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Given

$$ Y(t) = A X(t) \cos(\omega t + \phi) $$

with $X(t)$ is zero-mean WSS (wide-sense stationary) process, $\phi$ ~ Unif$(0,2\pi)$. Suppose $X(t)$ and $\phi$ are independent random variables. I want to compute the mean and autocorrelation functions of $Y(t)$.

My attempt is (using independence of $X(t)$ and $\phi$, and $E(X(t)) = 0$)

$$ E(Y(t))= E(A X(t) \cos(\omega t + \phi)) = A \cdot E(X(t)) \cdot E(\cos(\omega t + \phi)) = 0$$

Similarly we have

$$ R_Y(t,t+\tau)= E(A X(t) \cos(\omega t + \phi) \cdot A X(t+\tau) \cos(\omega (t + \tau) + \phi)) = 0$$

Is this a correct use of independence? If not, why?

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    $\begingroup$ Thanks. But this also means that the autocorrelation function is zero too which I don't think is right. Am I missing something? $\endgroup$ Commented Mar 27, 2023 at 18:49

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Your use of independence for computing $E[Y(t)]$ is correct. However, for the autocorrelation you get

$$\begin{align}E[Y(t)Y(t+\tau)]&=A^2E[X(t)X(t+\tau)\cos(\omega t+\phi)\cos(\omega(t+\tau)+\phi)]\\&=A^2E[X(t)X(t+\tau)]\cdot E[\cos(\omega t+\phi)\cos(\omega(t+\tau)+\phi)]\\&=A^2R_X(\tau)E[\cos(\omega t+\phi)\cos(\omega(t+\tau)+\phi)]\tag{1}\end{align}$$

Note that in general $X(t)$ and $X(t+\tau)$ are not independent, so $E[X(t)X(t+\tau)]\neq E[X(t)]E[X(t+\tau)]$.

You can compute the expectation in $(1)$ by using a trigonometric identity to write the product of the two cosines as a sum of two cosines. If you do it right, it should turn out that $E[Y(t)Y(t+\tau)]$ only depends on $\tau$ but not on $t$.

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