Finding the expected value for various functions of two independent random variables

Question

Let $X$ and $Y$ be independent random variables with uniform density functions on $[0, 1]$. Find:

$E(|X-Y|)$

$E(X)=E(Y)=1/2$, $f_{X,Y}(x,y)=1$

Integrating the region where $x > y$ and using symmetry for region $y > x$,

$$E(|X-Y|)=2\int_{0}^{1}\int_{0}^{x}(x-y)\,dy\,dx=2\int_{0}^{1}x^2/2\,dx=1/3$$

$E(\max(X,Y))$

Integrating over region where $y > x$ and symmetry

$$E(\max(X,Y))=2\int_{0}^{1}\int_{x}^{1}y\,dy\,dx=2\int_{0}^{1}1/2-x^2/2\,dx=2/3$$

$E(\min(X,Y))$

Integrating over region where $x < y$ and symmetry

$$E(\min(X,Y)) = 2\int_{0}^{1}\int_{0}^{y}x\,dx\,dy=2\int_{0}^{1}y^2/2\,dy=1/3$$

$E(X^2+Y^2)$

$$E(X^2+Y^2)=E(X^2)+E(Y^2)=2\int_{0}^{1}x^2dx=2/3$$

$E((X+Y)^2)$

$E((X+Y)^2)=E(X^2+2XY+Y^2)=2/3+2E(XY)=2/3+2E(X)E(Y)=7/6$

$E(X^Y)$

$$E(X^Y) = \int_{0}^{1}\int_{0}^{1}e^{y\log x}\,dy\,dx=\int_{0}^{1}\frac{e^{\log x}-1}{\log x}\,dx = \log 2$$ (using Wolfram Alpha)

Did I make any mistakes?

Answer 1

Or alternatively :

\begin{align*} \mathbb E[\max(X,Y)] &= \mathbb E\left[\frac{X+Y+|X-Y|}{2}\right]\\ &= \frac{1}{2} (\mathbb E[X]+\mathbb E[Y]+\mathbb E[|X-Y|])\\ &= \frac{2}{3}\\ \mathbb E[\min(X,Y)] &= \mathbb E[X+Y-\max(X,Y)]\\ &= \frac{1}{2}+\frac{1}{2}-\frac{2}{3}\\ &= \frac{1}{3} \end{align*}