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Conditioned local limit theorems for random walks defined on finite Markov chains

Abstract

Let \((X_n)_{n\geqslant 0}\) be a Markov chain with values in a finite state space \({\mathbb {X}}\) starting at \(X_0=x \in {\mathbb {X}}\) and let f be a real function defined on \({\mathbb {X}}\). Set \(S_n=\sum _{k=1}^{n} f(X_k)\), \(n\geqslant 1\). For any \(y \in {\mathbb {R}}\) denote by \(\tau _y\) the first time when \(y+S_n\) becomes non-positive. We study the asymptotic behaviour of the probability \({\mathbb {P}}_x \left( y+S_{n} \in [z,z+a],\, \tau _y > n \right) \) as \(n\rightarrow +\infty .\) We first establish for this probability a conditional version of the local limit theorem of Stone. Then we find for it an asymptotic equivalent of order \(n^{3/2}\) and give a generalization which is useful in applications. We also describe the asymptotic behaviour of the probability \({\mathbb {P}}_x \left( \tau _y = n \right) \) as \(n\rightarrow +\infty \).

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Appendix

Appendix

1.1 The non degeneracy of the Markov walk

In [14], it is proved that the statements of Propositions 2.12.3 hold under more general assumptions (see Hypotheses M1-M5 of [14]). We will link these assumptions to our Hypotheses M1M3. The assumptions M1-M3 in [14], with the Banach space \({\mathscr {C}}\), are well known consequences of Hypothesis M1 of this paper. Hypothesis M4 in [14] is also obvious with \(N=N_1 = \cdots = 0\). By Hypothesis M2, to obtain Hypothesis M5 of [14], it remains only to prove that \(\sigma \) defined by (2.2) is strictly positive. First we give a necessary and sufficient condition. Recall that the words path and orbit are defined in Sect. 4.

Lemma 10.1

Assume Hypothesis M1. The following statements are equivalent:

  1. 1.

    The Cesáro mean of f on the orbits is constant: there exists \(m \in {\mathbb {R}}\) such that for any orbit \(x_0,\dots ,x_n\) we have

    $$\begin{aligned} f(x_0) + \cdots + f(x_n) = (n+1)m. \end{aligned}$$
  2. 2.

    There exist a constant \(m \in {\mathbb {R}}\) and a function \(h \in {\mathscr {C}}\) such that for any \((x,x') \in {\mathbb {X}}^2\),

    $$\begin{aligned} {\mathbf {P}}(x,x') f(x') = {\mathbf {P}}(x,x') \left( h(x)-h(x')+m \right) . \end{aligned}$$
  3. 3.

    The following real \({{\tilde{\sigma }}}^2\) is equal to 0

    $$\begin{aligned} {{\tilde{\sigma }}}^2 = {\varvec{\nu }} \left( f^2 \right) - {\varvec{\nu }} \left( f \right) ^2 + 2 \sum _{n=1}^{+\infty } \left[ {\varvec{\nu }} \left( f {\mathbf {P}}^n f \right) - {\varvec{\nu }} \left( f \right) ^2 \right] = 0. \end{aligned}$$

Proof

The point 1 implies the point 2 Suppose that the point 1 holds. Fix \(x_0 \in {\mathbb {X}}\) and set \(h(x_0)= 0\). For any \(x \in {\mathbb {X}}\), we define h(x) in the following way: for any path \(x_0,x_1,\dots ,x_n,x\) in \({\mathbb {X}}\), we set

$$\begin{aligned} h(x) = -f(x)-f(x_n)-\cdots -f(x_1)+(n+1)m. \end{aligned}$$

We shall verify that h is well defined. By Hypothesis M1, we can find at least a path to define h(x). Now we have to check that this definition does not depend on the choice of the path. Let \(x_0,x_1,\dots ,x_p,x\) and \(x_0,y_1,\dots ,y_q,x\) be two paths. By Hypothesis M1, there exists a path \(x,z_1, \dots , z_n,x_0\) in \({\mathbb {X}}\) between x and \(x_0\). Since \(x_0,x_1,\dots ,x_p,x,z_1,\dots ,z_n\) and \(x_0,y_1,\dots ,y_p,x,z_1,\dots ,z_n\) are two orbits, by the point 1, we have

$$\begin{aligned}&-f(x)-f(x_p)-\cdots -f(x_1)+(p+1)m \\&\quad = f(x_0)+f(z_1)+\cdots +f(z_n)-(n+1)m \\&\quad = -f(x)-f(y_q)-\cdots -f(y_1)+(q+1)m \end{aligned}$$

and so the function h is well defined on \({\mathbb {X}}\). Now let \((x,x') \in {\mathbb {X}}^2\) such that \({\mathbf {P}}(x,x') > 0\). By Hypothesis M1, there exists \(x_0,x_1, \dots , x_n,x\) a path between \(x_0\) and x. Since

$$\begin{aligned} {\mathbf {P}}(x_0,x_1) \cdots {\mathbf {P}}(x_n,x){\mathbf {P}}(x,x') > 0, \end{aligned}$$

by the definition of h, we have

$$\begin{aligned} h(x)&= -f(x)-f(x_n)-\cdots -f(x_1)+(n+1)m \\ h(x')&= -f(x')-f(x)-f(x_n)-\cdots -f(x_1)+(n+2)m. \end{aligned}$$

In particular

$$\begin{aligned} h(x') = -f(x') + h(x) + m. \end{aligned}$$

The point 2 implies the point 1 Suppose that the point 2 holds and let \(x_0,\dots ,x_n\) be an orbit. Using the point 2,

$$\begin{aligned} h(x_0)= & {} h(x_n)-f(x_0)+m = \cdots = h(x_0)\\&-f(x_0)-f(x_n)-\cdots -f(x_1)+(n+1)m, \end{aligned}$$

and the point 1 follows.

The point 2 implies the point 3 Suppose that the point 2 holds. Denote by \({{\tilde{f}}}\) the \({\varvec{\nu }}\)-centred function:

$$\begin{aligned} {{\tilde{f}}}(x) = f(x) - {\varvec{\nu }}(f), \qquad \forall x \in {\mathbb {X}}. \end{aligned}$$
(10.1)

By the point 2, for any \(x\in {\mathbb {X}}\),

$$\begin{aligned} {\mathbf {P}} {{\tilde{f}}}(x) = h(x) - {\mathbf {P}} h(x) + m -{\varvec{\nu }}(f). \end{aligned}$$
(10.2)

Using the fact that \({\varvec{\nu }}\) is \({\mathbf {P}}\)-invariant, we obtain that \({\varvec{\nu }} \left( {{\tilde{f}}} \right) = 0 = m-{\varvec{\nu }}(f)\) and so,

$$\begin{aligned} m={\varvec{\nu }}(f). \end{aligned}$$
(10.3)

Consequently, by (10.2), \({\mathbf {P}}^n {{\tilde{f}}} = {\mathbf {P}}^{n-1} h - {\mathbf {P}}^n h\) for any \(n \geqslant 1\) and therefore,

$$\begin{aligned} \sum _{k=1}^{n} {\mathbf {P}}^{k} {{\tilde{f}}} = h - {\mathbf {P}}^n h. \end{aligned}$$
(10.4)

Let

$$\begin{aligned} {{\tilde{\varTheta }}} := \sum _{k=0}^{+\infty } {\mathbf {P}}^{k} {{\tilde{f}}} \end{aligned}$$

be the solution of the Poisson equation \({{\tilde{\varTheta }}} - {\mathbf {P}} {{\tilde{\varTheta }}} = {{\tilde{f}}}\), which by (2.1), is well defined. Taking the limit as \(n \rightarrow +\infty \) in (10.4) and using (2.1),

$$\begin{aligned} {\mathbf {P}} {{\tilde{\varTheta }}} = {{\tilde{\varTheta }}} - {{\tilde{f}}} = h - {\varvec{\nu }} (h). \end{aligned}$$

Therefore, for any \((x,x') \in {\mathbb {X}}^2\),

$$\begin{aligned} {{\tilde{\varTheta }}}(x') - {\mathbf {P}} {{\tilde{\varTheta }}}(x) = {{\tilde{f}}}(x') + {\mathbf {P}} {{\tilde{\varTheta }}}(x') - {\mathbf {P}} {{\tilde{\varTheta }}}(x) = {{\tilde{f}}}(x') + h(x') - h(x). \end{aligned}$$

Using the point 2 and (10.3), it follows that

$$\begin{aligned} {{\tilde{\varTheta }}}(x') - {\mathbf {P}} {{\tilde{\varTheta }}}(x) = 0, \end{aligned}$$
(10.5)

for any \((x,x') \in {\mathbb {X}}^2\) such that \({\mathbf {P}}(x,x') > 0\). Moreover,

$$\begin{aligned} {{\tilde{\sigma }}}^2= & {} {\varvec{\nu }} \left( {{\tilde{f}}}^2 \right) + 2 \sum _{n=1}^{+\infty } {\varvec{\nu }} \left( {{\tilde{f}}} {\mathbf {P}}^n {{\tilde{f}}} \right) = {\varvec{\nu }} \left( {{\tilde{f}}} \left( {{\tilde{f}}} + 2 {\mathbf {P}} {\tilde{\varTheta }} \right) \right) \\= & {} {\varvec{\nu }} \left( \left( {{\tilde{\varTheta }}} - {\mathbf {P}} {{\tilde{\varTheta }}} \right) \left( {{\tilde{\varTheta }}} + {\mathbf {P}} {{\tilde{\varTheta }}} \right) \right) . \end{aligned}$$

Since \({\varvec{\nu }}\) is \({\mathbf {P}}\)-invariant,

$$\begin{aligned} {{\tilde{\sigma }}}^2&= {\varvec{\nu }} \left( {\mathbf {P}} \left( {{\tilde{\varTheta }}}^2 \right) \right) - 2{\varvec{\nu }} \left( \left( {\mathbf {P}} {{\tilde{\varTheta }}} \right) ^2 \right) + {\varvec{\nu }} \left( \left( {\mathbf {P}} {{\tilde{\varTheta }}} \right) ^2 \right) \nonumber \\&= \sum _{(x,x') \in {\mathbb {X}}} \left[ {{\tilde{\varTheta }}}(x')^2 -2 {\tilde{\varTheta }}(x') {\mathbf {P}} {{\tilde{\varTheta }}}(x) + \left( {\mathbf {P}} {{\tilde{\varTheta }}}(x) \right) ^2 \right] {\mathbf {P}}(x,x') {\varvec{\nu }}(x) \nonumber \\&= \sum _{(x,x') \in {\mathbb {X}}} \left( {{\tilde{\varTheta }}}(x') - {\mathbf {P}} {\tilde{\varTheta }}(x) \right) ^2 {\mathbf {P}}(x,x') {\varvec{\nu }}(x). \end{aligned}$$
(10.6)

By (10.5), we conclude that \({{\tilde{\sigma }}}^2=0\).

The point 3 implies the point 2 Suppose that the point 3 holds. By (10.6), for any \((x,x') \in {\mathbb {X}}\) such that \({\mathbf {P}}(x,x')>0\) we have

$$\begin{aligned} {{\tilde{\varTheta }}}(x') - {\mathbf {P}} {{\tilde{\varTheta }}}(x) = 0. \end{aligned}$$

Let \(h = {\mathbf {P}} {{\tilde{\varTheta }}}\). Since \({{\tilde{\varTheta }}}\) is the solution of the Poisson equation,

$$\begin{aligned} {{\tilde{f}}}(x') + h(x') - h(x) = 0. \end{aligned}$$

By the definition of \({{\tilde{f}}}\) in (10.1), for any \((x,x') \in {\mathbb {X}}\) such that \({\mathbf {P}}(x,x')>0\),

$$\begin{aligned} f(x') = h(x) - h(x') + m, \end{aligned}$$

with \(m = {\varvec{\nu }}(f)\). \(\square \)

Note that under Hypothesis M2, Lemma 10.1 can be rewritten as follows.

Lemma 10.2

Assume Hypotheses M1 and M2. The following statements are equivalent:

  1. 1.

    The mean of f on the orbits is equal to zero: for any orbit \(x_0,\dots ,x_n\), we have

    $$\begin{aligned} f(x_0) + \cdots + f(x_n) = 0. \end{aligned}$$
  2. 2.

    There exists a function \(h \in {\mathscr {C}}\) such that for any \((x,x') \in {\mathbb {X}}^2\),

    $$\begin{aligned} {\mathbf {P}}(x,x') f(x') = {\mathbf {P}}(x,x') \left( h(x)-h(x') \right) . \end{aligned}$$
  3. 3.

    The real \(\sigma ^2\) is equal to 0:

    $$\begin{aligned} \sigma ^2 = {\varvec{\nu }} \left( f^2 \right) + 2 \sum _{n=1}^{+\infty } {\varvec{\nu }} \left( f {\mathbf {P}}^n f \right) = 0. \end{aligned}$$

Now we prove that the Hypothesis M3 (the “non-lattice” condition), implies that the Markov walk has non-zero asymptotic variance.

Lemma 10.3

Under Hypotheses M1M3, we have

$$\begin{aligned} \sigma ^2 = {\varvec{\nu }} \left( f^2 \right) + 2 \sum _{n=1}^{+\infty } {\varvec{\nu }} \left( f {\mathbf {P}}^n f \right) > 0 \end{aligned}$$

Proof

We proceed by reductio ad absurdum. Suppose that \(\sigma ^2 = 0\). By Lemma 10.2, for any orbit \(x_0,\dots ,x_n\), we have

$$\begin{aligned} f(x_0) + \cdots + f(x_n) = 0, \end{aligned}$$

which implies the negation of Hypothesis M3 with \(\theta = a = 0\). \(\square \)

1.2 Strong approximation

Let \((B_t)_{t\geqslant 0}\) be the standard Brownian motion on \({\mathbb {R}}\) defined on the probability space \((\varOmega , {\mathscr {F}}, {\mathbb {P}})\). Consider the exit time

$$\begin{aligned} \tau _y^{bm} := \inf \{ t \geqslant 0, \; y+\sigma B_t \leqslant 0 \}, \end{aligned}$$
(10.7)

where \(\sigma \) is defined by (2.2). It is proved in Grama, Le Page and Peigné [16] that there is a version of the Markov walk \((S_n)_{n\geqslant 0}\) and of the standard Brownian motion \((B_t)_{t\geqslant 0}\) living on the same probability space which are close enough in the following sense:

Proposition 10.4

There exists \(\varepsilon _0 >0\) such that, for any \(\varepsilon \in (0,\varepsilon _0]\), \(x\in {\mathbb {X}}\) and \(n\geqslant 1\), without loss of generality (on an extension of the initial probability space) one can reconstruct the sequence \((S_n)_{n\geqslant 0}\) with a continuous time Brownian motion\((B_t)_{t\in {\mathbb {R}}_{+} }\), such that

$$\begin{aligned} {\mathbb {P}}_x \left( \underset{0 \leqslant t \leqslant 1}{\sup } \left|S_{\left\lfloor tn\right\rfloor }-\sigma B_{tn}\right| > n^{1/2-\varepsilon } \right) \leqslant \frac{c_{\varepsilon }}{n^{\varepsilon }}. \end{aligned}$$

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Grama, I., Lauvergnat, R. & Le Page, É. Conditioned local limit theorems for random walks defined on finite Markov chains. Probab. Theory Relat. Fields 176, 669–735 (2020). https://doi.org/10.1007/s00440-019-00948-8

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Keywords

  • Markov chain
  • Exit time
  • Conditioned local limit theorem
  • Duality

Mathematics Subject Classification

  • 60J10
  • 60F05