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Let $X_1,\dots,X_n\geq 0$ be independent, continuous random variables, and assume that $X_i$ stochastically dominates $X_j$ if $i<j$. This means that for any $x>0$, \begin{align} \mathbb{P}(X_i\geq x) \geq \mathbb{P}(X_j\geq x). \end{align} In this case we clearly have $\mathbb{P}(X_i=X_{(1)})\geq\mathbb{P}(X_j=X_{(1)})$, where $X_{(1)}=max(X_1,\dots,X_n)$. In other words, $X_i$ is more likely than $X_j$ to be the largest random variable.

Can we show that for a given $\alpha>0$, \begin{align} \mathbb{P}(X_i=X_{(1)}|X_{(2)}\geq \alpha) \geq \mathbb{P}(X_j=X_{(1)}|X_{(2)}\geq \alpha). \end{align} Note that the event $X_i=X_{(1)}|X_{(2)}\geq\alpha$ means that given at least two of the random variables exceed the level $\alpha$, the random variable $X_i$ is the largest one.

P.S. $X_i$ stochastically dominating $X_j$ also means that $X_i$ has the same distribution as $X_j+Y$, where $Y\geq 0$ is a random variable independent of $X_j$.

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Slightly surprisingly, this is incorrect.

Let $n=2$.

$X_1=0.2$ w.p. $0.9$, and $X_1=1.2$ w.p. $0.1$.

$X_2=0$ w.p. $0.9$, and $X_2=1$ w.p. $0.1$.

$\alpha=0.1$ is a counterexample.

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  • $\begingroup$ The random variables are assumed to be continuous, but do you agree that your example can be extended to that setting? Specifically, by letting $X_1$ have a distribution that is greater than $\alpha$ with a high probability, and $X_2$ be a distribution that is greater than $\alpha$ with a low probability, we should have the same kind of counterexample? $\endgroup$
    – svonimir
    Commented Jul 2 at 14:38
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    $\begingroup$ Yes I agree! Continuity is not a main issue here. $\endgroup$
    – andy
    Commented Jul 2 at 16:26

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