Expectation of random variables ratio with positive weights Let $X_1, X_2, \dots, X_n$ be $n$ strictly positive iid random variables. Let $w_1, w_2, \dots, w_n$ be non-negative deterministic constants  such that $\sum_{i=1}^{n} w_i = 1$.
Then, can we say something on the following expectation? 
$$E\left(\frac{w_jX_j}{\sum_{i=1}^{n} w_i X_i}\right)$$
For the case where $w_i = \frac{1}{n}$, it is easy to show that the expectation is $\frac{1}{n}$ and have been asked numerous times here (see, e.g. this question). The main observation for this case is that since $X_i$'s are i.i.d, there is a symmetry that enables us to find the expectation exactly. 
Now, I was wondering whether by following the same logic we can say the expectation is $w_j$. If it helps, you can assume $X_i = \alpha+$Bernoulli($p$) for some deterministic constant $\alpha>0$.
Any comment greatly appreciated. 
 A: The conjecture is false.  Here is a simple counter-example with $n=2$.


*

*$X,Y$ are i.i.d. and $P(X=1)=P(X=2)=P(Y=1)=P(Y=2)=1/2$.

*$w_X = 1, w_Y=2$.  
Let $S=$ the weighted sum $w_X X + w_Y Y$.  Your conjecture is that 
$$E[{w_X X \over w_X X + w_Y Y}] = E[{X \over S}] = {w_X \over w_X + w_Y} = \frac13$$
There are only $4$ possible outcomes, shown in the table below:
X   Y     S      X/S
=   =   =====   =====
1   1   1+2=3    1/3
1   2   1+4=5    1/5
2   1   2+2=4    1/2
2   2   2+4=6    1/3

$$E[X/S] = (\frac13 + \frac15 + \frac12 + \frac13) / 4 \neq \frac13$$
Follow-up: Note that the same example shows that, in the case of equal weights, just being identically distributed (i.d.) is not enough but you also need independence (i.i.d.).  E.g. suppose $X,Y$ are as above and add $Z=Y$.  With all equal weights $w_X=w_Y=w_Z=1$ we have the same example:
X   Y   Z      S       X/S
=   =   =   =======    ===
1   1   1   1+1+1=3    1/3
1   2   2   1+2+2=5    1/5
2   1   1   2+1+1=4    1/2
2   2   2   2+2+2=6    1/3

And again $E[X/S] \neq 1/3$.  I haven't followed your links too much but any proof that claims dependence is acceptable must have a subtle error somewhere.
