Let $(V_{k})_{k\in \mathbb N}$ be IID RV's where $E[e^{\lambda V_{1}}]\leq 1$ and define $X_{n}:=\prod\limits_{i=1}^{n} e^{\lambda V_{i}}$. Further, let $X_{0}=1$ and $\mathbb F$ be the canonical filtration on $(V_{k})_{k\in \mathbb N}$.
I have shown that $\tau_{a}:=\inf\{n : \sum\limits_{i=1}^{n} V_{i} > a\} $ is a stopping time and further that $(X_{n})_{n\in \mathbb N}$ is a supermartingale.
I have also shown that $P(\tau_{a} < \infty)\leq e^{-\lambda a}$ but I used an assumption that I do not know whether I have justified, as the optimal stopping theorem cannot be used.
I want to state that $1=EX_{0}\geq EX_{\tau_{a}}$ from the supermartingale property. But I can only state this from the optimal stopping theorem if $\tau_{a}$ is finite a.s. And this does clearly not hold by the underlying proof
My idea: We know that $(X_{n \land \tau_{a}})_{n \in \mathbb N}$ is a supermartingale and hence:
$1=E[X_{0}]=E[X_{0 \land \tau_{a}}]\geq E[X_{n \land \tau_{a}}]$
and we thus have an upper bound $1=EX_{0}$ that is independent of $n$ and also by monotone convergence we know that:
$\lim\limits _{n \to \infty}E[X_{n\land \tau_{a}}]=E[X_{\tau_{a}}]$. This would then imply:
$E[X_{\tau_{a}}]\leq 1$. Is this justification correct?
Furthermore, if I am correct, this would imply that the necessity for any stopping time $\sigma$ to be finite a.s. is obsolete. Because I could simply - as above - define a supermartingale using $n \land \tau$ and then use monotone convergence to bound the limit, i.e. $EX_{\tau}$
The monotone convergence theorem does not apply since $X_{n \wedge \tau_a}$ is not increasing in $n$. For instance, if $V_i(\omega)$ is negative for some $\omega$ and $i<\tau_a(\omega)$, then $X_{(i-1) \wedge \tau_a}(\omega)$ is strictly larger than $X_{i \wedge \tau_a}(\omega)$.
You can get rid of this problem by applying Fatou's lemma instead of the monotone convergence theorem:
$$\mathbb{E}(X_{\tau_a}) = \mathbb{E}(\liminf_{n \to \infty} X_{n \wedge \tau_a}) \leq \liminf_{n \to \infty} \mathbb{E}(X_{n \wedge \tau_a})\leq 1.$$ Note that it is crucial that the process $(X_n)_{n \geq 1}$ is non-negative.