Computing Brownian Motion Cosine Expectations with Gaussian Methods

# How to calculate the expected value of a function of a standard brownian motion (Wiener process)


# How to calculate the expected value of a function of a standard brownian motion (Wiener process)












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?

## Answer by AFK (score 6, accepted)

https://quant.stackexchange.com/a/11006

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 $$ E[\cos(W_t)] = E[\Re(e^{iW_t})] = \Re(E[e^{iW_t}]) = \Re(e^{-t/2}) = e^{-t/2}. $$

## Answer by Damian Sowinski (score 3)

https://quant.stackexchange.com/a/83732

Sure this response is a decade late, but maybe it'll help a fellow traveller a decade from now? Hopefully by then all the SEs will have the same formatting for TeX. Until then, apologies for any non-lined-up equations that follow.

As almost stated by @mcisse, a Taylor expansion is a nice way to attack such expectation values. It does, however, require one to know the moments of a Wiener process:

$E[ W(t)^{n}]= \begin{cases} 0 & n\text{ is odd}\\ t^{n/2}(n-1)!! & n\text{ is even} \end{cases}$

and that the double factorial can be manipulated into

$\begin{align} (2n-1)!!&=(2n-1)(2n-3)(2n-5)\cdots 1\\ &=\frac{(2n-1)(2n-2)(2n-3)(2n-4)(2n-5)\cdots 1}{((2n-2)(2n-4)\cdots 2}\\ &=\frac{(2n-1)!}{2^{n-1}(n-1)!} \end{align} $

Then you can use the linearity of the expectation to find

$\begin{align} E[\cos W(t)]&=E[\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}W(t)^{2n}]\\ &=\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}E[W(t)^{2n}]\\ &=\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}t^n (2n-1)!!\\ &=\sum_{n=0}^\infty \frac{(-1)^n}{n!}\left(\frac{t}{2}\right)^n\\ &=e^{-\frac{1}{2}t} \end{align} $

Similarly, you can show that the expectation value of sine vanishes.

## Answer by mcisse (score -3)

https://quant.stackexchange.com/a/11011

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