Variance of correlated noise

I think this is an easy question but I am confused about the answer. Assume we have correlated noise distributed as $$n \sim \mathcal{N}(0,\sigma^2 \:A),$$ where $A$ is a given square matrix. If we multiply this noise by $A^{-1}$, i.e., $$w = A^{-1} n.$$ Then, the noise covariance matrix in $w$ is given as $$\mathbb{Var}\{w\} = \mathbb{Var}\{A^{-1} n\} = (A^{-1})^2 \mathbb{Var}\{n\} = (A^{-1})^2 A = A^{-1}.$$

Am I missing something?

• What do you mean by $Var(w)$? Do you want to compute the covariance matrix of $w$ i.e. $E[ww^T]$? If yes, you can just plug in $w=A^{-1}n$ and use the fact that $E[nn^T]=\sigma^2A$. – Atul Ingle Jan 4 '17 at 5:29
• What would you say is the "noise power" in $n$? Hint: it is neither $\sigma^2 A$ nor $A$ as you seem to be hinting at in your last displayed equation. – Dilip Sarwate Jan 4 '17 at 15:00
• @AtulIngle Yes, I mean the covariance matrix of $w$. So, I think my answer is fine. – Noor Jan 4 '17 at 15:30
• @DilipSarwate I mean the covariance matrix. I updated the question accordingly. – Noor Jan 4 '17 at 15:31

Even though your calculation yields the correct result (in this case), the steps are not completely correct.

First of all, the covariance matrix of a random variable $n$ is given by

$$C = E[nn^H].$$

The nomenclature of variance is more suitable for scalar random variables, in my opinion (or it refers to the diagonal of the covariance matrix, i.e. the variance of each element of $n$).

Now, you state the covariance of $n$ should be $\sigma^2A$. Hence, we have

$$E[nn^H] = \sigma^2A$$.

Now, you are interested in the covariance matrix of the random variable $y=A^{-1}n$, which is given by

$$D = E[A^{-1}nn^HA^{-H}]$$ Now, knowing that $A$ is constant, we can take it out of the expectation, to get

$$D = A^{-1}E[nn^H]A^{-H}=A^{-1}\sigma^2A A^{-H}=\sigma^2A^{-H}.$$

Now, knowing that $A$ is a covariance matrix and these are generally semi-positive definite and symmetric, we have $A^{-1}=A^{-H}$, so the final result is

$$D = \sigma^2 A^{-1}$$

which matches your guess. Note that in case you multiply $n$ with a different matrix $B$, the result is not $\sigma^2B^2A$ (as you might have guessed), but according to above calculation it is $\sigma^2BAB^H$.

I often refer to the following link, when it comes to expectations that involve matrices: http://www.psi.toronto.edu/matrix/expect.html