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Cheers, I have I am given the following signal $$A \cos ( 2 \pi f_o t + \Theta)$$ and $\Theta$ a random variable with pdf of $\frac{1}{2 \pi } ,0 \leq \theta \leq 2 \pi$ and 0 elsewhere and I am asked to find its mean. I know this is a simple exercise, but I cannot grasp why the computation would be $\frac{A}{2 \pi}\int_0^{2 \pi} \cos(2 \pi f_0 t + \theta) \operatorname{d}\theta$. I am getting confused to the fact that the random variable doesn't change the time but changes the phase of the signal, and thus I butcher the computation as well. Can anyone provide me with an intuitive explanation of how we go about thinking about this? Thanks a ton!

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    $\begingroup$ Since pdf is $\frac{1}{2 \pi}$ so $A \int_0^{2 \pi} \cos(2 \pi f_0 t + \theta) \operatorname{d}\theta$ = 0 in probability distribution. $\endgroup$ Commented Jun 23, 2022 at 14:26
  • $\begingroup$ Are you being asked to find the expected time average of the signal? What is the expected value of a time average of signal who's expected value at any point is known? What is the expected value of the time average of a signal who's instantaneous values all have the same expected value? What is the expected value of $A \cos ( 2 \pi f_o t + \Theta)$ at any arbitrary time $t$? $\endgroup$
    – TimWescott
    Commented Jun 23, 2022 at 14:42

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You have not been given a signal but rather a random process, a collection of random variables, one for each time instant $t$. The random variable for time $3.14908$ (call it $X(3.14908)$) is $A\cos(2\pi f_0\cdot 3.14908+\Theta)$ and is a function of the random variable $\Theta$. The random variable for time $5.12$ is $A\cos(2\pi f_0\cdot 5.12+\Theta)$ and thus is a different function of $\Theta$. How does one find the expected value of a function of a random variable? Well, the easiest way is to use a wonderful result called the Law of the Unconscious Statistician (LOTUS) which avers that you just have to multiply the function by the pdf and integrate over the real line (definite integral, no antiderivatives please). Thus, \begin{align}E[X(3.14908)] &= \int_{-\infty}^\infty A\cos(2\pi f_0\cdot 3.14908+u)\cdot f_{\Theta}(u) \,\mathrm du\\ &= \int_{0}^{2\pi}A\cos(2\pi f_0\cdot 3.14908+u)\cdot \frac{1}{2\pi}\,\mathrm du\\ &= 0. \end{align} Similarly, \begin{align}E[X(5.12)] &= \int_{-\infty}^\infty A\cos(2\pi f_0\cdot 5.12+u)\cdot f_{\Theta}(u) \,\mathrm du\\ &= \int_{0}^{2\pi}A\cos(2\pi f_0\cdot 5.12+u)\cdot \frac{1}{2\pi}\,\mathrm du\\ &= 0. \end{align} After a few more attempts like this, you might get the idea that the time has no effect on the answer you get, and that "$\Theta$ doesn't change the time, only the phase" is not the way to think about the question being asked.

You might want to look at this ancient answer for some understanding of the issues involved.

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