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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 $X \sim N(A, B)$.

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  • $\begingroup$ $Cov(A + BZ) = Cov(A) + B^TCov(Z)B$. 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. $\endgroup$ – Thoth Jun 30 '15 at 16:57
  • $\begingroup$ 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/…. $\endgroup$ – Thoth Jun 30 '15 at 17:07
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Since $X$ is a vector

$$\mathbb{E}X=\mathbb{E}[A+BZ]=A$$

and

$$Var(X)=\mathbb{E}[(X-\mathbb{E}X)(X-\mathbb{E}X)']=\mathbb{E}[BZ(BZ)']$$ $$=\mathbb{E}[BZZ'B']=B\mathbb{E}[ZZ']B'=BIB'=BB'$$

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