How to prove that $P[X_1=X^{(i)}]=\frac{1}{n}$?

Question

Let $X_1$, $X_2$ be two samples draw from a continuous distribution, then I think there is no reason to say that $X_1\leq X_2$ or $X_1\geq X_2$, so we may have $$P[X_1\leq X_2]=P[X_1>X_2]=\frac{1}{2}$$ more generally, let $X^{(1)},\ldots,X^{(n)}$ denote the $X_1,\ldots,X_n$ ordered, then $$P[X_1=X^{(i)}]=\frac{1}{n}$$ but I don't know how to prove this rigorously.

Answer 1

For any $k_1$ and $k_2$, by symmetry $\Pr(X_{k_1}=X^{(i)}) = \Pr(X_{k_2}=X^{(i)})$. Also, if $k_1 \ne k_2$, then $\Pr(X_{k_1} = X_{k_2}) = 0$ (since the distribution is continuous). So by the exclusion-inclusion formula $$ 1 = \sum_{k=1}^n \Pr(X_k = X^{(i)}) = n \Pr(X_1 = X^{(i)}) .$$