0
$\begingroup$

From my understanding, if you have a sample $x_{t_1},\dots,x_{t_n}$ of $X_{t_1},\dots,X_{t_n}$ which are iid $N(0,1)$, then $x_{t_1},\dots,x_{t_n}$ is a sample path of Gaussian white noise.

However, it is stated that the correlation function $c(s,t)$ of white noise is $\delta(s-t)$ where $\delta$ is the Dirac delta function. I understand that if $s\neq t$ then $c(s,t) = 0$. However, if $s=t$, you want $c(s-t)=1$. Why isn't the kronecker delta function used instead?

$\endgroup$
  • $\begingroup$ What is your definition of the Kronecker delta function? Be sure to identify all the symbols in what you write and tell us which are real numbers, which are integers, etc, Meanwhile, take a look at this answer. $\endgroup$ – Dilip Sarwate Mar 26 at 3:31
3
$\begingroup$

Dirac delta function has a continuous argument, but Kronecker delta function has a discrete argument. Your example is a discrete signal so Kronecker delta is used.

| improve this answer | |
$\endgroup$
  • $\begingroup$ But still, let's say that $X(t)$ is Gaussian white noise so that it is a continuous time stochastic process. If $c(s,t) = \delta(s-t)$, is the correlation function, does that mean that if $s=t$ then $c(s,t) \neq 1$ since $\delta(s-t)$ is the Dirac delta function? $\endgroup$ – user1237300 Mar 26 at 1:16
  • $\begingroup$ @user1237300 Yes, Dirac's delta function is not a well defined function, but surely its value at 0 can't be 1. $\endgroup$ – Mohammad M Mar 26 at 15:19
  • $\begingroup$ So the discrete Gaussian white noise has no continuous time equivalent? $\endgroup$ – user1237300 Mar 26 at 19:40
  • $\begingroup$ @user1237300 I'm not sure what did mean by equivalent, but continuous Gaussian white noise exist! $\endgroup$ – Mohammad M Mar 27 at 8:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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