Party Affiliation Example
Suppose you have undertaken a random sample of voters. The following table
shows the distribution of party affiliation by male voters:
Number of
Male Adults
Party
Affiliation
Democrat
200
Republican
300
Other
60
Total
560
f(D) = 200/560 = 0.357 f(R) = 300/560 = 0.536 f(O) = 60/560 = .107
f(D or R) = 200/560 + 300/560 = 0.357 + 0.536 = 0.883
The following joint distribution of gender and party affiliation was found to be the
following:
Party Affiliation
Gender
Total
Male
Female
Democrat
200
270
470
Republican
300
100
400
Other
60
70
130
Total
560
440
1000
Joint Probability Density Function [f(g,p)] of
Gender (G) and Party Affiliation
Party Affiliation
Gender
Total
Male
Female
Democrat
0.20
0.27
0.47
Republican
0.30
0.10
0.40
Other
0.06
0.07
0.13
Total
0.56
0.44
f(D,M) = 200/1000 = 0.20
Marginal Distributions
fD = f(D,M) + f(D,F) = 0.20 + 0.27 = 0.47
fO = F(O,M) + F(O,F) = 0.06 + 0.07 = 0.13
fM = F(D,M) + F(R,M) + F(O,M) = 0.20 + 0.30 + 0.06 = 0.56
Conditional Probability Density Functions
Conditional probability of party affiliation given that a person is female:
f(Pi | Gi = F) = f(Pi,G = F)/fF
Pi
f((Pi|G=F)
D
f(D,F)/fF = 0.27/0.44 = 0.614
R
f(R,F)/fF = 0.10/0.44 = 0.227
O
f(O,F)/fF = 0.07/0.44 = 0.159
Another Example of a Joint Distribution
Y
X
1
fY
3
9
fX
2
0.1250 0.0417 0.0833
0.25
4
0.2500 0.2500 0.0000
0.50
6
0.1250 0.0417 0.0833
0.25
0.50
0.3333 0.1667
Calculation of Expected Value of X and Y
E(X) = ∑ X f x (X) = 2(0.250) + 4(0.500) + 6(0.250) = 4.625
x
E(Y) = ∑ Y f Y (Y) = 1(0.50) + 3(0.333) + 9(0.167) = 3.000
Y
Calculation of Covariance Matrix
E(XY) = 2*1*(0.1250) + 2*3*(0.0417) + 2*9*(0.0833) +
4*1*(0.2500) + 4*3*(0.2500) + 4*9*(0.0000) +
6*1*(0.1250) + 6*3*(0.0417) + 6*9*(0.0833) = 12.0
Cov(XY) = E(XY) - E(X)E(Y) = 12.0 - (4.625)(3.000) = -1.875