Is there an explicit probability density function for $\frac{x_1}{\sum_{i=1}^{n}x_i}$ where $x_i \sim U(0,1)$ and are independent random variables Is there an explicit probability density function for $$\frac{x_1}{\sum_{i=1}^{n} x_i}$$ where $x_i$ are independent random variables and have a uniform probability density between 0 and 1.
 A: An explicit CDF was already worked out in the other answer, but I get the feeling you may be more interested in getting a good and simple approximation when $n$ is large. For this purpose, consider the random variable $Y$ which is the sum of $n-1$ iid uniform $[0,1]$ random variables. Then $Y$ has mean $\frac{n-1}{2}$ and variance $\frac{n-1}{12}$. You can get a decent approximation for $Y$ when $n$ is large by replacing $Y$ with a normal random variable $Z$ with the same mean and variance. Then, the distribution of this approximating distribution is obtained by computing the probability
$$
\mathbb P\bigl(\frac{U}{U+Z}<t\bigr),
$$
which can be computed explicitly.
Indeed, we can start by computing the conditional probability
$$
\mathbb P\bigl(\frac{U}{U+Z}<t\mid Z\bigr),$$
which equals $0$ if $Z$ is negative (which happens with tiny probability when $n$ is large). Thus
$$\mathbb P\bigl(\frac{U}{U+Z}<t\mid Z\bigr)=\frac{t}{1-t} Z\cdot 1[Z>0].
$$
Finally, taking the expectation results in 
$$
\mathbb P\bigl(\frac{U}{U+Z}<t\bigr)=\frac{t}{1-t}\mathbb E[Z;Z>0].
$$
If you want to make the $\mathbb E[Z;Z>0]$ more explicit, note that if we set $m=n-1$ for convenience of notation, then
$$
\mathbb E[Z;Z>0]=\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}xe^{-6(x-m/2)^2/m}\ dx
$$
is a constant depending only on $m$ (or equivalently, $n$).
