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
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