On Positive Recurrence of One-Dimensional Diffusions with Independent Switching

Abstract

Positive recurrence of one-dimensional diffusion with switching, with an additive Wiener process, and with one recurrent and one transient regime is established under suitable conditions on the drift in both regimes and on the intensities of switching. The approach is based on an embedded Markov chain with alternating jumps: one jump increases the average of the square norm of the process, while the next jump decreases it, and under suitable balance conditions this implies positive recurrence.

1 Introduction

On a probability space $(\varOmega ,\mathcal {F},(\mathcal {F}_t),\mathbb P)$ with a one-dimensional $(\mathcal {F}_t)$-adapted Wiener process $W=(W_{t})_{t\ge 0}$ on it, a one-dimensional SDE with switching is considered,

$$\begin{aligned} dX_{t} =b(X_{t}, Z_t)\, dt+ dW_{t}, \quad t\ge 0, \quad X_{0} =x, \; Z_0=z, \end{aligned}$$

where $Z_t$ is a continuous-time Markov process on the state space $S= \{0,1\}$ with (positive) intensities of respective transitions $\lambda _{01} =: \lambda _0, \, \& \, \lambda _{10} =: \lambda _1$; the process Z is assumed to be independent of W and adapted to the filtration $(\mathcal {F}_t)$. We assume that these intensities are constants; this may be relaxed. Under the regime $Z=0$ the process X is assumed positive recurrent, while under the regime $Z=1$ its modulus may increase in the square mean with the rate comparable to the decrease rate under the regime $Z=0$. This vague wording will be specified in the assumptions. Denote

$$ b(x,0) = b_-(x), \quad b(x,1) = b_+(x). $$

The problem addressed in this paper is to find sufficient conditions for the positive recurrence (and, hence, for convergence to the stationary regime) for solutions of stochastic differential equations (SDEs) with switching in the case where not for all values of the modulating process the SDE is recurrent, and where it is recurrent, this property is assumed to be “not very strong”. Earlier a similar problem was tackled in [2] in the exponential recurrent case; its method apparently does not work for the weaker polynomial recurrence. A new approach is offered. Other SDEs with switching were considered in [1, 4, 5, 7], see also the references therein. Neither of these works address exactly the problem which is attacked in this paper: some of them tackled an exponential recurrence, some other study the problem of a simple recurrence versus transience.

2 Main Result: Positive Recurrence

The existence and pathwise uniqueness of the solution follows easily from [9], or from [6], or from [8], although, neither of these papers tackles the case with switching. The next theorem is the main result of the paper.

Theorem 1

Let the drift b be bounded and let there exist $r_-, r_+,M>0$ such that

$$\begin{aligned} x b_-(x) \le - r_-, \quad x b_+(x)\le + r_+, \quad \forall \,|x|\ge M, \end{aligned}$$

(1)

and

$$ \begin{aligned} 2r_- > 1 \; \quad \& \quad \kappa _1^{-1}:= \frac{\lambda _0(2r_++1)}{\lambda _1(2r_- -1)} < 1. \end{aligned}$$

(2)

Then the process (X, Z) is positive recurrent; moreover, there exists $C>0$ such that for all $M_1$ large enough and all $x \in \mathbb R$

$$\begin{aligned} \mathbb E_x\tau _{M_1} \le C (x^2 + 1), \end{aligned}$$

(3)

where

$$ \tau _{M_1} := \inf (t\ge 0:\, |X_t|\le M_1). $$

Moreover, the process $(X_t,Z_t)$ has a unique invariant measure, and for each nonrandom initial condition x, z there is a convergence to this measure in total variation when $t\rightarrow \infty$.

3 Proof

Denote $\Vert b\Vert = \sup _x|b(x)|$. Let $M_1 \gg M$ (the value $M_1$ will be specified later); denote

$$ T_0 : = \inf (t\ge 0: Z_t = 0), $$

and

$$0 \le T_0< T_1< T_2 < \ldots , $$

where each $T_{n}$ is defined as the next moment of switch of the component Z; let

$$ \tau : = \inf (T_n\ge 0: \, |X_{T_n}|\le M_1). $$

It suffices to evaluate from above the value $\mathbb E_x\tau$ because $\tau \ge \tau _{M_1}$. Let us choose $\epsilon >0$ such that

$$\begin{aligned} \lambda _0(2r_++1 +\epsilon ) = q \lambda _1(2r_- -1 - \epsilon ) \end{aligned}$$

(4)

with some $q<1$ (see (2)). Note that for $|x|\le M$ there is nothing to prove; so assume $|x| > M$.

Lemma 1

Under the assumptions of the theorem for any $\delta >0$ there exists $M_1$ such that

$$\begin{aligned} \max \left[ \sup _{|x|>M_1}\mathbb E_x \left( \!\!\int _0^{T_1}1(\inf _{0\le s\le t}|X_s| \!\le \! M)dt|Z_0\!=\!0\!\right) \!, \right. \nonumber \\ \\ \nonumber \left. \! \sup _{|x|>M_1}\mathbb E_x \!\left( \!\int _0^{T_0}\!1(\inf _{0\le s\le t}|X_s| \!\le \! M)dt|Z_0\!=\!1\!\right) \right] <\delta . \end{aligned}$$

(5)

Proof

Let $X^i_t, \, i=0,1$ denote the solution of the equation

$$\begin{aligned} dX^i_{t} =b(X^i_{t}, i)\, dt+ dW_{t}, \quad t\ge 0, \quad X^i_{0} =x. \end{aligned}$$

Let $Z_0=0$; then $T_0=0$. The processes X and $X^0$ coincide a.s. on $[0,T_1]$ due to uniqueness of solution. Therefore, due to the independence of Z and W, and, hence, of Z and $X^0$, we obtain

$$\begin{aligned} \mathbb E_x \left( \int _0^{T_1}1(\inf _{0\le s\le t}|X_s| \le M)dt|Z_0=0\right) = \mathbb E_x \int _0^{T_1}1(\inf _{0\le s\le t}|X^0_s| \le M)dt \\ = \mathbb E_x \int _0^\infty 1(t<{T_1})1(\inf _{0\le s\le t}|X^0_s| \le M)dt = \int _0^\infty \mathbb E_x 1(t<{T_1})\mathbb P(\inf _{0\le s\le t}|X^0_s| \le M)dt \\ = \int _0^\infty \exp (-\lambda _0 t)\mathbb P(\inf _{0\le s\le t}|X^0_s| \le M)dt. \end{aligned}$$

Let us take t such that

$$ \int _t^\infty e^{-\lambda _0 s}ds <\delta /2. $$

Now, by virtue of the boundedness of b, it is possible to choose $M_1>M$ such that for this value of t we have

$$ t \, \mathbb P_x(\inf _{0\le s\le t}|X^0_s| \le M) <\delta /2. $$

The bound for the second term in (5) follows by using the process $X^1$ and the intensity $\lambda _1$ in the same way. QED

Lemma 2

If $M_1$ is large enough, then under the assumptions of the theorem for any $|x|>M_1$ for any $k=0,1,\ldots$

$$\begin{aligned} \mathbb E_x (X_{T_{2k+1}\wedge \tau }^2|Z_0=0, {\mathcal F}_{T_{2k}}) \le \mathbb E_x (X_{T_{2k}\wedge \tau }^2|Z_0=0, {\mathcal F}_{T_{2k}}) \nonumber \\ \nonumber \\ \nonumber - 1(\tau > T_{2k})\lambda _0^{-1}((2r_--1)- \epsilon ), \\ \end{aligned}$$

(6)

$$\begin{aligned} \mathbb E_x (X_{T_{2k+2}\wedge \tau }^2|Z_0=1, {\mathcal F}_{T_{2k+1}}) \le \mathbb E_x (X_{T_{2k+1}\wedge \tau }^2 |Z_0=1, {\mathcal F}_{T_{2k+1}}) \nonumber \\ \\ \nonumber + 1(\tau > T_{2k+1}) \lambda _1^{-1}((2r_-+1)+ \epsilon ). \end{aligned}$$

(7)

Proof

1. Recall that $T_0=0$ under the condition $Z_0=0$. We have,

$$ T_{2k+1} = \inf (t>T_{2k}: Z_t=1). $$

In other words, the moment $T_{2k+1}$ may be treated as “$T_{1}$ after $T_{2k}$”. Under $Z_0=0$ the process $X_t$ coincides with $X^0_t$ until the moment $T_1$. Hence, we have on $[0,T_1]$ by Ito’s formula

$$\begin{aligned} dX_t^2 - 2X_t dW_t = 2X_t b_-(X_t)dt + dt \le (- 2r_- + 1)dt, \end{aligned}$$

on the set $(|X_t|> M)$ due to the assumptions (1). Further, since $1(|X_t| > M) = 1 - 1(|X_t| \le M)$, we obtain

$$\begin{aligned} \int _0^{T_1\wedge \tau } 2X_t b_-(X_t)dt \\\\ = \int _0^{T_1\wedge \tau } 2X_t b_-(X_t) 1(|X_t|> M)dt +\int _0^{T_1\wedge \tau } 2X_t b_-(X_t)1(|X_t| \le M)dt \\\\ \le - 2r_- \int _0^{T_1\wedge \tau }1(|X_t| > M)dt +\int _0^{T_1\wedge \tau } 2M \Vert b\Vert 1(|X_t| \le M)dt \\\\ = - 2r_- \int _0^{T_1\wedge \tau }1dt +\int _0^{T_1\wedge \tau } (2M \Vert b\Vert + 2r_-) 1(|X_t| \le M)dt \\\\ \le - 2r_- \int _0^{T_1\wedge \tau }1dt +(2M \Vert b\Vert +2r_-) \int _0^{T_1\wedge \tau } 1(|X_t| \le M)dt. \end{aligned}$$

Thus, always for $|x|>M_1$,

$$\begin{aligned} \mathbb E_x \int _0^{T_1\wedge \tau } 2X_t b_-(X_t)dt \\\\ \le - 2r_- E\int _0^{T_1\wedge \tau }1dt + (2M \Vert b\Vert + 2r_-) E_x\int _0^{T_1\wedge \tau } 1(|X_t| \le M)dt \\\\ = - 2r_- \mathbb E\int _0^{T_1\wedge \tau }1dt + (2M \Vert b\Vert +2r_-)\mathbb E_x\int _0^{T_1\wedge \tau } 1(|X_t| \le M)dt \\\\ \le - 2r_- \mathbb E\int _0^{T_1\wedge \tau }1dt + (2M \Vert b\Vert + 2r_-) \mathbb E_x\int _0^{T_1} 1(|X_t| \le M)dt \\\\ \le - 2 r_- \mathbb E\int _0^{T_1\wedge \tau }1dt +(2M \Vert b\Vert +2 r_-) \delta . \end{aligned}$$

For our fixed $\epsilon >0$ let us choose $\delta = \lambda _0^{-1}\epsilon / (2M \Vert b\Vert + 2r_-)$. Then, since $|x|>M_1$ implies $T_1\wedge \tau = T_1$ on $(Z_0=0)$, we get

$$\begin{aligned} \mathbb E_x X_{T_1\wedge \tau }^2 - x^2 \le - (2r_--1)\mathbb E_x \int _0^{T_1} dt + \lambda _0^{-1}\epsilon = - \lambda _0^{-1}((2r_--1)- \epsilon ). \end{aligned}$$

Substituting here $X_{T_{2k}}$ instead of x and writing $\mathbb E_x(\cdot |{\mathcal F}_{T_{2k}})$ instead of $\mathbb E_x(\cdot )$, and multiplying by $1(\tau > T_{2k})$, we obtain the bound (6), as required.

2. The condition $Z_0=1$ implies the inequality $T_0>0$. We have,

$$ T_{2k+2} = \inf (t>T_{2k+1}: Z_t=0). $$

In other words, the moment $T_{2k+2}$ may be treated as “$T_{0}$ after $T_{2k+1}$”. Under $Z_0=1$ the process $X_t$ coincides with $X^1_t$ until the moment $T_0$. Hence, we have on $[0,T_0]$ by Ito’s formula

$$\begin{aligned} dX_t^2 - 2X_t dW_t = 2X_t b_+(X_t)dt + dt \le (2r_+ + 1)dt, \end{aligned}$$

on the set $(|X_t|> M)$ due to the assumptions (1). Further, since $1(|X_t| > M) = 1 - 1(|X_t| \le M)$, we obtain

$$\begin{aligned} \int _0^{T_0\wedge \tau } 2X_t b_+(X_t)dt \\\\ = \int _0^{T_0\wedge \tau } 2X_t b_+(X_t) 1(|X_t|> M)dt +\int _0^{T_0\wedge \tau } 2X_t b_+(X_t)1(|X_t| \le M)dt \\\\ \le 2r_+ \int _0^{T_0\wedge \tau }1(|X_t| > M)dt +\int _0^{T_0\wedge \tau } 2M \Vert b\Vert 1(|X_t| \le M)dt \\\\ = 2r_+ \int _0^{T_0\wedge \tau }1dt +\int _0^{T_1\wedge \tau } (2M \Vert b\Vert - 2r_+) 1(|X_t| \le M)dt \\\\ \le 2r_+ \int _0^{T_0\wedge \tau }1dt +2M \Vert b\Vert \int _0^{T_0\wedge \tau } 1(|X_t| \le M)dt. \end{aligned}$$

Thus, always for $|x|>M_1$,

$$\begin{aligned} \mathbb E_x \int _0^{T_0\wedge \tau } 2X_t b_+(X_t)dt \\\\ \le 2r_+ E\int _0^{T_0\wedge \tau }1dt + 2M \Vert b\Vert E_x\int _0^{T_0\wedge \tau } 1(|X_t| \le M)dt \\\\ = 2r_+ \mathbb E\int _0^{T_0\wedge \tau }1dt + 2M \Vert b\Vert \mathbb E_x\int _0^{T_1\wedge \tau } 1(|X_t| \le M)dt \\\\ \le 2r_+ \mathbb E\int _0^{T_0\wedge \tau }1dt + 2M \Vert b\Vert \mathbb E_x\int _0^{T_0} 1(|X_t| \le M)dt \\\\ \le 2 r_+ \mathbb E\int _0^{T_0\wedge \tau }1dt + 2M \Vert b\Vert \delta . \end{aligned}$$

For our fixed $\epsilon >0$ let us choose $\delta = \lambda _0^{-1}\epsilon / (2M \Vert b\Vert )$. Then, since $|x|>M_1$ implies $T_0\wedge \tau = T_0$ on $(Z_0=1)$, we get

$$\begin{aligned} \mathbb E_x X_{T_1\wedge \tau }^2 - x^2 \le - (2r_--1)\mathbb E_x \int _0^{T_1} dt + \lambda _0^{-1}\epsilon = - \lambda _0^{-1}((2r_--1)- \epsilon ). \end{aligned}$$

Substituting here $X_{T_{2k+1}}$ instead of x and writing $\mathbb E_x(\cdot |{\mathcal F}_{T_{2k+1}})$ instead of $\mathbb E_x(\cdot )$, and multiplying by $1(\tau > T_{2k+1})$, we obtain the bound (7), as required. QED

Lemma 3

If $M_1$ is large enough, then under the assumptions of the theorem for any $k=0,1,\ldots$

$$\begin{aligned} \mathbb E_x (X_{T_{2k+2}\wedge \tau }^2 |Z_0=0, {\mathcal F}_{T_{2k+1}}) \le \mathbb E_x (X_{T_{2k+1}\wedge \tau }^2 |Z_0=0, {\mathcal F}_{T_{2k+1}}) \nonumber \\ \\ \nonumber + 1(\tau > T_{2k+1}) \lambda _1^{-1}((2r_++1) + \epsilon )), \end{aligned}$$

(8)

and

$$\begin{aligned} \mathbb E_x (X_{T_{2k+1}\wedge \tau }^2 |Z_0=1, {\mathcal F}_{T_{2k}}) \le \mathbb E_x (X_{T_{2k}\wedge \tau }^2 |Z_0=1, {\mathcal F}_{T_{2k}}) \nonumber \\ \\ \nonumber - 1(\tau > T_{2k})\lambda _0^{-1}((2r_+-1) - \epsilon )). \end{aligned}$$

(9)

Proof

Let $Z_0=0$; recall that it implies $T_0=0$. If $\tau \le T_{2k+1}$, then (8) is trivial. Let $\tau > T_{2k+1}$. We will substitute x instead of $X_{T_{2k}}$ for a while, and will be using the solution $X^1_t$ of the equation

$$\begin{aligned} dX^1_{t} =b(X^1_{t}, 1)\, dt+ dW_{t}, \quad t\ge T_1, \quad X^1_{T_1} =X_{T_1}. \end{aligned}$$

For $M_1$ large enough, since $|x|\wedge |X_{T_1}|>M_1$ implies $T_2\le \tau$, and due to the assumptions (1) we guarantee the bound

$$\begin{aligned} 1(|X_{T_1}|>M_1) (\mathbb E_{X_{T_1}} X_{T_2\wedge \tau }^2 - X_{T_1\wedge \tau }^2) \\\\ \le 1(|X_{T_1}|>M_1)(\mathbb E_{X_{T_1}} (T_2 - T_1)((2r_++1) + \epsilon )) \\\\ = + 1(|X_{T_1}|>M_1)(\lambda _1^{-1}((2r_++1) + \epsilon )) \end{aligned}$$

in the same way as the bound (7) in the previous lemma. In particular, it follows that for $|x|>M_1$

$$\begin{aligned} (\mathbb E_{X_{T_1}} X_{T_2\wedge \tau }^2 - X_{T_1\wedge \tau }^2) \le 1(|X_{T_1}|>M_1)(\mathbb E_{X_{T_1}} (T_2 - T_1)((2r_++1) + \epsilon )) \\\\ = + 1(|X_{T_1}|>M_1)(\lambda _1^{-1}((2r_++1) + \epsilon )), \end{aligned}$$

since $|X_{T_1}|\le M_1$ implies $\tau \le T_1$ and $\mathbb E_{X_{T_1}} X_{T_2\wedge \tau }^2 - X_{T_1\wedge \tau }^2=0$. So, on the set $|x|>M_1$ we have,

$$\begin{aligned} \mathbb E_x (\mathbb E_{X_{T_1}} X_{T_2\wedge \tau }^2 - X_{T_1\wedge \tau }^2) \\\\ \le \mathbb E_x 1(|X_{T_1}|>M_1)(\lambda _1^{-1}((2r_++1) + \epsilon )) \le \lambda _1^{-1}((2r_++1) + \epsilon ). \end{aligned}$$

Now substituting back $X_{T_{2k}}$ in place of x and multiplying by $1(\tau > T_{2k+1})$, we obtain the inequality (8), as required.

For $Z_0=1$ we have $T_0>0$, and the bound (9) follows in a similar way. QED

Now we can complete the proof of the theorem. Consider the case $Z_0=0$ where $T_0=0$. Note that the bound (6) of the Lemma 2 together with the bound (8) of the Lemma 3 can be equivalently rewritten as follows:

$$\begin{aligned} \mathbb E_x X_{T_{2k+1}\wedge \tau }^2 - \mathbb E_x X_{T_{2k}\wedge \tau }^2 \le - ((2r_--1)- \epsilon ) \mathbb E_x (T_{2k+1}\wedge \tau - T_{2k}\wedge \tau ), \end{aligned}$$

(10)

and

$$\begin{aligned} \mathbb E_x X_{T_{2k}\wedge \tau }^2 - \mathbb E_x X_{T_{2k-1}\wedge \tau }^2 \le ((2r_++1)+ \epsilon ) \mathbb E_x (T_{2k}\wedge \tau - T_{2k-1}\wedge \tau ). \end{aligned}$$

(11)

We have the identity

$$ \tau \wedge T_n = T_0 + \sum _{m=0}^{n-1} ((T_{m+1}\wedge \tau ) - (T_{m}\wedge \tau )). $$

Therefore,

$$ \mathbb E_x(\tau \wedge T_n) = \mathbb E_xT_0 + \mathbb E_x\sum _{m=0}^{n-1} ((T_{m+1}\wedge \tau ) - (T_{m}\wedge \tau )), $$

Since $T_n\uparrow \infty$, by virtue of the monotonic convergence in both parts and due to Fubini theorem we obtain,

$$\begin{aligned} \mathbb E_x \tau = \mathbb E_xT_0 + \sum _{m=0}^{\infty } \mathbb E_x ((T_{m+1}\wedge \tau ) - (T_{m}\wedge \tau )) \nonumber \\ \nonumber \\ \nonumber = \mathbb E_xT_0 + \sum _{k=0}^{\infty } \mathbb E_x ((T_{2k+1}\wedge \tau ) - (T_{2k}\wedge \tau )) \\\nonumber \\ + \sum _{k=0}^{\infty } \mathbb E_x ((T_{2k+2}\wedge \tau ) - (T_{2k+1}\wedge \tau )). \end{aligned}$$

(12)

Due to (10) and (11) we have,

$$\begin{aligned} \mathbb E_x (T_{2k+1}\wedge \tau - T_{2k}\wedge \tau ) \le ((2r_--1)- \epsilon )^{-1}\left( \mathbb E_x X_{T_{2k+1}\wedge \tau }^2 - \mathbb E_x X_{T_{2k}\wedge \tau }^2\right) \end{aligned}$$

$$\begin{aligned} \mathbb E_x X_{T_{2m+2}\wedge \tau }^2 - x^2 \\\\ \le ((2r_++1)+ \epsilon ) \sum _{k=0}^{m} \mathbb E_x (T_{2k+2}\wedge \tau - T_{2k+1}\wedge \tau ) \\\\ - ((2r_--1)- \epsilon ) \sum _{k=0}^{m} \mathbb E_x (T_{2k+1}\wedge \tau - T_{2k}\wedge \tau ) \\\\ = \sum _{k=0}^{m} \left( - ((2r_--1)- \epsilon ) (\mathbb E_x (T_{2k+1}\wedge \tau - T_{2k}\wedge \tau ) \right. \\\\ \left. +((2r_++1)+ \epsilon ) \mathbb E_x (T_{2k+2}\wedge \tau - T_{2k+1}\wedge \tau ) \right) . \end{aligned}$$

By virtue of Fatou’s lemma we get

$$\begin{aligned} x^2 \ge ((2r_--1)- \epsilon ) \sum _{k=0}^{m} (\mathbb E_x (T_{2k+1}\wedge \tau - T_{2k}\wedge \tau ) \nonumber \\\\ \nonumber - ((2r_++1)+ \epsilon ) \sum _{k=0}^{m} \mathbb E_x (T_{2k+2}\wedge \tau - T_{2k+1}\wedge \tau ). \end{aligned}$$

(13)

Note that $1(\tau> T_{2k+1}) \le 1(\tau > T_{2k})$. So, $\mathbb P(\tau> T_{2k}) \ge \mathbb P(\tau > T_{2k+1})$. Hence,

$$\begin{aligned} \lambda _0 \mathbb E_x (T_{2k+1}\wedge \tau - T_{2k}\wedge \tau ) - \lambda _1 \mathbb E_x (T_{2k+2}\wedge \tau - T_{2k+1}\wedge \tau ) \\\\ = \lambda _0 \mathbb E_x (T_{2k+1}\wedge \tau - T_{2k}\wedge \tau )1(\tau \ge T_{2k}) \\\\ - \lambda _1 \mathbb E_x (T_{2k+2}\wedge \tau - T_{2k+1}\wedge \tau ) 1(\tau \ge T_{2k+1}) \\\\ = \lambda _0 \mathbb E_x 1(\tau> T_{2k}) \mathbb E_{X_{T_{2k}}}(T_{2k+1}\wedge \tau - T_{2k}\wedge \tau ) \\\\ - \lambda _1 \mathbb E_x 1(\tau> T_{2k+1}) \mathbb E_{X_{T_{2k+1}}}(T_{2k+2}\wedge \tau - T_{2k+1}\wedge \tau ) \\\\ = \lambda _0 \mathbb E_x 1(\tau> T_{2k}) \lambda _0^{-1} - \lambda _1 \mathbb E_x 1(\tau> T_{2k+1}) \lambda _1^{-1} \\\\ = \mathbb E_x 1(\tau> T_{2k}) - \mathbb E_x 1(\tau > T_{2k+1}) \ge 0. \end{aligned}$$

Thus,

$$ \mathbb E_x (T_{2k+2}\wedge \tau - T_{2k+1}\wedge \tau ) \le \frac{\lambda _0}{\lambda _1} \mathbb E_x (T_{2k+1}\wedge \tau - T_{2k}\wedge \tau ). $$

Therefore, we estimate

$$\begin{aligned} ((2r_++1)+ \epsilon ) \sum _{k=0}^{m} \mathbb E_x (T_{2k+2}\wedge \tau - T_{2k+1}\wedge \tau ) \\\\ \le ((2r_++1)+ \epsilon ) \frac{\lambda _0}{\lambda _1} \sum _{k=0}^{m} \mathbb E_x (T_{2k+1}\wedge \tau - T_{2k}\wedge \tau ) \\\\ = q ((2r_--1)- \epsilon ) \sum _{k=0}^{m} \mathbb E_x (T_{2k+1}\wedge \tau - T_{2k}\wedge \tau ). \end{aligned}$$

So, (13) implies that

$$\begin{aligned} x^2 \ge ((2r_--1)- \epsilon ) \sum _{k=0}^{m} (\mathbb E_x (T_{2k+1}\wedge \tau - T_{2k}\wedge \tau ) \\\\ - ((2r_++1)+ \epsilon ) \sum _{k=0}^{m} \mathbb E_x (T_{2k+2}\wedge \tau - T_{2k+1}\wedge \tau ) \\\\ \ge (1-q) ((2r_--1)- \epsilon ) \sum _{k=0}^{m} (\mathbb E_x (T_{2k+1}\wedge \tau - T_{2k}\wedge \tau ) \\\\ \ge \frac{1-q}{2}\, ((2r_--1)- \epsilon ) \sum _{k=0}^{m} (\mathbb E_x (T_{2k+1}\wedge \tau - T_{2k}\wedge \tau ) \\\\ + \frac{1-q}{2q}\, ((2r_++1)+ \epsilon ) \sum _{k=0}^{m} \mathbb E_x (T_{2k+2}\wedge \tau - T_{2k+1}\wedge \tau ). \end{aligned}$$

Denoting $\displaystyle c:= \min \left( \frac{1-q}{2q}\, ((2r_++1)+ \epsilon ), \frac{1-q}{2}\, ((2r_--1)- \epsilon )\right)$, we conclude that

$$\begin{aligned} x^2 \ge c \sum _{k=0}^{2m} \mathbb E_x (T_{k+1}\wedge \tau - T_{k}\wedge \tau ). \end{aligned}$$

So, as $m\uparrow \infty$, by the monotone convergence theorem we get the inequality

$$\begin{aligned} \sum _{k=0}^{\infty } \mathbb E_x (T_{k+1}\wedge \tau - T_{k}\wedge \tau ) \le c^{-1}x^2. \end{aligned}$$

Due to (12), it implies that (in the case $T_0=0$)

$$\begin{aligned} \mathbb E_x \tau \le c^{-1}x^2, \end{aligned}$$

(14)

as required. Recall that this bound is established for $|x|>M_1$, while in the case of $|x|\le M_1$ the left hand side in this inequality is just zero.

In the case of $Z_0=1$ (and, hence, $T_0>0$), we have to add the value $\mathbb E_x T_0 = \lambda _1^{-1}$ to the right hand side of (14), which leads to the bound (3), as promised.

In turn, this bound implies existence of the invariant measure, see [3, Section 4.4]. Convergence to it in total variation follows due to the coupling method in a standard way. So, this measure is unique. The details and some extensions of this issue will be provided in another paper. QED