Assuming that the price of the stock follows the model

$ S(t) = S(0) exp ( mt − (σ^2/ 2) t + σW(t) ) , $ where W(t) is a standard Brownian motion; σ > 0, S(0) > 0, m are some constants.

What is the expectation and variance of S(2t)?


$E[S(2t)]=E[S(0)exp(2mt-(t\sigma^2)+\sigma W(2t)] = $

$S(0)E[exp(2mt-(t\sigma^2)+\sigma W(2t))] = S(0)exp(2mt-\sigma^2 t)E[exp(\sigma W(2t)]$

using that $W(2t)$ is $N(0,2t)$ I get that $W(2t)=\sqrt{2t} Z$, where $Z$ is $N(0,1)$.

$S(0)exp(2mt-\sigma^2 t)E[exp(\sigma \sqrt{2t} Z)]$ =

$S(0)exp(2mt-\sigma^2 t)exp(\sigma \sqrt{2t})E[e^{Z}] =$

$S(0)exp(2mt-\sigma^2 t)exp(\sigma \sqrt{2t})$

Is this solution correct?

Variance: Assuming that the expectation is correctly solved I could just use that $Var(S(2t))= E[(S(2t))^2] - E[S(2t)]^2$ ?


1 Answer 1


No because $$ E(e^Z)=e^{\frac{1}{2}}\neq1 $$

More generally: $$ N \sim \mathcal N(\mu,\sigma^2)\\ E(e^{Nt})=MGF_{\mathcal N(\mu, \sigma^2)}(t)=e^{\mu t+\frac{1}{2}\sigma^2t^2} $$

The last lines should be: $$ S(0)exp(2mt-\sigma^2 t)exp(\sigma \sqrt{2t})E[e^{Z}] =\\ S(0)exp(2mt-\sigma^2 t)exp(\sigma \sqrt{2t})e^{\frac{1}{2}} $$

For the rest it is correct.


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.