Lecture_16

Joint Probability Distributions

• Often we are not interested in the behavior of an individual random variable.

• Instead, we may be interested in the relationship between two components of a bivariate random variable.

Example: X = number of hours study per week

Joint and Marginal PMF

The joint probability mass function (or joint PMF) of the jointly discrete random variable X and Y is

\[ p(x,y)=P(X=x,Y=y) \]

• If \(S=\{(x_1,y_1),(x_2,y_2)....\}\) is the sample space of (X,Y), then

\[ p(x_i,y_i)\geq0~for~all~i\\\ \sum_{(x_i,y_i)\in S}p(x_i,y_i)=1\\ p(x<X\leq b,c<Y\le d)=\sum_{i:a,x_i\le b,c<y_i\leq d}p(x_i,y_i) \]

• The distributions of the individual random variables are called marginal distributions. The marginal PMFs can be found using the joint PMF.

\[ p_X(x)=\sum_{y\in S_Y}p(x,y)\\ p_Y(y)=\sum_{x\in S_X}p(x,y) \]

Example: Measurements for the length and width of plastic covers for CDs are rounded to the nearest mm.

X = length of a randomly selected CD cover

Y = width of a randomly selected CD cover

The possible values of X are 129, 130, and 131 mm. The possible values of Y are 120 and 121 mm. The joint PMF for X and Y is

  y
  120 121

x 129 0.12 0.08 130 0.42 0.28 131 0.06 0.04

1. Verify that this is a valid joint PMF

The sum of probability is 1, or each item is greater than 0.

2. Find \(P(129.5 \le X \le 131.5, Y \geq 121)\).

\[ =P(X=130,Y=121)+P(X=131,Y=121)=0.28+0.06=0.32 \]

3. Find \(P(129.5\le X\le131.5)\)

\[ =P(X=130)+P(X=131)=0.42+0.28+0.06+0.04=0.8 \]

4. Find the marginal PMF of Y.

y	120	121
p(y)	0.6	0.4