# Can deleting a random entry from an iid sequence destroy the iid property?

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?

• 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 $1/\sqrt{8} + 1/\sqrt{2} \ne 1$. Oct 4 at 12:26

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.
• @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. Oct 3 at 14:47
• @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? Oct 3 at 19:12