# Gaussian white noise

I have two questions about a Gaussian white noise $\xi_t$:

1. First of all, $\xi_t$ is a white noise. So, it's variance should be infinitive. But if the process is Gaussian then it has finite variance. How can Gaussian white noise exist?
2. Say, we have to compute the function $u(t)= \sigma \xi_t$ at the moment $t=t_i$ during an integration of another complex system. If $\xi_t$ is a standard Gaussian white noise then we could simulate $\xi_t$ as a random number of standard normal distribution. And we get $u(t_i)=\sigma \hat{\xi_t},\, \hat{\xi_t} \in N(0,1)$. But in case the process isn't a Gaussian one we could write as follows: $$u(t) dt =\sigma dW,$$ where W - a Weiner process. Then we get $u(t_i)=\frac{\sigma}{\sqrt{\Delta t}}\hat{\xi_t}, \, \hat{\xi{_t}} \in N(0,1).$
Where do I make mistake?
• I'm not sure what you mean by your second question. How is your example not a Gaussian process? And what contradiction are you trying to point out? In general, switching freely between continuous in real time is probably what's screwing you up here since thr heart of this is all about subtleties of the continuum limit. – spaceisdarkgreen Sep 27 '17 at 18:36
• In the second question I want to realize how to simulate a white noise which is not Gaussian. As a variance of a white noise is infinite, the second approach looks like quite natural: if $\Delta t =t_{i+1}-t_i \longrightarrow 0$ then at the moment $t=t_i$ the variable $u(t_i)$ goes to infinite. But I'm still confused about it. Both question, probably, are about the same problem. – Artem Zefirov Sep 27 '17 at 18:56
• well that last part comports well with your first question where $"\xi(t)"$ has infinite variance. If you read my answer you will see why there is no such thing as $\xi(t)$. – spaceisdarkgreen Sep 27 '17 at 19:08

• I say it does not exist because we cannot do that and it literally does not exist. If it existed, it would be the derivative at $t$ of a nowhere differentiable function... can't happen. I did not say we can use an arbitrary distribution to get gaussian wn, I said an arbitrary dist would add up to a (gaussian) Brownian motion. It's imprecise as I said it, but the idea is that if you take a discrete random walk with any finite variance step distribution, at long timescales it will look like Brownian motion because the central limit says the sum converges in distribution to a normal. – spaceisdarkgreen Sep 27 '17 at 23:57