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Let $A$ and $B$ be independent, positive random variables. Why must $E(\min(A, B)) < \min(E(A), E(B))$, where $\min(X, Y)$ is the minimum of $X$ and $Y$?

I would think the opposite, that $E(A, B) > \min(E(A), E(B))$ because $E(\min(A, B))$ weights all possible values.

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2 Answers

up vote 10 down vote accepted

$\min(A,B) \le A$, so $E[\min(A,B)] \le E[A]$. Similarly $E[\min(A,B)] \le E[B]$.
Thus $E[\min(A,B)] \le \min(E[A],E[B])$.

But the statement with $<$ is not true. For example, if $A < B$ in all outcomes, $\min(A,B) = A$ and $E[\min(A,B)] = E[A] = \min(E[A],E[B])$.

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This indicates that if F is a function, then F(min(a, b))≤min(F(a), F(b)). So one could respond to the question by saying "because E is a function." –  Doug Spoonwood Dec 12 '12 at 1:22
Or rather, because $E$ is a nondecreasing function. –  Robert Israel Dec 12 '12 at 2:12
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Robert's answer is great, but to gain some more intuition, imagine that the expected value behaves like a mean and compare the following inequalities.

\begin{align} \mathbb{E} \min(A,B) &\leq \min(\mathbb{E} A, \mathbb{E} B) \\ \\ \frac{\min(a_1,b_1) + \min(a_2,b_2)}{2} &\leq \min\left(\frac{a_1+a_2}{2}, \frac{b_1+b_2}{2}\right) \end{align}


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