Expectation value of random walk on $\mathbb{R}$ [closed]

Question

Given iid random variables $\xi_1,\dots,\xi_n,\dots$ with probability $P(\xi_1=-\log 3)=\frac{1}{2}=P(\xi_1=\log 5-\log 3)$. Let $S_n=\sum_{i=1}^n \xi_i$. This is a random walk on $R$. If let a stopping time $T:=\inf\{n: S_n\le -4\log 3\}$. I want to get $\mathbb{P}(\tau<\infty)$.

I try to follow the solution in https://math.stackexchange.com/questions/1108134/unbounded-stopping-time

1. The stopped process $M_n := e^{\xi_1+\ldots+\xi_{n \wedge \tau}}$ is also a martingale; hence, $$\mathbb{E}M_n = \mathbb{E}M_1=E(e^{\xi_1})=(1/3)*(1/2)+(5/3)*(1/2)=1.$$ Calculate $\mathbb{E}M_1$.
2. By the strong law of large numbers, $$\frac{\xi_1+\ldots+\xi_n}{n} \to \mathbb{E}\xi_1 <0.$$ Thus, $$\xi_1+\ldots+\xi_n \to - \infty \quad \text{almost surely as} \, \, n \to \infty.$$
3. Conclude that $$\begin{align*} M_n &= e^{\xi_1+\ldots+\xi_n} 1_{\{ \tau=\infty\}} + e^{\xi_1+\ldots+\xi_{n \wedge \tau}} 1_{\{\tau<\infty\}}\\ &\to 0 + e^{-4log 3} 1_{\{\tau<\infty\}}. \end{align*}$$
4. Deduce from step 1 and 3 that $$ 1=3^{-4}P(\tau<\infty) $$ $$\mathbb{P}(\tau<\infty) = 3^4.$$ But this is impossible? Where am I wrong?

Answer 1

1. In the MSE setting, there could be no overshoot: there, $X_1+ \dots+X_\tau$ is exactly $1$ (on the event $\{\tau<\infty\}$). In contrast, in your setting there will be, with a nonzero probability, an undershoot: $P(S_T<-4\ln3)>0$. So, the convergence you claimed in Step 3 does not actually take place.

2. More importantly, in the MSE setting, the (positive) martingale $(M_n)$ is bounded (by $3$), whereas your martingale $(M_n)$ is unbounded. So, even if your claim in Step 3 were correct, you cannot use the dominated convergence theorem to go from Step 3 to Step 4.

Mainly for the latter reason, you get the absurd value, $3^4$, for $P(\tau<\infty)$ (you apparently use two different symbols, $T$ and $\tau$, to denote the same thing). In fact, in your setting, by the strong law of large numbers, $P(T<\infty)=1$.