(I hope this question may be too easy for some experts here). It is well-known that if $X$ is normal rv and $Y$ is a chi-square rv, then $Z=X/\sqrt{Y/n}$ follows student's $t$ distribution, where $n$ is the degrees of freedom of $Y$, and $X$ and $Y$ need to be independent.

However, what comes if $X$ and $Y$ are dependent? More specifically, $X_1$ and $X_2$ are independent normal rvs, then $Z=X_1/\sqrt{X_1^2+X_2^2}$ would follow what kind of distribution?

  • 2
    $\begingroup$ The initial assertion is incorrect unless you specify that $X$ has zero mean and unit variance (although obviously changing the variance only rescales $Z$). Upon adopting the same assumption for $X_1$ and $X_2$--namely, that they have zero means and their variances are equal, integration gives the PDF of $Z$ as $dz/(\pi\sqrt{1-z^2})$ (a rescaled Beta distribution). If they have different variances and $\sigma^2$ is the ratio of the variance of $X_2$ to that of $X_1$, the PDF generalizes to $\frac{\sigma }{\pi \sqrt{1-z^2} \left(1+\left(\sigma^2-1\right)\right)z^2}$ (which can be trimodal). $\endgroup$
    – whuber
    Commented Mar 17, 2013 at 16:20

2 Answers 2


Let $X = (X_1, \ldots, X_n)$ be independently and Normally distributed with common mean $0$ and common standard deviation $\sigma$. The ratio

$$Z = \frac{X_1}{\sqrt{X_1^2 + \cdots + X_n^2}} = \frac{X\cdot \nu}{|X||\nu|}$$

for $\nu = (1, 0, \ldots, 0)$ is the correlation coefficient of $X$ and $\nu$. We could just as well replace $\nu$ with an independent multinormal vector $Y$ having the same distribution as $X$, because the correlation coefficient is the cosine of the angle and the distribution of the angle between independent $X$ and $Y$ is the distribution of the angle between $X$ and any fixed vector (we may rotate $Y$ into that fixed vector without changing the distribution of $X$). If you have any trouble believing this, consider the case $n=2$, where the claim is that the distribution of the angle made between a random point on the circle and the y-axis is the same as the distribution of the angle made between two random points on the circle.

Consequently, $Z$ has the same distribution as the sample correlation coefficient for a sample of size $n+1$ from a Binormal distribution with correlation parameter $\rho=0$. (The sample size is one greater than $n$ because in computing the sample correlation, one degree of freedom is lost through the centering process.) We may therefore avail ourselves of the well known result that this distribution has pdf

$$f(z) = \frac{(1-z^2)^{(n+1-4)/2}}{B(\frac{1}{2}, \frac{n+1-2}{2})} = \frac{(1-z^2)^{(n-3)/2}}{2^{n-2} B(\frac{n-1}{2},\frac{n-1}{2})}.$$

This is a Beta$(\frac{n-1}{2},\frac{n-1}{2})$ distribution scaled to the interval $[-1, 1]$.

If instead of $X_1$ we use any nonzero linear combination $Z_a = a_1 X_1 + \cdots + a_n X_n$, upon writing this as $|a|(\alpha_1 X_1 + \cdots + \alpha_n X_n)$ with $\alpha_1^2 + \cdots + \alpha_n^2=1$, the same argument applies with the same result, showing that $Z_a$ has the same distribution as $|a|Z$: it is a Beta distribution rescaled to the interval $[-|a|, |a|]$.


It would depend* on the form of the dependence. In many (almost all) cases, it won't be any well-known distribution.

* pun unintended

For the specific question, an answer may be possible.

Note that if the $X_i$ are iid $N(0,\sigma^2)$, then $Z^2$ would have a beta distribution.

  • $\begingroup$ Thx a lot. And, so if $Z=(a*X_1+b*X_2)/\sqrt{X_1^2+X_2^2}$, a and b are constant, then the distribution of $Z$ is generally unknown? $\endgroup$
    – Jingjings
    Commented Mar 17, 2013 at 8:33
  • $\begingroup$ Well, I can't say that I know. $\endgroup$
    – Glen_b
    Commented Mar 17, 2013 at 9:00
  • $\begingroup$ The distribution of that $Z$ can readily be worked out (with just simple algebra) when the $X_i$ both have zero mean and equal variances by applying the results of my comment (anent the question) to the independent Normal variates $Y_1=a X_1 + b X_2$ and $Y_2=b X_2 - a X_1$. $\endgroup$
    – whuber
    Commented Mar 17, 2013 at 16:28
  • $\begingroup$ @whuber. Thanks for your comments. However, I don't understand it very clearly. So, let $X_i$ are i.i.i. $N(0,1)$ rvs, one can calculate the PDF of Z according to the integration.(Using convolution?) When there are constants, $Y_1$ and $Y_2$ are independent, what if there are more than 2 rvs, that is $Z=a_1*X_1+...a_d*X_d/\sqrt(X_1^2+...+X_d^2)$? $\endgroup$
    – Jingjings
    Commented Mar 18, 2013 at 11:21
  • $\begingroup$ The same approach works. The integration is perhaps most easily done using Fisher's geometric methods. (More formally, change to spherical coordinates and exploit the symmetry.) $\endgroup$
    – whuber
    Commented Mar 18, 2013 at 14:56

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