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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!

Don P.
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    As another issue: for each $i$, is ${Y = X_i}$ measurable with respect to $\sigma{X_1, X_2, \ldots, X_n}$? Or maybe they're independent? Both give answers, but they're not equivalent. (for instance, in the first case you could have an answer of any fixed order statistic $X_{(i)}$ while in the latter case you would have an answer of the sample mean $\bar{X}$) – Brian Moehring Feb 25 '22 at 06:29
  • But in this case ${Y=X_i}$ could it be independent to $X_i$?, I was thinking to given a borelian $B$ is it true that $Y^{-1}(B)=\bigcup_{i=1}^n X^{-1}(B)$?. The thing is I don't know the answer of your question, this is a problem to understand bootstrapping, and the problem says: Let $Y$ a bootstrap sample of distintic observations $X_1,..,X_n$ compute $\mathbb{E}[Y|X_1,..,X_n]$. And I would say thank you for your answers and advices. – Don P. Feb 25 '22 at 06:42
  • Your Question has been closed for lack of context. Your Comment above points to some of the context, but the facts "this is a problem to understand bootstrapping," and "Let $Y$ [be] a bootstrap sample of" $X_1,\ldots,X_n$ ought to be incorporated into the body of the Question. – hardmath Feb 27 '22 at 04:13

1 Answers1

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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.

Brian Moehring
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  • I have a question: In the case when $N$ is independet to $X_1,...,X_n$, then $E[X_N|X_1,...,X_n]=E[X_N]=\bar{X}$ but this is a random variable, right?, and again I really appreciate your time! Thanks!. – Don P. Feb 25 '22 at 07:09
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    Your intermediate step of $\mathbb{E}[X_N]$ is wrong. Even if $N$ is independent of $X_1, \ldots, X_n$, it doesn't mean $X_N$ is (in fact, except in degenerate cases where $\bar{X}$ is almost surely constant, this shows $X_N$ cannot be independent of $X_1, \ldots, X_n$) – Brian Moehring Feb 25 '22 at 07:12
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    @DonP. See my edit to see how to show it's the sample mean in that case. – Brian Moehring Feb 25 '22 at 07:23
  • Now I understand my mistake in the step $\mathbb{E}[X_N]$. Thank you very much!!! – Don P. Feb 25 '22 at 07:28