We would be looking at the first question, all parts here as:
a) The probabilities for each combination of values of X, Y and Z here are computed as:
x | y | z | p(x, y, z) |
1 | 2 | 0 | c |
1 | 2 | 1 | 4c |
1 | 3 | 0 | c |
1 | 3 | 1 | 5c |
2 | 2 | 0 | 4c |
2 | 2 | 1 | 8c |
2 | 3 | 0 | 4c |
2 | 3 | 1 | 9c |
Sum of all probabilities should be 1. Therefore, we have
here:
c + 4c + c + 5c + 4c + 8c + 4c + 9c = 1
36c = 1
Therefore 1/36 is the required value of c here.
b) The PDF here is obtained from the above table as:
p(x = 1, y = 2) = p(x = 1, y = 2, z = 0) + p(x = 1, y = 2, z = 1) = 5c = 5/36
Similarly all other probabilities are obtained and PDF here is given as:
p(1,2) = 5/36
p(1,3) = 6/36 = 1/6
p(2,2) = 12/36 = 1/3
p(2,3) = 13/36
This is the required joint PDF for XY
c) The expected value of xy2 here is computed as:
E(xy2) = 4*(5/36) + 9*(1/6) + 8*(1/3) + 18*(13/36) = (5/9) + (3/2) + (8/3) + (13/2) = 8 + (29/9) = (101)/9
Therefore (101)/9 is the required expected value here.
Consider 3 jointly discrete andom variables X, Y, Z whose joint PMF is given by p(x,...
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