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$z \mapsto E(e^zB_t)$ where $B_t$, $z\in \mathbb{C}$ is a Brownian motion is analytic. I am looking at the proof that for a one dimensional Brownian motion $B_t$, $E e^{zB_t}=e^{tz^2/2}$ for all $z\in \mathbb{C}$. The proof uses the fact that $E^{izB_t}=e^{-tz^2/2}$ and $e^{ay}e^{-y^2/2}$ is integrable on $\mathbb{R}$, and analytic functions agreeing on a real line must agree on the whole complex plane. However, I do not understand why $z \mapsto E(e^zB_t)$ is analytic in the first place. I would greatly appreciate any help.

nomadicmathematician
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  • Well, the mapping is holomorphic, isn't it? There is no trouble differentiating with respect to $z$... – saz Jun 04 '17 at 04:53
  • @saz what justifies the interchange of the integral and differentiation here? Lebesgue's differentiation theorem doesn't seem to apply here as we're differentiating over $z$ and not real number $t$. – nomadicmathematician Jun 04 '17 at 09:25
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    Have a look at this: https://math.stackexchange.com/q/842942/36150 (... you have to replace the integral wrt Lebesgue measure $\int_0^1 \dots , dt$ by an integral wrt a probability measure $\int_{\Omega} \dots , d\mathbb{P}$). – saz Jun 04 '17 at 10:20
  • @saz Thank you that clears me up. Can you help me with this question as well? https://math.stackexchange.com/questions/2309217/proof-of-the-markov-property-of-bm-using-intersection-stable-generators – nomadicmathematician Jun 04 '17 at 11:11

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