# Understanding the White noise in $\Bbb{R}^2$

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 $z_{(0,1) \times (0,1)}$ ?

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 $[0,1)^2$.