P1 (15 points). Show that E[X] is finite (|E[X]| < ) if and only if the following holds: For all > 0, there exits 8 > 0 such that E[|X|I] < for all A with P(A) < . Here, I is the indicator function. Hint: First, we note that |E[X]| < if and only if E|X| < . One direction of this is easy. If E|X| < , then |E[X]| E|X| < . = . Next, suppose |E[X]| < . Since X = X+ X_, E[X] = E[X+]E[X_]. In the definition of general integral, E[X] is undefined when E[X+] = and E[X] = . E[X] is defined when (i) E[X+] < and E[X_] < , (ii) E[X+] = and E[X] < , or (iii) E[X+] < and E[X_] In (i), clearly, E[X]| < ; in (ii), E[X] = ; in (iii), E[X] Therefore, under the assumption of |E[X]| < , we are in case (i). Since |X| = X+ + X_, we have E|X| = E[X+]+E[X_] < . = -8. We now return to the original question. For the 'only if' part, suppose |E[X]| < . But, it is more convenient to work with E|X| < . Then, note that E|X| = E[|X|I{xy}]+E[|X|I{\x\>y}]. Using the Monotone Conver- gence Theorem on E[|X|I|{[|X| y}] = 0. You need to show: For any > 0, you can find 8 > 0 such that E[|X|IA] < for any A with P(A) < 8. To do that, let By = {|X|> y}. Any set A can be split into two disjoint parts An By and An Br. Now, E[|X|IA] = E[|X|IA_By]+E[|X|IABg]. Show that you can choose y and to make each of the above two terms less than /2. For the 'if' part, again consider E|X| = E[|X|IB]+E[|X|IB]. Use the condition in the 'if' part of the statement and argue why the right hand side is finite. You will need to argue P(Bx) 0 as x and use the result.