A transformation is not a distributional assumption

Topic: Estimation and inference for the multivariate normal · more questions on this topic
‹ Normal conditioning and whitening

Let \mathbf{X}\sim\operatorname{N}_2(\boldsymbol{\mu},\Sigma) with \boldsymbol{\mu}=(0,2)^\top and

\Sigma=\begin{pmatrix}2&1\\1&3\end{pmatrix}.

  1. Derive the mean and variance of Y=X_1-2X_2 and state its distribution.

  2. Explain how the covariance matrix determines the orientation and relative axis lengths of equal-density contours.

  3. A different, non-normal dataset is whitened to have sample covariance I. Does this establish independent standard normal observations? Explain.

A linear image of a jointly normal vector is normal. Its mean and variance identify the resulting univariate distribution. Whitening asks a separate question about a whole coordinate system.

  1. With \mathbf{a}=(1,-2)^\top,

\mathbb{E}(Y)=0-2\cdot 2=-4, \qquad \operatorname{Var}(Y)=2+4\cdot 3-4\cdot 1=10.

The cross term in the variance is 2 a_1 a_2\sigma_{12}=2(1)(-2)(1)=-4. Linear combinations of a jointly normal vector are normal, so Y\sim\operatorname{N}(-4,\ 10).

  1. Orthogonal eigenvectors give the principal axes. An ellipsoid (\mathbf{x}-\boldsymbol{\mu})^\top\Sigma^{-1}(\mathbf{x}-\boldsymbol{\mu})=c has semi-axis lengths \sqrt{c\lambda_j}. Larger eigenvalues mean more spread. The mean translates the centre and does not change the orientation.

  2. Whitening controls empirical first and second moments. It does not determine the shape of the population distribution, establish independence between coordinates, or make observations independent across rows. Even whitened jointly normal population coordinates would need the population transformation for that exact population claim.