# Probability of an event involving the maximum of (stochastically ordered) random variables

Let $$X_1,\dots,X_n\geq 0$$ be independent, continuous random variables, and assume that $$X_i$$ stochastically dominates $$X_j$$ if $$i. This means that for any $$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 $$X_j$$.

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.

• 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? Commented Jul 2 at 14:38
• Yes I agree! Continuity is not a main issue here.
– andy
Commented Jul 2 at 16:26