Not sure this is the best place to ask for verification, but I can't seem to find a derivation anywhere else. I want to calculate $\mathbb{E}[e^{\sigma(W_t + W_s)}]$, where $W_t$ and $W_s$ are two values of a standard Brownian motion and $\sigma$ is some positive constant. WLOG assume $s < t$ and let $\mathcal{F}$ be the filtration generated by the Brownian motion. Then using standard results I get

$$ \begin{align*} \mathbb{E}[e^{\sigma(W_t + W_s)}] & = \mathbb{E}[\mathbb{E}[e^{\sigma(W_t - W_s)}e^{2\sigma W_s}| \mathcal{F}_s]] \\ & = \mathbb{E}[\mathbb{E}[e^{\sigma(W_t - W_s)}]e^{2\sigma W_s}] \\ & = \mathbb{E}[e^{2\sigma W_s}]e^{\frac{1}{2}\sigma^2(t-s)} \\ & = e^{2\sigma^2 s + \frac{1}{2}\sigma^2(t-s)} \\ & = e^{\frac{1}{2}\sigma^2(3s + t)}. \end{align*} $$


  • $\begingroup$ This seems to use "Brownian motion" for "normal random variable". $\endgroup$ – Did May 27 '14 at 19:18
  • $\begingroup$ I've used properties of Brownian motion in the derivation, namely independence and known distribution of its increments. $\endgroup$ – bcf May 27 '14 at 19:21
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    $\begingroup$ Then $W_s$ and $W_t$ would be two values of the same Brownian motion $W$ at times $t$ and $s$--in which case you need to seriously revise the question. $\endgroup$ – Did May 27 '14 at 19:25
  • $\begingroup$ Seriously revised. $\endgroup$ – bcf May 27 '14 at 19:34
  • $\begingroup$ Much better now... :-) If every $1/2 \sigma^2$ in your answer is actually $\sigma^2/2$, then the answer is correct. $\endgroup$ – Did May 27 '14 at 19:42

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