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:
- If $\alpha_1=...=\alpha_N=\alpha$, then $\mathbb{E}[X_i|\sum_{k=1}^N\alpha_kX_k]=\mathbb{E}[X_j|\sum_{k=1}^N\alpha_kX_k]=\frac{1}{N\alpha}\sum_{k=1}^NX_k,\ \forall i,j=1,...,N$;
- 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.