# Distribution of a stopped random sum, with subexponential stopping time

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 $$\mathbb{E}[e^{\sum_{n=1}^\tau X_n}] \leq 1$$.

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.
• 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. – Clement C. May 29 '20 at 23:09
• 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. – Yuval Peres May 29 '20 at 23:47