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Let $X=(X_1,\ldots,X_n)$ be an iid sequence of random variables, and let $\nu$ be a uniformly random integer in the range $1,\ldots,n$. Then $\xi_\nu$ is a random entry of $X$. Is it always true that after deleting such a random entry from $X$, the remaining sequence is still iid?

Due to the uniform randomness of the index $\nu$, it is tempting to answer that the leftover sequence always remains iid. This is correct, if we also assume that $\nu$ is chosen independently of $X$.

What happens, however, if no such independence is assumed? For example, consider a 0-1 valued iid sequence. Remove a random 1, chosen uniformly at random among all 1s. If there is no 1, then remove a random 0. In this situation, the position of the removed entry is still uniformly random, but not independent of the original sequence. Can such a removal destroy the iid property in the remaining sequence?

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    $\begingroup$ Your concluding example can be calculated by hand when $n=3$. The probability of getting $11$ is $1/8$ and the probability of getting $00$ is $1/2$. If the sequence were i.i.d. then we would have to have $\Pr(1) = 1/\sqrt{8}$ and $\Pr(0) = 1/\sqrt{2}$ , but $1/\sqrt{8} + 1/\sqrt{2} \ne 1$. $\endgroup$
    – Timothy Chow
    Commented Oct 4, 2022 at 12:26

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The independence will be then in general lost. E.g., let $X_1,\dots,X_n$ be independent random variables each uniformly distributed on $[0,1]$. Let $M:=\max(X_1,\dots,X_n)=X_\nu$, so that $\nu$ is uniformly distributed on $[n]:=\{1,\dots,n\}$. Let $(Y_1,\dots,Y_{n-1})$ be the leftover sequence, after the removal of $X_\nu$. Then, conditionally on $M$, the $Y_i$'s are iid uniformly distributed on $[0,M]$.

So, for $n\ge2$ and $i\in[2]$ we have $E(Y_i|M)=M/2$ and $E(Y_1Y_2|M)=E(Y_1|M)E(Y_2|M)=(M/2)^2$. So, $$EY_1Y_2=E(M/2)^2>(EM/2)^2=EY_1\,EY_2,$$ with the inequality taking place because $Var\,M>0$. Thus, $Y_1$ and $Y_2$ are not independent.

Intuitively, it can be expected that the $Y_i$'s are positively dependent. Indeed, if $M$ is small, then all $Y_i$'s will be small. So, if $Y_1$ turns out to be small, a reason for that may be that $M$ is small, and then $Y_2$ will be small. Thus, the smallness of $Y_1$ seems to make $Y_2$ tend to be small.

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  • $\begingroup$ Nice example! Could the argument be carried over to the example mentioned in the last paragraph of the original question? $\endgroup$
    – Andras Farago
    Commented Oct 3, 2022 at 14:31
  • $\begingroup$ @AndrasFarago : Yes, of course. However, the expressions there are messy, because of ties, as the index $\nu$ of the maximum is not unique in that case. In that example, I suggest you consider $Y_1$ and $Y_2$ for $n=3$, which is quite elementary. $\endgroup$
    – Iosif Pinelis
    Commented Oct 3, 2022 at 14:47
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    $\begingroup$ I love this question and answer! Very interesting! $\endgroup$
    – Stanley Yao Xiao
    Commented Oct 3, 2022 at 18:44
  • $\begingroup$ @IosifPinelis The intuitive argument at the end of your answer does not seem to work for the 0-1 case, because the maximum is almost always the largest possible value (1). The only exception is when all terms are 0, but if the entries take value 1, say, with probability 1/2, and $n$ is large, then this happens with exponentially small probability. We may say that let's start with a conditionally iid sequence, where the condition is that not all terms are 0. Is it true that leaving out a random 1, the leftover will not be iid even in this case? $\endgroup$
    – Andras Farago
    Commented Oct 3, 2022 at 19:12
  • $\begingroup$ @StanleyYaoXiao : Thank you for your comment. $\endgroup$
    – Iosif Pinelis
    Commented Oct 3, 2022 at 21:06

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