# $E(X(t)X(t))=\sigma^2\delta(\tau)$ or $E(X(t)X(t))=\sigma^2$

(Question already asked on Math StackExchange)

Let's say we have a white noise process $$x(t)$$ such that:

$$E(X(t)X(t+\tau))=N\delta(\tau)$$

$$E(X(t))=0$$

In particular, with $$\tau=0$$, $$E(X(t)X(t))=E(X^2(t))$$ is infinite.

Now, I want $$X(t)$$ at each time $$t$$ to have a normal distribution of 0 mean and $$\sigma^2$$ variance. That is:

$$E(X^2(t))=\sigma^2$$

This is not consistent. I guess the expectations mean something different in both cases, but I don't find an explanation. This prevents me from moving ahead in a study of the mean, variance and autocorrelation of a process $$y(t)$$ defined as $$y(t)=1$$ if $$a and 0 otherwise.

• Is this what you are looking for dsp.stackexchange.com/questions/8629/…? – AlexTP Dec 23 '18 at 12:14
• It is not considered good practice to post the same question on two different groups. Be that as it may, you might want to read this question (which I asked) and its answers on math.SE. Basically, the answer I received is that while what you would like to have is desirable from your perspective and is something that mathematicians have studied, it is not at all useful for engineering applications such as the one you have in mind. – Dilip Sarwate Dec 23 '18 at 15:02
• To complement other comments, this is the way I see this problem. The white noise process does not have a variance. The good news is that we never observe white noise directly; we always observe it through an instrument that has finite bandwidth. The output of that instrument does have a variance, which can be calculated from the white noise PSD and the device's bandwidth and gain. – MBaz Dec 23 '18 at 20:16