Given n uniformly distributed r.v's, what is the PDF for one r.v. divided by the sum of all n r.v's? I'm interested in the following type of case: there are 'n' continuous random variables which must sum to 1. What then would be the PDF for any one individual such variable? So, if $n=3$, then I am interested in the distribution for $\frac{X_1}{X_1+X_2+X_3}$, where $X_1, X_2$, and$ X_3 $are all uniformly distributed. The mean of course, in this example, is $1/3$, as the mean is just $1/n$, and though it is easy to simulate distribution in R, I do not know what the actual equation for the PDF or CDF is.
This situation is related to the Irwin-Hall distribution (https://en.wikipedia.org/wiki/Irwin%E2%80%93Hall_distribution). Only Irwin-Hall is the distribution of the sum of n uniform random variables, whereas I would like the distribution for one of n uniform r.v's divided by the sum of all $n$ variables.
 A: The breakpoints in the domain make it somewhat messy. A simple but tedious approach is to build up to the final result. For $n=3,$ let $Y=X_2 + X_3,$ $W = {{X_2 + X_3} \over X_1},$ and $T = 1 + W.$ Then $Z = {{1} \over {T}}={{X_1} \over {X_1 + X_2 + X_3}}.$
The breakpoints are at 1 for $Y,$ 1 and 2 for $W,$ 2 and 3 for $T,$ and $1/3$ and $1/2$ for $Z.$ I found the complete pdf to be 
$$f(z) = \begin{cases} \ \ \ \ \ {{1} \over {(1-z)^2}} \ , & \text{if} \ {0} \leq z \leq {1/3} \\\\ {{3z^3-9z^2+6z-1} \over {3z^3(1-z)^2}} \ , & \text{if} \ {1/3} \leq z  \leq {1/2} \\\\ \ \ \ \ \ \ \ {{1-z} \over {3z^3}} \ , & \text{if} \ {1/2} \leq z \leq {1} \end{cases}$$
The cdf can then be found as
$$F(z) = \begin{cases} \ \ \ \ \ \ \ \ \ \ \ {{z} \over {(1-z)}} \ , & \text{if} \ {0} \leq z \leq {1/3} \\\\ {{1} \over {2}}+{{-18z^3+24z^2-9z+1} \over {6z^2(1-z)}} \ , & \text{if} \ {1/3} \leq z  \leq {1/2} \\\\ \ \ \ \ \ \ \ \ {{5} \over {6}} + {{2z-1} \over {6z^2}} \ , & \text{if} \ {1/2} \leq z \leq {1} \end{cases}$$
A: Let $Y=\sum_{i=2}^n X_i$. We can find the cdf of $X_1/\sum_{i=1}^n X_i$ by calculating
\begin{align*}
P(\frac{X_1}{\sum_{i=1}^n X_i} \leq t)
&= P(X_1 \leq t\sum_{i=1}^n X_i) \\
&= P((1-t)X_1 \leq t\sum_{i=2}^n X_i) \\
&= P(X_1 \leq \frac t{1-t}Y)\\
&= \int_0^1 P(x_1 \leq \frac t{1-t}Y)\ dx_1\\
&= \int_0^1 (1-F_Y(\frac{1-t}{t}x_1))\ dx_1\\
&= 1-\int_0^1 F_Y(\frac{1-t}{t}x_1)\ dx_1\\
\end{align*}
We then differentiate and substitute the Irwin-Hall pdf to obtain the desired pdf:
\begin{align*}
f(t) &= \int_0^1 f_Y(\frac{1-t}{t}x_1)\cdot \frac{x_1}{t^2}\ dx_1\\
&= \frac{1}{t^2}\int_0^{1\wedge \frac{(n-1)t}{1-t}} \sum_{k=0}^{\lfloor \frac{1-t}{t}x_1\rfloor}\frac1{(n-2)!}(-1)^k\binom{n-1}k(\frac{1-t}{t}x_1-k)^{n-1} x_1\ dx_1
\end{align*}
From here it gets a little messy, but you should be able to interchange the integral and summation and then perform a substitution (e.g, $u=\frac{tx_1}{1-t}-k$) to evaluate the integral and hence obtain an explicit formula for the pdf.
A: Assuming 

"the N uniform distributions don't sum to 1."

This is how I started(it's incomplete):
Consider $Y = \sum_{i=1}^n X_i$ and let $X=X_i$ by a slight abuse of notation.
Consider, $U = \frac{X}{Y}$ and $V =Y$:
$$
X=UV\\
Y=V
$$
Then following transformation of variables:
$$
J = \begin{bmatrix}
V & U\\
0 & 1
\end{bmatrix}
$$
The joint probability function of $(U,V)$ is given by:
$f_{U,V}(u,v) = f_{X,Y}(uv,v)|J|$
Where $X \sim U(0,1)$ and $Y \sim IrwinHall$
$$
f_X(x) = \begin{cases}
1 & 0 \leq x\leq 1\\
0 & otherwise
\end{cases}
$$
And, 
$$
f_Y(y) = \frac{1}{2(n-1)!}\sum_{k=0}^n(-1)^k {n\choose k}(x-k)^{n-1} sign(x-k)
$$
Thus,
$$
f_{U,V}(u,v) = \begin{cases}
\frac{1}{2(n-1)!}\sum_{k=0}^n(-1)^k {n\choose k}(uv-k)^{n-1} sign(uv-k) & 0 \leq uv \leq 1\\
0 & otherwise
\end{cases}
$$
and $f_U(u) = \int f_{U,V}(u,v) dv$
A: Suppose we already know sum of $U(0,1)$ has a Irwin-Hall distribution.
Now your question changes to find the pdf (or CDF) of $\frac{X}{Y}$ when X had a $U(0,1)$ distribution and $Y$ has a Irwin-Hall distribution.
First we need to find he joint pdf of $X$ and $Y$.
Let $Y_1=X_1\\Y_2=X_1+X_2\\Y_3=X_1+X_2+X_3$
Then
$X_1=Y_1\\X_2=Y_2-Y_1\\X_3=Y_3-Y_2-Y_1$
$\therefore$
$J=\begin{vmatrix}
\frac{\partial X_1}{\partial Y_1} & \frac{\partial X_1}{\partial Y_2} &\frac{\partial X_1}{\partial Y_3} \\ 
 \frac{\partial X_2}{\partial Y_1} & \frac{\partial X_2}{\partial Y_2} &\frac{\partial X_2}{\partial Y_3} \\ 
\frac{\partial X_3}{\partial Y_1} & \frac{\partial X_3}{\partial Y_2} &\frac{\partial X_3}{\partial Y_3}
\end{vmatrix}=-1$
Since $X_1, X_2, X_3$ are i.i.d with $U(0,1),$ therefore, $f(x_1,x_2,x_3)=f(x_1)f(x_2)f(x_3)=1$
The joint distribution with $y_1,y_2,y_3$ is
$g(y_1,y_2,y_3)=f(y_1,y_2,y_3)|J|=1$
Next let us integrate out the $Y_2$ and we can get the joint distribution of $Y_1$ and $Y_3$ i.e the joint distribution of $X_1$ and $X_1+X_2+X_3$
As suggested by whuber now I changed the the limits
$$h(y_1,y_3)=\int_{y_1+1}^{y_3-1} g(y_1,y_2,y_3)dy_2=\int_{y_1+1}^{y_3-1} 1 dy_2=y_3-y_1-2 \tag{1}$$
Now, we know the joint pdf of $X,Y$ i.e joint pdf $X_1$ and $X_1+X_2+X_3$ is   $y_3-y_1-2$.
Next let find the pdf of $\frac{X}{Y}$
We need another transformation:
Let $Y_1=X\\Y_2=\frac{X}{Y}$
Then $X=Y_1\\Y=\frac{Y_1}{Y_2}$
Then
$J=\begin{vmatrix}
\frac{\partial x}{\partial y_1} & \frac{\partial x}{\partial y_2}\\ 
\frac{\partial y}{\partial y_1}  & \frac{\partial y}{\partial y_2}
\end{vmatrix}=
\begin{vmatrix}
1 & 0\\ 
\frac{1}{y_2}  & -\frac{y_1}{y_2^2}
\end{vmatrix}=-\frac{y_1}{y_2^2}$
we already the joint distribution of $X,Y$ from above steps ref (1).
$\therefore$
$g_2(y_1,y_2)=h(y_1,y_3)|J|=(y_3-y_1-2)\frac{y_1}{y_2^2}$
Next, we integrate the $y_1$ out we get the pdf of $y_2$ then we get the pdf of $\frac{X}{Y}$
$$h_2(y_2)=\int_0^1(y_3-y_1-2)\frac{y_1}{y_2^2}dy_1=\frac{1}{y_2^2}(\frac{y_3}{2}-\frac{1}{3}-1)\tag{2}$$
This is the pdf of $X/Y$ i.e $\frac{X_1}{X1+X_2+X_3}$
We are not finish yet, what is $y_3$ in (2) then?
We know that $Y_3=X_1+X_2+X_3$ from the first transformation.
So at least we know $Y_3$ has a Irwin-Hall distribution.
I wonder can we plug  the Irwin-Hall for $n=3$ pdf  to (2) to get a explicit formula? or can we do some simulations from here as Glen suggested?
