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.
Best Answer
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:
$$E[X/S] = (\frac13 + \frac15 + \frac12 + \frac13) / 4 \neq \frac13$$
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:
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.