I know $E[W_T-W_t]=0$ but I have a solution which implies this is wrong.
Question
Answer
1
-
1$\begingroup$ This question was originally closed for being too basic. I'd like to point to the subtlety of expectations as beautifully explained out by the accepted answer. I think this is important and could confuse other students in the field. Also, this site has a popular question on annualizing the Sharpe Ratio. If that question is sophisticated enough, so is knowing that, "E(X)=μ doesn't necessarily imply E[f(X)]=f(μ)." $\endgroup$– conaCommented Apr 1, 2021 at 1:26
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
$E(X)=\mu$ 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 $$E[e^{\sigma\gamma(W_T-W_t)}]=e^{\frac{1}{2}\sigma^2\gamma^2(T-t)}$$
