Question 4 (28 points): The chart below shows the joint probability func- tion o

Question

Question 4 (28 points): The chart below shows the joint probability func- tion of two discrete random variables X and Y, each having possible outcomes 1, 2 or 3 X 2 1/8 1/8 1/8 X-3 1/8 1/8 0 (a) Find the (discrete) probability functions of X and of Y by listing all their values (or by a table) (b) Find E(X) and E(Y) (c) Find Var(X) and Var(Y) (d) Find all possible outcornes of the random variable Z = XY (e) Find the probability function of Z and E(Z) () Find Cov(x,y) (g) Find the correlation coefficient (X,Y) (h) Can you give an intuitive explanation of the sign of the correlation by only arguing with the values in the above chart of the joint probability function? In other words, could you have predicted the sign of (X.Y) before starting to work any of the parts of this problem?

Explanation / Answer

a) The marginal PDF for X here is obtained as:

P(X = 1) = 3/8 = 0.375
P(X = 2) = 3/8 = 0.375
P(X = 3) = 2/8 = 0.250

The marginal PDF for Y here is obtained as:

P(Y = 1) = 3/8 = 0.375
P(Y = 2) = 3/8 = 0.375
P(Y = 3) = 2/8 = 0.250

b) The expected values of X and Y here are obtained as:

E(X) = E(Y) = 1*0.375 + 2*0.375 + 3*0.25 = 1.875

Therefore E(X) = E(Y) = 1.875

c) The second moments of X and Y here are obtained as:

E(X2) = E(Y2) = 12*0.375 + 22*0.375 + 32*0.25 = 4.125

Therefore, we get here:

Var(X) = Var(Y) = E(X2) - E2(X) = 4.125 - 1.8752 = 0.609375

Therefore Var(X) = Var(Y) = 0.609375

d) All the outcomes for Z here are obtained as:

1, 2, 3, 4, 6, 9

e) The PDF for Z here is obtained as:

P(Z = 1) = 1/8 = 0.125
P(Z = 2) = 2/8 = 0.250
P(Z = 3) = 2/8 = 0.250
P(Z = 4) = 1/8 = 0.125
P(Z = 6) = 2/8 = 0.250

The expected value of Z here is computed as:

E(Z) = E(XY) = 0.125*1 + 0.25*2 + 0.25*3 + 0.125*4 + 0.25*6 = 3.375

Therefore E(Z) = 3.375

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