Expected value of max(x, y) We are given a square of unit length which has its centre at $(0,0)$ and its edges are parallel to the axes. How to find the expected value of $\max(x, y)$ where $(x,y)$ is a point in the square.
Edit:
Firstly I thought about just $x$ axis. $x$ axis varies in $[-0.5, 0.5]$. And $y$ will always be $0$. Therefore, $\max(x, y)$ in $-0.5 \le x \lt 0$ will be equal to $0$ and so will be the expected value. For $0 \le x \lt 0.5, \space \max(x, y) = x$. Therefore, expected value will be $\int_0^{0.5} x^2 dx $. But I don't know how to account for $2$ dimensional case.
 A: Transform the problem to have $X$ and $Y$ independent and uniformly distributed
on $[0,1]$. Let $Z = \max\{X,Y\}$ and note that $0 \leq Z \leq 1$. Then, for
any $\alpha \in [0,1]$, 
$$F_Z(\alpha) = P\{\max\{X,Y\} \leq \alpha\} = P\{X \leq \alpha, Y \leq \alpha\} = \alpha^2$$ and
so $f_Z(\alpha) = 2\alpha\mathbf 1_{\alpha \in [0,1]}$. This density has
a triangular shape and so the center of mass is $\frac{2}{3}$ for those
who remember their high-school physics, while those who don't, write
$$E[Z] = \int_0^1 \alpha \cdot 2\alpha \, \mathrm d\alpha = \frac{2}{3}.$$
Transforming the problem back gives the answer to the original problem as
$\frac{2}{3}$ $-\frac{1}{2} = \frac{1}{6}$ which matches the result obtained
from the double integral in @kaine's answer.
PS. changed $E[X]$ to $E[Z]$, but site doesn't accept single letter edits, hence this note.
A: Draw a picture of the square and overlay the line $y=x$.  Anything above $y=x$ has y greater than x and the problem is exactly symmetrical about this line.  The value is, therefore, $$2\int_{-\frac{1}{2}}^{\frac{1}{2}} \int_x^\frac{1}{2} y dy dx$$ which would be divided by the area but the area is 1.  I am looking for a way to do this without integrals but don't know one off hand.
