I have the following question given in Communication Systems by Dr Sanjay Sharma :- "Show that the random process $X(t) = A cos(\omega t + \theta)$ where $\theta$ is a random variable uniformly distributed in range $(0, 2 \pi )$ , is a wide sense stationary process." In the solution while calculating the mean, the author writes, $\mu _X(t) = \int_{-\infty}^{\infty}X f_X(x,t) dx$ and $f_X(x,t) = f_{\theta}(\theta) = \frac{1}{2\pi} U(0,2\pi)$. But while calculating mean of functions (before introducing random process) the book used the formula as $\mu _X = \int_{-\infty}^{\infty}x f_X(x) dx$. I am not able to get the meaning of the mean/expectation in random process (which one is random variable, which one is distribution function). according to me it should have been $\mu _X(t) = \int_{-\infty}^{\infty}\theta f_{\theta}(\theta) d\theta$.

PS. In Simon Haykins the formulae for mean is $\mu _X(t) = \int_{-\infty}^{\infty}x f_{X(t)}(x) dx$ that means the integration has to be performed wrt the same varible that is being multiplied to $f$.

  • 1
    $\begingroup$ Depending on how you try to understand it, the expression "$\mu _X(t) = \int_{-\infty}^{\infty}X f_X(x,t) dx$" is either nonsensical or wrong. The right hand side needs to be $ \int_{-\infty}^{\infty}x f_X(x,t) dx$. Is this what you are asking about--a typographical error? $\endgroup$
    – whuber
    Apr 19, 2017 at 13:15

1 Answer 1


$\mu_X(t)$ is a conditional expectation, which means it is a function of $t$ rather than a number as is the case for a regular expectation. Here $\theta$ is a random variable and $t$ is some variable (possibly to be made random at some later time) and $\omega$ is a fixed parameter. So $\mu_X(t)$ represents the mean value of $X$ at $t$, having integrated out the random variable $\theta$.

I would personally read this whole apparatus as $X$ being a family of functions of a random variable $\theta$ and some parameters $t,A, \omega$ so we could index a member of the family as $X_{t,A,\omega}$. We can (apprarently) obtain the expectation $E_{f(\theta)}[X_{t,A,\omega}(\theta)]$ for all members of the family in a closed form.

  • $\begingroup$ what exactly is meant by X(t) = Acos(wt + theta)? is it a distribution, I read in Haykins that X(t_1) is a random variable. Then shouldn't X(t_1) be equal to theta (which is a random variable) $\endgroup$
    – euler16
    Apr 19, 2017 at 12:29
  • $\begingroup$ Given that the question concerns the concepts underlying the notation, I am concerned that characterizing $\mu_X(t)$ as a "conditional" expectation might further confuse the issue by (incorrectly) suggesting $t$ is a random variable. Later you refer to $t$ as a parameter. This means $\mu_X$ is indeed a function but it is not "conditional" in the sense of being conditioned on values of a random variable. Furthermore, although it is intended that many values of $t$ be considered in any application, typically $A$ and $\omega$ are fixed: maybe they shouldn't all be lumped as "parameters." $\endgroup$
    – whuber
    Apr 19, 2017 at 13:10
  • $\begingroup$ @euler16 $X(t)$ is a random variable, because (at least) $\theta$ is random and $X(t)$ is a function of $\theta$. $X(t)$ could not be a distribution as need not integrate to one. To take its expectation we need to know its distribution, but we don't. However we do know the distribution of $\theta$ and one could potentially express the density of $X$ transformed into $\theta$ (except that the relationship isn't straightforwardly invertible because $cos(-y)=cos(y)$)... blah, blah. Really this is just saying look at $\int_0^{2\pi} (1/2\pi)A\cos(\omega t + \theta) d\theta$. $\endgroup$ Apr 19, 2017 at 15:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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