1
$\begingroup$

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!

$\endgroup$

2 Answers 2

1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.