Home page for accesible maths 8.1 Joint probability mass functions 8.1 Joint probability mass functions 8.2 Independence

Style control - access keys in brackets

Font (2 3) - + Letter spacing (4 5) - + Word spacing (6 7) - + Line spacing (8 9) - +

8.1.1 Properties of $p_{X,Y}(x,y)$ :

1.

For all $x$ and $y$ , $0\leq p_{X,Y}(x,y)\leq 1$ ,
2.

$\sum_{\mbox{all }x,y}p_{X,Y}(x,y)=1$ ,
3.

${\rm P}((X,Y)\in A)=\sum_{(x,y)\in A}p_{X,Y}(x,y).$

Example 8.2.

The joint pmf of $X$ and $Y$ is

p_{X,Y}(x,y)=(x+y)/18

for $x,y=0,1,2$ .

a.

Write out the joint probability table.
b.

Show this is a valid joint pmf.
c.

Evaluate (i) ${\rm P}(X=2)$ , (ii) ${\rm P}(X=Y)$ , (iii) ${\rm P}(X+Y\geq 2)$ .

Solution.

a.

$\begin{array}[]{cc|ccc}&&&y&\\ &p_{X,Y}(x,y)&0&1&2\\ \hline&0&0&1/18&2/18\\ x&1&1/18&2/18&3/18\\ &2&2/18&3/18&4/18\end{array}$
b.

$p_{X,Y}(x,y)\geq 0$ for all $x, y$ , and $\sum_{\mbox{all }(x,y)}p_{X,Y}(x,y)=(0+1+2+1+2+3+2+3+4)/18=1$ .
c.

(i) ${\rm P}(X=2)={\rm P}((X,Y)\in\{(2,0),(2,1),(2,2)\})=2/18+3/18+4/18=9/18$
(ii) ${\rm P}(X=Y)={\rm P}((X,Y)\in\{(0,0),(1,1),(2,2)\})=0/18+2/18+4/18=6/18$
(iii) ${\rm P}(X+Y\geq 2)={\rm P}((X,Y)\in\{(0,2),(1,1),(1,2),(2,0),(2,1),(2,2)\})=(2% +2+3+2+3+4)/18=16/18$

Note that each of $X$ and $Y$ still have their own probability mass functions $p_{X}$ and $p_{Y}$ . In the context of jointly distributed random variables, these are called the marginal probability mass functions.

p_{X}(x)={\rm P}(X=x)={\rm P}((X,Y)\in\{(x,0),(x,1),(x,2),\ldots\})=\sum_{y=0}% ^{\infty}p_{X,Y}(x,y).

Similarly,

p_{Y}(y)=\sum_{x=0}^{\infty}p_{X,Y}(x,y).

Exercise 8.3.

Bivariate random variables ${X}$ and ${Y}$ have joint pmf

		${y}$				$p_{X}(x)$
		0	1	2	3
${x}$	1	5/60	8/60	2/60	1/60	16/60
	2	12/60	7/60	3/60	2/60	24/60
	3	4/60	8/60	6/60	2/60	20/60
$p_{Y}(y)$		21/60	23/60	11/60	5/60	1

Fill in the marginal pmfs in the final row and column.

Style control - access keys in brackets

8.1.1 Properties of pX,Y⁢(x,y):

Example 8.2.

Solution.

Exercise 8.3.

8.1.1 Properties of $p_{X,Y}(x,y)$ :