Given a signal

$x[n] = A + w[n]$

where $A$ is a Gaussian random variable and $w[n]$ is Gaussian white noise, then the covariance matrix of the signal is given by


I understand this definition, but where does the reduced expression


come from?

  • $\begingroup$ Have you tried expanding the term inside the expectation operator, and then apply the expectation to the individual terms (which is allowed because of the linearity of the expectation operator)? Do you know the definition of $\sigma_A^2$? Do you know what it means for a random process to be white? $\endgroup$ – Matt L. Aug 15 '18 at 20:41
  • $\begingroup$ By the way, is this homework? $\endgroup$ – Matt L. Aug 15 '18 at 20:41
  • $\begingroup$ I have tried expanding, reaching $E[A^2] + E[Aw[j-1]] + E[w[i-1]A] + E[w[i-1]w[j-1]]$. This is where I get stuck, because my intuition in interpreting and reducing expectations isn't too keen yet. I know the definition of $\sigma^2_A$ is the variance of Gaussian RV A, and that white processes are uncorrelated and zero mean (but I don't think x[n] is white since it has the added term A, unless A has mean zero. And no, this isn't homework-- I'm actually trying to work through the examples in Kay's Statistical Signal Processing book to better understand the subject. $\endgroup$ – ROS Aug 15 '18 at 21:24
  • $\begingroup$ Hi:The first term above gives you the $\sigma^2_A$. The next two drop out because of the independence of $A$ and $w$, The last one is only not zero when $i = j$ which gives you the $\sigma^2 \delta_{ij}$. $\endgroup$ – mark leeds Aug 15 '18 at 21:56
  • $\begingroup$ and, yes, first term is $\sigma^2_{A}$ only if $A$ has mean zero. $\endgroup$ – mark leeds Aug 15 '18 at 22:04

It is important to add that $w[n]$ is uncorrelated with $A$. This means that


because we may assume that $w[n]$ is zero mean: $E\{w[n]\}=0$. In this case all cross-terms in the expectation are zero, and we're left with


Since $w[n]$ is white, you have $E\{w[i-1]w[j-1]\}=\sigma^2\delta_{ij}$. The expectation of $A^2$ is given by


where $\mu_A=E\{A\}$ is the mean of $A$.

So we finally have


The expression in your question is only valid if $\mu_A=E\{A\}=0$, i.e., if $A$ is a zero mean random variable.


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