Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I found a contradiction I couldn't resolve by my self. It's about a "Uniform White Noise".

Let ${x}_{t}$ be a "White Noise" i.i.d. Random Process:

$ \forall t \in \mathbb{R}, \ {x}_{t} \sim U[-1, \ 1] $

If we chose to go by the PSD definition of "White Noise" (Constant all over the Frequencies) we get:

$ {R}_{xx}( \tau ) = var({x}_{t}) \delta ( \tau ) $

Yet, Clearly:

$ E[{x}_{t} {x}_{t + \tau}] \underset{ \tau = 0}{=}E[{x}_{t} {x}_{t}]= \frac{1}{3} $

Intuitively, a Process with bounded variance and values can't be "White Noise".
Please mind this is a Continuous Random Process. We don't have such problem in the Discrete case.

What am I missing here? Either there's no such "White Noise" (Why?) or There's a good explanation (Could someone derive it Mathematically) how to get the Delta in The Variance.


share|cite|improve this question

Apparently, saying that $(x_t)$ is a continuous white noise process simply refers to the fact that $(x_t)$ is a continuous time process, that is $(x_t)$ is indexed by a continuous parameter set. That is, the sample paths of $(x_t)$ are not assumed to be continuous, and in fact may be expected to be discontinuous at every fixed point (almost surely). Indeed, consider a continuous Gaussian white noise process. Then, $E[x_s x_t]=0$, $s \neq t$, implies that $x_s$ and $x_t$ are independent, and hence the sample paths must be discontinuous at every point (almost surely). The case of "Uniform White Noise" is essentially the same.

share|cite|improve this answer
Hello Shai. I don't care about how smooth are the paths. Continuous as I meant means t can get any real value. I want to know how to explain the contradiction between calculating the Variance using the PSD to calculating it by using the Distribution of the Variable at certain time. – Drazick Dec 14 '10 at 4:20
By the PSD definition (DC all over the Frequencies) E[Xt Xt] = Variance * Delta -> Infinite Energy. Yet using the Uniform Distribution you get 1/3, with no Delta, Finite Energy. – Drazick Dec 14 '10 at 8:26
No, there's a big difference. Using the delta notation means it has infinite amount of variance. A function f(x) which equals to 1 at x = 0 and zero everywhere else isn't a delta. A delta means that at zero its value is infinite. You are using Delta just like Indicator Function, Where it means something else. – Drazick Dec 14 '10 at 13:48
I now see your point. – Shai Covo Dec 14 '10 at 14:12
So probably you cannot assume $X_t \sim U[-1,1]$. Maybe the following example can help, in the setting of Gaussian white noise. If $Z_1,\ldots,Z_n$ are i.i.d. ${\rm N}(0,1)$, then their sum is ${\rm N}(0,n)$, which has mean zero but variance which tends to $\infty$. – Shai Covo Dec 14 '10 at 14:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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