M1 - HOMEWORK 1 SOLUTION Northwestern University Fall 2009...

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HOMEWORK 1 SOLUTION Northwestern University Marciano Siniscalchi Fall 2009 Econ 331-0 1. The Peter Lynch Problem This was pretty straightforward. For the first part, we first compute the probability of k successes in n = 13 trials, for k = 11 , 12 , 13, where the probability of a success is p = 1 2 ; then we sum these three numbers to get the probability of at least 11 successes in 13 trials. This is approximately 1 . 123% (i.e. 0 . 01123), a small number. For the second part, the trick is to note that the probability that at least one fund manager outperforms Peter Lynch equals 1 minus the probability that no one does. The latter probability is easy to calculate: the probability that one particular manager does not beat Peter Lynch is 1 - 0 . 01123, so the number we need is (1 - 0 . 01123) 500 0 . 003528 (i.e. 0 . 3538%). Therefore, the probability that at least one manager outperforms Peter Lynch is approximately 1 - 0 . 003528 = 0 . 9965 = 99 . 65%. Bottom line: maybe Peter Lynch was a really clever guy, but then again, maybe he was just lucky—even if all fund managers chose stocks at random over theperiod in question, someone just “had” to outperform Peter Lynch! 2. Coin Flips! A) State space S = { HHHHH, HHHHT, ..., TTTTT } will do. Notice that there are 2 5 = 32 states. Let probability P be defined by P ( { HHHHH } ) = P ( { HHHHT } ) = ... = P ( { TTTTT } ) = 1 32 , and we are done. B) X = 5 for s = HHHHH 3 for s ∈ { HHHHT, HHHTH, HHTHH, HTHHH, THHHH } . . . - 5 for s = TTTTT C) Probability distribution is a tuple ( - 5 , 1 32 ; - 3 , 5 32 ; - 1 , 10 32 ; 1 , 10 32 ; 3 , 5 32 ; 5 , 1 32 ) where the first element is the outcome and the second probability. How do you get probabilities? Well you can write out the whole table and count the number of states for which X = x holds. For this problem this is still feasible. Subtle version is using the binomial distribution.

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