Probability: determine if $Z[n] = X[n] + Y[n]$ is WSS when $X[n]$ and $Y[n]$ are WSS

Question

Determine if $Z[n] = X[n] + Y[n]$ is WSS when $X[n]$ and $Y[n]$ are WSS and every sample of $X[n]$ is independent of every sample of $Y[n]$.

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Since both $X[n]$ and $Y[n]$ are WSS, $\mu_X[n] = \mu_1$, $\mu_Y[n] = \mu_2$, $E[X[n_1]X[n_2]] = g(\lvert n_1 - n_2\rvert)$ and $E[Y[n_1]Y[n_2]] = h(\lvert n_1 - n_2\rvert)$.

Then $$ E[Z[n]] = E[X[n]] + E[Y[n]] = \mu_1 + \mu_2 $$ which is a constant so condition one is met. For condition two, we have \begin{align} E[Z[n_1]Z[n_2]] &= \cdots\\ &= E[X[n_1]X[n_2]] + E[Y[n_1]Y[n_2]] + E[X[n_1]Y[n_2]] + E[Y[n_1]X[n_2]]\\ &= g(\lvert n_1 - n_2\rvert) + h(\lvert n_1 - n_2\rvert) + E[X[n_1]Y[n_2]] + E[Y[n_1]X[n_2]] \end{align} I am not really sure with what to do with $E[X[n_1]Y[n_2]] + E[Y[n_1]X[n_2]]$. I am guessing the statement that $X$ and $Y$ are independent will play a role here but not sure how.

Answer 1

Since $X$ and $Y$ are independent, then:

$$E[XY] = E[X]E[Y]$$

Then:

$$E[Z[n_1]Z[n_2]] = g(\lvert n_1 - n_2\rvert) + h(\lvert n_1 - n_2\rvert) + 2\mu_1 \mu_2 = f(|n_1 - n_2|)$$

This met the WSS requirements on second order moment.