One can define the Gaussian white noise in $\Bbb{R}^d$ as a distribution on test functions

$$\xi(\phi) \sim N(0, \|\phi\|_{L^2}) \quad \forall \, \phi \in S(\Bbb{R}^d)$$

This definition if very abstract and does not seem to be unique one. I mean, I suspect that there is another way do construct a space white noise.

Let's begin in $\Bbb{R}$. Here we can think of the White noise as the derivative of the Brownian motion. If you fix a certain $\phi$ smooth the object you are interested in is $\int \phi(t) \, dB_t$. This is as if you had a strategy $\phi$ for the amount of a certain stock on the market ( the value of such a stock is modelled by the Brownian motion). In this case the value $\xi(\phi)$ is the value of the wealth you will have after following this strategy. This is OK.

We can think that in each interval $ I = (a,b)$ there is a normal random variable $z_I$ with variance $b-a$. These normals are independent if the intervals are disjoints

How do we carry this way of thinking to $\Bbb{R}^2$?

My trouble is the following. One could similarly just say that in each Block $(a,b)\times(c,d)$ of the plane one has a normal random variable of variance $(b-a)(d-c)$. These normals are independent if the regions are disjoints. However this is not as clear as the previous one dimensional case, since in $\Bbb{R}$ one can from a normal variable $z_{(0,1)}$ defined in $(0,1)$ say, construct independent random variables $z_{(0,1/2)}$ and $z_{(1/2,1)}$ on $(0,1/2)$, $(1/2,1)$ such that $$z_{(0,1/2)} +z_{(1/2,1)}= z_{(0,1)}$$

This procedure allows us to refine the normals distributed in the space.

Is there an analogous way to construct normals in the $(0,1) \times (0,1)$ that are independent and that summed together yield $z_{(0,1) \times (0,1)}$ ?

enter image description here


Yes, there is.

It suffices to use the same idea in the Paul-Levy construction:

divide the interval $[0, 1)$ in the disjoint intervals $[0,1/4)$,$[1/4,2/4)$,$[2/4,3/4)$,$[3/4, 1)$.

Then remember the construction as explained in the book Brownian motion of Yuval and Morters:

Note that we have already done this for $\mathcal{D}_0=\{0,1\}$. Proceeding inductively we may assume that we have succeeded in doing it for some $n-1$. We then define $B(d)$ for $d\in\mathcal{D}_n\backslash\mathcal{D}_{n-1}$ by $$B(d)=\dfrac{B(d-2^{-n})+B(d+2^{-n})}2+\dfrac{Z_d}{2^{(n+1)/2}}.$$ Note that the first summand is the linear interpolation of the values of $B$ at the neighbouring points of $d$ in $\mathcal{D}_{n-1}$. Therefore $B(d)$ is independent of $(Z_t:t\in\mathcal{D}\backslash\mathcal{D}_n)$ and the second property is fulfilled.

Moreover, as $\frac12[B(d-2^{-n})+B(d+2^{-n})]$ depends only on $(Z_t:t\in\mathcal{D}_{n-1})$, it is independent of $Z_d/2^{(n+1)/2}$. Both terms are normally distributed with mean zero and variance $2^{-(n+1)}$. Hence their sum $B(d)-B(d-2^{-n})$ and their difference $B(d+2^{-n})-B(d)$ are independent and normally distributed with mean zero and variance $2^{-n}$ by Corollary II.3.4.

So you can obtain the random variables you are looking for and define for every Dyadic set $C$ (i.e. for every set $C$ of the form $C = \cup_i D_i$ where $D_i = [d_i, \tilde d_i) \times [d'_i, \tilde d'_i)$ $D_i \cap D_j = \emptyset$ )

$$W(C) = \sum_{i: D_i \subset C} z_i$$

where $z_i$ are independent normal random variables with mean $0$ and variance $\operatorname{Vol}(D_i)$.

This construction is well defined since we have a compatibility in the decomposition of the diadic blocks that can be obtained by analogous reasoning as we did in the simple case of the block $[0,1)^2$.

| cite | improve this answer | |

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.