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.

  • 7
    $\begingroup$ 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
    Commented Mar 3, 2014 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
    Commented Mar 4, 2014 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
    Commented Mar 4, 2014 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
    Commented Mar 4, 2014 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
    Commented Mar 5, 2014 at 17:04

1 Answer 1


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.


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)
  • $\begingroup$ I am curious. I have been delving into ratios of quadratic forms, which seems to be the type of structure that this problem question has. There's a considerable literature on the topic (e.g. sciencedirect.com/science/article/pii/S016794730200213X), but it doesn't seem like people use this approach to solve those problems. I don't see why this approach shouldn't apply to that large class of problems, but I don't think these people would have missed it. So, am I missing a reason why this approach wouldn't generalize to those problems? thanks $\endgroup$
    – dherrera
    Commented Mar 6, 2023 at 18:43
  • 1
    $\begingroup$ @dherrera This thread concerns iid Normal variables with zero means. The general problem with zero means but differing variances is a sum of Gamma distributions, addressed at stats.stackexchange.com/questions/72479 (where you can see what complications ensue). When the means are nonzero, you're in the realm of non-central chi-squared distributions and worse. $\endgroup$
    – whuber
    Commented Mar 6, 2023 at 19:06
  • $\begingroup$ thanks for the explanation $\endgroup$
    – dherrera
    Commented Mar 6, 2023 at 19:51

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