0
$\begingroup$

Given volatility $\sigma_1$, $\sigma_2$ and time $t_1$, $t_2$, how to calculate the expectation $E(e^{\sigma_1 W_{t_1} + \sigma_2 W_{t_2}})$ where $W_{t}$'s are Wiener processes?

I am starting from the probability density function of normal distribution and calculating the integral, but get stuck a little in the correlation of Wiener processes. Thanks.

$\endgroup$
  • $\begingroup$ Suppose that $t_1<t_2$. Write $W_{t_2} = (W_{t_2}-W_{t_1})+W_{t_1}$ and use the independence of $W_{t_2}-W_{t_1}$ and $W_{t_1}$ to split up the expectation. After that you will have to use the stationarity of the increments and the fact that $W_r \sim N(0,r)$ for any $r>0$. $\endgroup$ – saz Jun 8 '18 at 16:41
  • $\begingroup$ @saz It is truly great. Let me follow this. $\endgroup$ – Yi Bao Jun 8 '18 at 17:47
1
$\begingroup$

Inspired by saz's comment, I am answering this question myself.

Assume $t_2 > t_1$, and notice $W_{t_2}-W_{t_1}\sim N(0, t_2 - t_1)$.

$E(e^{\sigma_1 W_{t_1} + \sigma_2 W_{t_2}}) = E(e^{(\sigma_1 + \sigma_2)W_{t_1} + \sigma_2 (W_{t_2} - W_{t_1})}) = E(e^{(\sigma_1 + \sigma_2)W_{t_1}})E(e^{\sigma_2(W_{t_2} - W_{t_1})}) = e^{\frac{1}{2}(\sigma_1 + \sigma_2)^2 t_1}e^{\frac{1}{2}\sigma_2^2(t_2-t_1)}$.

That $E(e^X) = e^{\mu+\frac{1}{2}\sigma^2}$ where $X \sim N(\mu, \sigma^2)$ is used above.

Similarly the solution of the case where $t_1 < t_2$ can be derived.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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