Expectation value of exponential of a function - first two moments of function known

Question

I have a function $f(t)$ and know that $\langle f(t)\rangle=0$ and $\langle f(t)f(t')\rangle=C(t-t')$;

Now i want to calculate:

$$\left\langle\exp\left(\int\limits_{0}^t f(t') \, \mathrm{d}t'\right)\right\rangle$$

I tried to look at the sum definition of the exponential but could do anything...

Any help?

Note that proper notation is $\langle f(t)\rangle,$ not $<f(t)>. \qquad$ — Michael Hardy, Jul 04 '17 at 20:02
Thanks for the answer! $\left\langle f(t)\right\rangle$ dentoes the expectationvalue — Martin, Jul 04 '17 at 20:03
Hmm. Must be nonstandard notation. I'm used to $E(X)$ but hey, we learn something new! — Sean Roberson, Jul 04 '17 at 20:04
yes, it's a real function where only these two moments are known — Martin, Jul 04 '17 at 20:04
yeah...as a physicist the notation may anger the common mathematician :D — Martin, Jul 04 '17 at 20:05

score 0 · Answer 1 · answered Jul 04 '17 at 22:06

0

I think what you mean is that $f$ is a stochastic process (rather than a function) of mean $0$ and covariance $C(t - t')$. If it's a Gaussian process, that's all you need to determine it, but if it's non-Gaussian you really don't know enough to say anything about expectations of exponentials. They might not even exist.

answered Jul 04 '17 at 22:06

Robert Israel

448,999

Thanks for the answer, I know assumed that $\left\langle f(t_1)f(t_2)....f(t_n)\right\rangle=0$ for $ n \ge 3$, which helps evaluating the integral easily! – Martin Jul 05 '17 at 09:33

Expectation value of exponential of a function - first two moments of function known

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