Axes of a contour

Topic: Covariance and the multivariate normal · more questions on this topic
‹ Uncorrelated but dependent Normal marginals without joint normality ›

For

\Sigma=\begin{pmatrix}2&1\\1&2\end{pmatrix}

and \boldsymbol{\mu}=\mathbf{0}, find the eigenvalues and eigenvectors of \Sigma, the directions and lengths of the semi-axes of the contour \mathbf{x}^\top\Sigma^{-1}\mathbf{x}=C, and the value of C for which the ellipse contains probability 0.95.

Worked answer (click to show)

Diagonalisation identifies the directions. The contour level determines how far to travel along them. The probability content is chosen after those two geometric steps.

The characteristic equation is (2-\lambda)^2-1=0, so \lambda_1=3 and \lambda_2=1. Solving (\Sigma-\lambda I)\mathbf{v}=\mathbf{0} gives

\boldsymbol{\gamma}_1=(1,1)^\top/\sqrt{2}, \qquad \boldsymbol{\gamma}_2=(1,-1)^\top/\sqrt{2}.

The contour \mathbf{x}^\top\Sigma^{-1}\mathbf{x}=C has semi-axes of length \sqrt{C\lambda_j} along \boldsymbol{\gamma}_j.

The squared Mahalanobis distance of a \operatorname{N}_2(\mathbf{0},\Sigma) vector is \chi^2_2, so probability 0.95 requires the upper 5\% point

C=\chi^2_{2;\,0.05}=-2\log 0.05=5.991.

The semi-axes are then \sqrt{5.991\times 3}=4.24 along (1,1) and \sqrt{5.991\times 1}=2.45 along (1,-1).