multivariate prob. question, help?

Question

p(x,y)=(4 x)(3 y)(2 3-x-y)/(9 3)
where (4 x) means number of comb. of x chosen from 4 items.

domain 0<=x<=3, 0<=y<=3, 1<=x+y<=3

find the cov(x, y)
please show me your solution with clear explanations
thank you. really want to learn

Answer 1

First, we will need the probabilities of each possible outcome (there are only a small number of them, so computing them individually is feasible). These probabilities are:

P(1, 0) = (4!/(1!3!) 3!/(0!3!) 2!/(2!0!))/(9!/(3!6!)) = 4*1*1/84 = 1/21
P(0, 1) = (4!/(0!4!) 3!/(1!2!) 2!/(2!0!))/(9!/(3!6!)) = 1*3*1/84 = 1/28
P(2, 0) = (4!/(2!2!) 3!(0!3!) 2!/(1!1!))/(9!/(3!6!)) = 6*1*2/84 = 1/7
P(1, 1) = (4!/(1!3!) 3!(1!2!) 2!/(1!1!))/(9!/(3!6!)) = 4*3*2/84 = 2/7
P(0, 2) = (4!/(0!4!) 3!(2!1!) 2!/(1!1!))/(9!/(3!6!)) = 1*3*2/84 = 1/14
P(3, 0) = (4!/(3!1!) 3!(0!3!) 2!/(0!2!))/(9!/(3!6!)) = 4*1*1/84 = 1/21
P(2, 1) = (4!/(2!2!) 3!(1!2!) 2!/(0!2!))/(9!/(3!6!)) = 6*3*1/84 = 3/14
P(1, 2) = (4!/(1!3!) 3!(2!1!) 2!/(0!2!))/(9!/(3!6!)) = 4*3*1/84 = 1/7
P(0, 3) = (4!/(0!4!) 3!(3!0!) 2!/(0!2!))/(9!/(3!6!)) = 1*1*1/84 = 1/84

Checking our result, the sum of all probabilities is:

1/21 + 1/28 + 1/7 + 2/7 + 1/14 + 1/21 + 3/14 + 1/7 + 1/84 = 1

So indeed, this is a valid probability distribution and our calculations are accurate.

Now, we calculate some derived probabilities. Specifically, we will need the probabilities for each of the possible values of X, each possible value of Y, and each possible value of XY:

P(X=0) = P(0, 1) + P(0, 2) + P(0, 3) = 1/28 + 1/14 + 1/84 = 5/42
P(X=1) = P(1, 0) + P(1, 1) + P(1, 2) = 1/21 + 2/7 + 1/7 = 10/21
P(X=2) = P(2, 0) + P(2, 1) = 1/7 + 3/14 = 5/14
P(X=3) = P(3, 0) = 1/21

P(Y=0) = P(1, 0) + P(2, 0) + P(3, 0) = 1/21 + 1/7 + 1/21 = 5/21
P(Y=1) = P(0, 1) + P(1, 1) + P(2, 1) = 1/28 + 2/7 + 3/14 = 15/28
P(Y=2) = P(0, 2) + P(1, 2) = 1/14 + 1/7 = 3/14
P(Y=3) = P(0, 3) = 1/84

P(XY=0) = P(X=0) + P(Y=0) = 5/42 + 5/21 = 5/14 (since X and Y cannot both be zero simultaneously)
P(XY=1) = P(1, 1) = 2/7
P(XY=2) = P(1, 2) + P(2, 1) = 1/7 + 3/14 = 5/14

Now, we compute the expected values of X, Y, and XY:

E(X) = 0*P(X=0) + 1*P(X=1) + 2*P(X=2) + 3*P(X=3) = 0 + 10/21 + 10/14 + 3/21 = 4/3

E(Y) = 0*P(Y=0) + 1*P(Y=1) + 2*P(Y=2) + 3*P(Y=3) = 0 + 15/28 + 6/14 + 3/84 = 1

E(XY) = 0*P(XY=0) + 1*P(XY=1) +2*P(XY=2) = 0 + 2/7 + 10/14 = 1

Finally, we calculate the covariance:

Cov(X, Y) = E(XY) - E(X)E(Y) = 1 - 4/3 = -1/3

And we are done.