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If I have two variables $X$ and $Y$ which randomly take on values uniformly from the range $[a,b]$ (all values equally probable), what is the expected value for $\max(X,Y)$?

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3 Answers

up vote 8 down vote accepted

Here are some useful tools:

  1. For every nonnegative random variable $Z$, $$\mathrm E(Z)=\int_0^{+\infty}\mathrm P(Z\geqslant z)\,\mathrm dz.$$
  2. As soon as $X$ and $Y$ are independent, $$\mathrm P(\max(X,Y)\leqslant z)=\mathrm P(X\leqslant z)\,\mathrm P(Y\leqslant z).$$
  3. If $U$ is uniform on $(0,1)$, then $a+(b-a)U$ is uniform on $(a,b)$.

If $(a,b)=(0,1)$, items 1. and 2. together yield $$\mathrm E(\max(X,Y))=\int_0^1(1-z^2)\,\mathrm dz=\frac23.$$ Then item 3. yields the general case, that is, $$\mathrm E(\max(X,Y))=a+\frac23(b-a)=\frac13(2b+a).$$

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did's excellent answer proves the result. The picture here enter image description here

may help your intuition. This is the "average" configuration of two random points on a interval and, as you see, the maximum value is two-thirds of the way from the left endpoint.

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-1 Sorry for the down vote. I know what you mean, but I hate such examples, as it confuses the situation for those who do not fully understand what an "average" configuration means. –  Calvin Lin Jul 1 '13 at 2:57
I don't follow. In what sense is this an average configuration? –  Jonah Jul 9 '13 at 18:02
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Here is another way to prove it, for $a=0$ and $b=1$ which should not be too hard to extend to whatever values of $a$ and $b$

\begin{align*} E(\max\{X,Y\}) & = \int_0^1 \int_0^1 \max\{x,y\} f_{X,Y}(x,y) dx dy\\ & = \int_0^1 \int_0^1 \max\{x,y\} dx dy\\ & = \int_0^1 \int_0^x y~~ dxdy + \int_0^1 \int_0^y x~~ dydx\\ & = \int_0^1 \frac{x^2}{2} dx + \int_0^1 \frac{y^2}{2} dy\\ & = \frac{1}{3} + \frac{1}{3} \\ & = \frac{2}{3} \end{align*}

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