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I am trying to find a reference (or, if it's false, a counterexample) for the following sort-of-intuitive fact: if $\tau$ is a stopping time with a subexponential probability distribution, and $(X_n)_{n\geq 1}$ are independent r.v.'s, also subexponential, then $\sum_{n=1}^\tau X_n$ aso has a subexponential distribution.

Specifically, I would like to know if the following statement is known:

Let $(X_n)_{n\geq 1}$ be independent random variables satisfying $\mathbb{E}[e^{X_n}] \leq 1$ for all $n$, and $\tau$ be a stopping time. Suppose $\mathbb{E}[e^{\alpha \tau}] \leq e^\beta$ for some $\alpha >0$ and $\beta<\infty$. Then $\mathbb{E}[e^{\sum_{n=1}^\tau X_n}] \leq 1$.

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The hypothesis implies that $M_k=[e^{\sum_{n=1}^k X_n}]$ is a supermartingale, with $M_0=1$. Then the optional stopping theorem for positive supermartingales implies the requested inequality. (see e.g. Williams' book "Probability with martingales"). Note that no moment conditions on the stopping time $\tau$ are needed, just that it is an almost surely finite stopping time. Alternatively, look up "Wald's third identity"and apply it to the independent variables $e^{X_n}/[\mathbb{E}e^{X_n}]$ and the given stopping time. Note that the case of a general stopping time follows from the case of a bounded stopping time via Fatou's lemma.

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  • $\begingroup$ Thank you! I can't believe it was so simple -- I got stuck at the almost sure boundedness assumption for $\tau$ in Wald's third identity. If I understand correctly, the assumption on $\tau$ can even be relaxed: all that's needed is almost sure finiteness. $\endgroup$ – Clement C. May 29 '20 at 23:09
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    $\begingroup$ You need some assumption on $\tau$ for Wald's identity to hold. But what you asked is an inequality, so you could apply Wald to $\min\{\tau,m\}$ and then let $m \to \infty$ using Fatou. $\endgroup$ – Yuval Peres May 29 '20 at 23:47

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