Correlation of Proportions To introduce my question, here is a small simplification for consideration:
Let $X,Y$ be independent random variates, each with finite mean and variance. Interestingly,
$$\text{Corr}\big(\frac{X}{X+Y},\frac{Y}{X+Y} \big) = \text{Corr}\big(\frac{X}{X+Y},1-\frac{X}{X+Y} \big) = -1$$
The question becomes, if I have a set of $X_i$ which are independent with finite mean and variance, is there a possible simplification (much like the relationship above) which can be made to the following for some selection of $j,k$?
$$\text{Corr}\big(\frac{X_j}{\sum_i{X_i}},\frac{X_k}{\sum_i{X_i}} \big)$$
If the variates are 'iid' this problem is likely simplified considerably. And while I would like to see something of that form, the case I'm most interested in is when $X_i$ are all from the same 'family' of distribution, but with different parameters, $\mu_i$ as a varying mean, for example. 
I attempted looking at the fact that
$$\frac{X_j}{\sum_i{X_i}} = 1 - \frac{\sum_{i\neq j}X_i}{\sum_i{X_i}} = 1 - \frac{X_k}{\sum_i X_i} - \frac{\sum_{i\notin\{j,k\}}X_i}{\sum_i X_i}, \ k\neq j$$
But using this just gives us
$$\text{Corr}\big(\frac{X_j}{\sum_i{X_i}},\frac{X_k}{\sum_i{X_i}} \big) = $$
$$\text{Corr}\big(1 - \frac{X_k}{\sum_i X_i} - \frac{\sum_{i\notin\{j,k\}}X_i}{\sum_i X_i}, 1 - \frac{X_j}{\sum_i X_i} - \frac{\sum_{i\notin\{j,k\}}X_i}{\sum_i X_i}\big)$$
Which if I understand correctly is very similar to:
$$\text{Corr}\big(1 - \frac{X_k}{\sum_i X_i},1 - \frac{X_j}{\sum_i X_i}\big)$$
So it didn't really take me anywhere
 A: Given iid random variables $X_i, i = 1, \ldots, n$. Define $Z_j = \frac{X_j}{\sum_{i=1}^n X_i}.$ Clearly, $$\sum_{j=1}^n Z_j = 1.$$ Take variance on both sides, we get
$$
\begin{split}
Cov(\sum_{j=1}^n Z_j, \sum_{j=1}^n Z_j) 
&= 
0.
\end{split}
$$
We know
$$
\begin{split}
Cov(\sum_{j=1}^n Z_j, \sum_{j=1}^n Z_j) 
&= 
\sum_{j=1}^n Var(Z_j) + 2\sum_{j < k} Cov(Z_j, Z_k) 
\\ 
&= 
n Var(Z_1) + n(n-1) Cov(Z_1, Z_2). 
\\ 
\end{split}
$$
We get to second step using symmetry. In particular, $Var(Z_j) = Var(Z_k)$ and $Cov(Z_j, Z_k) = Cov(Z_1, Z_2)$.
Using the last equation, we get $\frac{Cov(Z_1, Z_2)}{Var(Z_1)} = -\frac{1}{n-1}$. Since,
$$
\frac{Cov(Z_1, Z_2)}{Var(Z_1)} 
= 
\frac{Cov(Z_1, Z_2)}{\sqrt{Var(Z_1)Var(Z_2)}} 
= 
\rho_{Z_1, Z_2},
$$
we get correlation is $-1/(n-1)$.
A: Well a related formula for random variables that sum to a constant $c$ is the following: Suppose $Y_1, \ldots, Y_N$ are random variables that satisfy $\sum_{i=1}^N Y_i = c$.  Then: 
\begin{align} 
c &= \sum_{i=1}^NE[Y_i] \\
c^2 &= \sum_{i,  j} E[Y_i]E[Y_j] \: \: (*)
\end{align} 
Also: 
\begin{align} 
c^2 &= \sum_{i,  j} Y_i Y_j\\
c^2 &= \sum_{i, j} E[Y_iY_j] \: \: (**)
\end{align}
Subtracting equation (*) from (**) gives: 
$$  0 = \sum_{i, j} (E[Y_iY_j]-E[Y_i]E[Y_j]) = \sum_{i=1}^NVar(Y_i) + \sum_{i\neq j} Cov(Y_i,Y_j)  $$
In particular: 
$$ \sum_{i \neq j} Cov(Y_i,Y_j) = -\sum_{i=1}^NVar(Y_i) $$
Assuming $Var(Y_i)>0$ for at least one $i \in \{1, \ldots, N\}$, we get: 
$$ \boxed{\frac{\sum_{i\neq j} Cov(Y_i,Y_j)}{\sum_{i=1}^NVar(Y_i)} = -1} $$
