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?



Expectation value of random walk on $\mathbb{R}$
1. In the [MSE setting](https://math.stackexchange.com/questions/1108134/unbounded-stopping-time), 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](https://math.stackexchange.com/questions/1108134/unbounded-stopping-time), 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, $P(T<\infty)=1$.


