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In an exercise I am asked to prove the following Wald's identities: let $S_n$ be a simple random walk and $T$ a stopping time. Then for all $\lambda \in \mathbb R,$ $$ \mathbb E(e^{\lambda S_1}) = 1 \Rightarrow \mathbb E(e^{\lambda S_T})=1. $$ I can easily prove this from the property of martingales.

However (although not asked in the exercise), I think we should be able to deduce the moment generating function of $S_T$ from this. However things does not quite work.

If $$ \mathbb E(e^{zS_1})=e^{\gamma}, $$

then define $\tilde S_n = S_n - n\gamma/z.$ We have $$ \mathbb E(e^{z \tilde S_1}) = 1, $$ and so $\mathbb E(e^{z \tilde S_T}) = 1.$ So $$ \mathbb E(e^{zS_T -T\gamma}) =1. $$

However, it is hard to separate $T$ from $S_T.$ They may not be independent.

Can we obtain the moment generating function (or characteristic function) of $S_T$ in this way?

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  • $\begingroup$ The moment generating function of $S_T$ uniquely determines the distribution of $S_T$, so obviously it cannot be determined uniquely without using some information about how $T$ is defined. Moreover, if it were Brownian motion not RW, then Skorokhod's embedding theorem says that $S_T$ can be distributed in arbitrary way (say, with finite second moment). $\endgroup$
    – Kostya_I
    Commented Apr 30, 2021 at 17:31
  • $\begingroup$ @Kostya_I I mean, we can include some information $T$ and combine it with the result above. $\endgroup$
    – Ma Joad
    Commented Apr 30, 2021 at 17:32

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