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  $\frac{\sum{x_i}}{\sqrt{\sum{x_i^2}}}$

Hint: you can rewrite $\sum x_i^2 = \bar{x}^2 + \sum (x_i -\bar{x})^2$.
Note that the terms $n\bar{x}^2$ and $\sum (x_i -\bar{x})^2$ are independent chi-squared distributed variables.
Then you can rewrite $Y = \frac{\sum{x_i}}{\sqrt{\sum{x_i^2}}}$ as
$$\begin{array}{}
Y^2 &= &\frac{\left(\sum{x_i}\right)^2}{{\sum{x_i^2}}}\\
&=& \frac{n^2\bar{x}^2}{\bar{x}^2+{\sum{(x_i-\bar{x})^2}}}\\
&=& \frac{n}{1/n+{\left(\sum{(x_i-\bar{x})^2}\right)/(n\bar{x}^2)}}\\
\end{array}$$
And this term in the denominator ${\left(\sum{(x_i-\bar{x})^2}\right)/(n\bar{x}^2)}$ follows an F-distribution
So we have that $n/Y^2-1/n$ follows a F-distribution.
Possibly this can be simplified further (edit: and as Henry noted in the comments you can 'turn it into' a beta distribution. Or... $Y^2$ follows a beta distribution.)
