Math 426: Probability MWF 1pm, Gasson 310 Homework 4 Selected Solutions
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1 Exercses from Ross, 3, : Math 26: Probablty MWF pm, Gasson 30 Homework Selected Solutons 3, p. 05 Problems 76, 86 3, p. 06 Theoretcal exercses 3, 6, p. 63 Problems 5, 0, 20, p. 69 Theoretcal exercses 2, 3, 3, Problem 86. Let S {,..., n}m and suppose that A and B are, ndependently, equally lkely to be any of the 2 n subsets ncludng and S of S. a Show that P A B 3/ n. b Show that P A B 3/ n. Soluton a. Followng the hnt n the book, we condton on the number of elements n B: let NB ndcate the number of elements n B. Evdently, NB {0,..., n}. Because the whole sample space s the dsjont unon of the events n whch B has k elements as k goes from to n, by Bayes Theorem we have P A B P A B NB P NB. Moreover, gven that NB, there are 2 subsets of B, and 2 n subsets n total. Thus P A B NB 2 2 n. Fnally, there are n subsets of S of sze, so we have Puttng these together, we have P NB n 2 n. P A B P A B NB P NB n 2 n 2 n n n 2 2 n n 2 n + n 2n n 3. Soluton b. The condton A B s equvalent to A B c. The probablty P A B c can be computed wth the same method as above: Bayes Theorem agan apples condtonng on NB c, the
2 number of elements n B c, and P NB c P NB n. Thus we have P A B c P A B c NB c P NB c n 2 n 2 n n j0 n j n 2 n n n. 2 n n j 2 n j + n 2n 3 n 2 n n 3, Theoretcal exercse 3. A communty conssts of m famles, where n of them have chldren, for each,..., k, so that n m. Consder the followng two methods for choosng a random chld: Choose one of the m famles at random, then randomly choose a chld from that famly. 2 Choose one of the n chldren at random. Show that method s more lkely than method 2 to result n the choce of a frstborn chld. Soluton. Let F ndcate the event that a frstborn chld s chosen. We compute P F for each method, startng wth method. Let E ndcate the event that we start by choosng a famly wth chldren. Snce we are choosng one of m famles randomly and n of them have chldren, we have P E n /m. Snce E,..., E k are dsjont sets whose unon s the whole sample space, by Bayes Theorem we have P F P F E P E. Moreover, gven that we ve chosen a famly wth chldren, the probablty that a randomly chosen chld s frstborn s /. Thus P F E /. Puttng ths altogether, we have P F P F E P E n m m Call ths probablty f. Now we compute P F for method 2: because there are n chldren and m frstborn, we have P F m. n Call ths probablty f 2. We clam that f f 2. Ths nequalty s equvalent to n mf n mf 2, and snce m n the latter s equvalent to n 2 n n. Introducng a second ndex for convenence, ths s equvalent to n jn j n n j. j j 2 n.
3 It remans to demonstrate ths nequalty. We wll wrte LHS and RHS for the lefthand and rghthand sdes, respectvely, of the last nequalty above. By expandng these products, we have RHS n n j n n j n n j + 2n n j j j j <j n 2 + 2n n j <j n j LHS jn j n n j j n n j + j + j n n j j j j <j n 2 + j + j n n j. <j Evdently, we wll have LHS RHS provded we show that + j 2, for all postve ntegers, j. j Because j s an nteger, we have j 2 0, whch when expanded and rearranged gves 2 +j 2 2j. Snce and j are postve, ths mples that 2 + j 2 /j 2, whch s precsely + j 2 as desred. j, Problem 20. A gamblng book recommends the followng wnnng strategy for the game of roulette: Bet $ on red. If red appears whch has probablty 8/38, then take the $ proft and qut. If red does not appear and you lose ths bet whch probablty 20/38 of occurng, make addtonal $ bets on red on each of the next two spns of the roulette wheel and then qut. Let X denote your wnnngs when you qut. a Fnd P X > 0. b Are you convnced that the strategy s ndeed wnnng? Explan. c Fnd E[X]. Soluton a. Let R ndcate the event that red appears on the frst spn of the wheel. If red appears on the frst spn, than our strategy produces a proft of $, so that P X > 0 R. Let R ndcate the event that there are reds on the second and thrd spns of the roulette wheel, for 0,, 2. Snce R, R c R 0, R c R, R c R 2 are dsjont sets whose unon s the whole sample space, by Bayes Theorem we have Our strategy dctates: P X > 0 P X > 0 RP R + P X > 0 R c R 0 P R c R 0 + P X > 0 R c R P R c R + P X > 0 R c R 2 P R c R 2 f ω R 3 f ω R c R 0 Xω f ω R c R f ω R c R 2 Thus P X > 0 R P X > 0 R c R 2 and P X > 0 R c R 0 P X > 0 R c R 0, and we conclude that P X > 0 P R + P R c R
4 Soluton b. The answer here depends a bt on what s meant by wnnng. The gamblng book s correct n the sense that P X > 0 > /2, so that we are more lkely than not to make money. However, f we play ths game over and over, repeatedly, whether we make money n the long run whch many readers would farly nterpret wnnng to mean depends on how much we we wn or lose as well. That s, a reasonable nterpretaton of wnnng depends on E[X], and not on P X > 0. Snce we compute ths quantty below, and E[X] < 0, one could farly nterpret ths strategy to be losng. Soluton c. The possble values for X are recorded above, and they are, 3, and. Thus we have E[X] P X 3 P X 3 P X P R R c R 2 3P R c R 0 P R c R If we perform ths strategy 000 tmes, we should expect fnal wnnngs to be approxmately ,.e. we expect to lose $08., Theoretcal exercse. The random varable X s sad to have the Yule-Smons dstrbuton f P X n, for n. nn + a Show that the precedng s actually a probablty mass functon,.e. P X n. b Show that E[X] 2. c Show that E[X 2 ]. Soluton a. We frst smplfy the expresson for P X n: nn + nn + n n n + 2 Now the partal sums can be computed drectly: P X n nn + n 2 n + n N N n N 2 n + 2 N + 2 N N + 2. N n n. + 2 N + N + 2
5 To fnsh, the nfnte sum s gven by the lmt of the partal sums: P X n lm N P X n lm N 2 N N + 2. Soluton b. We compute the expected value drectly: E[X] n P X n n + lm N lm N N lm N n + 2 N + 2 n nn + n + lm N 2 2 N Soluton c. For the second moment, we have: E[X 2 ] n 2 P X n n n +. n 2 nn + Snce n + 2n and 3n for all n, we have n + 6n 2. Because the seres n 6n 2 dverges by the p-test, or because the harmonc seres dverges, the comparson 2 3 n test now mples that n n + dverges. Thus E[X2 ]. The followng problem appears as Theoretcal exercse n a prevous edton of the textbook:, Theoretcal exercse. For a nonnegatve nteger-valued random varable N, show that E[N] P N. Soluton. Because the event N s the dsjont unon of the events N j, for j, +,..., we have P N P N j and j P N P N j. j 5
6 For a fxed value of j, P N j appears n the double sum once for each value of from to j. That s, nterchangng the order of summaton gves j P N P N j j P N j. j j Because N s non-negatve and nteger-valued, ths last sum s the same as x P N x,.e. E[N]. x:px>0 6
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