$X(t) = cos(2\pi f_o t + \phi), f_o > 0 $ is a constant, $\phi$ is a random variable with: $$p_\phi (\varphi) = \frac{1}{4}[\delta (\varphi) + \delta (\varphi - \pi /2) + \delta(\varphi - \pi) + \delta(\varphi - 3\pi /2)]$$
How do I calculate $\mu _X (t)$ ? I know that $\mu _X$ is the expected value with respect to $p_X (x)$, but I'm having trouble manipulating this pdf of X and using that of $\phi$ instead of something explicitly in x (?). I don't know if anyone can understand my issue if they haven't been there, but maybe you'll resolve it by answering the question.
Thanks!
I think the answer is $\mu_X(t) = 0$. \begin{align*} \mu_X(t) &= E_p[X(t,\varphi)] \\ &= \int_{-\infty}^{\infty}X(t, \varphi)p_\phi(\varphi) d\varphi \\ &= \frac{1}{4} \left( \cos (2\pi f_0 t) + \cos \left(2\pi f_0 t + \frac{\pi}{2}\right) + \cos (2\pi f_0 t + \pi ) + \cos \left(2\pi f_0 t + \frac{3\pi}{2}\right)\right) \\ &= \frac{1}{4} \left( \cos (2\pi f_0 t) + \cos \left(2\pi f_0 t + \frac{\pi}{2}\right) - \cos (2\pi f_0 t) - \cos \left(2\pi f_0 t + \frac{\pi}{2}\right)\right) \\ &= 0 \end{align*}
I made use of the fact that $\cos(u+\pi)=-\cos(u)$. For clarification, I wrote $X$ as a function of $t$ and $\varphi$.