Prof

Prof. Bryan Caplan

bcaplan@gmu.edu

http://www3.gmu.edu/departments/economics/bcaplan

Economics 345

Fall, 1998

Problem Set #1 Answer Key
Wallace and Silver Problems.

1.1.

a. Since the problem specifies that there is no replacement, you have to be careful when calculating the probability of different events. The best approach is to make a table (or tree) showing all possibilities and their probabilities:

1st draw	2nd draw	3rd draw	Prob
O	Y	Y	(3/5)(2/4)(1/3)=6/60
Y	O	Y	(2/5)(3/4)(1/3)=6/60
Y	O	Y	(2/5)(1/4)(3/3)=6/60
Y	O	O	(2/5)(3/4)(2/3)=12/60
O	Y	O	(3/5)(2/4)(2/3)=12/60
O	O	Y	(3/5)(2/4)(2/3)=12/60
O	O	O	(3/5)(2/4)(1/3)=6/60

Reading off the table, P(X=1)=18/60; P(X=2)=36/60; P(X=3)=6/60;

P(Y=0)=6/60; P(Y=1)=36/60; P(Y=2)=18/60

Y/X	1	2	3	P(Y)
0	0	0	6/60	6/60
1	0	36/60	0	36/60
2	18/60	0	0	18/60
P(X)	18/60	36/60	6/60

c. E(X)=1.80; E(Y)=1.20; Var(X)=.36; Var(Y)=.36

d. Cov(X,Y)=-.36

e. P(1)=54/60; P(2)=0; P(3)=6/60

Z/X	1	2	3	P(Z)
1	18/60	36/60	0	54/60
2	0	0	0	0
3	0	0	6/60	6/60
P(X)	18/60	36/60	6/60

f. If X=1, then Y=2, so |X-Y|=|1-2|=1 with probability 1.

g. E(Z)=1.2; V(Z)=.360; Cov(X,Y)=.24

1.8.

a. 8

b. V(X-Y)=Var(X)+Var(Y)+2*Cov(X,Y)=6

c. 44

d. (a) 8; (b) 0; (c) 80

1.9. X must be a constant.

1.16. X has a normal distribution with mean of 66 and SD=10 (Since Var(X)=100). Define Z=(X-66/10)~N(0,1). Thus P(X<50) equals P(Z<(50-66)/10)=.0548.

Freedman and Lane Problems.

p.30, #5.

a. T. Mathematically, , so it can't be negative. Intuitively, you can have a big spread, a small spread, or no spread, but you can't have negative spread!