Order statistics: probability of second highest order statistics conditional on highest order value

Question

I need to calculate a probability of being second highest order statistics for some value, given that this value is a highest order statistics: $$ Pr(X_{n-1,n}\leq x|X_{n,n}=x) $$ I guess that the probability is either zero or one, but can not prove it. Thanks in advance.

Answer 1

The random set $\{X_{1,n},X_{2,n},\ldots,X_{n-1,n}\}$ is distributed like a sample of size $n-1$ from the conditional distribution of $X$ conditionally on $X\leqslant x$. Thus, for every $y\leqslant x$, $$ P[X_{n-1,n}\leqslant y\mid X_{n,n}=x]=\left(\frac{P[X\leqslant y]}{P[X\leqslant x]}\right)^{n-1}. $$

Answer 2

Is the probability 1 by definition of order statistics?