Explicit formula to $\mathbb{E}[Y|X_1,..,X_n]$?

Question

Let $X_1,..,X_n$ random variables i.i.d. with distribution function $F$, with $\mathbb{P}[X_i=X_j]=0$ for $i\neq j$ and $Y$ a r.v. such that $\mathbb{P}[Y=X_i]=\frac{1}{n}$. Is there a explicit formula to compute: \begin{equation} \mathbb{E}[Y|X_1,..,X_n] \end{equation} I really appreciate any help. Thanks!

Answer 1

In essence you're defining $Y = X_N$ where $N$ is a uniform random variable on $\{1, 2, \ldots, n\}$.

Here's the problem: None of what you've written tells us anything how the distribution of $N$ relates to the distribution of the random vector $(X_1, \ldots, X_n)$ so in particular, we cannot find $\mathbb{E}[N | X_1, \ldots, X_n],$ which would be necessary to answer the question.

Here are some examples:

• If $N = \operatorname{argmin}_iX_i,$ then as $\mathbb{P}(X_i = X_j) = 0$ for all $i \neq j$, this defines $Y = X_{(1)}$ which satisfies the assumptions. However, as $X_{(1)}$ is measurable with respect to $\sigma(X_1,\ldots, X_n)$, we see $\mathbb{E}[Y\mid X_1,\ldots, X_n] = X_{(1)}$
• More generally, whenever $N$ is measurable with respect to $\sigma(X_1, \ldots, X_n)$, it follows that $\mathbb{E}[Y \mid X_1, \ldots, X_n] = Y.$
• If $N$ is independent of $(X_1, \ldots, X_n)$, then $$\begin{align*}E[X_N \mid X_1,\ldots, X_n] &= \mathbb{E}\left[\sum_{i=1}^nX_i\mathbf{1}_{N=i}\middle| X_1, \ldots, X_n\right] \\ &= \sum_{i=1}^nX_i\mathbb{P}[N=i\mid X_1,\ldots, X_n] \\ &= \sum_{i=1}^n X_i \mathbb{P}[N=i] \\ &= \sum_{i=1}^n X_i \cdot (1/n) = \bar{X}\end{align*}$$ is the sample mean.

That is, we can certainly find explicit expressions for the conditional expectation given some more information, but as it stands, I don't see how we may infer that information from the given problem.