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UCLA STATS 100A - 100aHW8Soln

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Statistics 100AHomework 8 SolutionsRyan RosarioPart 1: Chapter 71. A player throws a fair die and simultaneously flips a fair coin. If the coin lands heads, thenshe wins twice, and if tails, the one-half of the value that appears on the die. Explain herexpected winnings.First, this problem is poorly worded. If the coin lands heads, then she wins twice the valueon the die, otherwise she wins12the value that appears on the die.Let X represent the number of pips that show on the die. Let Y be the result of the toss ofthe fair coin, such that Y = 1 if the coin lands heads, and Y = 0 otherwise.Then,E(XY ) =2X, Y = 112, Y = 0Each value of Y occurs with probability12. So,E(XY ) =12(2X) +1212X=12216+ 2 ×16+ . . . + 6 ×66+121216+ 2 ×16+ . . . + 6 ×66= 4.3755. The county hospital is located at the center of a square whose sides are 3 miles wide. Ifan accident occurs within this square, then the hospital sends out an ambulance. The roadnetwork is rectangular so the travel distance from the hospital, whose coordinates are (0,0),to the point (x, y) is |x|+ |y|. If an accident occurs at a point that is uniformly distributed inthe square, find the expected travel distance of the ambulance.First, let Z = |X| + |Y | for simplicity. We want to find E(Z). Note that −32< X <32and−32< Y <32since the sides of the square have length 3. Thus, |X| <32and |Y | <32.E(Z) = E (|X| + |Y |)= E|X| + E|Y |=Z32−32x32−−32dx +Z32−32y32−−32dy= 2Z320x3dx + 2Z320y3dy=3216. A fair die is rolled 10 times. Calculate the expected sum of the 10 rolls.This is a sum X of 10 independent random variables Xi.E(X) = E 10Xi=1Xi!=10Xi=1E(Xi)=10Xi=116× 1 +16× 2 + . . . +16× 6=10Xi=1216= 10216= 357. Suppose that A and B each randomly, and independently, choose 3 of 10 objects. Find theexpected number of objects(a) chosen by both A and B.Let X be the number of objects that are selected by both A and B. To further simplifythe problem we use indicator variables Xi. Let Xi= 1 if object i is selected by both Aand B, and Xi= 0 otherwise, where 1 ≤ i ≤ 10. Then,E(X) = E 10Xi=1Xi!=10Xi=1E(Xi)Now we must find E(Xi). We know that Xionly takes on one of two values, Xi= 1 orXi= 0. So, for the case of a sum of independent random indicator variables, E(Xi) =P (Xi= 1).Each person can choose 3 of the 10 items. There are 3 ways to choose the item ofinterest, since a person can draw 3 objects. Since person A and B draw independently,P (Xi= 1) =3102Then,E(X) =10Xi=1E(Xi) =10Xi=13102= 103102= 0.92(b) not chosen by either A or B.The principle is similar to part (a). Let Xi= 1 if object i is not chosen by A and is notchosen by B. P (Xi= 1) =7102, because the probability that an arbitrary person doesnot choose object i is710and person A and person B draw independently. Then,E(X) =10Xi=1E(Xi) =10Xi=17102= 107102= 4.9(c) Chosen by exactly one of A or BIn this case, either person A draws object i and person B does not, or person B drawsobject i and person A does not. Again, let Xi= 1 if exactly one of A or B draws objecti, Xi= 0 otherwise. The person that eventually draws object i had probability310ofdrawing the object and the person that does not draw object i had probability710ofnot drawing object i. But there are two ways to arrange this situation. A can draw theobject, and B does not, or B draws the object and A does not. Thus,E(Xi) = P (Xi= 1) = 2 ×310×710andE(X) = 102 ×310×710= 4.29. A total of n balls, numbered 1 through n, are put into n urns, also numbered 1 through n insuch a way that ball i is equally likely to go into any of the urns 1, 2, . . . , i. Find(a) the expected number of urns that are empty.Let X be the number of empty urns. Define an indicator variable Xi= 1 if urn i isempty, Xi= 0 otherwise. We must find E(Xi) = P (Xi= 1).The ith urn remains empty as the first i −1 balls are deposited into the other urns. Onthe ith drop, urn i must remain empty. Since a ball can land in any of the i urns withequal probability, the probability that the ith ball will not land in urn i is 1 −1i. Toremain empty, the urn must not receive a ball on the i + 1st drop etc. so the probabilitythat the i + 1st ball will not land in urn i, is 1 −1i+1. So,E(Xi) =1 −1i1 −1i + 11 −1i + 2. . .1 −1n=i − 1iii + 1. . .n − 1n=i − 1n3So,E(X) =nXi=1E(Xi)=nXi=1i − 1n=1nnXi=1i − 1=1nn−1Xj=0j=1n·n(n − 1)2=n−12(b) the probability that none of the urns is empty.For all of the urns to have at least one ball in them, the nth ball must be dropped intothe nth urn, which has probability1n. The n − 1st ball must be placed in the n − 1sturn which has probability1n−1and so on. So, the probability that none of the urns willbe empty is1n×1n − 1×1n − 2× . . . ×12×11=1n!17. A deck of n cards, numbered 1 through n, is thoroughly shuffled so that all possible n! orderingscan be assumed to be equally likely. Suppose you are to make n guesses sequentially, wherethe ith one is a guess of the card in position i. Let N denote the number of correct guesses.(a) If you are not given any information about your earlier guesses show that, for anystrategy, E(N ) = 1.Proof. Let Nibe an indicator variable such that Ni= 1 if the ith guess is correct. Theprobability that the guess is correct is simply1n. Then,E(N) =nXi=1E(Ni) =nXi=1P (Ni= 1) =nXi=11n= n ×1n= 1(b) Suppose that after each guess you are shown the card that was in the position in question.What do you think is the best strategy? Show that under this strategyE(N) =1n+1n − 1+ . . . + 1 ≈Zn11xdx = log nOnce the card is shown, and assuming the person is not a complete idiot, the player hasn −1 cards remaining to guess from. Therefore, the best strategy (or is it just commonsense?) is to not guess any cards that have already been shown.4Proof. After a card is shown, there is one less card to guess from given our elegantstrategy, so the denominator in the probability dereases by one at each guess. So,E(N) =nXi=1E(Ni) =1n+1n − 1+1n − 2+ . . . +11which is a discrete sum. For large n, the sum can be approximated as an integral.Zn11xdx = log n(c) Suppose that you are told after each guess whether you are right or wrong. In this caseit can be shown that the strategy that maximizes E(N ) is one which keeps on guessingthe same card until you are told you are correct and then changes to a new card. Forthis strategy show thatE(N) = 1


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UCLA STATS 100A - 100aHW8Soln

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