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What is the Hilbert transform of a white noise $\xi(t)$?

By the Hilbert transform I mean: http://mathworld.wolfram.com/HilbertTransform.html

Thank you.

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Just a thought, use Karhunen-Loeve decomposition of Brownian motion, differentiate w.r.t. $t$ to obtain white noise, and find the Hilbert transform term-wise. But will the sum converge, I am not sure... –  Sasha Sep 1 '11 at 6:03
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Let me explain the background. For a real-valued signal $f(t)$ (oscillatory but noisy), by applying the Hilbert transform $H$ one can determine its phase $\phi(t)$: $f(t) + H[f(t)]i =A(t)e^{i\phi(t)}$. I would like to know how noise in the original data $f(t)$ can affect the precision of determination of $\phi(t)$. –  pharmine Sep 1 '11 at 15:41
    
Thank you Sasha for your comment. –  pharmine Sep 2 '11 at 11:59
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1 Answer 1

up vote 3 down vote accepted

The Hilbert transform of white noise is white noise.

Let $(\Omega,\mathcal{F},P)$ be a probability space. A map $\xi:L^2(\mathbb{R})\to L^2(\Omega)$ is called white noise on $\mathbb{R}$ if

  1. $(\xi(g_1),\ldots,\xi(g_n))$ has a mean zero, multivariate Gaussian distribution, for all $g_1,\ldots,g_n\in L^2(\mathbb{R})$, and

  2. $E[\xi(g)\xi(h)] = \langle g,h\rangle$, for all $g,h\in L^2(\mathbb{R})$.

Other notation for $\xi(g)$ includes $\xi(g)=\langle \xi,g\rangle=\int_\mathbb{R}\xi(t)g(t)\,dt$.

Let $H:L^2(\mathbb{R})\to L^2(\mathbb{R})$ denote the Hilbert transform and $H^*$ its dual. The Hilbert transform of white noise can be defined in the usual way by duality, so that $$ \langle H\xi,g\rangle = \langle \xi,H^*g\rangle. $$ Or, in other words, $H\xi=\xi\circ H^*:L^2(\mathbb{R})\to L^2(\Omega)$.

Since $(H\xi(g_1),\ldots,H\xi(g_n))=(\xi(H^*g_1),\ldots,\xi(H^*g_n))$, it follows that $H\xi$ satisfies (1) above. Also, $H^*=-H$ and $H^2=-I$, which gives $HH^*=I$. Therefore, \begin{align*} E[\langle H\xi,g\rangle\langle H\xi,h\rangle] &= E[\langle \xi,H^*g\rangle\langle \xi,H^*h\rangle]\\ &= \langle H^*g,H^*h\rangle\\ &= \langle g,HH^*h\rangle\\ &= \langle g,h\rangle. \end{align*} This verifies (2) and shows that $H\xi$ is a white noise.

Edit:

Although $\xi$ and $H\xi$ are both white noise, they are not the same white noise, and one may wonder about the relationship between the two. One way to study this relationship is to consider their integrals, which are two-sided Brownian motions.

A two-sided Brownian motion is a process $B$ of the form $$ B(t) = \begin{cases} B_1(t) &\text{if $t\ge 0$},\\ B_2(-t) &\text{if $t<0$}, \end{cases} $$ where $B_1,B_2$ are independent, standard Brownian motions. A two-sided Brownian motion can be constructed from a white noise $\xi$ on $\mathbb{R}$ by using $B_1(t)=\xi(1_{[0,t]})$ and $B_2(t)=\xi(1_{[-t,0]})$. We then have $$ \xi(g) = \int_{\mathbb{R}}g(t)\,dB(t), $$ for all $g\in L^2(\mathbb{R})$, where the above is the Itô integral.

Let $W$ be the two-sided Brownian motion constructed from $H\xi$ in the above manner. Then for $t>0$, $$ W(t) = H\xi(1_{[0,t]}) = \xi(H^*1_{[0,t]}) = \int_{\mathbb{R}} H^*1_{[0,t]}(s)\,dB(s). $$ Using $H^*=-H$ and the formula from the Wikipedia page, we have $$ W(t) = -\frac1\pi\int_{\mathbb{R}}\log\left|{\frac{s}{s-t}}\right|\,dB(s). $$ A similar formula can be derived for $t<0$. Together, these formulas give an explicit relationship between the two integrated white noises.

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Thanks so much! This gives me the motivation to learn more about the theory of probability. –  pharmine Sep 4 '11 at 12:15
    
I do have a problem with the accepted answer. I don't claim it's not correct or whatever, but I don't get something, if anyone could give more details that would be great. Here is my point: $\xi: L^2 \mapsto L^2$ (with the correct sets)... well why not. I m actually quite fine with that. Now, I start getting confused with the notations. –  Jean-Luc Bouchot Feb 27 '12 at 15:18
    
> Other notation for $\xi(g)$ includes $\xi(g)=\langle \xi,g\rangle= \int_\mathbb{R}ξ(t)g(t)dt.$ I really don't get that point. Given $g\in L^2$, the definition of $\xi$ (or at least how it was introduced), then $\xi(g)$ should be in $L^2$ too. I first thought this notation would be justified by using Riesz representation theorem but anyway, we need a linear functional, which we, at the moment, do not have. Am I missing something? –  Jean-Luc Bouchot Feb 27 '12 at 15:18
    
Now assuming my notation problems are not actually troubling. I don't get (this might be due to a lack of knowledge!) the conditions for $\xi$ to be white noise. The first one sounds strange to me. How can it be zero mean? I would have expected something like $\mathbb{E}[\xi(g)] = \mathbb{E}[g], \forall g$ –  Jean-Luc Bouchot Feb 27 '12 at 15:18
    
Is there any further references on that topic? I feel a bit lost with that. I have indeed a more practical background and am not familiar with those theoretical aspects. Assume I have a signal $s \in L^2$ and a noisy measurement of it $\widetilde{s} = s+n$ with $n$ being my white noise. What I normally assume, is that $\mathbb{E}[n] = 0$. In that case, how can I find the $\xi$ as introduced above and write my $\widetilde{s} = \xi(s) =? \langle\xi,s\rangle$ ? –  Jean-Luc Bouchot Feb 27 '12 at 15:18
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