Given two stochastic processes, $X[n]$ and $Y[n]$, both being WSS (wide state stationary) and independents. What would be the Average and Autocorrelation function of $Z[n] = Y[n] X[n]$?

Is the resulting $Z[n]$ process also WSS? If so, can the PSD (power spectral density) be calculated?

Thank you.


1 Answer 1


Let $(X_n)$ and $(Y_n)$ be independent wide-sense stationary (WSS) processes with averages $\mu_X=EX_n$ and $\mu_Y=EY_n$, covariance functions $C_X(m)=E(X_{n+m}-EX_{n+m})(X_{n}-EX_{n})$ and $C_Y(m)=E(Y_{n+m}-EY_{n+m})(Y_{n}-EY_{n})$, correlation functions $R_X(m)=C_X(m)/C_X(0)$ and $R_Y(m)=C_Y(m)/C_Y(0)$, and power spectral densities (PSDs) $f$ and $g$, so that $R_X(m)=\int_{-\infty}^\infty e^{2\pi i mx} f(x)dx$ and $R_Y(m)=\int_{-\infty}^\infty e^{2\pi i mx} g(x)dx$. Let $Z_n=X_n Y_n$.

Then the process $(Z_n)$ is also WSS, with the average $$\mu_Z=EX_n Y_n=EX_n\,EY_n=\mu_X \mu_Y,$$ covariance function \begin{align*} C_Z(m)&=E(X_{n+m}Y_{n+m}-\mu_X\mu_Y)(X_{n}Y_{n}-\mu_X\mu_Y) \\ &=EX_{n+m}X_{n}\,EY_{n+m}Y_n-\mu_X^2\mu_Y^2\\ &=(C_X(m)+\mu_X^2)(C_Y(m)+\mu_Y^2)-\mu_X^2\mu_Y^2\\ &=C_X(m)C_Y(m)+\mu_Y^2C_X(m)+\mu_X^2C_Y(m), \end{align*} correlation function \begin{equation*} R_Z(m)=\frac{C_Z(m)}{C_Z(0)} =a\,R_X(m)R_Y(m) +b\,R_X(m) +c\,R_Y(m) \end{equation*} with $a:=\frac{C_X(0)C_Y(0)}{C_Z(0)}$, $b:=\frac{C_X(0)\mu_Y^2}{C_Z(0)}$, $c:=\frac{C_Y(0)\mu_X^2}{C_Z(0)}$, and PSD \begin{equation*} h=a f*g+bf+cg, \end{equation*} where $f*g$ is the convolution of $f$ and $g$.

  • $\begingroup$ Hi Iosif. Thank you so much for your answer. One question: What is the 'm' in the Covariance and Correlation functions meaning? Thanks again. $\endgroup$
    – Sergio
    Mar 19, 2018 at 18:09
  • $\begingroup$ @Sergio : If in your question $n$ stands for any integer, then in my answer $m$ also denotes any integer. If in your question $n$ stands for any nonnegative integer (say), then in my answer $m$ also denotes any nonnegative integer. $\endgroup$ Mar 19, 2018 at 22:07

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