Derive the distribution of the fraction ${\sum{x_i}}\Big/{\sqrt{\sum{x_i^2}}}$ Suppose ${X_i}$ follows the standard normal distributon N(0,1), what is the distribution of  $\frac{\sum{x_i}}{\sqrt{\sum{x_i^2}}}$

Some empirical assertions if there are $n$ terms and $Y_n = \frac{\sum_1^n{X_i}}{\sqrt{\sum_1^n{X_i^2}}}$  with the $X_i$ iid $N(0,1)$:

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*$Y_n$ has expectation $0$ and variance $1$


*$Y_n$ has support on $[-\sqrt{n},\sqrt{n}]$ (unless $n=1$, in which case the support is just $\{-1,1\}$)


*The density for $Y_n$ on its support is proportional to $(n-y^2)^{(n-3)/2}$


*In particular if $n=3$ this is a uniform distribution; if $n=4$ it is a Wigner semicircle distribution; if $n=2$ it is a version of the arcsine distribution


*$\frac{Y_n}{\sqrt{n}}+\frac12$ has a Beta distribution with parameters $\alpha=\frac{n-1}{2}$ and $\beta=\frac{n-1}{2}$; perhaps $Y_n$ could be called a standardised symmetric Beta distribution


*As $n$ increases, the distribution of $Y_n$ converges to a standard normal distribution


*If $Z_{i,n} = \frac{{X_i}}{\sqrt{\sum_1^n{X_i^2}}}$  with the $X_i$ iid $N(0,1)$ then $Y_n =\sum_1^n Z_{i,n}$ and, as StubbornAtom commented,
the vector $\mathbf Z_{n}$ is uniformly distributed on the surface of a unit hypersphere. $Z_{i,n}$ has the same distribution as $\frac{Y_n}{\sqrt{n}}$, so with mean $0$ and variance $\frac1n$ supported on $[-1,1]$
