Question

Let X,Y and Z be the face value of three independent dice, i.e. X, Y, Z are i.i.d. Define the following variables:

U = X + Y, V = X + Z, W = X + Y + Z.

Use appropriate definitions and properties to calculate the followings:

(a) E(U),E(V/2) and E(W/3).
(b) Var(U),SD(2V) and SD(3W).
(c) E(X/(YZ)).
(d) Cov(U,V) and Corr(U,W).

Answer 1

The expected values are E(U) =7, E(V/2) = 3.5 and E(W/3) = 3.5

How can you solve for the expected values?

(a) Expected Values

E(U) = E(X + Y) = E(X) + E(Y) = 3.5 + 3.5 = 7 (since X and Y are i.i.d. with mean 3.5)

E(V/2) = E(X/2 + Z/2) = 1/2 * E(X) + 1/2 * E(Z) = 1/2 * 3.5 + 1/2 * 3.5 = 3.5

E(W/3) = E((X + Y + Z)/3) = 1/3 * E(X) + 1/3 * E(Y) + 1/3 * E(Z) = 1/3 * 3.5 + 1/3 * 3.5 + 1/3 * 3.5 = 3.5

(b) Variances and Standard Deviations

Var(U) = Var(X + Y) = Var(X) + Var(Y) = 35/12 + 35/12 = 35/6 (since X and Y are independent)

SD(2V) = SD(2(X + Z)) = 2 * SD(X + Z) = 2 *
`\sqrt`(Var(X) + Var(Z)) = 2 *
`\sqrt`(35/12 + 35/12) = 5/√3

SD(3W) = SD(3(X + Y + Z)) = 3 * SD(X + Y + Z) = 3 *
`\sqrt`(Var(X) + Var(Y) + Var(Z)) = 3 *
`\sqrt`(35/12 + 35/12 + 35/12) = 5√3

(c) E(X/(YZ))

E(X/(YZ)) = E(X) * E(1/(YZ)) ≈ 3.5 * 0.054 ≈ 0.19 (assuming independence of X, Y, and Z, and using the fact that E(1/(YZ)) ≈ 1/(E(Y) * E(Z)) ≈ 0.054)

(d) Covariance and Correlation

Cov(U, V) = Cov(X + Y, X + Z) = Var(X) + Cov(X, Z) + Cov(Y, X) + Cov(Y, Z) = Var(X) = 35/12 (since X, Y, and Z are independent)

Corr(U, W) = Cov(U, W) / (SD(U) * SD(W)) = Cov(X + Y, X + Y + Z) / (SD(X + Y) * SD(X + Y + Z)) = Var(X + Y) / (SD(X + Y) * SD(X + Y) *
`\sqrt`(3)) = 35/6 / (5/√3 * 5/√3 *
`\sqrt`(3)) = 7/5