Stopping time of a martingale $Y_n=e^{S_n}$

Question

Let $X_1,X_2$, . . . be independent, identically distributed random variables with \begin{equation} \mathbb{P}\{X_j = 1\} = q, \mathbb{P}\{X_j = −1\} = 1 − q. \end{equation} Let $S_0 = 0$ and for $n \geq 1$, $S_n = X_1 + X_2 + · · · + X_n$. Let $Y_n = e^{S_n}$.

Use the optional sampling theorem to determine the probability that $Y_n$ ever attains a value greater than 100.

I have found that $q=\frac{1}{e+1}$ so that $Y_n$ is a martingale. And it is then assumed.

Then I tried to define the stopping time $T=\min \{n:Y_n>100\}$, and $T_J=\min \{n:Y_n>100\ \text{or}\ Y_n=e^{-J}\}$. To solve the question, I need to find $\mathbb{P}\{T<\infty\}$. Hence, I may find $\mathbb{P}\{T=\infty\}=\lim_{J\to\infty}\mathbb{P}\{T_J=\infty\}$. And this value is 0, so that answer to the question is 1. Am I correct?

Answer 1

If I understand correctly, you assume that $q=\frac{1}{1+e}$ so $Y_n$ is a martingale? Then, note that $T = \inf\{n : Y_n > 100\} = \inf\{ n : Y_n = e^5\}$.

Write $T_j = \min\{j,T\}$ and note that by Doob's theorem $$1 = \mathbb EY_0 = \mathbb EY_{T_j} = \mathbb E[Y_{T_j} 1_{\{T < \infty\}}] + \mathbb E[Y_{T_j} 1_{\{T = \infty\}}].$$ On the set where $T < \infty$, we have by Dominated Convergence Theorem $$\mathbb E[Y_{T_j} 1_{\{T < \infty\}}] \to e^5 \mathbb P(T<\infty).$$ Moreover, $Y_n = e^{n \cdot \frac{X_1 + ... + X_n}{n}} \to 0$ since $\frac{X_1 + ... +X_n}{n} \to \mathbb EX_1 = \frac{1}{e+1} - (1-\frac{1}{e+1}) = \frac{2}{e+1} - 1 < 0$ hence (also by DCT) $$ \mathbb E[Y_{T_j} 1_{\{T=\infty\}}] = \mathbb E[Y_j 1_{\{T = \infty\}}] \to 0.$$ In particular, $1 = e^5 \cdot \mathbb P(T < \infty)$, i.e. $\mathbb P(T<\infty) = e^{-5}$.