Generating correlated random numbers: Why does Cholesky decomposition work?

Let's say I want to generate correlated random variables. I understand that I can use Cholesky decomposition of the correlation matrix to obtain the correlated values. If $C$ is the correlation matrix, then we can do the cholesky decomposition:

$LL^{T}=C$

Then I can easily generate correlated random variables:

$LX=Y$,

where $X$ are uncorrelated values and $Y$ are correlated values. If I want two correlated random variables then $L$ is:

$L = \left[ {\begin{array}{*{20}c} 1 & 0 \\ \rho & {\sqrt {1 - \rho ^2 } } \\ \end{array}} \right]$

I understand that this works, but I don't really understand why... My question is: Why does this work?

• Isn't C here the covariance matrix ? It is mentioned as Correlation matrix – ThePredator Jul 25 at 6:36

The co-variance matrix of any random vector $Y$ is given as $\mathbb{E} \left(YY^T \right)$, where $Y$ is a random column vector of size $n \times 1$. Now take a random vector, $X$, consisting of uncorrelated random variables with each random variable, $X_i$, having zero mean and unit variance $1$. Since $X_i$'s are uncorrelated random variables with zero mean and unit variance, we have $\mathbb{E} \left(X_iX_j\right) = \delta_{ij}$. Hence, $$\mathbb{E} \left( X X^T \right) = I$$ To generate a random vector with a given covariance matrix $Q$, look at the Cholesky decomposition of $Q$ i.e. $Q = LL^T$. Note that it is possible to obtain a Cholesky decomposition of $Q$ since by definition the co-variance matrix $Q$ is symmetric and positive definite.

Now look at the random vector $Z = LX$. We have $$\mathbb{E} \left(ZZ^T\right) = \mathbb{E} \left((LX)(LX)^T \right) = \underbrace{\mathbb{E} \left(LX X^T L^T\right) = L \mathbb{E} \left(XX^T \right) L^T}_{\text{ Since expectation is a linear operator}} = LIL^T = LL^T = Q$$ Hence, the random vector $Z$ has the desired co-variance matrix, $Q$.

• But how do you specify whether the correlations of those variables are going to be positive or negative? – f1r3br4nd Jul 3 '14 at 12:23
• So it should work for any decomposition of $Q=LL^T$, right? I mean, $L$ does not necessarily have to be triangular, as long as it satisfies that equation. Am I right? – Michael Apr 1 '16 at 17:50
• Why do you state that covariance is given by $\mathbb{E} \left(YY^T \right)$? What happen to the $\mathbb{E} \left(Y\right)\mathbb{E} \left(Y\right)^T$ term? – Confounded Nov 7 '17 at 11:47
• @Confounded He probably meant centered variables (zero mean) – Michael Nov 13 '18 at 20:42

The specified correlation, C, can have negative element, i.e. Negative correlation. But, C, must Be positive definite to be a proper corr. matrix. E.g. [ 1 -0.9 0.5 -0.9 1 -0.1 0.5 -0.1 1 ] Is ok, but 0.9,-0.5,0.1 (in all places) produces a negative definite matrix, so not ok. Actually, for all values positive will have the effect that the sign of the second random variable changes.

protected by SaadOct 3 '18 at 14:13

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