I am fairly new to signal processing and the concept of noise. I recently started working on a project related to data gathered from some cheap sensors. I have therefore been trying to gather some knowledge about white noise. I have understood that we can not necessarily assume much about the distribution of white noise, however it is (from wikipedia:)a discrete signal whose samples are regarded as a sequence of serially uncorrelated random variables with zero mean and finite variance. Would this imply that a white noise signal is symmetric around 0, in the sense that given a white noise signal $X = (X_1, \ldots, X_m)$ would the expected value $$ \mathbb{E}\Bigg(\frac{|\max_i(X_i) - \min_j(X_j)|}{2}\Bigg) $$ equal $0$?


1 Answer 1


No, that would not imply that; it really just implies that the expectation is zero.

But as a "crazy" example, you can have noise that takes the value -4 with probability 1/3, and +2 with probability 2/3, then the expectation is 1/3·(-4) + 2/3·(+2) = (-4 + 2·2)/3 = 0. Such noise, however, is rarely what we model; it's more common that noise is Gaussian (and if it's white, hence central Gaussian), uniform or exponential (one or two sided). If anything squaring appears in the physical model, things like $\chi^2$ and Rice distributions happen – and of course, combinations of these. (Mixtures of Gaussian distributed random variables are especially popular when observing sums of individually noisy non-central phenomena.)

For example, the most common model is Gaussian white noise, because that happens in electronics (and generally, due to forms of the Central Limit Theorem, always when you add independent but finite-variance random things); when you look at the square of that, you get a $\chi^2$ distributrion (for example, when you square your observed signal, to know whether a transmitter was emitting power or not). If you do things like measuring the distance in an X- and an orthogonal Y-direction with a sensor with white Gaussian noise, and then divide the two values (e.g. to feed it into an arctan later on to estimate an angle of a puck on a hockey field), you get Cauchy-distributed noise (which isn't white any more by your definition, because its variance isn't limited).

So, it really pays to do what you are currently doing: read the part of anything where noise plays a role precisely, and look which noise model they are really using.

  • $\begingroup$ Thanks for the reply. Would the expectation I provided equal 0 in any case where the distribution is symmetric, e.g. gaussian? $\endgroup$
    – mNugget
    Commented Jun 14 at 14:19
  • 1
    $\begingroup$ yes, from "distribution symmetric around 0" follows "zero expection" (but not the other way around). That's really just looking at the very definition of expectation for continuously distributed variables. It's a very simple integral, I'd strongly recommend you fresh up your stochastics basics :) $\endgroup$ Commented Jun 14 at 14:25
  • $\begingroup$ Would it also hold if we have a white noise signal + a constant offset. That is $X_i = Y_i + c$ where $c$ is the constant offset. Then, $\max(X_i) = \max(Y_i + c) = \max(Y_i)+c$, same holds for $\min(X_j)$. Hence the expectation would offset invariantt? $\endgroup$
    – mNugget
    Commented Jun 14 at 14:43
  • $\begingroup$ expectation is not offset-invariant. it's linear. Seriously, these are basic questions, all answered by the wikipedia article on expectation. $\endgroup$ Commented Jun 14 at 14:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.