I am looking at some practise questions and stumped on one. Likely just been sat down for too long, but I tried two days in a row and not sure how to proceed.
I was given a Covariance matrix M: \begin{matrix} a^2+b^2 & 0 & 2ab \\ 0 & 4 & 0 \\ 2ab & 0 & a^2+b^2 \end{matrix}
My tutor in a video stated the below result for M^(1/2) ... is there some method or trick to get there?
\begin{matrix} a & 0 & b \\ 0 & 2 & 0 \\ b & 0 & a \end{matrix}
and how would one proceed from here to get the brownian motions? would you simply state a column vector has components of independent Brownian Motions? I have a feeling it's something to do with Spectral Decomposition (M=QDQ^T) but really not sure.
I had a look at Missing a trick: 3D Brownian Motion from a covariance matrix but couldnt follow...
Thanks for any help!
In this case $M^{1/2}$ can be easily guessed and verified. In general, $M^{1/2}$ can be obtained from diagonalizing $M$: If $M=QDQ^T$ Then $M^{1/2}=QD^{1/2}Q^T$.
Let $B=(B_1,B_2,B_3)^T$ be standard Brownian motion in three dimensions, that has covariance matrix $E[B(t)B^T(t)]=tI$ at time $t$. Then $W(t)= M^{1/2} B(T)$ is a Brownian motion with covariance matrix $$E[W(t)W^T(t)]= M^{1/2} E(B(t) B^T(t))M^{1/2} =M^{1/2} \cdot t \cdot M^{1/2} =tM \,$$ at time $t$.