# Expectation of Brownian motion increment and exponent of it

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!

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

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.