Two factors of one covariance

Topic: Estimation and inference for the multivariate normal · more questions on this topic
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Take

\Sigma=\begin{pmatrix}4&2\\2&3\end{pmatrix}.

Find a lower-triangular L with LL^\top=\Sigma (the Cholesky factor), and also the matrix B=\Gamma\Lambda^{1/2} from the spectral decomposition. Show that both give \operatorname{Var}(B\mathbf{Z})=\Sigma for \mathbf{Z}\sim\operatorname{N}_2(\mathbf{0},I), although L\neq B.

Worked answer (click to show)

A simulation factor need only satisfy the required matrix product. Different valid factors can map the same normal draw to different points without changing the resulting distribution.

The Cholesky factor has the form

L=\begin{pmatrix}l_{11}&0\\ l_{21}&l_{22}\end{pmatrix}

with l_{11}^2=4, l_{11}l_{21}=2 and l_{21}^2+l_{22}^2=3. So l_{11}=2, l_{21}=1 and l_{22}=\sqrt{2}:

L=\begin{pmatrix}2&0\\1&1.414\end{pmatrix}, \qquad LL^\top=\begin{pmatrix}4&2\\2&1+2\end{pmatrix}=\Sigma.

For the spectral factor, the eigenvalues solve \lambda^2-7\lambda+8=0, so

\lambda_1=(7+\sqrt{17})/2=5.562, \qquad \lambda_2=(7-\sqrt{17})/2=1.438,

with unit eigenvectors \boldsymbol{\gamma}_1=(0.788,\ 0.615)^\top and \boldsymbol{\gamma}_2=(-0.615,\ 0.788)^\top. Then

B=\Gamma\Lambda^{1/2}=\begin{pmatrix}1.859&-0.738\\1.451&0.945\end{pmatrix}.

For either factor M\in\{L,B\},

\operatorname{Var}(M\mathbf{Z})=MIM^\top=MM^\top=\Sigma.

The factors differ because MM^\top=\Sigma determines M only up to a right multiplication by an orthogonal matrix Q: (MQ)(MQ)^\top=MM^\top. The simulated vectors differ draw by draw and have the same distribution.