non-central chi-square distribution.

#1
If \( X_i\) is distributed as normal with mean \(\mu\) and variance-covariance matrix \(D\),where \(D\) is the diagonal matrix, then show that \(X^TD^{-1}D\) is distributed as noncentral \(\chi^2\) with k degrees of freedom and noncentrality parameter \(\mu^TD^{-1}\mu=\lambda.\)
 

BGM

TS Contributor
#4
By independence (and as indicated by the question \( D \) is a diagonal matrix),

\( D = \begin{bmatrix} \sigma_1^2 & 0 & \ldots & 0 \\
0 & \sigma_2^2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \sigma_k^2 \end{bmatrix} \)

and thus its inverse is

\( D^{-1} = \begin{bmatrix}
\displaystyle \frac {1} {\sigma_1^2} & 0 & \ldots & 0 \\
0 & \displaystyle \frac {1} {\sigma_2^2} & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \displaystyle \frac {1} {\sigma_k^2} \end{bmatrix} \)

Now you merely need to use basic matrix multiplication to verify that

\( X^TD^{-1}X = \begin{bmatrix} X_1 & X_2 & \ldots & X_k \end{bmatrix}
\begin{bmatrix}
\displaystyle \frac {1} {\sigma_1^2} & 0 & \ldots & 0 \\
0 & \displaystyle \frac {1} {\sigma_2^2} & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \displaystyle \frac {1} {\sigma_k^2} \end{bmatrix}
\begin{bmatrix} X_1 \\ X_2 \\ \vdots \\ X_k \end{bmatrix} \) \(
= \sum_{i=1}^k \left(\frac {X_i} {\sigma_i}\right)^2
\)

which by definition follows a non-central \( \chi^2 \) distribution with \( k \) degrees of freedom. The method to verify the non-centrality parameter \( \lambda \) is similar.