# Show that two random variables are equal in distribution

I found this exercise at the end of a chapter about Brownian motion. Let $(X_j)_{j=1}^{2^M}$ be independent standard Gaussian random variables, where $M$ is a integer. Define $B_k:=2^{\frac{-M}{2}}\sqrt{T}\sum_{j=1}^kX_j$, where $0<T\in\mathbb{R}$. Furthermore let $(W_{k2^{-M}T})_{k=1}^{2^M}$ be a Brownian motion. Then $(W_{k2^{-M}T})_{k=1}^{2^M}=(B_k)_{k=1}^{2^M}$ in distribution. There is a hint: I should find a matrix $A$, which transform the $(X_k)_{k=1}^{2^M}$ into $(B_k)_{k=1}^{2^M}$.

So far I did the following:

For the matrix $2^{\frac{-M}{2}}\sqrt{T}A$ I've chosen: $$\begin{pmatrix} 1 & 0 & \cdots \\ 1 & 1 & 0 & \cdots \\ \vdots & \vdots & \vdots& 0 \\ 1 & \cdots & \cdots & 1 \end{pmatrix}$$

Then clearly $2^{\frac{-M}{2}}\sqrt{T}AX=B$, seen as vectors. Hence I know that $B$ is normal distributed with mean $0$ and covariance matrix $A\cdot \mathrm{Id}\cdot A^T$. For the product $A\cdot A^T$ I got:

$$\begin{pmatrix} 1 & 1 & \cdots \\ 1 & 2 & 2 & \cdots \\ \vdots & 2 & 3&3&\cdots \\ \vdots & 2 & 3 & 4& \cdots\\ \vdots & \vdots&\vdots&\vdots \end{pmatrix}$$ with the factor $2^{-M}T$ in front. Clearly $(W_{k2^{-M}T})_{k=1}^{2^M}$ has expectation zero and is normally distributed. But I do not see why the covariance matrices are the same. I hope someone could help me. Thanks in advance.

hulik

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The covariance matrices are the same. Indeed, if $j\leq k$, then \begin{align}\operatorname{Cov}(W_{k2^{-m}T},W_{j2^{-m}T})&=\operatorname{Cov}(W_{k2^{-m}T}-W_{j2^{—m}},W_{j2^{-m}T})+\operatorname{Cov}(W_{j2^{-m}T},W_{j2^{-m}T})\\ &=\operatorname{Cov}(W_{j2^{-m}T},W_{j2^{-m}T})\\ &=j2^{-m}T\\ &=2^{-m}T\min\{j,k\}, \end{align} which is exactly what was obtained.