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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.

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1 Answer 1

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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.

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  • $\begingroup$ Thanks. I have a question @antkam. In the link I posted in the question, it is stated independent assumption is not required and we only need identically distributed. If that is the case, can't we break $2Y$ to two $Y$'s and essentially get the case with equal weights? $\endgroup$
    – abolfazl
    Sep 28, 2019 at 8:00
  • $\begingroup$ Unsurprisingly, I think I am right and the other guy is wrong. :) See new example above. Not exactly sure where his error is, but my new example is rock solid. $\endgroup$
    – antkam
    Sep 28, 2019 at 14:50
  • $\begingroup$ Yes, your example makes total sense. BTW, do you have any simple way to bound the expectation or the variance of this random variable? At least for the Bernoulli case? $\endgroup$
    – abolfazl
    Sep 30, 2019 at 3:28
  • $\begingroup$ Those are great questions! The variance seems hard - indeed even in the all weights equal case, the symmetry argument only helps with the mean, not with the variance (at least, AFAIK). But the question about bounding the mean seems just difficult enough to be fun, but not impossible... Nothing comes to mind right away but that does seem like a fun problem. In fact, why dont you post a new question asking that? (Pls dont just modify this question, but post a new one instead.) Or if you dont want to, may I do that in a day or two? $\endgroup$
    – antkam
    Sep 30, 2019 at 5:17
  • $\begingroup$ For variance, I was thinking of Popoviciu's inequality. But that is not as sophisticated as I'd like. I'll post a new question tmrw. $\endgroup$
    – abolfazl
    Sep 30, 2019 at 6:18

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