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.

  • $\begingroup$ 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

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.


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