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While reading a proof of a theorem I stumbled upon the following derivation which I failed to replicate myself. Let $\mu$ be a constant and $B(t)$ be a standard Brownian motion with $t > s$. Show that

$$ E\left( (B(t)−B(s))e^{−\mu (B(t)−B(s))} \right) = - \frac{d}{d\mu}(e^{\mu^2(t-s)/2})$$

Many thanks!

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The increments $B(t)-B(s)$ have a Gaussian distribution with mean zero and variance $t-s$, for $t>s$. Hence $$E\left( (B(t)−B(s))e^{−\mu (B(t)−B(s))} \right) =\int_{-\infty}^\infty xe^{-\mu x}e^{-\frac{x^2}{2(t-s)}}\,dx$$ $$=-\mu(t-s)e^{\mu^2(t-s)/2}=- \frac{d}{d\mu}(e^{\mu^2(t-s)/2}).$$

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Let $m:=\mu$ and $X:=B(t)-B(s)$, so that $X\sim N(0,t-s)$ and hence $Ee^{-mX}=e^{m^2(t-s)/2}$. So, in view of the Leibniz_integral_rule, the expectation in question is $$EXe^{-mX}=-E\frac d{dm}e^{-mX}=-\frac d{dm}Ee^{-mX}=-\frac d{dm}e^{m^2(t-s)/2},$$ by as desired.

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