Recitation 20 Problem 2. I'm God. Seriously. So, I know everything that everybody thinks In particu lar, I know who each one of the 250,000,000 Americans want to vote for in the upcoming elections. I know that a fraction p=0.52 of them want to vote for the current president. You are mortal. An insignificant dot in space-time. But a quite significant dot among dots. You work close to the president and, within a week, you must answer his agonizing question: " Am I win\ing?"Or, in math jargon(but with the same agony :"Isp>1p"8 Your first idea is to ask me (I'm God). But you haven t talked to me for a long time, so you know I won' t tell you. Your second idea is to call every American, ask them, then divide the yes's by 250 million. But you soon realize there is not enough time(there is a reason or representative democracy). Your third idea. You have no third idea! In your panic as the week is almost over, you start picking Americans at random, call them, and ask Amazingly, thats the correct approach. But you should be careful what you are going to say to the president! Let's see (a) In you first phone call, you pick 1 American uniformly at random, call, and ask whether he/she will vote for the president. What is the probability that the answer is going to be"yes".(i) from my perspective? (ii) from your perspective? How would you model this in terms of coin flips Solution. From my perspective, it's 0.52. From your perspective, it is also 0.52. The only problem is that you dont know that, so you just call it p. Clearly, from your perspective, the first phone call is the same as flipping a coin with an unknown bias, which you call p(and I know is 0.52 (b) In your second phone call, you again pick 1 American uniformly at random, call, and ask whether he/ she will vote for the president. But wait! When selecting the second voter, shouldn 't you exclude the guy that you asked in the first phone call? No What's bad if you exclude him/her? Solution. If you do this, you alter the coin that you are flipping. The bias will decrease or increase, depending on whether the first guy said"yes"or"no",respec tively. The analysis gets messy, so you dont want to do this (c) So, in each one of n phone calls, you pick 1 American uniformly at random and ask. Your plan is to eventually divide the number M of positive answers by n to get P=M. An MIT friend tells you that, as the numerical outcome of a random experiment, this P is a random variable; and that, according to his calculations, Pr(P-p≤0.03)≥0.95 When you are done calling people, you divide to get P, and it's 0. 53. You call the president up and... what do you say? Mr President, p=0.53 Mr President, with probability at least 95%, p is within 0.03 of 0.53!Recitation 20 2 Problem 2. I’m God. Seriously. So, I know everything that everybody thinks. In particu￾lar, I know who each one of the 250,000,000 Americans want to vote for in the upcoming elections. I know that a fraction p = 0.52 of them want to vote for the current president. You are mortal. An insignificant dot in space­time. But a quite significant dot among dots. You work close to the president and, within a week, you must answer his agonizing question: “Am I winning?” Or, in math jargon (but with the same agony): “Is p > 1 2 ?” Your first idea is to ask me (I’m God). But you haven’t talked to me for a long time, so you know I won’t tell you. Your second idea is to call every American, ask them, then divide the yes’s by 250 million. But you soon realize there is not enough time (there is a reason for representative democracy). Your third idea. . . You have no third idea! In your panic as the week is almost over, you start picking Americans at random, call them, and ask! Amazingly, that’s the correct approach. But you should be careful what you are going to say to the president! Let’s see. (a) In you first phone call, you pick 1 American uniformly at random, call, and ask whether he/she will vote for the president. What is the probability that the answer is going to be “yes”. . . (i) from my perspective? (ii) from your perspective? How would you model this in terms of coin flips? Solution. From my perspective, it’s 0.52. From your perspective, it is also 0.52. The only problem is that you don’t know that, so you just call it p. Clearly, from your perspective, the first phone call is the same as flipping a coin with an unknown bias, which you call p (and I know is 0.52). (b) In your second phone call, you again pick 1 American uniformly at random, call, and ask whether he/she will vote for the president. But wait! When selecting the second voter, shouldn’t you exclude the guy that you asked in the first phone call? No. What’s bad if you exclude him/her? Solution. If you do this, you alter the coin that you are flipping. The bias will decrease or increase, depending on whether the first guy said “yes” or “no”, respec￾tively. The analysis gets messy, so you don’t want to do this. (c) So, in each one of n phone calls, you pick 1 American uniformly at random and ask. Your plan is to eventually divide the number M of positive answers by n to get P = n M . An MIT friend tells you that, as the numerical outcome of a random experiment, this P is a random variable; and that, according to his calculations, Pr (| | P − p ≤ 0.03) 0 ≥ .95. (1) When you are done calling people, you divide to get P, and it’s 0.53. You call the president up and. . . what do you say? • Mr. President, p = 0.53! • Mr. President, with probability at least 95%, p is within 0.03 of 0.53!