Question

24.

Let S be the educational attainment of individuals in a town, with values S=0 for less than high school and S=1 for high school or above. Also, let Y be their individual annual income with values Y=0 for less than $20,000, Y=1 for between $20,000 and $40,000, and Y=2 for above $40,000. Consider now the following joint probabilities:

S\Y         Y=0 (less than $20K)       Y=1 ($20K-$40K)              Y=2 (more than $40K)

S=0 (Less than HS)            0.05        0.03        0.01

S=1 (HS or more)              0.22        0.36        0.33

Determine the expected value (mean) of the conditional expected level of educational attainment (given the various levels of individual annual income).

Answer 1

Consider the two events

S : Level of educational attainment.

Y : Levels of individual annual income.

		Y			Total
		Y = 0 ( Less than $ 20 K)	Y=1 ( $ 20 K - $ 40 K)	Y=2 ( more than $ 40 K)
S	S =0 ( less than High School)	0.05	0.03	0.01	0.09
	S=1 ( High school or more)	0.22	0.36	0.33	0.91
Total		0.27	0.39	0.34	1

The Marginal probability distribution of S is

S	0	1	Total
P(S=s)	0.09	0.91	1

The marginal probability distribution of Y is

Y	0	1	2	Total
P(Y=y)	0.27	0.39	0.34	1

We have to find

E ( S / Y=0) , E(S / Y=1) and E(S / Y=2).

by definition of conditional distribution

P ( S / Y=y) = P ( S=s, Y=y) / P ( Y=y).

Hence the conditional distribution of S given Y = 0 is

S	0	1	Total
P(S/ Y=0)	0.1852	0.8148	1.0000

by definition of conditional expectation

E ( S / Y=0) = 0 * 0.1852 + 1 * 0.8148

E(S / Y = 0) = 0.8148.

The conditional distribution of S given Y =1 is

S	0	1	Total
P(S / Y=1)	0.0769	0.9231	1.0000

E ( S / Y=1) = 0 * 0.0769 + 1 * 0.9231

E ( S / Y=1) = 0.9231.

The conditional distribution of S given Y =2 is

S	0	1	Total
P(S / Y=2)	0.0294	0.9706	1.0000

E ( S / Y=2) = 0* 0.0294 + 1 * 0.9706

E(S/ Y=2) = 0.9706.