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Let us have identically independently distributed random variables $x_1, x_2, \dots, x_{10}$. Now let us pick indices $\alpha, \beta$ uniformly independently from $1,2,\dots,10$. Are variables $x_{\alpha}$ and $x_{\beta}$ independent?

My intuition says that they are not, since there is a chance $\frac{1}{10}$ that they are the same. But how can I formally prove this? (or prove that they are independent in case I am wrong)

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Can we assume $\alpha,\beta$ independent from x's? – ved Jul 5 '14 at 16:19
@ved yes, independent – Martin Drozdik Jul 5 '14 at 16:20
up vote 5 down vote accepted

To summarize:

With replacement (the question), $(\alpha,\beta)$ is independent and $(x_\alpha,x_\beta)$ is not. Without replacement, $(\alpha,\beta)$ is not independent and $(x_\alpha,x_\beta)$ is.

To show the case with replacement, consider $$P(x_\alpha\in A,x_\beta\in A)=P(x_\alpha\in A,x_\beta\in A,\alpha\ne\beta)+P(x_\alpha\in A,\alpha=\beta), $$ thus, $$ P(x_\alpha\in A,x_\beta\in A)=\left(1-\frac1{10}\right)p^2+\frac1{10}p, $$ where $p=P(x_1\in A)$. On the other hand, $$ P(x_\alpha\in A)P(x_\beta\in A)=p^2. $$ If $p$ is not $0$ or $1$ these are different.

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Independence has nothing to do with whether the answers are the same. They can still be independent if answers are the same. Independence has more to do with conditional probability. Two random variables are independent if knowing about outcome of one rv tells you no information about the other rv (i.e. $P(X_\alpha \mid X_\beta=x)=P(X_\alpha)$. Now one way this situation could not be independent is if your numbers from uniform $1,2,\ldots,10$ where selected without replacement.

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Also to prove independence just use definition which is outcome A and B are independent if $P(A\cap B)=P(A)P(B)$ – Kamster Jul 5 '14 at 16:28
"Without replacements" means that the $x_i$s are not independent, contrary to the assumption in the question. – Henning Makholm Jul 5 '14 at 16:28
x @user159813: But your example is not an example of the situation in the question. As Did's computation shows, $x_\alpha$ and $x_\beta$ are actually not independent with any nontrivial probability distribution for the $x_i$s. – Henning Makholm Jul 5 '14 at 16:31
Without replacement, $X_\alpha$ and $X_\beta$ ARE independent. – Did Jul 5 '14 at 16:35
@MartinDrozdik Of course (what else?). The current last sentence of the post is ambiguous: without replacement, $\alpha$ and $\beta$ are not independent but this is not the point, rather one asks whether $x_\alpha$ and $x_\beta$ are independent (and they are). – Did Jul 5 '14 at 17:20

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