Verifying Property of Constructed Brownian Motion using Haar Basis

35 Views Asked by At

This is an alternative proof to the construction of BM (to the one using Kolmogorov Extension), called Lévy construction. Let us define the Haar basis as $H_{k,l}$ and $Z_{k,l}\sim N(0,1)$ as IID random variables. Then define $G_k(t)=\sum_{l=0}^{2^{n-1}}Z_{k,l}\int_0^t H_{k,l}(s)ds$, and let our construction of Brownian motion be:

$$W_t=\sum_{k-1}^\infty G_k(t)$$

Recall that the Haar basis is $H_{k,l}(t)=2^{k/2}$ or $-2^{k/2}$ on alternating $2^{-k-1}$-length sub-intervals on the interval $[0,1]$.

What I don't understand is how this respects the property of BM, that is, $(B_t - B_s)$ is independent, mean 0, normally distributed of variance $(t-s)$. If we take, for example, $t=1/2$ and $s=0$, the above construction would include both the $G_2$ path of $N(0,1)/\sqrt{2}=N(0,1/2)$ and well as half of the $G_1$ path of $N(0,1)$ (the remaining paths vanishing to 0). Any clarification would be appreciated (and lmk if any is needed from my end).

1

There are 1 best solutions below

0
On

This is shown in The Haar functions and the Brownian motion and Construction of Brownian Motion using Haar wavelets. For $t>s$ we have

$$W_t-W_s=\sum_{k-1}^\infty G_k(t)-G_k(s)=\sum_{k-1}^\infty \sum_{l=0}^{2^{n-1}}Z_{k,l}\int_{s}^{t} H_{k,l}(r)dr.$$

Let $\Phi_{k,l}(s,t):=\int_{s}^{t} H_{k,l}(r)dr$. We study the correlation $t>s, v>u$

$$\begin{align*} \mathbb{E}((W(t)-W(s))(W(v)-W(u))) &\stackrel{1}{=} \lim_{N \to \infty} \mathbb{E}((W_N(t)-W_N(s))(W_N(v)-W_N(u))) \\ &\stackrel{2}{=} \lim_{N \to \infty} \mathbb{E} \left( \sum_{m,n=1}^N\sum_{z,l} \Phi_{m,z}(s,t)\Phi_{n,l}(u,v) Z_{m,z}Z_{n,l} \right) \\ &\stackrel{3}{=} \sum_{n=1}^{\infty}\sum_{l} \Phi_{n,l}(s,t)\Phi_{n,l}(u,v) \\ &\stackrel{4}{=} \int_0^1 1_{[s,t)}(x) 1_{[u,v)}(x) \, dx. \end{align*}$$

For step 3 we used iid and for step 4 we used the orthonormality of the Haar basis. The right-hand side equals $0$ if $[s,t) \cap [u,v) = \emptyset$; this shows that $W(t)-W(s)$ and $W(v)-W(u)$ are uncorrelated; hence, independent, as these form a Gaussian vector.

Reference: Schilling & Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Section 3.1.