# How to show that $|X|^2$ has the chi-squared distributions with r degrees of freedom?

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?

• $D$ 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.) Feb 2, 2022 at 21:28
• How do you take this transformation $Y$ here? Feb 2, 2022 at 21:32

## 1 Answer

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 $$|Y|^2=|X|^2$$.