Question

Suppose the joint probability distribution of X and Y is given by the following table.

Y=>3 6 9 X

1 0.2 0.2 0

2 0.2 0 0.2

3 0 0.1 0.1

The table entries represent the probabilities. Hence the outcome [X=1,Y=6] has probability

0.2.

a) Compute E(X), E(X2), E(Y), and E(XY). (For all answers show your work.) b) Compute E[Y | X = 1], E[Y | X = 2], and E[Y | X = 3].

c) In this case, E[Y | X] is linear, given by E[Y | X] = β0 + β1X where β0 and β1 are constants. Make a plot with E[Y | X] on the vertical axis and X on the horizontal. Can you use your plot to deduce the values of β0 and β1?

d) When E[Y | X] is linear, a formula for β1 is β1 = Cov(X,Y)/Var(X).

And given β1, a formula for β0 is β0 = E(Y) – β1E(X).

Does applying these formulas yield the same answers that you deduced in part c?

e) Let u = Y – (β0 + β1X). It so happens that u can take four possible values: 1.5, -1.5, 3, and -3. Find the joint distribution of u and X. The first row is done for you.

u=>-3 -1.5 1.5 3 X

1 0 0.2 0.2 0

2 _____ _____ _____ _____

3 _____ _____ _____ _____

Does E(u) = 0? Does Cov(X,u) = 0?

Answer 1