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Suppose $X_1,X_2, \dots, X_N$ are i.i.d. random variables with $\mathbb{E}[X_1]=\mu\neq 0$. Extending the analysis for Wald's equation to where $\tau$ is a stopping time gives their partial sum subject to a stopping time $\tau$ to be $$ \mathbb{E}[\sum_{i=1}^{\tau} X_i] = \mathbb{E}[\sum_{i=1}^{\tau} \mathbb{E}[X_1]] = \mathbb{E}[\sum_{i=1}^{\tau} \mu]. $$ I am interested in the partial product $$ \prod_{i=1}^{\tau} X_i, $$ where $\{\tau=i\}$ is dependent on $X_1,X_2,\dots,X_i$ and the empty product equals one. Ruben states that such a partial product satisfies $$ \mathbb{E}[\prod_{i=1}^{\tau} X_i\mu^{-\tau}] = 1. $$ Can we say anything more about the properties of $\mathbb{E}[\prod_{i=1}^{\tau} X_i]$? Alternatively, if we knew that $\mu>1$, can we say that $$ \mathbb{E}[\prod_{i=1}^{\tau} X_i] > 1? $$

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I'm not sure what an empty product is defined to be, but let's assume an empty product is just $1$. I'm also assuming that the $X_i$ are independent of $\tau$, which is needed for Wald's formula anyway. In that case, we have \begin{align} \mathbb{E}[\prod_{i=1}^{\tau} X_i]&=\sum_{n=0}^{\infty}\Pr[\tau = n]\mathbb{E}[\prod_{i=1}^{n} X_i\vert \tau = n ]\\ &=\sum_{n=0}^{\infty}\Pr[\tau = n] \mu^n\\ &=\Pr[\tau=0]+\sum_{n=1}^{\infty} \Pr[\tau=n]\mu^n\\ &>\Pr[\tau=0]+\sum_{n=1}^{\infty} \Pr[\tau=n]\\ &=1, \end{align} where the last inequality follows from the assumption $\mu>1$.

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  • $\begingroup$ I did mean $\tau$ is dependent on $X_i$'s. I have updated the question and corrected some of the equations to reflect that. $\endgroup$ – thisisenfield Aug 12 '19 at 22:04

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