# Cholesky decomposition and variance

Well, I've been reading about simulating correlated data and I've come across Cholesky decomposition. Everything seemed clear until I found a couple of posts on this site and Cross-Validated that showed a way to alter mean and variance of simulated data. The proposed solution is as follows:

Let $Z$ be a set of uncorrelated random variables normally distributed with mean 0 and variance 1, i.e.

$$Z \sim N(0, I)$$ Then if we make an affine transformation

$$X \equiv A + BZ$$ $X$ will have a distribution $$X \sim N(A, B{B}^{T})$$ So given a covariance matrix $\Sigma$ we can find $B$ using the Cholesky decomposition $\Sigma = B{B}^{T}$.

So, I don't feel like I understand why the application of an affine transformation of form $X \equiv A + BZ$ results in $X \sim N(A, B{B}^{T})$ instead of $X \sim N(A, B)$.

• $Cov(A + BZ) = Cov(A) + B^TCov(Z)B$. This is a direct result of the linearity of expectation. However it should be clear right away that it definitely can't be just $B$, since the covariance matrix must be positive semi-definite. – Thoth Jun 30 '15 at 16:57
• It actually may be better to write it as just $Cov(A+BZ)=B^TCov(Z)B$, since of course adding a constant has no effect on variance and so $Cov(A)=0$, a reference can be found here: en.wikipedia.org/wiki/…. – Thoth Jun 30 '15 at 17:07

Since $X$ is a vector
$$\mathbb{E}X=\mathbb{E}[A+BZ]=A$$
$$Var(X)=\mathbb{E}[(X-\mathbb{E}X)(X-\mathbb{E}X)']=\mathbb{E}[BZ(BZ)']$$ $$=\mathbb{E}[BZZ'B']=B\mathbb{E}[ZZ']B'=BIB'=BB'$$