Question

For the following function

x                 1                      1                   2                      4

y                  3                      4                   5                      6          

f_XY(x,y)      3/7                    1/7                2/7                   1/7    

1. Determine the covariance and the correlation

b) Are X and Y independent? Explain why or why not.

Answer 1

x                 1                      1                   2                      4

y                  3                      4                   5                      6          

fXY(x,y)      3/7                    1/7                2/7                   1/7    

a.

E(x) = 1*3/7 + 1*1/7 + 2*2/7 + 4*1/7 = 1.7143

E(y) = 3*3/7 + 4*1/7 + 5*2/7 + 6*1/7 = 4.1429

covariance :

cov (x,y) = (1 - 1.7143)*(3 - 4.1429)*(3/7) + (1 - 1.7143)*(4 - 4.1429)*(1/7) + (2 - 1.7143)*(5 - 4.1429)*(2/7) + (4 - 1.7143)*(6 - 4.1429)*(1/7)

cov (x,y) = 1.0408

correlation :

std dev. of x = s(x) = [ (1 - 1.7143)^2*(3/7) + (1 - 1.7143)^2*(1/7) + (2 - 1.7143)^2*(2/7) + (4 - 1.7143)^2*(1/7) ]^0.5

s(x) = 1.0302

std dev. of y = s(y) = [ (3 - 4.1429)^2*(3/7) + (4 - 4.1429)^2*(1/7) + (5 - 4.1429)^2*(2/7) + (6 - 4.1429)^2*(1/7) ]^0.5

s(y) = 1.1249

correlation coeff. = r = cov(x,y) / (s(x)*s(y))

= 1.0408 / (1.0302*1.1249)

= 0.8981

correlation coeff. = 0.8981

b.

for independence :

P(x,y) = P(x)*P(y)

P(x=1) = P(x=1,y=3) + P(x=1,y=4) = 3/7 + 1/7 = 4/7

P(y=3) = P(x=1,y=3) = 3/7

P(x=1)*P(y=3) = (4/7)*(3/7) = 12/7

P(1,3) = 3/7

we can see ,

P(1,3) ≠ P(x=1)*P(y=3)

so they are not independent