-\infty }^{0} f_{X,2}(x-u) \mathrm{d}x \right) f_{Y,2}(u)\mathrm{d}u\right| \\&\quad = \left| \int _{-\infty }^{\infty } \left( \int _{-\infty }^{-u} f_{X,1}(v) \mathrm{d}v \right) f_{Y,1}(u)\mathrm{d}u - \int _{-\infty }^{\infty } \left( \int _{-\infty }^{-u} f_{X,2}(v) \mathrm{d}v \right) f_{Y,2}(u)\mathrm{d}u\right| \\&\quad = \left| \int _{-\infty }^{\infty } \left( \int _{-\infty }^{-u} f_{X,1}(v) \mathrm{d}v \right) f_{Y,1}(u)\mathrm{d}u - \int _{-\infty }^{\infty } \left( \int _{-\infty }^{-u} \right. \right. \\&\qquad \quad \left. \left. [f_{X,1}(v) + g_X(v)] \mathrm{d}v \right) [f_{Y,1}(u) + g_Y(u)]\mathrm{d}u\right| \\&\quad = |K_1 + K_2 + K_3|, \end{aligned}$$

where

$$\begin{aligned}&K_1 := \int _{-\infty }^{\infty } \left( \int _{-\infty }^{-u} f_{X,1}(v) \mathrm{d}v \right) g_Y(u) \mathrm{d}u,\\&K_2 := \int _{-\infty }^{\infty } \left( \int _{-\infty }^{-u} g_X(v) \mathrm{d}v\right) f_{Y,1}(u) \mathrm{d}u,\\&K_3 := \int _{-\infty }^{\infty } \left( \int _{-\infty }^{-u} g_X(v) \mathrm{d}v\right) g_Y(u) \mathrm{d}u. \end{aligned}$$

We have

$$\begin{aligned} K_1&= b\int _{-\infty }^{\infty } \left( \int _{-\infty }^{-u} \psi (v) \mathrm{d}v \right) J(-Tu +c) \mathrm{d}u \\&= b\int _{-\infty }^{\infty } \left( \int _{u}^{\infty } \psi (v) \mathrm{d}v \right) J(-Tu +c) \mathrm{d}u \\&= -b\int _{-\infty }^{\infty } \left( \int _{u}^{\infty } \psi (v) \mathrm{d}v \right) J(Tu - c) \mathrm{d}u\\&= -b\int _{-\infty }^{\infty } \left( \int _{-\infty }^{v} J(Tu -c) \mathrm{d}u \right) \psi (v) \mathrm{d}v \\&= -\frac{b}{T}\int _{-\infty }^{\infty } \left( \int _{-\infty }^{Tv-c} J(w) \mathrm{d}w \right) \psi (v) \mathrm{d}v = -\frac{b}{T}\int _{-\infty }^{\infty } \widetilde{J}(Tv-c) \psi (v) \mathrm{d}v, \\ K_2&= b\int _{-\infty }^{\infty } \left( \int _{-\infty }^{-u} J(Tv) \mathrm{d}v\right) \psi \left( -u + \frac{c}{T}\right) \mathrm{d}u \\&= b\int _{-\infty }^{\infty } \left( \int _{-\infty }^{u} J(Tw) \mathrm{d}w\right) \psi \left( -u + \frac{c}{T}\right) \mathrm{d}u \\&= \frac{b}{T}\int _{-\infty }^{\infty } \left( \int _{-\infty }^{Tu} J(v) \mathrm{d}v\right) \psi \left( -u + \frac{c}{T}\right) \mathrm{d}u \\&= \frac{b}{T}\int _{-\infty }^{\infty } \left( \int _{-\infty }^{-Tw+c} J(v) \mathrm{d}v\right) \psi (w) \mathrm{d}w \\&= \frac{b}{T}\int _{-\infty }^{\infty } \left( \int _{-\infty }^{Tw+c} J(u) \mathrm{d}u\right) \psi (w) \mathrm{d}w = \frac{b}{T}\int _{-\infty }^{\infty } \widetilde{J}(Tw+c) \psi (w) \mathrm{d}w . \end{aligned}$$

Now we prove that $K_1 = -K_2$. Indeed, put

$$\begin{aligned}\varphi (c):= \int _{-\infty }^{\infty } \widetilde{J}(Tv-c) \psi (v) \mathrm{d}v.\end{aligned}$$

Then,

$$\begin{aligned} \phi _{\varphi }(t)&= \int _{-\infty }^{\infty } \varphi (c) \mathrm {e}^{\mathrm {i}tc} \mathrm{d}c = \int _{-\infty }^{\infty } \left( \int _{-\infty }^{\infty } \widetilde{J}(Tv-c) \psi (v) \mathrm{d}v \right) \mathrm {e}^{\mathrm {i}tc} \mathrm{d}c \\&= \int _{-\infty }^{\infty } \left( \int _{-\infty }^{\infty } \widetilde{J}(Tv-c) \mathrm {e}^{\mathrm {i}tc} \mathrm{d}c \right) \psi (v) \mathrm{d}v\\&= \int _{-\infty }^{\infty } \left( \int _{-\infty }^{\infty } \widetilde{J}(w) \mathrm {e}^{\mathrm {i}t(Tv-w)} \mathrm{d}w \right) \psi (v) \mathrm{d}v \\&= \phi _{\widetilde{J}}(-t) \phi _{\psi }(Tt). \end{aligned}$$

Hence,

$$\begin{aligned} \varphi (c)&= \frac{1}{2\pi } \int _{-\infty }^{\infty } \phi _{\varphi }(t) \mathrm {e}^{-\mathrm {i}tc} \mathrm{d}t = \frac{1}{2\pi } \int _{-\infty }^{\infty } \phi _{\widetilde{J}}(-t) \phi _{\psi }(Tt) \mathrm {e}^{-\mathrm {i}tc} \mathrm{d}t \\&= \frac{1}{2\pi } \int _{-\infty }^{\infty } \overline{\phi _{\widetilde{J}}(t)} \phi _{\psi }(Tt) \mathrm {e}^{-\mathrm {i}tc} \mathrm{d}t = \frac{1}{2\pi } \int _{-\infty }^{\infty } \phi _{\widetilde{J}}(t) \phi _{\psi }(-Tt) \mathrm {e}^{-\mathrm {i}tc} \mathrm{d}t= \varphi (-c), \end{aligned}$$

where we have used the fact that $\phi _{\widetilde{J}}(t)$ is a real-valued function. Since $K_1 = - b\varphi (c)/T$, $K_2 = b\varphi (-c)/T$ and $\varphi (-c) = \varphi (c)$, we deduce $K_1 + K_2 = 0$. Therefore,

$$\begin{aligned} |F_{X+Y,1}(0) - F_{X+Y,2}(0)|&= |K_3| = \left| \int _{-\infty }^{\infty } \left( \int _{-\infty }^{-u} g_X(v) \mathrm{d}v\right) g_Y(u) \mathrm{d}u\right| \\&= b^2\left| \int _{-\infty }^{\infty } \left( \int _{-\infty }^{-u} J(Tv) \mathrm{d}v\right) J(-Tu+c) \mathrm{d}u\right| \\&= b^2\left| \int _{-\infty }^{\infty } \left( \int _{-\infty }^{u} J(Tw) \mathrm{d}w\right) J(Tu-c) \mathrm{d}u\right| = \frac{b^2}{T^2}|J_{*}(c)|. \end{aligned}$$

Now, we let $b:= T^{(1-2\alpha )/2}$ to give $b^2 T^{2\alpha -1} = 1$ and

$$\begin{aligned} |F_{X+Y,1}(0) - F_{X+Y,2}(0)| \ge |J_{*}(c)| T^{-(2\alpha +1)}. \end{aligned}$$

(5.10)

It follows from (5.9) and (5.10) that

$$\begin{aligned} \mathcal {R}[\widehat{F}_{X+Y}; \mathcal {S}(\alpha , L), \mathcal {S}(\alpha , L)] \ge C |J_{*}(c)| T^{-(2\alpha +1)} \end{aligned}$$

(5.11)

whenever the distance $\chi ^2(\mathbf {P}_{\mathbf {X}',1} \times \mathbf {P}_{\mathbf {Y}',1}, \mathbf {P}_{\mathbf {X}',2} \times \mathbf {P}_{\mathbf {Y}',2})$ is bounded above.

Now we bound $\chi ^2(\mathbf {P}_{\mathbf {X}',1} \times \mathbf {P}_{\mathbf {Y}',1}, \mathbf {P}_{\mathbf {X}',2} \times \mathbf {P}_{\mathbf {Y}',2})$. Indeed, we have (see, e.g., [17, page 86])

$$\begin{aligned}\chi ^2(\mathbf {P}_{\mathbf {X}',1} \times \mathbf {P}_{\mathbf {Y}',1}, \mathbf {P}_{\mathbf {X}',2} \times \mathbf {P}_{\mathbf {Y}',2}) = [1+\chi ^2(\mathbf {P}_{\mathbf {X}',1}, \mathbf {P}_{\mathbf {X}',2})][1+\chi ^2(\mathbf {P}_{\mathbf {Y}',1}, \mathbf {P}_{\mathbf {Y}',2})] - 1.\end{aligned}$$

Note that

$$\begin{aligned}&\chi ^2(\mathbf {P}_{\mathbf {X}',1}, \mathbf {P}_{\mathbf {X}',2})\\&\quad = \int _{-\infty }^{\infty }\cdots \int _{-\infty }^{\infty } \frac{\left( \prod _{j=1}^{n} (f_{X,1} *f_{\zeta })(x_j) - \prod _{j=1}^{n} (f_{X,2} *f_{\zeta })(x_j)\right) ^2}{\prod _{j=1}^{n} (f_{X,1} *f_{\zeta })(x_j)} \mathrm{d}x_1 \cdots \mathrm{d}x_n \\&\quad = \int _{-\infty }^{\infty }\cdots \int _{-\infty }^{\infty } \frac{\prod _{j=1}^{n} ((f_{X,2} *f_{\zeta })(x_j))^2}{\prod _{j=1}^{n} (f_{X,1} *f_{\zeta })(x_j)} \mathrm{d}x_1 \cdots \mathrm{d}x_n -1 \\&\quad = \left( \int _{-\infty }^{\infty } \frac{((f_{X,2} *f_{\zeta })(x))^2}{(f_{X,1} *f_{\zeta })(x)} \mathrm{d}x\right) ^n -1 = (1+ Q_{\zeta })^n -1, \end{aligned}$$

where

$$\begin{aligned}Q_{\zeta }:= \int _{-\infty }^{\infty } \frac{\left( (f_{X,1} *f_{\zeta })(x) - (f_{X,2} *f_{\zeta })(x)\right) ^2}{(f_{X,1} *f_{\zeta })(x)} \mathrm{d}x = \int _{-\infty }^{\infty } \frac{|(g_X *f_{\zeta })(x)|^2}{(\psi *f_{\zeta })(x)} \mathrm{d}x.\end{aligned}$$

Similarly,

$$\begin{aligned}\chi ^2(\mathbf {P}_{\mathbf {Y}',1}, \mathbf {P}_{\mathbf {Y}',2}) = (1+ Q_{\eta })^m -1 \end{aligned}$$

with

$$\begin{aligned}Q_{\eta }:= \int _{-\infty }^{\infty } \frac{\left( (f_{Y,1} *f_{\eta })(x) - (f_{Y,2} *f_{\eta })(x)\right) ^2}{(f_{Y,1} *f_{\eta })(x)} \mathrm{d}x = \int _{-\infty }^{\infty } \frac{|(g_Y *f_{\eta })(x)|^2}{(f_{Y;1} *f_{\eta })(x)} \mathrm{d}x .\end{aligned}$$

Hence,

$$\begin{aligned} \chi ^2(\mathbf {P}_{\mathbf {X}',1} \times \mathbf {P}_{\mathbf {Y}',1}, \mathbf {P}_{\mathbf {X}',2} \times \mathbf {P}_{\mathbf {Y}',2})&= (1+Q_{\zeta })^{n} (1+Q_{\eta })^{m} - 1 \nonumber \\&= \mathrm {e}^{n\ln (1+Q_{\zeta }) + m\ln (1+Q_{\eta })} - 1 \nonumber \\&\le \mathrm {e}^{nQ_{\zeta } + mQ_{\eta }} - 1. \end{aligned}$$

(5.12)

Next, we bound $Q_{\zeta }$. In fact, by the Lebesgue dominated convergence theorem, we have $\lim _{d \rightarrow \infty } \int _{-d}^{d} f_{\zeta }(u) \mathrm{d}u =1$, so there exists a constant $d_0 >0$ depending on $f_{\zeta }$ such that $\int _{-d_0}^{d_0} f_{\zeta }(u) \mathrm{d}u \ge 1/2$. Then,

$$\begin{aligned} (\psi *f_{\zeta })(x)&= \int _{-\infty }^{\infty } \frac{f_{\zeta }(u)}{\pi [1+(x-u)^2]} \mathrm{d}u \ge \int _{-d_0}^{d_0} \frac{f_{\zeta }(u)}{\pi (1+2x^2 + 2u^2)} \mathrm{d}u \\&\ge \frac{1}{2\pi (1+x^2)(1+d_0^2)} \int _{-d_0}^{d_0} f_{\zeta }(u) \mathrm{d}u \ge \frac{1}{4\pi (1+x^2)(1+d_0^2)}, \end{aligned}$$

which gives

$$\begin{aligned} Q_{\zeta }&\le \mathcal {O}\left( \int _{-\infty }^{\infty } (1+x^2) |(g_X *f_{\zeta })(x)|^2 \mathrm{d}x \right) \\&\le \mathcal {O}\left( \int _{-\infty }^{\infty } |(g_X *f_{\zeta })(x)|^2 \mathrm{d}x + \int _{-\infty }^{\infty } |x(g_X *f_{\zeta })(x)|^2 \mathrm{d}x\right) . \end{aligned}$$

By the Parseval identity and the inequality $(u+v)^2 \le 2(u^2 + v^2)$ (for $u,v\in \mathbb {R})$,

$$\begin{aligned}&\int _{-\infty }^{\infty } |(g_X *f_{\zeta })(x)|^2 \mathrm{d}x \\&\quad = \frac{1}{2\pi } \int _{-\infty }^{\infty } |\phi _{g_X *f_{\zeta }}(t)|^2 \mathrm{d}t\\&\quad = \frac{1}{2\pi } \int _{-\infty }^{\infty } |\phi _{g_X}(t)|^2 |\phi _{\zeta }(t)|^2 \mathrm{d}t ,\\ \int _{-\infty }^{\infty } |x(g_X *f_{\zeta })(x)|^2 \mathrm{d}x&= \frac{1}{2\pi } \int _{-\infty }^{\infty } |(\phi _{g_X} \phi _{\zeta })'(t)|^2 \mathrm{d}t \\&\quad \le \frac{1}{\pi } \int _{-\infty }^{\infty } (|\phi _{g_X}(t)|^2|\phi _{\zeta }'(t)|^2 + |\phi _{g_X}'(t)|^2|\phi _{\zeta }(t)|^2) \mathrm{d}t . \end{aligned}$$

Thus,

$$\begin{aligned}Q_{\zeta } \le \mathcal {O}\left( \int _{-\infty }^{\infty } (|\phi _{g_X}(t)|^2 + |\phi _{g_X}'(t)|^2)(|\phi _{\zeta }(t)|^2 + |\phi _{\zeta }'(t)|^2) \mathrm{d}t \right) .\end{aligned}$$

Because $\phi _{g_X}(t) = \frac{b}{T}\phi _{J}\left( \frac{t}{T}\right) = \frac{\mathrm {i}b}{T}H\left( \frac{t}{T}\right)$, we deduce

$$\begin{aligned} Q_{\zeta }&\le \mathcal {O}\left( \int _{-\infty }^{\infty } \left( \frac{b^2}{T^2}\left| H\left( \frac{t}{T}\right) \right| ^2 + \frac{b^2}{T^4}\left| H'\left( \frac{t}{T}\right) \right| ^2 \right) (|\phi _{\zeta }(t)|^2 + |\phi _{\zeta }'(t)|^2) \mathrm{d}t \right) \nonumber \\&\le \mathcal {O}\left( \frac{b^2}{T}\int _{1}^{2} (|H(s)|^2 + \left| H'(s)\right| ^2) (|\phi _{\zeta }(Ts)|^2 + |\phi _{\zeta }'(Ts)|^2) \mathrm{d}s \right) . \end{aligned}$$

(5.13)

We now bound $Q_{\eta }$. First,

$$\begin{aligned} (f_{Y;1} *f_{\eta })(x)&= \int _{-\infty }^{\infty } f_{Y;1}(x-u) f_{\eta }(u) \mathrm{d}u = \int _{-\infty }^{\infty } \psi \left( u-x + \frac{c}{T}\right) f_{\eta }(u) \mathrm{d}u \\&= \int _{-\infty }^{\infty } \psi \left( x - \frac{c}{T} -u \right) f_{\eta }(u) \mathrm{d}u = (\psi *f_{\eta })\left( x - \frac{c}{T}\right) . \end{aligned}$$

Similar to the estimate of $(\psi *f_{\zeta })(x)$, there exists a constant $d_1>0$ depending on $f_{\eta }$ such that

$$\begin{aligned}(\psi *f_{\eta })(x) \ge \frac{1}{4\pi (1+x^2)(1+d_1^2)},\end{aligned}$$

which yields

$$\begin{aligned}(f_{Y;1} *f_{\eta })(x) \ge \frac{1}{4\pi \left[ 1+\left( x - \frac{c}{T}\right) ^2\right] (1+d_1^2)},\end{aligned}$$

thus

$$\begin{aligned} Q_{\eta }&\le \mathcal {O}\left( \int _{-\infty }^{\infty } |(g_Y *f_{\eta })(x)|^2 \mathrm{d}x + \int _{-\infty }^{\infty } \left( x - \frac{c}{T}\right) ^2 |(g_Y *f_{\eta })(x)|^2 \mathrm{d}x\right) \\&\le \mathcal {O}\left( \int _{-\infty }^{\infty } |(g_Y *f_{\eta })(x)|^2 \mathrm{d}x + \int _{-\infty }^{\infty } |x(g_Y *f_{\eta })(x)|^2 \mathrm{d}x\right) . \end{aligned}$$

By the Parseval identity and the inequality $(u+v)^2 \le 2(u^2 + v^2)$ (for $u,v\in \mathbb {R}$),

$$\begin{aligned}\int _{-\infty }^{\infty } |(g_Y *f_{\eta })(x)|^2 \mathrm{d}x = \frac{1}{2\pi } \int _{-\infty }^{\infty } |\phi _{g_Y}(t)|^2 |\phi _{\eta }(t)|^2 \mathrm{d}t ,\end{aligned}$$

$$\begin{aligned} \int _{-\infty }^{\infty } |x(g_Y *f_{\eta })(x)|^2 \mathrm{d}x&= \frac{1}{2\pi } \int _{-\infty }^{\infty } |(\phi _{g_Y} \phi _{\eta })'(t)|^2 \mathrm{d}t \\&\le \frac{1}{\pi } \int _{-\infty }^{\infty } (|\phi _{g_Y}(t)|^2|\phi _{\eta }'(t)|^2 + |\phi _{g_Y}'(t)|^2|\phi _{\eta }(t)|^2) \mathrm{d}t . \end{aligned}$$

Thus,

$$\begin{aligned}Q_{\eta } \le \mathcal {O}\left( \int _{-\infty }^{\infty } (|\phi _{g_Y}(t)|^2 + |\phi _{g_Y}'(t)|^2)(|\phi _{\eta }(t)|^2 + |\phi _{\eta }'(t)|^2) \mathrm{d}t \right) .\end{aligned}$$

Since $\phi _{g_Y}(t) = \mathrm {e}^{\mathrm {i}tc/T} \phi _{g_X}(-t) = -\frac{b\mathrm {i}}{T} \mathrm {e}^{\mathrm {i}tc/T} H\left( \frac{t}{T}\right)$, we infer that

$$\begin{aligned} Q_{\eta }&\le \mathcal {O}\left( \int _{-\infty }^{\infty } \left( \frac{b^2}{T^2}\left| H\left( \frac{t}{T}\right) \right| ^2 + \frac{b^2}{T^4}\left| H'\left( \frac{t}{T}\right) \right| ^2 \right) (|\phi _{\eta }(t)|^2 + |\phi _{\eta }'(t)|^2 ) \mathrm{d}t \right) \nonumber \\&\le \mathcal {O}\left( \frac{b^2}{T}\int _{1}^{2} (|H(s)|^2 + |H'(s)|^2) (|\phi _{\eta }(Ts)|^2 + |\phi _{\eta }'(Ts)|^2 ) \mathrm{d}s \right) . \end{aligned}$$

(5.14)

(a) For $b \equiv T^{(1-2\alpha )/2}$, it follows from (5.13), (5.14) and (A.8) that

$$\begin{aligned}&Q_{\zeta } \le \mathcal {O} \left( \frac{b^2}{T}\mathrm {e}^{-2k_{\zeta }T^{\upsilon _{\zeta }}}\right) = \mathcal {O}\left( T^{-2\alpha } \mathrm {e}^{-2k_{\zeta }T^{\upsilon _{\zeta }}}\right) ,\\&Q_{\eta } \le \mathcal {O} \left( \frac{b^2}{T}\mathrm {e}^{-2k_{\eta }T^{\upsilon _{\eta }}}\right) = \mathcal {O}\left( T^{-2\alpha } \mathrm {e}^{-2k_{\eta }T^{\upsilon _{\eta }}}\right) .\end{aligned}$$

Put

$$\begin{aligned}T_{*,1}:= \left( \frac{1}{2\min \{k_{\zeta }; k_{\eta }\}} \ln (\max \{n;m\}) \right) ^{1/\min \{\upsilon _{\zeta }; \upsilon _{\eta }\}}.\end{aligned}$$

Choosing $T = T_{*,1}$ yields $Q_{\zeta }, Q_{\eta } \le \mathcal {O}((\max \{n;m\})^{-1})$. From this and (5.12), we conclude that $\chi ^2(\mathbf {P}_{\mathbf {X}',1} \times \mathbf {P}_{\mathbf {Y}',1}, \mathbf {P}_{\mathbf {X}',2} \times \mathbf {P}_{\mathbf {Y}',2})$ is bounded above. Then, inserting $T=T_{*,1}$ into (5.11) leads to the desired result of this part.

(b) For $b \equiv T^{(1-2\alpha )/2}$, it follows from (5.13), (5.14) and (A.9) that $Q_{\zeta } \le \mathcal {O}(T^{-(2\alpha +2\widetilde{\beta }_{\zeta })})$ and $Q_{\eta } \le \mathcal {O}(T^{-(2\alpha +2\widetilde{\beta }_{\eta })})$. After that, letting $T = T_{*,2} := (\max \{n;m\})^{1/(2\alpha + 2\min \{\widetilde{\beta }_{\zeta }; \widetilde{\beta }_{\eta }\})}$ leads to $Q_{\zeta }, Q_{\eta } \le \mathcal {O}((\max \{n;m\})^{-1})$. Combination of this and (5.12) gives that $\chi ^2(\mathbf {P}_{\mathbf {X}',1} \times \mathbf {P}_{\mathbf {Y}',1}, \mathbf {P}_{\mathbf {X}',2} \times \mathbf {P}_{\mathbf {Y}',2})$ is bounded above. Then, inserting $T=T_{*,2}$ into (5.11) leads to the desired result of this part. $\square$