The ratio of the dependent normal rv and chi rv is? (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?
 A: 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 $[-|a|, |a|]$.
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.
