Examples

Expectation of X squared plus Y from joint PMF chart

Suppose the joint PMF of $X$ and $Y$ is given by this chart:

$Y\downarrow X\to$	$1$	$2$
$-1$	0.2	0.2
$0$	0.35	0.1
$1$	0.05	0.1

Define $W=X^2+Y$. Find the expectation $E[W]$.

Solution

First compute the values of $W$ for each pair $(X,Y)$ in the chart:

$Y\downarrow X\to$	$1$	$2$
$-1$	0	3
$0$	1	4
$1$	2	5

Now take the sum, weighted by probabilities:

$$\begin{array}{c}0\cdot (0.2)+3\cdot (0.2)+1\cdot (0.35)\\ +4\cdot (0.1)+2\cdot (0.05)+5\cdot (0.1)\end{array}\gg \gg 1.95=E[W]$$

Expectation of Y two ways and Expectation of XY from joint density

Suppose $X$ and $Y$ are random variables with the following joint density:

$$f_{X,Y}(x,y)=\{\begin{array}{cc}\frac{3}{16}x{y}^2& x,y\in [\mathrm{0,2}]\\ 0& \text{ otherwise }\end{array}$$

(a) Compute $E[Y]$ using two methods.

(b) Compute $E[XY]$.

Solution

(a)

(1) Method One: via marginal PDF $f_Y(y)$:

$$f_Y(y)={\int }_0^2\frac{3}{16}x{y}^{2}dx\gg \gg \{\begin{array}{cc}\frac{3}{8}{y}^2& y\in [\mathrm{0,2}]\\ 0& \text{otherwise}\end{array}$$

Then expectation:

$$E[Y]={\int }_0^{2}y{f}_Y(y)dy\gg \gg {\int }_0^2\frac{3}{8}{y}^{3}dy\gg \gg 3/2$$

(2) Method Two: directly, via two-variable formula:

$$E[Y]={\int }_0^2{\int }_0^{2}y\cdot \frac{3}{16}x{y}^{2}dydx\gg \gg {\int }_0^2\frac{3}{4}xdx\gg \gg 3/2$$

(b) Directly, via two-variable formula:

$$\begin{array}{c}E[XY]={\int }_0^2{\int }_0^{2}xy\cdot \frac{3}{16}x{y}^{2}dydx\\ \\ \gg \gg {\int }_0^2\frac{3}{4}{x}^{2}dx\gg \gg 2\end{array}$$