Cholesky decomposition & Correlated Normals

I was talking to a friend of mine the other day, about, financial science of course, and the question came up: “I know that by transforming a vector of uncorrelated normals with the cholesky-factoring the correlation matrix will result in a vector of correlated normals… but why does that work?”. I was going to rig up the argument in a hurry, in a bar, on the back of a napkin, but I couldn’t recall it. I promised my friend I’d give him a very simple one-liner argument and the intention here is to fulfill the promise.

First, recall that the correlation matrix of a vector  \overrightarrow{y} of unit normals is given by

\displaystyle \Big\langle \overrightarrow{y} \cdot \overrightarrow{y}^\textrm{T} \Big\rangle = \left( \begin{array}{cccc}1 & \rho_{12} & \cdots & \rho_{1N} \\ \rho_{21} & 1 & \vdots & \vdots \\ \vdots & \cdots & \ddots & \rho_{N-1,N} \\ \rho_{N1} & \cdots & \rho_{N,N-1} & 1 \\ \end{array} \right) = \rho

(Notice by the way, that there is an easy fix to the stupid 038 appearing in the matrix, fix explained here but for some reason I don’t have write access to the file /wp-content/plugins/latexrender/latex.php so we’ll have to wait for this to be legible.)

Anyway, that was just to recall what the correlation matrix was. Now assume we Cholesky factorize it as follows:

\displaystyle C^\textrm{T} \cdot C = \rho

and consider the outcome of an uncorrelated vector normal transformed by the cholesky factor:

\displaystyle \overrightarrow{z} = C \cdot \overrightarrow{x}

We know that the this must be a unit vector normal, the only thing left to determine is its correlation matrix:

\displaystyle \Big\langle \overrightarrow{z} \cdot \overrightarrow{z}^\textrm{T} \Big\rangle  = \Big\langle C \cdot \overrightarrow{x} \cdot \overrightarrow{x}^\textrm{T} \cdot C^\textrm{T} \Big\rangle

But the cholesky matrix is deterministic and comes out of the expectation, and using the uncorrelated property of x we get:

\displaystyle \Big\langle \overrightarrow{z} \cdot \overrightarrow{z}^\textrm{T} \Big\rangle  = C \cdot \Big\langle \overrightarrow{x} \cdot \overrightarrow{x}^\textrm{T} \Big\rangle \cdot C^\textrm{T} = C \cdot C^\textrm{T} = \rho^\textrm{T} = \rho

And since the only thing which prevented the last two equations to fit in one line is the unusually narrow margins here, the argument has been presented as a one-liner.

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