Expectation conditional on a linear combination of i.i.d random variables

Suppose $$X_1,X_2,...,X_N$$ are i.i.d random variables. I want to calculate the conditional expectation $$\mathbb{E}[X_i|\sum_{k=1}^N\alpha_k X_k].$$

For now, my idea is as the following:

Denote $$Y_k=\alpha_kX_k$$ and let $$\pi=(\pi_1,...,\pi_N)$$ be any permutation of $$(1,...,N)$$. It follows from the pairwise independence of $$(Y_1,...,Y_N)$$ that $$Y_1,...,Y_N$$ are exchangeable, i.e., for any two permutations $$\pi$$ and $$\pi'$$, $$(Y_{\pi_1},...,Y_{\pi_N})$$ and $$(Y_{\pi_1'},...,Y_{\pi_N'})$$ share the same joint distribution. Thus, function $$f(Y_{\pi_1},...,Y_{\pi_N})$$ and $$f(Y_{\pi_1'},...,Y_{\pi_N'})$$ have the same distribution. Applying this fact to function $$f(Y_{\pi_1},...,Y_{\pi_N})=(Y_{\pi_1},\sum_{k=1}^NY_k)$$ implies $$(Y_i,\sum_{k=1}^NY_k)$$ and $$(Y_j,\sum_{k=1}^NY_k)$$ share the same joint distribution.

By definition, the conditional expectation $$\mathbb{E}[X_i|\sum_{k=1}^NY_k]$$ is given by $$\mathbb{E}[X_iZ]=\mathbb{E}\left[Z \mathbb{E}[X_i|\sum_{k=1}^NY_k]\right],\ Z\in\mathbf{M}\left(\sum_{k=1}^NY_k\right)$$ where $$\mathbf{M}\left(\sum_{k=1}^NY_k\right)$$ is the closed subspace of $$L^2$$ consists of all Borel function of $$\sum_{k=1}^NY_k$$. Multiplying both sides of the above equation by $$\alpha_i$$, we can also write $$\mathbb{E}[Y_iZ]=\mathbb{E}\left[Z \mathbb{E}[Y_i|\sum_{k=1}^NY_k]\right].$$ Then we have $$\mathbb{E}[Y_iZ]=\mathbb{E}[Y_jZ]=\mathbb{E}\left[Z \mathbb{E}[Y_j|\sum_{k=1}^NY_k]\right] \Longrightarrow \mathbb{E}[Y_i|\sum_{k=1}^NY_k] = \mathbb{E}[Y_j|\sum_{k=1}^NY_k],$$ since $$(Y_i,\sum_{k=1}^NY_k)$$ and $$(Y_j,\sum_{k=1}^NY_k)$$ share the same joint distribution and $$Z$$ is a function of $$\sum_{k=1}^NY_k$$. Thus, $$\alpha_i\mathbb{E}[X_i|\sum_{k=1}^NY_k]=\alpha_j\mathbb{E}[X_j|\sum_{k=1}^NY_k].$$ Summing up both sides with respect to $$j$$ gives $$N\alpha_i\mathbb{E}[X_i|\sum_{k=1}^NY_k]=\mathbb{E}[\sum_{j=1}^N\alpha_jX_i|\sum_{k=1}^NY_k]=\sum_{j=1}^N\alpha_jX_i.$$ Therefore, we have $$\mathbb{E}[X_i|\sum_{k=1}^NY_k]=\frac{1}{N\alpha_i}\sum_{j=1}^N\alpha_jX_i,\ \alpha_i\ne0.$$ For $$\alpha_i=0$$, it follows from the independence that $$\mathbb{E}[X_i|\sum_{k=1}^N\alpha_kX_k]=\mathbb{E}[X_i|\sum_{k\ne i}\alpha_kX_k]=0$$.

Implications:

1. If $$\alpha_1=...=\alpha_N=\alpha$$, then $$\mathbb{E}[X_i|\sum_{k=1}^N\alpha_kX_k]=\mathbb{E}[X_j|\sum_{k=1}^N\alpha_kX_k]=\frac{1}{N\alpha}\sum_{k=1}^NX_k,\ \forall i,j=1,...,N$$;
2. In general, for $$\alpha_i\ne \alpha_j$$ with $$\alpha_i\alpha_j\ne 0$$, $$\mathbb{E}[X_i|\sum_{k=1}^N\alpha_kX_k]\ne\mathbb{E}[X_j|\sum_{k=1}^N\alpha_kX_k]$$, although $$X_i$$ and $$X_j$$ are i.i.d.

Does it make sense? Any hint is appreciated!

EDIT 1: Thank @Leander Tilsted Kristensen for pointing out the mistake in the comment. Now I know $$(Y_{\pi_1},...,Y_{\pi_N})$$ and $$(Y_{\pi_1'},...,Y_{\pi_N'})$$ do not necessarily have the same distribution. Then, if $$\alpha_i$$s are not identical, can I make a statement that $$\mathbb{E}[X_i|\sum_{k=1}^N\alpha_kX_k]\ne\mathbb{E}[X_j|\sum_{k=1}^N\alpha_kX_k]$$ in general?

EDIT 2: Thank @Paresseux Nguyen for the examples and comments. It turns out that in the EDIT 1, "not equal" is still a strong conclusion and we should say the two conditional expectation are "not necessarily" equal.

• The claim that $(Y_1, \dots ,Y_N)\sim (Y_{\pi_1} , \dots ,Y_{\pi_N})$ for any permutation $\pi$ is not true. You are allowed to permute the $X$ values, since they are i.i.d., but you are not allowed to permute the constants $\alpha_1,\dots , \alpha_N$. Nov 6, 2020 at 16:55
• @Leander Tilsted Kristensen Thanks for the comment. Is it because that $Y_1,...,Y_N$ are independent, but not idendical distributed?
– HXW
Nov 7, 2020 at 0:17
• Yes, that is exactly why. Nov 7, 2020 at 11:29

As showed by Kristensen, you kicked off your approach with a not true observation.
In fact, I don't think you can do anything for it in general. I think I can dive a little bit in analysis of this problem to show you where things get complicated.
But I guess it'll be better if I give some examples.
Case 1
Let's say $$X=\alpha_1 X_1+\alpha_2X_2+...+\alpha_n X_n$$
If $$\alpha_1=\alpha_2=...=\alpha_n=1$$,
$$\mathbb{E}( X_1|X)= \frac{X}{n}$$
Case 2
If $$(X_k, k \in [n])$$ are all standard normal distribution, then
$$\mathbb{E}( X_1|X)= \frac{\alpha_1X}{\sum_{k=1}^n \alpha_k^2}$$
Case 3
If $$(X_k, k \in [n])$$ are all Bernoulli variable $$\mathcal{B}(p) (0, $$a_k=2^{-k}$$, then:
$$\mathbb{E}( X_1|X)= [2X]$$
$$\mathbb{E}( X_2|X)= [4X]-2[2X]$$
etc.
**Disclaimer **: I may have mistaken somewhere but the generality stays the same.

• I see. Thanks! These are really helpful!
– HXW
Nov 7, 2020 at 0:26
• Zut! I accidentally delete my comment. Anyway, for your updated question, you also cannot have that a.s. Look at my third case, when $X=0$, but you can show they are different variable by calculate the expectation of $X( \mathbb{E}(X_1|X) - \mathbb{E}(X_2|X) )$ (assuming $E(X)=0)$. Nov 7, 2020 at 0:36
• Got it. Then should I say that they are "not necessarily" equal?
– HXW
Nov 7, 2020 at 0:42
• Yeah, I think we can say that Nov 7, 2020 at 0:44
• Ok. Thanks again!
– HXW
Nov 7, 2020 at 0:45