Question

Let the random variable X and Y have the joint pmf f(x, y) = , c xy 2 where x = 1, 2, 3; y = 1, 2, x + y ≤ 4 , that is, (x, y) are {(1, 1),(1, 2),(2, 1),(2, 2),(3, 1)} .

(a) Find c > 0 .

(b) Find μ . X

(c) Find μ . Y

(d) Find σ . 2 X

(e) Find σ . 2 Y

(f) Find Cov (X, Y ) .

(g) Find ρ , Corr (X, Y ) .

(h) Are X and Y independent?

Please show work/explanation if you can, thank you!

Answer 1

We would be looking at the first 4 parts of the question here as:

a) The value of C can be computed by summing up the probability values on X, Y range as 1.

P(1, 1) = c,
P(1, 2) = 4c,
P(2, 1) = 2c,
P(2,2) = 8c,
P(3,1) = 3c

Total probability = 18c = 1

c = 1/18 is the required value of c here.

b) The marginal PDF for X is first obtained here as:
P(X = 1) = P(1, 1) + P(1,2) = c + 4c = 5c = 5/18
P(X = 2) = P(2, 1) + P(2, 2) = 10c = 10/18
P(X = 3) = 3/18

Therefore the mean of X here is obtained as:

= E(X) = 1*(5/18) + 2*(10/18) + 3*(3/18) = 17/9

Therefore 17/9 = 1.2143 is the required mean value of X here.

c) Now similar to the above method, we obtain the marginal PDF for Y here as:
P(Y = 1) = c + 2c + 3c = 6c = 1/3
P(Y = 2) = 12c = 2/3

E(Y) = 1*(1/3) + 2*(2/3) = 5/3

Therefore 5/3 = 1.6667 is the required mean of Y here.

d) The second moment of X here is obtained as:
E(X²) = 1²*(5/18) + 2²*(10/18) + 3²*(3/18) = 4

Therefore, the variance now is computed here as:

= E(X²) - [E(X)]² = 4 - (17/9)² = 0.4321

therefore 0.4321 is the required variance of X here.