Hilbert transform of white noise 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.
 A: 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


*

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

*$E[\xi(g)\xi(h)] = \langle g,h\rangle$, 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.
