Solved: People are arriving at a party one at a time. While waiting for more people to

Chapter 4, Problem 54

(choose chapter or problem)

People are arriving at a party one at a time. While waiting for more people to arrive they entertain themselves by comparing their birthdays. Let X be the number of people needed to obtain a birthday match, i.e., before person X arrives there are no two people with the same birthday, but when person X arrives there is a match. Assume for this problem that there are 365 days in a year, all equally likely. By the "result of the birthday problem from Chapter 1, for 23 people there is a 50.7% chance of a birthday match (and for 22 people there is a less than 50% chance). But this has to do with the median of X (defined below); we also want to know the mean of X, and in this problem we will find it, and see how it compares with 23. (a) A median of an r.v. Y is a value m for which P(Y m) 1/2 and P(Y m) 1/2 (this is also called a median of the distribution of Y ; note that the notion is completely determined by the CDF of Y ). Every distribution has a median, but for some distributions it is not unique. Show that 23 is the unique median of X. (b) Show that X = I1 +I2 ++I366, where Ij is the indicator r.v. for the event X j. Then find E(X) in terms of pj s defined by p1 = p2 = 1 and for 3 j 366, pj = 1 1 365 1 2 365 ... 1 j 2 365 . (c) Compute E(X) numerically. In R, the pithy command cumprod(1-(0:364)/365) produces the vector (p2,...,p366). (d) Find the variance of X, both in terms of the pj s and numerically. Hint: What is I2 i , and what is IiIj for j? or i