Quantitative Finance Stack Exchange is a question and answer site for finance professionals and academics. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Have a problem regarding the expected value of the Wiener process inside a function, namely:

Compute $E[cos(W_t)]$.

To extend my question, what is the general method of computing these E´s when it is wrapped up inside some function? For this I have a hunch of having to use some Taylor series for the cosine but how do I know? When do I need some special method apart from just using Ito´s?

share|improve this question
up vote 4 down vote accepted

In this particular case, the simplest way to compute the expected value is to write $\cos(x) = \Re(e^{ix})$ and use the formula for the characteristic function of a Gaussian variable: if $Z \sim \mathcal{N}(\mu,\sigma^2)$, $E[e^{iuZ}] = e^{iu\mu - \frac{1}{2}u^2 \sigma^2 }$ (simply write the expected value as an integral $\int_{\mathbb{R}} e^{iuz} \frac{1}{\sqrt{2\pi \sigma^2}} e^{\frac{(z-\mu)^2}{2\sigma^2}} dz$, regroup the exponentials and "complete the square").

So, since $W_t \sim \mathcal{N}(0,t)$, we get $$ E[\cos(W_t)] = E[\Re(e^{iW_t})] = \Re(E[e^{iW_t}]) = \Re(e^{-t/2}) = e^{-t/2}. $$

share|improve this answer
How do you mean I should use the char. function? Would this way work? $E[cos(W_t)]=\sum (-1)^k\frac{1}{(2k)!}E[W^{2k}]$? @YBL – user2069136 Apr 20 '14 at 11:05
Edited the answer. Using the Taylor expansion can work but it is much more complicated. The exponential of a Gaussian variable is really easy to work with and appears a lot: exponential martingales, geometric brownian motion (Black-Scholes process), Girsanov theorem etc... – AFK Apr 20 '14 at 22:39
If the OP is not comfortable with using $\cos x = \Re \{ e^{i x} \} $, let $\cos x = \frac{e^{i x} + e^{-i x}}{2}$ and proceed from there. – wsw Apr 21 '14 at 15:36
Thanks y´all! Anyone knows any good reading about expectation values of brownian motions and normal distributions? @YBL – user2069136 Apr 22 '14 at 11:07

Yes, I was thinking Taylor series approximation. Another possibility is to use bootstrapping or Jacknife, which is a linear approximation of bootstrapping.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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