Let $$w=X^TV^{-1}X$$ Where $X$ is a vector of Normally distributed variables with covariance matrix $V$. Then $w$ is chi-squared.

I have read (without an explanation) that w has degrees of freedom equal to the rank of $V$, rather than its dimension. This seems intuitive, but is there an easy proof to show this?

Ps. I think that $V^{-1}$ should be the Moore-Penrose generalized inverse, since if $V$ has not full rank, it cannot be inverted. Is that correct?


1 Answer 1


Let $X \sim \mathcal N(0, V)$ with $V \in \mathbb R^{p\times p}$.

First assume $V$ is invertible so $V^{-1/2}X \sim \mathcal N(0, I)$ and therefore $$ (V^{-1/2}X)^T(V^{-1/2}X) = X^TV^{-1}X \sim \chi^2_p. $$ $V$ being invertible means it is full rank so in this case $\text{rank } V = p$ and the two coincide.

Now suppose $V$ is not invertible, i.e. it is just PSD. Let $V = Q\Lambda Q^T$ be the eigendecomposition of $V$ and let $r =\text{rank } V < p$ so there are $p-r > 0$ eigenvalues that are exactly $0$.

Consider $$ X^TV^+X = X^TQ\Lambda^+Q^TX $$ where $V^+ = Q\Lambda^+ Q^T$ is the pseudoinverse of $V$ and $(\Lambda^+)_{ii}$ is $1/\lambda_i$ if $\lambda_i \neq 0$ and $0$ otherwise.

Partition $\Lambda$ as $$ \Lambda = \left(\begin{array}{c|c} L & 0\\ \hline0& 0\end{array}\right) $$ so $L \in \mathbb R^{r\times r}$ is diagonal with all non-zero entries, and the lower right block is $(p-r)\times (p-r)$. Then $$ \Lambda^+ = \left(\begin{array}{c|c} L^{-1} & 0\\ \hline0& 0\end{array}\right). $$ Let $Y = \sqrt{\Lambda^+}Q^TX$, where $\lambda_i \geq 0$ makes $\sqrt{\Lambda^+}$ well-defined, and note that $Y^TY = X^TV^+X$. Then $$ \text{Var}(Y) = \sqrt{\Lambda^+}Q^TQ\Lambda Q^T\sqrt{\Lambda^+} $$ $$ = \left(\begin{array}{c|c} I & 0\\ \hline0& 0\end{array}\right) := H. $$ Thus $Y_{r+1} = \dots=Y_p = 0$ so $$ X^TV^+X = Y^TY = \sum_{i=1}^r Y_i^2 \sim \chi^2_r. $$

Here's the proof that this $V^+$ really is the Moore-Penrose inverse. For a matrix $A$, the Moore-Penrose inverse $A^+$ needs to satisfy the following properties:

  1. $AA^+A=A$
  2. $A^+AA^+=A^+$
  3. $AA^+$ and $A^+A$ are symmetric

For (1) and (2), since $\Lambda^+\Lambda = \Lambda\Lambda^+ = H$, and $H$ acts as the identity on $\Lambda$ and $\Lambda^+$, we have $$ VV^+V = Q\Lambda \Lambda^+\Lambda Q^T = Q\Lambda Q^T = V $$ and $$ V^+VV^+ = Q\Lambda^+ \Lambda\Lambda^+ Q^T = V^+. $$

For (3), $$ V^+V = Q\Lambda^+\Lambda Q^T = QHQ^T = Q\Lambda \Lambda^+Q^T = VV^+ $$ so $VV^+ = V^+V$ and both are symmetric.

This proves that $V^+$ is indeed the Moore-Penrose inverse of $V$.


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