# Expected Value of a Random Process

$X(t) = cos(2\pi f_o t + \phi), f_o > 0$ 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!

• Well, by definition of $p_\phi$, $$E(X(t))=\frac14\left(\cos\left(2\pi f_0t\right)+\cos\left(2\pi f_0t+\frac\pi2\right)+\cos\left(2\pi f_0t+\pi\right)+\cos\left(2\pi f_0t+\frac{3\pi}2\right)\right)$$ Can you simplify this?
– Did
Oct 31, 2017 at 22:01
• Hint: use symmetry Oct 31, 2017 at 22:22
• @Did Cab you further clarify how you applied the definition of E[X(t)]? This is the answer to the problem, but I don't understand why we got here. Oct 31, 2017 at 23:15
• If $X$ is discrete with $P(X=x_i)=p_i$ then $E(X)=\sum\limits_ip_ix_i$, right?
– Did
Oct 31, 2017 at 23:38
• @Did Yup, I got my old statistics book and reviewed the definitions rigorously. I'm still having trouble understanding how the random process function of t, is distributed according to $p_\phi$ , so the random process is a function of phi, but also of t, yet we're only considering $p_\phi$. It makes more sense if we would take $p_(t, \phi )$ for example, but I still wouldn't be comfortable doing that. Nov 1, 2017 at 20:27

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 $\varphi$.
• Thanks! If you could please elaborate more on one little thing: I still don't get is considering $X(t, \phi )$ instead of $X(t)$. I don't feel comfortable making this transition. Maybe it's related to X being a random process and not variable? That's why $X(t)$ is according to $p_\phi$? Nov 1, 2017 at 20:25
• Is used $X(t,\phi)$ in stead of $X(t)$, just to make clear that $X$ depends on $t$ and $\phi$. So purely for clarification, no deep maths here. We want to have the expectation of $X$, given the variation of $\varphi$, that is why the integral is performed over $\varphi$.