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Consider the random walk where $X_t=\sum_{i=1}^t Y_i$, $Y_i$s are iid and take $\pm$1 with probabilities $p$ and $1-p$ respectively, where $0<p<0.5$. Define stopping time $\tau_k=\min\left\{t:X_t=k\right\}$. Suppose we know $P(\tau_1<\infty)=\frac{p}{1-p}=q$, is there a way to argue $P(\tau_k<\infty)=q^k$?

I tried to use an inductive argument. Suppose $P(\tau_{k-1}<\infty)=q^{k-1}$, $$P(\tau_k<\infty)=P(\tau_k<\infty\mid \tau_{k-1}<\infty)P(\tau_{k-1}<\infty)+P(\tau_k<\infty\mid \tau_{k-1}=\infty)P(\tau_{k-1}=\infty)$$

The second of the probabilities is zero, and I only need to argue $P(\tau_k<\infty\mid \tau_{k-1}<\infty)=q$. I can intuitively see a regeneration basis for this, but is there a clear way to argue this? Also is there a cleaner way to proceed with this proof?

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For $p = \frac 23$, we get $q = \frac p{1-p} = \frac{2 \cdot 3}{3 \cdot 1} = 2$?! –  martini Jun 8 '13 at 11:44
    
@martini: Thanks, I missed out $0<p<0.5$. –  Bravo Jun 8 '13 at 11:50

1 Answer 1

Let $k \in \mathbb{N}$. Denote by $$Z_s := X_{s+\tau_k}-X_{\tau_k} \qquad (s \in \mathbb{N}_0)$$ the restarted random walk. Then $\tau_k$ is independent of $(Z_s)_{s \geq 0}$. This is intuitively clear, for a proof see below. Now note that

$$\begin{align*} \tau_{k+1} &= \tau_k + \min \{t-\tau_k; X_{(t-\tau_k)+\tau_k} - X_{\tau_k} = 1\} \\ &= \tau_k + \underbrace{\min\{s \geq 0; Z_s=1\}}_{=:\sigma_1} \end{align*}$$

Since $\tau_k$ is independent of $(Z_s)_{s \geq 0}$, we see that $\tau_k$ and $\sigma_1$ are also independent. Moreover, the increments $Y_j$ are supposed to be identically distributed, so $\sigma_1 \sim \tau_1$. Thus,

$$\mathbb{P}(\tau_{k+1}<\infty) = \mathbb{P}(\tau_k < \infty, \sigma_1 < \infty) = \mathbb{P}(\tau_k<\infty) \cdot \underbrace{\mathbb{P}(\sigma_1<\infty)}_{\mathbb{P}(\tau_1<\infty)}.$$

By induction,

$$\mathbb{P}(\tau_{k+1}<\infty) = q^k \cdot q = q^{k+1}$$


$\tau_k$ and $(Z_s)_{s \geq 0}$ are independent:

$$\begin{align*} \mathbb{E} \exp \left(\imath \, \xi \cdot \tau_k + \imath \sum_{j=0}^n \eta_j \cdot Z_{s_j} \right) &= \sum_{m \geq 1} e^{\imath \, \xi \cdot m} \cdot \mathbb{E} \bigg( \underbrace{1_{[\tau_k=m]}}_{\in \mathcal{F}_m} \cdot \underbrace{\exp \bigg(\imath \sum_{j=0}^n \eta_j \cdot \underbrace{(X_{s_j+m}-X_m)}_{\sim X_{s_j}} \bigg)}_{\text{independent of} \, \mathcal{F}_m} \bigg) \\ &= \mathbb{E}e^{\imath \, \xi \cdot \tau_k} \cdot \mathbb{E}\exp \left(\imath \, \sum_{j=0}^n \eta_j \cdot X_{s_j} \right) \end{align*}$$ where $\xi, \eta_j \in \mathbb{R}$, $s_j \in \mathbb{N}_0$ and $\mathcal{F}_m := \sigma(X_1,\ldots,X_m)$ denotes the canonical filtration. For $\xi=0$ this shows that the finite dimensional distributions of $X$ and $Z$ coincide. Hence,

$$\begin{align*} \mathbb{E} \exp \left(\imath \, \xi \cdot \tau_k + \imath \sum_{j=0}^n \eta_j \cdot Z_{s_j} \right)&= \mathbb{E}e^{\imath \, \xi \cdot \tau_k} \cdot \mathbb{E}\exp \left(\imath \, \sum_{j=0}^n \eta_j \cdot Z_{s_j} \right) \end{align*}$$

which implies that $\tau_k$ and $(Z_s)_{s \geq 0}$ are independent, see here.

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