# Variance of white noise and the STFT

I hope someone can help me understand the meaning of variance in this case. I am reading an old research paper by Ephraim and Malah (for my DSP project) and it says: Knowing that $D_k$ is the FFT of white noise, what I am trying to find out is $\lambda_d(k)$. Can anyone explain to me the meaning of $E\lbrace|D_k|^2\rbrace$? I think it is the variance of a particular $D_k$ with respect to the mean, but since this is white noise, shouldn't it be zero?

$E\{|D_k|^2\}$ is the expected value of the squared magnitude of the noise in the $k^{th}$ frequency bin. Since they assume that the noise is zero mean, this equals the noise variance. Remember that the variance of a random variable $X$ is defined as $\sigma_X^2=E\{|X-\mu_X|^2\}$, where $\mu_X=E\{X\}$ is the mean of $X$. So if $\mu_X=0$ (as assumed here for the noise) then $\sigma^2_X=E\{|X|^2\}$. Note that $E\{|D_k|^2\}$ is also the noise power in the $k^{th}$ bin, so it can't be zero unless there is no noise at all in that bin.
• @Mona: No, $E\{|D_k|^2\}$ is the average noise power, which needs to be estimated. To keep things simple, you could assume that your input data contain a segment with only noise from which you can obtain an estimate of $E\{|D_k|^2\}$ by averaging $|D_k|^2$ over all noise frames. Obviously, this assumes that the noise is stationary, i.e. its characteristics don't change over time. In a more realistic setting you would use a noise tracking algorithm such as this one. – Matt L. Nov 30 '14 at 20:13