Help for HW1 ECE662 Spring 2012
{ Summary: To generate "colored" samples $ \tilde{x}\in\mathbb{R}^n \sim \mathcal{N}(\mu,\Sigma) $ from "white" samples $ x $ drawn from $ \mathcal{N}(\vec{0},I_n) $, simply let $ \tilde{x} = Ax + \mu $, where $ A $ is the Cholesky decomposition of $ \Sigma $, i.e. $ \Sigma = AA^T $}
Consider generating samples $ \tilde{x}\in\mathbb{R}^n \sim \mathcal{N}(\mu,\Sigma) $. Many platforms (e.g. Matlab) have a random number generator to generate iid samples from (white) Gaussian distribution. If we seek to "color" the noise with an arbitrary covariance matrix $ \Sigma $, we must produce a "coloring matrix" $ A $. Let us consider generating a colored sample $ \tilde{x} = [\tilde{x}_1,\tilde{x}_2,\ldots,\tilde{x}_n]^T $ from $ x = [x_1,x_2,\ldots,x_n]^T $, where $ x_1, x_2, \ldots, x_n $ are iid samples drawn from $ \mathcal{N}(0,1) $. (Note: Matlab has a function, mvnrnd.m, to sample from $ \mathcal{N}(\mu,\Sigma) $, but I discuss here the theory behind it). Relate $ \tilde{x} $ to $ x $ as follows:
$ \begin{align} \tilde{x}_1 &= a_{11} x_1 \\ \tilde{x}_2 &= a_{21} x_1 + a_{22} x_2 \\ &... \\ \tilde{x}_n &= \sum_{i=1}^n a_{ni}x_i \\ \end{align} $
We can rewrite this in matrix form as $ \tilde{x} = Ax $, where matrix $ A $ is lower triangular. We have, then, that
$ E[\tilde{x}_n] = \sum_{i=1}^n a_{ni}E[x_i] = 0 $, and
$ Cov[\tilde{x}_n,\tilde{x}_m] = E\left[\left(\sum_{i=1}^na_{ni}x_i\right)\left(\sum_{j=1}^m a_{mj}x_j\right)\right] = \sum_{i=1}^n\sum_{j=1}^m a_{ni}a_{mj}E[x_ix_j] \Rightarrow $
$ Cov(\tilde{x}_n,\tilde{x}_m) $ = $ \sum_{i=1}^{\min(m,n)}a_{ni}a_{mi} $, since $ x_i $'s are independent, $ E[x_i] = 0 $ and $ Var[x_i] = 1 $.
We are now left with the problem of defining $ a_{ni} $'s so that the form of $ Cov(\tilde{x}_n,\tilde{x}_m) $ follows the form of $ \Sigma_{nm} $: i.e.
$ \Sigma_{nm} $ = $ Cov(\tilde{x}_n,\tilde{x}_m) $ = $ \sum_{i=1}^{\min(m,n)}a_{ni}a_{mi} $
$ \Rightarrow \Sigma = AA^T $, where $ A $ is lower triangular, and $ \Sigma $ is positive definite. Therefore, $ A $ follows the form of what is called the Cholesky decomposition of $ \Sigma $.
