Supplementary Solutions to HW 5

H&G 2.11

X/Y	1	3	9
2	1/8	1/24	1/12
4	1/4	1/4	0
6	1/8	1/24	1/12

a) Find marginal of Y, f(y):
P(Y=1) = f(1) = 4/8 = 1/2
f(3) = 8/24 = 1/3
f(9) = 2/12 = 1/6

b) f(y|2)
f(1|2) = f(1,2)/P(X=2) = (1/8) / (6/24) = 1/2
f(3|2) = (1/24)/ (1/4) = 1/6
f(9|2) = (1/12)/(1/4) = 1/3

c) COV(X,Y) = E(X - mu_x)*(Y-mu_y)
First we need mu_x = E(X) = 2*(6/24) + 4*(2/4) + 6*(6/24) = 4
mu_y = E(Y) = 1*(1/2) + 3*(1/3) + 9*(1/6) = 3

E(X-mu_x)(Y-mu_y) = (2-4)*(1-3)*(1/8) + (2-4)*(3-3)*(1/24) + (2-4)*(9-3)*(1/12) +
(4-4)*(1-3)*(1/3) + (4-4)*(3-3)*(1/4) + (4-4)*(9-3)*0 +
(6-4)*(1-3)*(1/8) + (6-4)*(3-3)*(1/24) + (6-4)*(9-3)*(1/12) = 0

d)  WARNING:  THIS IS THE WRONG ANSWER: "THEY ARE INDEPENDENT BECAUSE COV = 0."

THIS IS WRONG BECAUSE YOU CAN'T CONCLUDE INDEPENDENCE JUST BECAUSE COV=0.  IF X, Y ARE INDEPENDENT, THEN THEIR COVARIANCE WILL BE 0. HOWEVER, SOME PAIRS OF RANDOM VARIABLES  HAVE COVARIANCE 0 AND YET ARE  DEPENDENT.  

So we're forced to check.  There are three approaches.  You can choose whichever one you want.  They will all give the same result:
1) Does f(x,y) = f(x) * f(y)?  If so, they're indpt.
2) Does f(x|y) = f(x)?  If so, indpt.
3) Does f(y|x) = f(y)?  If so, indpt.

We've already got a head-start on method (3).  In part (a) we found f(y).  In part (b) we found f(y|2) and, in fact, f(3|2) = 1/6 but f(3) = 1/3. These are not equal, so they are DEPENDENT.