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Fall 2011 Strategic Practice 9: Section 3 (Conditional Expectation) - Question 1
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You get to choose between two envelopes, each of which contains a check for some positive amount of money. Unlike in the two envelope paradox from class, it is not given that one envelope contains twice as much money as the other envelope. Instead, assume that the two values were generated independently from some distribution on positive real numbers, with no information given about what that distribution is. After picking an envelope, you can open it and see how much money is inside (call this value x), and then you have the option of switching. As no information has been given about the distribution, it may seem impossible to have better than a 50% chance of picking the better envelope. Intuitively, we may want to switch if x is "small" and not switch if x is "large," but how do we define "small" and "large" in the grand scheme of all possible distributions? [The last sentence was a rhetorical question.] Consider the following strategy for deciding whether to switch. Generate a "threshold" T ~ Expo(1), and switch envelopes if and only if the observed value x is less than the value of T. Show that this strategy succeeds in picking the envelope with more money with probability strictly greater than 1/2. *Hint: Let t be the value of T (generated by a random draw from the Expo(1) distribution). First explain why the strategy works very well if t happens to be in between the two envelope values, and does no harm in any case (i.e., there is no case in which the strategy succeeds with probability strictly less than 1/2).*

Solution: Let a be the smaller value of the two envelopes and b be the larger value (assume a < b since in the case a = b it makes no difference which envelope is chosen!). Let G be the event that the strategy succeeds and A be the event that we pick the envelope with a initially. Then P(G|A) = P(T > a) = 1-(1-e^-a) = e^-a, and P(G|Ac) = P(T =< b) = 1 − e^-b. Thus, the probability that the strategy succeeds is (1/2)e^−a + (1/2)(1 - e^−b) > 1/2.