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 $\max(X,Y)$?
 A: 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, $$\mathrm E(\max(X,Y))=a+\frac23(b-a)=\frac13(2b+a).$$
A: 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: 

A: 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.     
A: 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;
$$\iint_{A} Z \times P(X=x,Y=y)=\int_{0}^1\int_{0}^x x\frac{dydx}{1}+\int_{0}^1\int_{x}^1 y\frac{dydx}{1}=\int_{0}^1 x^2dx+\int_{0}^1 \frac{1}{2}\times(1-x^2)dx= \frac{2}{3} $$
A: I find the approach described in https://www.probabilitycourse.com/chapter4/4_1_3_functions_continuous_var.php easy to follow and applicable for this problem. It's a 3 step process

*

*Get the range of the required distribution, in this case, max(X, Y)

*Find the CDF of this distribution as a function of the known
distributions

*Find the PDF of the distribution by differentiating the
CDF

Let's say our new distribution is denoted by Z, it takes values in the range [0,1]
$P(Z <= z) = P(Max(X, Y) <= z)$
= $P((X,Y) <= z)$
= $P(X <= z,Y <= z)$
= $P(X <= z) . P(Y <= z)$
= $(z-a) \over (b-a) $ . $(z-a) \over (b-a) $
It can be used to get the CDF for [0,1] uniform distribution, PDF is differential of the CDF and E[Z] will be straightforward integral over [0,1].
PS: There is another approach described in the book for generic order static problems but the proof is relatively involved when compared to this approach.
