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I read a statement in a book and I can't realize what exactly it means, and why it is true.

a set of random variables are given. in general, the expected value of the maximum of random variables is not equal to the maximum of the expected values of these variables.

I would like to know what exactly it means and why it is true.

thanks

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Take two independent coins $c_1,c_2$ which have equal probability of heads or tails.

Assign a value of $1$ for a head and $0$ for a tail.

Then $E c_1 = E c_2 = {1 \over 2}$ and $\max (E c_1, E c_2) = {1 \over 2}$.

Now consider $\max(c_1,c_2)$, that is, toss both coins and take the 'best' result. Then $E [\max(c_1,c_2)] = {3 \over 4}$ (just look at all 4 possibilities).

Hence $\max (E c_1, E c_2) \neq E [\max(c_1,c_2)]$.

Addendum: Note that in the first case, heads and tails are equally likely so $E c_1 = {1 \over 2} 0 + {1 \over 2} 1 = {1 \over 2}$.

In the second case, each pair of possibilities is equally likely so \begin{eqnarray} E [\max(c_1,c_2) &=& {1 \over4} \max(0,0) + {1 \over4} \max(0,1) +{1 \over4} \max(1,0) +{1 \over4} \max(1,1)\\ &=& {1 \over4} 0 + {1 \over4} 1 +{1 \over4} 1 +{1 \over4} 1 \\ &=& {3 \over 4} \end{eqnarray}

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  • $\begingroup$ Excuse me if my question is very obvious, it has been 9 years I didn't study statistics. $E[max(c_1,c_2)]$ equals to the expectation of head for both coins, right? then it should be $$x_1 \times P(x_1) + x_2 \times P(x_2) = 1/2 \times1+1/2\times1=1 $$ then why it is $3/4$? $\endgroup$ – M a m a D Nov 17 '15 at 17:19
  • $\begingroup$ No. It is the expected value of the $\max$. The possible values for $(c_1,c_2)$ are $(0,0), (0,1),(1,0), (1,1)$ and the corresponding $\max$ are $0,1,1,1$ hence the expected value is ${3 \over 4}$. $\endgroup$ – copper.hat Nov 17 '15 at 17:22
  • $\begingroup$ Note that $\max$ takes the 'best' result. If you toss two coins then intuitively the chances of getting at least one head must be greater than getting a head with a single toss. $\endgroup$ – copper.hat Nov 17 '15 at 17:30
  • $\begingroup$ best result occurs when the result of tossing both coins be head, am I right? $\endgroup$ – M a m a D Nov 17 '15 at 17:32
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    $\begingroup$ Yes, I added something to that effect in the answer. Be careful though, you are looking at the expectation over all possibilities, not just a restricted expectation over a subset of 'best possibilities'. $\endgroup$ – copper.hat Nov 17 '15 at 17:41

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