Find expected value of function of a random variable

Question

I have $U_t = cos(\sigma W_t)$ where $W_t$ is Brownian Motion, that is $f(w) = \frac{1}{t \sqrt{2\pi}} \exp(-\frac{1}{2} (\frac{w}{t})^2)$ or in words $W_t$ is distributed as $N(0,t)$ I can calculate $dU_t$ like so:

$$ dU_t = -\sigma sin(\sigma W_t) dW_t - \frac{\sigma}{2} cos(\sigma W_t) dt $$

Then I need to find $E[U_t]$. However, to do this, I need:

To get $g(u)$, the pdf of $U_t$
To get $g(u)$ I need the function linking $U_t$ and $W_t$ to be one-to-one and it is not

I'm learning about Ito Calculus and it feels to me I'm missing something - I'm not sure I should be using the method of getting the pdf and expected value of a random variable which is a function of another random variable.

Does this answer your question? Evaluate $\frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^\infty \cos (x)e^{-\frac{1}{2} \left( \frac{x}{\sigma} \right)^2 }dx$ — NCh, Jan 12 '20 at 17:40
@NCh Yeah - I figured but the question first says, find $dU_t$ and hence find $E[U_t]$ — s5s, Jan 12 '20 at 17:53
Related: https://math.stackexchange.com/questions/2222976/ito-integral-representation-of-cosine-of-brownian-motion-and-expected-value — s5s, Jan 12 '20 at 18:16

score 0 · Answer 1 · answered Jan 12 '20 at 17:51

0

Note

$$E[U_t] = \int_{-\infty}^{\infty} U(x) f(x)dx =\frac{1}{t \sqrt{2\pi}} \int_{-\infty}^{\infty} \cos(\sigma x)e^{-\frac{x^2}{2t^2}}dx =e^{-\frac{(\sigma t)^2}{2}}$$

answered Jan 12 '20 at 17:51

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Find expected value of function of a random variable

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