Based on this question Expectation of product of iid random variables limited by stopping time. Let $X_1, X_2, \cdots$ be i.i.d. such that $X_i > 0$ and $\mathbb E[X_i]=1$ ($P(X_i=1/3)=1/2=P(X_i=5/3)$) and consider $\mathbb F = \{\mathcal F_n\}_{n\ge 1}$ to be the discrete filtration. Denote $Y_n = \prod\limits_{i=1}^n X_i$. For any bounded $\tau \in \mathbb F$, we have $\mathbb E[Y_{\tau}]=1$.
Stopping time product of iid random variables
analysisprobability
Best Answer
After the discussion in the comments, I see that your question is whether this conflicts with the Optional Stopping Theorem. The answer is no. Your $\tau$ is not bounded, so there's no contradiction.
Note also that your $\tau$ is not only bounded, it doesn't even seem to be almost surely finite; that is, $\mathbb P(\tau = \infty) > 0$. (In fact, I think this probability is one.) This results in a number of other problems -- in particular, it's not clear how you would define $Y_{\infty}$, which means that $\mathbb E[Y_{\tau}]$ is also not even clearly defined.
EDIT: From other comments, I think you intended for your real stopping time to be $\tau := \inf\{n: Y_n \color{red}{\leq} 1/18\}$ instead. This stopping time would still not be bounded.
For $\tau$ to be bounded, there must exist some $k$ such that $\mathbb P(\tau < k) = 1$. However, for any $k$, there is a positive probability (specifically, a probability of $1/2^k$) that the first $k$ steps are all upward; that is, $X_1 = \dots = X_k = 5/3$). Hence, $\tau$ cannot be bounded.