HMMT 二月 2019 · 冲刺赛 · 第 29 题

HMMT February 2019 — Guts Round — Problem 29

[ 20 ] Yannick picks a number N randomly from the set of positive integers such that the probability − n that n is selected is 2 for each positive integer n . He then puts N identical slips of paper numbered 1 through N into a hat and gives the hat to Annie. Annie does not know the value of N, but she draws one of the slips uniformly at random and discovers that it is the number 2. What is the expected value of N given Annie’s information?

解析

[ 20 ] Yannick picks a number N randomly from the set of positive integers such that the probability − n that n is selected is 2 for each positive integer n . He then puts N identical slips of paper numbered 1 through N into a hat and gives the hat to Annie. Annie does not know the value of N, but she draws one of the slips uniformly at random and discovers that it is the number 2. What is the expected value of N given Annie’s information? Proposed by: Yuan Yao 1 Answer: 2 ln 2 − 1 1 Let S denote the value drawn from the hat. The probability that 2 is picked is if n ≥ 2 and 0 if n n = 1 . Thus, the total probability X that 2 is picked is ∞ − k ∑ 2 P ( S = 2) = . k k =2 − n P ( N = n,S =2) 2 /n By the definition of conditional probability, P ( N = n | S = 2) = = if n ≥ 2 and 0 if P ( S =2) X n = 1 . Thus the conditional expectation of N is ∞ ∞ ∞ − n ∑ ∑ ∑ 2 /n 1 1 − n E [ N | S = 2] = n · P ( N = n | S = 2) = n · = 2 = . X X 2 X n =1 n =2 n =2 ∑ ∞ 1 k It remains to compute X . Note that x = for | x | < 1 . Integrating both sides with respect k =0 1 − x to x yields ∞ k ∑ x = − ln(1 − x ) + C k k =1 1 for some constant C, and plugging in x = 0 shows that C = 0 . Plugging in x = shows that 2 ∑ − k ∞ 2 1 = ln 2 . Note that X is exactly this summation but without the first term. Thus, X = ln 2 − , k =1 k 2 1 1 so = . 2 X 2 ln 2 − 1