11
$\begingroup$

Assume that $ X = X_1 + X_2+\cdots+ X_n $ where $X_i \sim N(0,\sigma^2)$ are independent.

My question is, what distribution does

$$ Z = \frac{X^2}{X_1^2 + X_2^2 + \cdots + X_n^2}$$

follow? I know from here that the ratio of two chi-squared random variables expressed as $\frac{W}{W + Y}$ follows a Beta distribution. I think that this assumes independence between $W$ and $Y$. In my case though, the denominator of $Z$ contains the components of $X$ squared.

I think $Z$ must also follow a variation of the Beta distribution but I am not sure. And if this assumption is correct, I don't know how to prove it.

$\endgroup$
  • 6
    $\begingroup$ Because the distribution of the denominator is invariant under rotations, you can rotate $X$ to equal $\sqrt{n}X_1$, which reduces your question to something familiar :-). $\endgroup$ – whuber Mar 3 '14 at 22:23
  • 1
    $\begingroup$ I'm pretty sure @whuber means exactly what was typed there. When you say 'nominator' do you mean 'numerator'? $\endgroup$ – Glen_b Mar 4 '14 at 4:34
  • 3
    $\begingroup$ When you rotate anything you (by definition) preserve its length. Therefore the variance of any rotated version of $X$ must equal the variance of $X$, which is $1+1+\cdots+1=n$: that's where the $\sqrt{n}$ term comes from. $\endgroup$ – whuber Mar 4 '14 at 15:37
  • 1
    $\begingroup$ @whuber Your answer seems very interesting indeed but I have some doubts about it. When you say that I can rotate $X$ to become equal to $\sqrt nX_1$, this basically means that I can rewrite the numerator of $Z$ as $nX_1^2$ and consequently, $Z$ itself turns into $n\frac{X_1^2}{X_1^2+X_2^2+\cdots+X_n^2}$. Now, if I assume $W=X_1^2$ and $Y=X_2^2+\cdots+X_n^2$ and since $W$ and $Y$ are independent, I can assume that $Z=n\frac{W}{W+Y}$ has a $\beta$ distribution and so forth. Am I getting your point up to now? So, here is my confusion. Before using the concept of rotational invariance and modifyi $\endgroup$ – ssah Mar 4 '14 at 17:45
  • 2
    $\begingroup$ @ssah You err in your application of my reasoning: without the $X_1^2$ in the denominator, its distribution is no longer invariant to arbitrary rotations of $(X_1,\ldots, X_n),$ and so the conclusions no longer hold. $\endgroup$ – whuber Mar 5 '14 at 17:04
7
$\begingroup$

This post elaborates on the answers in the comments to the question.


Let $X = (X_1, X_2, \ldots, X_n)$. Fix any $\mathbf{e}_1\in\mathbb{R}^n$ of unit length. Such a vector may always be completed to an orthonormal basis $(\mathbf{e}_1, \mathbf{e}_2, \ldots, \mathbf{e}_n)$ (by means of the Gram-Schmidt process, for instance). This change of basis (from the usual one) is orthogonal: it does not change lengths. Thus the distribution of

$$\frac{(\mathbf{e}_1\cdot X)^2}{||X||^2}=\frac{(\mathbf{e}_1\cdot X)^2}{X_1^2 + X_2^2 + \cdots + X_n^2} $$

does not depend on $\mathbf{e}_1$. Taking $\mathbf{e}_1 = (1,0,0,\ldots, 0)$ shows this has the same distribution as

$$\frac{X_1^2}{X_1^2 + X_2^2 + \cdots + X_n^2}.\tag{1} $$

Since the $X_i$ are iid Normal, they may be written as $\sigma$ times iid standard Normal variables $Y_1, \ldots, Y_n$ and their squares are $\sigma^2$ times $\Gamma(1/2)$ distributions. Since the sum of $n-1$ independent $\Gamma(1/2)$ distributions is $\Gamma((n-1)/2)$, we have determined that the distribution of $(1)$ is that of

$$\frac{\sigma^2 U}{\sigma^2 U + \sigma^2 V} = \frac{U}{U+V}$$

where $U = X_1^2/\sigma^2 \sim \Gamma(1/2)$ and $V = (X_2^2 + \cdots + X_n^2)/\sigma^2 \sim \Gamma((n-1)/2)$ are independent. It is well known that this ratio has a Beta$(1/2, (n-1)/2)$ distribution. (Also see the closely related thread at Distribution of $XY$ if $X \sim$ Beta$(1,K-1)$ and $Y \sim$ chi-squared with $2K$ degrees.)

Since $$X_1 + \cdots + X_n = (1,1,\ldots,1)\cdot (X_1, X_2, \cdots, X_n) = \sqrt{n}\,\mathbf{e}_1\cdot X$$

for the unit vector $\mathbf{e}_1=(1,1,\ldots,1)/\sqrt{n}$, we conclude that $Z$ is $(\sqrt{n})^2 = n$ times a Beta$(1/2, (n-1)/2)$ variate. For $n\ge 2$ it therefore has density function

$$f_Z(z) = \frac{n^{1-n/2}}{B\left(\frac{1}{2}, \frac{n-1}{2}\right)} \sqrt{\frac{(n-z)^{n-3}}{z}}$$

on the interval $(0,n)$ (and otherwise is zero).


As a check, I simulated $100,000$ independent realizations of $Z$ for $\sigma=1$ and $n=2,3,10$, plotted their histograms, and superimposed the graph of the corresponding Beta density (in red). The agreements are excellent.

Figure

Here is the R code. It carries out the simulation by means of the formula sum(x)^2 / sum(x^2) for $Z$, where x is a vector of length n generated by rnorm. The rest is just looping (for, apply) and plotting (hist, curve).

for (n in c(2, 3, 10)) {
  z <- apply(matrix(rnorm(n*1e5), nrow=n), 2, function(x) sum(x)^2 / sum(x^2))
  hist(z, freq=FALSE, breaks=seq(0, n, length.out=50), main=paste("n =", n), xlab="Z")
  curve(dbeta(x/n, 1/2, (n-1)/2)/n, add=TRUE, col="Red", lwd=2)
}
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.