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 $\max(X,Y)$?
Expected value of maximum of two random variables from uniform distribution
4 Answers
Here are some useful tools:
- For every nonnegative random variable $Z$, $\mathrm E(Z)=\int_0^{+\infty}\mathrm P(Z\geqslant z)\,\mathrm dz=\int_0^{+\infty}(1-\mathrm P(Z\leqslant z))\,\mathrm dz.$
- As soon as $X$ and $Y$ are independent, $\mathrm P(\max(X,Y)\leqslant z)=\mathrm P(X\leqslant z)\,\mathrm P(Y\leqslant z).$
- 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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0This is totally true...You can think $Z= \max{(x,y)}$ – 2016-08-08
I very much liked Martin's approach but there's an error with his integration. The key is on line three. The intution here should be that when y is the maximum, then x can vary from 0 to y whereas y can be anything and vice-versa for when x is the maximum. So the order of integration should be flipped:
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0E(max(x,y)) should read E(max(X,Y)). – 2015-08-22
did's excellent answer proves the result. The picture 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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5I don't follow. In what sense is this an average configuration? – 2013-07-09