On the conditioned tail of a maximum of a sequence of random variables Suppose we have a finite sequence $X_1, X_2, \cdots X_n$ of i.i.d real-valued random variables. Let $X=\max_{i} X_i$. Let $C$ be a fixed (large) constant. Does it hold for all $r$ that $$P(X>r|X>C) \leq P(X>r | X_i > C \ \forall i)$$
?
Intuitively, it seems this should be the case, since the latter condition asserts that the random variables are simultaneously large whereas the former condition only asserts that one of them is large.
EDIT: Note that the answer by angryavian has reduced the problem to that of solving an inequality.
 A: [This is a partial answer that attempts to express each side in terms of the CDF of $X_1$.]
If $r \le C$, then both sides equal $1$ and the inequality holds.
Let's assume $r > C$. The left-hand side is
$$\frac{P(X > r)}{P(X > C)} = \frac{1-P(X_1 \le r)^n}{1-P(X_1 \le C)^n}.$$
The right-hand side is
$$\begin{align}1 - P(X \le r \mid X_i > C \forall i)
&= 1- \frac{P(X\le r, X_1 > C, \ldots, X_n > C)}{P(X_1 > C, \ldots, X_n > C)}
\\
&= 1-\frac{P(C < X_1 \le r)^n}{P(X_1 > C)^n}
\\
&= 1-\left(1 - \frac{P(X_1 > r)}{P(X_1 > C)}\right)^n.
\end{align}$$
[Sanity check: both expressions are equal when $n=1$.]
From here, one can either try to prove the inequality, or instead produce some CDF for $X_1$ that falsifies the inequality.
A: We remark that this result can be generalized to the case where the random variables are independent but not identically distributed. The details are just a lot of algebra, but basically in the inequality discussed in angryavian's answer, powers like $x^n, c^n$ become products $x_1x_2 \cdots x_n, c_1c_2 \cdots c_n$, and the inequality is the generalization I have in my own answer to this.
