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I am aware there are similar questions about the subject, but my question is probably much more simple.

I want to know why is the expectation of the second moment of continuous time white-noise equal to the some constant times the delta function, or:

$E[w(t)w(t+\tau)] = Q \delta(\tau)$

I am aware of the difficulties involving the definition of a function that is discontinuous in all points, but I just wish to know why engineers defined it like this. I appreciate on-line references to the subject. Thank you.

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That is because the PSD of the white noise process is by definition a constant $Q$. So, by Wiener-Khinchin theorem the auto-correlation function is $$r_{ww}(\tau):=\mathbb{E}\left(w(t)w(t+\tau)\right)=\mathcal{F}^{-1}(Q)=Q\delta(\tau)$$

The motivation for the definition comes from the frequency domain description of random processes. White noise is a pure theoretical model which simply assumes that the spectral power density is same for the noise process in all frequency bands.

A practical example which has a description close to white noise is the Thermal noise in electrical circuits which has a constant power spectral density over a large band of frequencies. See here for more details.

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See this answer (and the link cited therein) on dsp.SE for why engineers define white noise the way they do. On the other hand, as the lack of answers to this question seems to indicate, there appears to be no mathematical difficulty in defining a continuous-time white noise as one for which $$E[X(t)X(t+\tau)]= \begin{cases}\sigma^2, & \tau = 0,\\ 0, & \tau \neq 0.\end{cases}$$ where $\sigma^2$ is the variance of all the random variables comprising the process.

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  • $\begingroup$ I like your answer, but I guess what I wanted to know was this Wiener–Khinchin theorem. $\endgroup$ Commented Jul 25, 2013 at 20:38
  • $\begingroup$ Just be careful and read beyond the Wikipedia page on the Wiener-Khinchin theorem. As that page says, the theorem holds if $E[X(t)X(t+\tau)]$ is finite for all $\tau$, and so applying the theorem to the Dirac delta requires some additional machinery. $\endgroup$ Commented Jul 25, 2013 at 21:15

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