Let $X_1, X_2, \cdots$ be i.i.d. such that $X_i > 0$ and $\mathbb E[X_i]=1$ and consider $\mathbb F = \{\mathcal F_n\}_{n\ge 1}$ to be the discrete filtration. Denote $Y_n = \prod\limits_{i=1}^n X_i$. I need to show that:
i. For any bounded $\tau \in \mathbb F$ we have $\mathbb E[Y_{\tau}]=1$.
and the for any $\tau \in \mathbb F$ including , if $\mathbb E[ sup_{1\le n\le\tau} Y_n] < \infty$, then $E[Y_{\tau}]=1$.
I thought I need to use insights from Martingale theory to find an $\alpha$ such that $e^{\alpha Z_i} = X_i$ and hence $\mathbb E[e^{\alpha Z_i}]=1$. Then $Y_n = e^{\alpha S_n}$ where $S_n = \sum\limits_{i=1}^n X_i$ and assuming the bounded $\tau = \min\{\tau_{-a},\tau_b\}$ for some $a,b>0$ that satisfy $\mathbb E[e^{\alpha S_{\tau}}]=1$. Then I will have the two following equation:
$\mathbb E[e^{\alpha S_{\tau_{-a} \wedge \tau_b}}] = e^{-\alpha a} P(\tau_{-a} < \tau_b) + e^{\alpha b} P(\tau_{-a} > \tau_b)$
and
$1 = P(\tau_{-a} < \tau_b) + P(\tau_{-a} > \tau_b)$
But then I don't know how to extend this to derive the required results. I wonder if you could please share your thoughts on this.
Sketch:
$$E[Y_\tau]=E[E[Y_\tau|\tau]]=E[E[X_1]^\tau]=1$$
For more details, write $Y_\tau=\sum_nY_n1_{\tau=n}$ and note that the sum is finite if $\tau$ is bounded, so the expectation of $Y_\tau$ will commute with the sum.