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A 8-sided die is rolled twice. Let X be the sum of the two rolls, Y be the first minus the second, and Z be the 2 times the first, minus the second. (a) Compute Cov(X, Y): Cov(X, Y) = (b) Compute Cov(X, Z): Cov(X, Z) =

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Final answer:

Cov(X, Y) = -27, Cov(X, Z) = -31.5.

Step-by-step explanation:

Cov(X, Y):

Let's start by calculating the covariance between X and Y.

Cov(X, Y) = E[(X - E[X])(Y - E[Y])]

Since X and Y are defined as X = first roll minus second roll, and Y = sum of the two rolls:

X = X1 - X2

Y = X1 + X2

Substituting these values into the covariance formula:

Cov(X, Y) = E[(X1 - X2 - E[X1 - X2])(X1 + X2 - E[X1 + X2])]

Cov(X, Y) = E[(X1 - X2 - 0)(X1 + X2 - E[2X1])]

Cov(X, Y) = E[(X1 - X2)(X1 + X2 - 2E[X1])]

Cov(X, Y) = E[(X1 - X2)(X1 + X2 - 2(4.5))]

Cov(X, Y) = E[(X1 - X2)(X1 + X2 - 9)]

Since the two rolls are independent, we can use the linearity of the expectation:

Cov(X, Y) = E[X1^2 - X1X2 + X1X2 - X2^2 - 9X1 + 9X2]

Cov(X, Y) = E[X1^2 - X2^2 - 9X1 + 9X2]

Using the fact that X1 and X2 are each rolled on an 8-sided die, their expected values are both equal to (1 + 2 + ... + 8) / 8 = 4.5:

Cov(X, Y) = E[X1^2 - X2^2 - 9X1 + 9X2]

Cov(X, Y) = E[X1^2] - E[X2^2] - 9E[X1] + 9E[X2]

Since the expected value of a roll on an 8-sided die is 4.5, we have:

Cov(X, Y) = E[X1^2] - E[X2^2] - 9(4.5) + 9(4.5)

Since X1 and X2 are independent identically distributed random variables, we have:

Cov(X, Y) = E[X^2] - E[X^2] - 9(4.5) + 9(4.5)

Cov(X, Y) = Var(X) - 0 - 9(4.5) + 9(4.5)

Since X is the sum of two rolls on an 8-sided die, its variance can be calculated as follows:

Var(X) = Var(X1 + X2) = Var(X1) + Var(X2) = (1/3) + (1/3) = 2/3

Substituting this value back into the covariance formula:

Cov(X, Y) = Var(X) - 0 - 9(4.5) + 9(4.5)

Cov(X, Y) = 2/3 - 0 - 9(4.5) + 9(4.5)

Cov(X, Y) = -27

Cov(X, Z):

Now, let's calculate the covariance between X and Z.

Cov(X, Z) = E[(X - E[X])(Z - E[Z])]

Since X and Z are defined as X = first roll minus second roll, and Z = 2 times the first roll, minus the second roll:

X = X1 - X2

Z = 2X1 - X2

Substituting these values into the covariance formula:

Cov(X, Z) = E[(X1 - X2 - E[X1 - X2])(2X1 - X2 - E[2X1 - X2])]

Cov(X, Z) = E[(X1 - X2 - 0)(2X1 - X2 - E[2X1])]

Cov(X, Z) = E[(X1 - X2)(2X1 - X2 - 2E[X1])]

Cov(X, Z) = E[(X1 - X2)(2X1 - X2 - 2(4.5))]

Cov(X, Z) = E[(X1 - X2)(2X1 - X2 - 9)]

Using the linearity of the expectation:

Cov(X, Z) = E[2X1^2 - X1X2 - 2X2X1 + X2^2 - 9X1 + 2X2]

Cov(X, Z) = E[2X1^2 - 2X2X1 - 9X1 + 2X2]

Using the same logic as before, the expected value of a roll on an 8-sided die is 4.5, so we have:

Cov(X, Z) = E[2X1^2 - 2X2X1 - 9X1 + 2X2]

Cov(X, Z) = E[2X^2 - 2X2 - 9X1 + 2X2]

Since X1 and X2 are independent identically distributed random variables, we have:

Cov(X, Z) = E[2X^2 - 2X^2 - 9X1 + 2X2]

Cov(X, Z) = E[-9X1 + 2X2]

Again, since X1 and X2 are independent identically distributed random variables, we have:

Cov(X, Z) = -9E[X1] + 2E[X2]

Substituting the expected values of X1 and X2:

Cov(X, Z) = -9(4.5) + 2(4.5)

Cov(X, Z) = -40.5 + 9

Cov(X, Z) = -31.5

User Artur Iwan
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