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Proof Assignment #7 1. You and the bank play the following game: You flip n coins. If X of them come up heads, you receive 2X dollars. (a)[1 point] You have to buy a ticket to play this game. What is the fair price of a ticket? The fair price of a ticket is its expected value. We can think of this as a Bernoulli trials process, where “success” is when the coin lands heads. Let Yi = 2i (i.e. the amount that we expect to receive if the coin lands heads i times). Then, if Y = 2X , n n−i n n i n−i X X X 1 n i 1 n i 1 = 1 . E(Y ) = E(Yi ) = 2 2 2 i 2 i i=0 i=0 i=0 According to the Binomial Theorem, this last sum is equivalent to (1 + 12 )n . Hence, a fair price for the ticket would be ( 23 )n dollars. (b)[1 point] Prove: The probability that you break even (i.e. receive at least your ticket’s worth) is exponentially small. (Hint: At least how many “heads” do you need to break even?) In order to break even, we need to have the number of times i that the coin lands heads satisfy 2i ≥ ( 32 )n (i.e. “the payoff is at least as large as the amount that we pay for the ticket”). Taking logs of both sides of the equation yields: log 2i ≥ log( 32 )n , so i log 2 ≥ n log 3 − n log 2. Hence, i ≥ n(log 3/ log 2 − 1). Using a calculator, we see that this implies that i ≥ .585n. We can use the Central Limit Theorem to determine the probability of getting at least .585n heads. We note that p = 1/2 and µ = np = n/2. Hence, .585n − n/2 .085n ∗ ∗ √ √ P (Sn ≥ .585n) = P Sn ≥ = P Sn ≥ . 1/2 n 1/2 n It’s clear that n → ∞. .058n √ 1/2 n approaches infinity as n → ∞, so P (Sn∗ ≥ .085n √ ) 1/2 n must tend to 0 as Another approach that many people took in solving this problem used Chebyshev’s Inequality in lieu of the Central Limit Theorem. Both are valid approaches. (c)[1 point] Calculate the standard deviation of the variable 2X . Your answer should be a simple formula. To find the standard deviation of 2X , we first must find its variance. Recall that, if Y = 2X , then V (Y ) = E(Y 2 ) − (E(Y ))2 . We have already computed E(Y ) in part (a), so we just need to find E(Y 2 ). Using the same idea as in part (a), we have i n−i X n−i n n X 1 1 1 2 2i i E(Y ) = 2 = 2 . 2 2 2 i=0 i=0 According to the Binomial Theorem, this last sum is equivalent to (2 + 21 )n = ( 52 )n . Hence, q V (Y ) = ( 25 )n − ( 32 )2n = ( 25 )n − ( 94 )n . Therefore, SD= ( 52 )n − ( 49 )n . 1 Proof Assignment #7 (d)[1 point] Show that your answer to (c) is asymptotically equal to an even simpler formula. Make it as simple as possible. q The simplest choice is ( 52 )n . It is easy to see that this is asymptotically equal to the expression that we obtained in part (c). Namely, q s s s n ( 52 )n − ( 94 )n ( 25 )n − ( 94 )n ( 52 )n ( 94 )n 9 n q lim = = lim − 5 n = lim 1 − . 5 5 n n n→∞ n→∞ n→∞ 10 ( ) ( ) ( ) 5 n 2 2 2 ( ) 2 Observing that (9/10)n approaches 0 as n → ∞ allows us to conclude that the limit above is 1, i.e. our solution to part (d) is, in fact, asymptotically equal to our solution to part (c). (e)[1 point] State what the (Weak) Law of Large Numbers would say about the variable 2X . The Weak Law of Large Numbers would say that, as n tends to infinity, the average winnings (i.e. 2X ) get closer and closer to the price of the ticket (i.e. ( 23 )n ). Of course, as we’ll see in part (f), the Weak Law of Large Numbers is not applicable in this situation. (f )[Bonus! +2 points] Prove that the (Weak) Law of Large Numbers does NOT hold for this variable. There were a number of reasons that the Weak Law of Large Numbers should fail. Here is one possible answer: The Weak Law of Large Numbers says that, as n → ∞, the probability of winning back the price that you paid for a ticket should approach 1. However, we showed in part (b) that, as n → ∞, the probability of winning back the price that you paid for the ticket actually approaches 0. Thus, the Weak Law of Large Numbers cannot hold in this scenario! 2