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2. The law of iterated expectations is a very useful property of conditional expectations operators. In the Wooldridge book this is property CE.4 in Appendix B, which says that if X, Y, and Z are random variables then E [Y|X] = E [E [Y|X, Z] [X]. Essentially, this means that we can take a conditional expectation in two steps. First, over X and Z and then over X alone. In this question, you'll work through an example to see that this is true using random variables X, W, and Y. Start with the following joint distribution: .2 if x = 1, w = -2, y =5 .1 if x =0, w = -2, y = 10 .1 if x =1, w = -2, y = 10 f (x, w, y) = .1 if x = 0, w = 2. y =5 .1 if x = 1, w =2, y =5 4 if x = 0, w = 2, y = 10 0 otherwise (a) (3 points) This is what the conditional distribution of f (W|X, Y) would look like, except that I have left out some values. What are the correct values for A, B, C, D and E? f ( W | = = 0 , y = 5 ) = 1 So if w = -2 1 if w = 2 f (W|x = 1,y =5) = if w = -2 if w = 2 f ( WIX, Y ) = f (W|x = 0, y = 10) = JA if w = -2 B if w = 2 f ( W | x = 1, y = E) = C if w = -2 D if w = 2 (b) (4 points) Use the conditional distribution from part (a) to find the conditional expectation E [W|X, Y].(c) (4 points) Then, using E [W|X, Y] that you found in part (b) find E [E [W|X, Y] [X]. Notice that now the outer conditioning set is just X. Show your work. (d) (5 points) Now use either the joint distribution in the question set up or the con- ditional distribution in part (a) to find the conditional distribution of f (W|X) (e) (2 points) Using f (W|X) from part (d), find E [W|X]. Show your work. (f) (1 point) How does your answer in part (e) compare to your answer in part (c)