Stopping time with $\mathbb{P}(\tau<\infty)=1$

Question

Let $\tau=\min \left\{t: X_t=3\right\}$. Using the the fact that $\tau$ is a stopping time satisfying $\mathbb{P}(\tau<\infty)=1$, prove the following relation: $$ \mathbb{E}\left[\alpha^\tau\right]=\left(\frac{1-\sqrt{1-\alpha^2}}{\alpha}\right)^3, \text { for any constant } \alpha \in(0,1) $$

where $X_0=0$ and $$ X_t=Z_1+\cdots+Z_t . $$ and $Z_1, Z_2, \ldots$ be a sequence of i.i.d. (independent and identically distributed) random variables with $$ \mathbb{P}\left(Z_i=1\right)=\mathbb{P}\left(Z_i=-1\right)=1 / 2, $$

Assume: $\mathcal{F}_0=\{\emptyset, \Omega\}$ and $\forall t \in \mathbb{N}^*, \mathcal{F}_t=\sigma\left(Z_1, \ldots, Z_t\right)$.

I'm stuck on this problem that I can't solve, any help is welcome. I think I should probably use Optional Sampling Theorem but I dont know how to proceed.

Answer 1

We should use the Doob's optional sampling theorem that you mentionned.

First, we need to construct a martingale $S_t$, I suggest $$\color{red}{S_t=\underbrace{c^{-X_t}}_{\text{(1)}}\cdot \underbrace{\alpha^t}_{(2)}}$$

• The first term $(1)$ is a function $f(X_t)$ because we will use the property $X_\tau = 3$ and I choose $f(X_t) = c^{-X_t}$ with $c>0$. We will determine $c$ later.
• The second term $(2)$ because the question asks for $\alpha^\tau$.

$S_t$ is martingale if and only if $$\begin{align} \mathbb{E}(S_t |\mathcal{F}_{t-1}) = S_{t-1} &\iff \mathbb{E}(c^{-Z_t}\cdot\alpha |\mathcal{F}_{t-1})=1 \\ &\iff \frac{\alpha}{2}\left( c+ \frac{1}{c} \right)=1 \\ &\iff c = \frac{1}{\alpha} \pm \sqrt{\frac{1}{\alpha^2}-1} \end{align}$$ So, it suffices to choose $c = \frac{1}{\alpha} \color{red}{-} \sqrt{\frac{1}{\alpha^2}-1}$ so that $S_t$ is martingale. Besides, with this choice, it's easy to find that $0<c<1$ and then $$S_{t\wedge\tau} = c^{-X_{t\wedge\tau}}\underbrace{\alpha^{t\wedge\tau}}_{<1}<c^{-X_{t\wedge\tau}} \le \left(\frac{1}{c}\right)^{3} \hspace{1cm} \text{ because }\frac{1}{c}>1$$ and so the condition $(c)$ of Doob's optional sampling theorem is satisfied.

Applying the theorem, we have:

$$\begin{align} 1 = S_0 = \mathbb{E}(S_\tau) &= \mathbb{E}(c^{-X_\tau}\alpha^\tau) =\mathbb{E}(c^{-3}\alpha^\tau) \end{align}$$

or $$\color{red}{\mathbb{E}(\alpha^\tau) = c^3 = \left(\frac{1-\sqrt{1-\alpha^2}}{\alpha}\right)^3}$$

Q.E.D

Answer 2

Note that $$\frac{a^{S_{n}}}{\frac{1}{2^{n}}(a+\frac{1}{a})^{n}}$$ is a martingale wrt the natural filtration.

(How did I come up with it:- Usually if $X_{n}$'s are independent positive random variables , then $\frac{\prod_{i=1}^{n}X_{n}}{\prod_{i=1}^{n}E(X_{n})}$ is a martingale . This is an useful trick to keep in mind. I am precisely using this when I set $X_{n}=a^{Z_{n}}$).

Now apply the optional sampling theorem to see that $$E(\frac{a^{S_{T}}}{\frac{1}{2^{T}}(a+\frac{1}{a})^{T}})=1$$

as $a^{S_{T}}=a^{3}$ we have:-

$$E\left[\left(\frac{1}{\frac{a}{2}+\frac{1}{2a}}\right)^{T}\,\right]=a^{-3}$$

Now just replace $\displaystyle\alpha=\frac{1}{\frac{a}{2}+\frac{1}{2a}}$ and verify that $\left(\frac{1-\sqrt{1-\alpha^2}}{\alpha}\right)^3=\frac{1}{a^{3}}$ .

Just note that for $a\mapsto\frac{1}{2}(a+\frac{1}{a})$ is a surjection from $(0,\infty)\to(1,\infty)$ and thus $a\mapsto \frac{2}{(a+\frac{1}{a})}$ is a surjection from $(0,\infty)\to(0,1)$.