# Expected value of maximum of two random variables from uniform distribution

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 $\max(X,Y)$?

• The answers all assume that $X$ and $Y$ are independent. Without this assumption the question cannot be answered. If you meant to imply this assumption, please add it to the question. – joriki Jan 11 at 9:24
• math.stackexchange.com/q/1874340/321264 – StubbornAtom May 19 at 15:12

Here are some useful tools:

1. For every nonnegative random variable $Z$, $$\mathrm E(Z)=\int_0^{+\infty}\mathrm P(Z\geqslant z)\,\mathrm dz=\int_0^{+\infty}(1-\mathrm P(Z\leqslant z))\,\mathrm dz.$$
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, $$\mathrm E(\max(X,Y))=a+\frac23(b-a)=\frac13(2b+a).$$

• Hi, would you please tell me what theorem step 2 comes from? – Austin Jan 26 '16 at 1:24
• @Larry Theorem? Rather, the observation that $\{\max(X,Y)\leqslant z\}=\{X\leqslant z,Y\leqslant z\}$, plus independence. – Did Jan 26 '16 at 1:57
• I feel really dumb for asking this, but why is this true? I can't remember my professor ever mentioning the min or max functions in my previous probability course. – Austin Jan 26 '16 at 2:22
• @Larry ?? $\max(x,y)\leqslant z\iff (x\leqslant z \land y\leqslant z)$. Not a probabilistic result... – Did Jan 26 '16 at 8:19
• This is totally true...You can think $Z= \max{(x,y)}$ – Xiaonan Aug 8 '16 at 6:18

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:

• Thanks for pointing that out. As your answer corrects the mistake, I'll simply delete mine. – Martin Van der Linden Nov 28 '14 at 16:00
• E(max(x,y)) should read E(max(X,Y)). – Did Aug 22 '15 at 22:24

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.

• -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

Since x and y are independent random variables, we can represent them in x-y plane bounded by x=0, y=0, x=1 and y=1. Also we can say that choosing any point within the bounded region is equally likely. So, if were to choose a small area around a set of value (x,y)- probability of that, i.e., ***P(X=x,Y=y)= dx.dy/(A)***. Where A is the total area where (x,y) might belong, Hence A=1*1= 1. Also note that $$\iint_{A} P(X=x,Y=y)=\iint_{A}\frac{(dx)(dy)}{1}= 1$$ Hence, P(X=x,Y=y) is indeed a probability density function. Please see the Image of random variables in xy plane.

Now, let Z= max(x,y). Note that when (x,y) is below the line y=x (i.e, x>y); Z=x When (x,y) is above the line y=x (i.e, y>x), Z=y. So if we compute the expected value over the whole region it would be; $$\iint_{A} Z \times P(X=x,Y=y)=\int_{0}^1\int_{0}^x x\frac{dydx}{1}+\int_{0}^1\int_{x}^1 y\frac{dydx}{1}=\int_{0}^1 x^2dx+\int_{0}^1 \frac{1}{2}\times(1-x^2)dx= \frac{2}{3}$$