# What is the expectation of a change in Brownian motion? [closed]

I know $$E[W_T-W_t]=0$$ but I have a solution which implies this is wrong.

## Question

$$E(X)=\mu$$ doesn't necessarily imply $$E[f(X)]=f(\mu)$$. In this case, if $$X \sim N(\mu,\sigma^2)$$ then $$e^X \sim lnN(\mu,\sigma^2)$$ (lognormal distribution) and $$E(e^X)=e^{\mu+\frac{\sigma^2}{2}}$$. We know that $$W_T-W_t \sim N(0,T-t)$$ therefore $$\sigma\gamma(W_T-W_t) \sim N(0,\sigma^2\gamma^2(T-t))$$. It is easy to see that $$E[e^{\sigma\gamma(W_T-W_t)}]=e^{\frac{1}{2}\sigma^2\gamma^2(T-t)}$$