Examples

Covariance from 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

Find $\mathrm{Cov}[X,Y]$.

Solution

We need $E[X]$ and $E[Y]$ and $E[XY]$.

$$E[X]=1\cdot (0.2+0.35+0.05)+2\cdot (0.2+0.1+0.1)\gg \gg 1.4$$ $$\begin{array}{c}E[Y]=-1\cdot (0.2+0.2)+0\cdot (0.35+0.1)+1\cdot (0.05+0.1)\\ \\ \gg \gg -0.25\end{array}$$ $$E[XY]=-1\cdot (0.2)-2\cdot (0.2)+0+1\cdot (0.05)+2\cdot (0.1)\gg \gg -0.35$$

Therefore:

$$\begin{array}{c}\mathrm{Cov}[X,Y]=E[XY]-E[X]E[Y]\\ \\ \gg \gg -0.35-(1.4)(-0.25)\gg \gg 0\end{array}$$

Variance of sum of indicators

An urn contains 3 red balls and 2 yellow balls.

Suppose 2 balls are drawn without replacement, and $X$ counts the number of red balls drawn.

Find $\mathrm{Var}[X]$.

Solution

Let $X_1$ indicate (one or zero) whether the first ball is red, and $X_2$ indicate whether the second ball is red, so $X=X_1+X_2$.

Then $X_1{X}_2$ indicates whether both drawn balls are red; so it is Bernoulli with success probability $\frac{3}{5}\frac{2}{4}=\frac{3}{10}$. Therefore $E[X_1{X}_2]=\frac{3}{10}$.

We also have $E[X_1]=E[X_2]=\frac{3}{5}$.

The variance sum rule gives:

$$\begin{array}{cc}\mathrm{Var}[X]& =\mathrm{Var}[X_1]+\mathrm{Var}[X_2]+2\mathrm{Cov}[X_1,X_2]\\ & \\ & \gg \gg E[X_1^2]-E{[X_1]}^2+E[X_2^2]-E{[X_2]}^2+2(E[X_1{X}_2]-E[X_1]E[X_2])\\ & \\ & \gg \gg \frac{3}{5}-{\left(\frac{3}{5}\right)}^2+\frac{3}{5}-{\left(\frac{3}{5}\right)}^2+2(\frac{3}{10}-\frac{3}{5}\cdot \frac{3}{5})\gg \gg \frac{9}{25}\end{array}$$