Question

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

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

B) Determine the expected level of educational attainment.

Answer 1

A)

S\Y	Y=0	Y=1	Y=2	Total
S = 0	0.05	0.03	0.01	0.09
S = 1	0.22	0.36	0.33	0.91
Total	0.27	0.39	0.34	1

P(Y=0) = total under ( Y=0 )

= 0.27

P(Y=1) = total under ( Y=1 )

= 0.39

P(Y=2) = total under ( Y=2 )

= 0.34

Expected value formula in common is ,

  

We have to find the conditional expected level of educational attainment given Y =0 , Y=1 , Y = 2

First we can find with Y = 0

P ( S / Y =0) this will have two values with S =0 and S = 1

P(S=0/ Y=0) = P(S=0,Y=0)/ P (Y=0)

= 0.05/0.27

= 0.19

P(S=1/Y=0) = P( S=1 , Y=0)/ P(Y=0)

= 0.22/ 0.27

= 0.81

Then expected value of P( S/Y=0)

= 0×0.19 + 1×0.81

= 0+ 0.81

= 0.81

For given Y = 1

P ( S/Y = 1) , S =0 and S= 1

P(S=0/Y=1) = P(S=0,Y=1)/ P(Y=1)

= 0.03 / 0.39

= 0.08

P(S=1/Y=1) = P(S=1,Y=1)/P(Y=1)

= 0.36/0.39

= 0.92

Expected value of P(S/Y=1) is,

= 0×0.08 + 1×0.92

= 0 + 0.92

= 0.92

For given Y=2

P(S/Y=2) , for S=0 and S=1

P(S=0/Y=2) = P(S=0,Y=2) / P(Y=2)

= 0.01 / 0.34

= 0.03

P(S=1/Y=2) = P(S=1,Y=2) / P(Y=2)

= 0.33 / 0.34

= 0.97

Expected value of P(S/Y=2) is

= 0×0.03 + 1×0.97

= 0 + 0.97

= 0.97

Mean value of these 3 expected values are given as,

Mean = ( 0.81 + 0.92 + 0.97 ) / 3

= 2.7/3

= 0.9

Answer :

0.9

B)

We are asked to determine the expected level of the education attainment .

We know that educational attainment is indicated by two methods one is S= 0 and S=1

So we have to find the probability of S=0 and S=1 to get the expected value . From the table we can find P(S=0) and P(S=1) by adding the values in the same row and it is added and written in the total column in the solution of A which is given above this problem. Then ,

P(S=0) = 0.05 + 0.03 + 0.01

= 0.09

P(S=1) = 0.22 + 0.36 + 0.33

= 0.91

Then finding the expected value ,

E(S) = 0×0.09 + 1×0.91

= 0 + 0.91

= 0.91

Answer :

0.91