Question

) Let X be the minimum of the two numbers obtained by rolling a die twice and Y the maximum.

a) Compute E(X).

b) Compute Var(X).

c) Compute E(Y).

Answer 1

The values of X and Y are obtained for different combinations of the 2 dice throw values are obtained as:

	1	2	3	4	5	6
1	X = 1, Y = 1	X = 1, Y = 2	X = 1, Y = 3	X = 1, Y = 4	X = 1, Y = 5	X = 1, Y = 6
2	X = 1, Y = 2	X = 2, Y = 2	X = 2, Y = 3	X = 2, Y = 4	X = 2, Y = 5	X = 2, Y = 6
3	X = 1, Y = 3	X = 2, Y = 3	X = 3, Y = 3	X = 3, Y = 4	X = 3, Y = 5	X = 3, Y = 6
4	X = 1, Y = 4	X = 2, Y = 4	X = 3, Y = 4	X = 4, Y = 4	X = 4, Y = 5	X = 4, Y = 6
5	X = 1, Y = 5	X = 2, Y = 5	X = 3, Y = 5	X = 4, Y = 5	X = 5, Y = 5	X = 5, Y = 6
6	X = 1, Y = 6	X = 2, Y = 6	X = 3, Y = 6	X = 4, Y = 6	X = 5, Y = 6	X = 6, Y = 6

From above distribution, we have the PDF for X as:
P(X = 1) = 11/36
P(X = 2) = 9/36
P(X = 3) = 7/36
P(X = 4) = 5/36
P(X = 5) = 3/36
P(X = 6) = 1/36

Therefore the expected value of X here is computed as:

Therefore E(X) = 2.5278

The second moment first is computed here as:

Now the variance of X is computed here as:

Var(X) = E(X²) - [E(X)]² = 8.3611 - 2.5278² = 1.9715

Therefore 1.9715 is the required variance of X.

c) The PDF for Y here is obtained as:
P(Y = 1) = 1/36
P(Y = 2) = 3/36
P(Y = 3) = 5/36
P(Y = 4) = 7/36
P(Y = 5) = 9/36
P(Y = 6) = 11/36

The expected value of Y here is obtained as:

Therefore 4.4722 is the expected value of Y here.