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I am studying White Noise. But I am really beginner level, so I have a confusion with its construction. White Noise is usually defined as a Wide Sense Stationary process $N=\{N_t\}_{t\in T}$ (for $T$ a time index set), that have a constant PSD, say $S_{NN}(f)=\sigma^2$. Since Correlation function $R_{NN}(t)=\mathcal{F}[S_{NN}(f)]$, we have also that $R_{NN}(t)=\sigma^2\delta(t)$ where $\delta$ is the Dirac delta. The thing is that $\mathbb{E}\{|N_t|^2\}=R_{NN}(0)=\sigma^2\delta(0)$. Indeed, if we assume that $N$ is ergodic we have that Power is also the Variance of the Process [$Var(N)=R_{NN}(0)-\mu_N(t)=R_{NN}(0)=\lim_{\tau\to\infty}\int_{-\infty}^{\infty}\frac{\mathbb{E}\{|\hat{N}^\tau(f)|^2\}}{2\tau}df$, where $\hat{N}^\tau$ is the Fourier transform of windowing of $N$ on a interval of lenght $2\tau$]. So, what I do not understand is how $N$ can be WSS if it has an infinite power and variance; because $\int_{-\infty}^{\infty}\sigma^2df=\infty.$

As you can see, I am requiring a process to have finite second moment for being WSS. Also, I have that this WN has identical distribution with zero mean and variance $\sigma^2$.

In short, my confusion lies on the fact that for a WN, $\sigma^2\delta(0)=R_{NN}(0)=\mathbb{E}\{N_t^2\}$ for every $t$. By hypothesis, $\sigma^2=\mathbb{E}\{N_t^2\}<\infty$. So, I see a contradiction where we have $\sigma^2\delta(0)=\sigma^2$.

I feel that I have a big misconception, but I do not know where the problem is. I really appreciate any help.

C David Reinach
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  • but it doesn't have infinite variance nor infinite power. It has infinite energy, but finite variance $\sigma^2$. Maybe you want to spell out why you think it has infinite energy! – Marcus Müller Dec 19 '21 at 15:00
  • Okay, maybe that is where my confusion lies. Why do you say that it have infinite energy from the construction? – C David Reinach Dec 19 '21 at 15:07
  • well, if it has nonzero constant power and you integrate that (over infinite time) or nonzeroconstant power spectral density and you integrate that (over infinite bandwidth), then you get an infinite value. – Marcus Müller Dec 19 '21 at 15:08
  • Exactly, but having a nonzero constante PSD means that power is the total area over infinite interval, thus having an infinite power. Indeed, in various text I have read that for a WN the power is infinite (that's the reason why they are not more than an ideal process). – C David Reinach Dec 19 '21 at 15:12
  • @CDavidReinach Strictly speaking, white noise has no variance (even though it's often defined as infinite). Personally, I think of white noise as not WSS -- however, any actual observation of white noise (i.e. through a finite-bandwidth filter) is WSS. – MBaz Dec 19 '21 at 16:18
  • @MarcusMüller put your comment as an answer— it’s short but to the point and clears up the OP’s confusion. I will upvote – Dan Boschen Dec 19 '21 at 16:27
  • I understand what you say, but my confusion lies on the fact that for a WN, $\sigma^2\delta(0)=R_{NN}(0)=\mathbb{E}{N_t^2}$ for every $t$. By hypothesis, $\sigma^2=\mathbb{E}{N_t^2}<\infty$. So, I see a contradiction where we have $\sigma^2\delta(0)=\sigma^2$. – C David Reinach Dec 19 '21 at 17:16
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    Does this answer your question? Is white noise WSS by nature or not? – Dilip Sarwate Dec 19 '21 at 17:45

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