Expectation of random variables ratio with positive weights

conditional-expectationexpected valueprobabilityprobability distributionsrandom variables

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:

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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