Expected value of $\sin(x)$

Question

I have a uniformly distributed random variable $ \omega $ in the range $[\frac\pi2, \frac\pi{-2}]$. Then I have the function $ s = \sin(\omega) $ I want to calculate the expected value of this function $ s $.

So far I know that the uniformly distributed random variable can be written as $$ \omega = \frac1{\frac\pi2 - - \frac\pi2} = \frac 1\pi $$

Then I don't know if the correct way of calculating the expected value is $$ E = \int_{-\frac\pi2}^{\frac\pi2} \frac1\pi \sin(x) dx $$ or if I'm completely off.

Answer 1

Your working seems fine.

And by the fact that sine is an odd function and $\omega$ is uniformly distributed symmetrically about $0$, the integral is evaluated to be $0$.

Answer 2

The expected value of any random variable $s(\omega)$ where $\omega$ is having the probability distribution function $f(\omega)$ is given by:

$$ E(s(\omega)) = \int_{-\infty}^{\infty} s(\omega)f(\omega)d\omega$$ since $\omega$ is distributed uniformly in the interval $[-\pi/2,\pi/2]$ we have $$f(\omega) = \frac{1}{(\pi/2-(-\pi/2))} = \frac{1}{\pi}$$ Now, $$E(s(\omega))=\int_{-\infty}^{\infty} \sin(\omega)f(\omega)d\omega$$ or, $$E(s(\omega))=\int_{-\infty}^{\infty} \sin(\omega)\frac{1}{\pi}d\omega$$ The limits for $\omega$ is from $[-\pi/2,\pi/2]$ so the integral is $$E(s(\omega))=\int_{-\pi/2}^{\pi/2} \sin(\omega)\frac{1}{\pi}d\omega$$ The answer of this integral is $0$