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Two envelopes, each containing a check, are placed in front of you. You are to choose one of the envelopes, open it, and see the amount of the check. At this point, either you can accept that amount or you can exchange it for the check in the unopened envelope. What should you do? Is it possible to devise a strategy that does better than just accepting the first envelope? Let A and B, A<B, denote the (unknown) amounts of the checks, and note that the strategy that randomly selects an envelope and always accepts its check has an expected return of (A+B) / 2. Consider the following strategy: Let F()F(\cdot) be any strictly increasing (that is, continuous) distribution function. Choose an envelope randomly and open it. If the discovered check has the value x, then accept it with probability F(x) and exchange it with probability 1-F(x).

a. Show that if you employ the latter strategy, then your expected return is greater than (A+B) / 2.

b. Hint: Condition on whether the first envelope has the value A or B.

c. Now consider the strategy that fixes a value x and then accepts the first check if its value is greater than x and exchanges it otherwise.

a. Show that for any x, the expected return under the x -strategy is always at least (A+B) / 2 and that it is strictly larger than (A+B) / 2 if x lies between A and B.

b. Let X be a continuous random variable on the whole line, and consider the following strategy: Generate the value of X, and if X=x, then employ the x-strategy of part (b). Show that the expected return under this strategy is greater than (A+B) / 2.