The question is Suppose that the random vector $X$ has a centered k-dimensional normal distribution whose covariance matrix has 1 as an eigenvalue of multiplicity $r$ and $0$ as an eigenvalue of multiplicity $k-r$. Show that $|X|^2$ has the chi-squared distributions with r degrees of freedom?

For this question, I need to show that $|X|^2=\sum_{I=1}^r z_i^2$ where $\{z_i\}$ are $i.i.d$ $N(0,1)$.

Now, there exist an orthogonal matrix $U$ and a diagonal matrix $D$ such that $U^t \Sigma U= D$ where $\Sigma$ is the given covariance matrix.Then,

$\Sigma=AA^t$, where $A=UD_0$ and $D_0=\sqrt{D}$.

I am not sure how do I make conclusion here. Can anyone give me some hints how do I solve this question?

  • $\begingroup$ $D$ will have $r$ ones in the diagonal and $k-r$ zeroes. So when you do the classical transformation $Y = \Sigma^{-\frac{1}{2}} X$ you will find the $Y$ is a vector with $r$ entries that are independent $\mathsf{Norm}(0,1)$ and $k-r$ zeroes. (I believe this is what will happen. Try it.) $\endgroup$
    – William M.
    Feb 2, 2022 at 21:28
  • $\begingroup$ How do you take this transformation $Y$ here? $\endgroup$
    – User124356
    Feb 2, 2022 at 21:32

1 Answer 1


By construction, the diagonal $D$ consists of the eigenvalues of the covariance matrix of $X$.

Define the random vector $Y:=U^tX$. Argue that:

  1. $Y$ has multivariate normal distribution.

  2. $Y$ has mean vector zero.

  3. $Y$ has covariance matrix $D$.

Since $D$ is diagonal with $r$ ones and $r-k$ zeros, conclude that the vector $Y$ has $r$ components which are iid standard normal variables, with the remaining $k-r$ components equal to zero. Thus the sum of the squares of the components in $Y$ has chi-squared distribution with $r$ degrees of freedom.

Finally, argue that $|Y|^2=|X|^2$.


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