0
$\begingroup$

$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!

$\endgroup$
5
  • 1
    $\begingroup$ 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? $\endgroup$
    – Did
    Oct 31, 2017 at 22:01
  • $\begingroup$ Hint: use symmetry $\endgroup$
    – stochastic
    Oct 31, 2017 at 22:22
  • $\begingroup$ @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. $\endgroup$
    – Rami Awar
    Oct 31, 2017 at 23:15
  • 1
    $\begingroup$ If $X$ is discrete with $P(X=x_i)=p_i$ then $E(X)=\sum\limits_ip_ix_i$, right? $\endgroup$
    – Did
    Oct 31, 2017 at 23:38
  • $\begingroup$ @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. $\endgroup$
    – Rami Awar
    Nov 1, 2017 at 20:27

1 Answer 1

0
$\begingroup$

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 $\varphi$.

$\endgroup$
3
  • $\begingroup$ 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 $p_\phi$? $\endgroup$
    – Rami Awar
    Nov 1, 2017 at 20:25
  • $\begingroup$ 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 $\varphi$. $\endgroup$
    – EdG
    Nov 2, 2017 at 1:12
  • $\begingroup$ Alright, thanks! $\endgroup$
    – Rami Awar
    Nov 2, 2017 at 9:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.