Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $E[\min(A,B)] = E[A] = \min(E[A],E[B])$.

share|cite|improve this answer
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

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}


share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.