Are random selections from i.i.d. random variables independent? 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)
 A: 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.
A: 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. 
