UC Berkeley Department of Electrical Engineering and Computer Sciences. EE126: Probability and Random Processes

Transcription

1 UC Berkeley Departmet of Electrical Egieerig ad Computer Scieces EE26: Probability ad Radom Processes Problem Set Fall 208 Issued: Thursday, August 23, 208 Due: Wedesday, August 29, 207 Problem. (i Sho that the probability that exactly oe of the evets A ad B occurs is P(A + P(B 2P(A B. (ii If A is idepedet of itself, sho that P(A = 0 or. Solutio.. The probability of the evet that exactly oe of A ad B occur is Pr(A B c + Pr(A c B = Pr(A Pr(A B + Pr(B Pr(A B = Pr(A + Pr(B 2 Pr(A B. 2. Pr(A A = Pr(A Pr(A, so Pr(A = Pr(A 2 ; this implies that Pr(A {0, }. Alteratively, suppose for the sake of cotradictio that 0 < Pr(A <. The, Pr(A A = Pr(A, hich cotradicts the supposed idepedece of A ith itself. Hece, Pr(A {0, }. Problem 2. Alice ad Bob have 2 + fair cois, each coi ith probability of heads equal to /2. Bob tosses + cois, hile Alice tosses the remaiig cois. Assumig idepedet coi tosses, sho that the probability that after all cois have bee tossed, Bob ill have gotte more heads tha Alice is /2 (Hit: use symmetry. Solutio 2. If e let Ω be the sample space cosistig of all possible 2 + tosses, the Ω is a uiform probability space by assumptio. Defie the evets A = {there are more heads i the first + tosses tha the last tosses}, B = {there are more tails i the first + tosses tha the last tosses}. By symmetry, Pr(A = Pr(B, ad e ote that A B = sice it is impossible for the first + tosses to have more heads ad more tails tha the last tosses, ad A B = Ω. So, Pr(A + Pr(B = ad hece Pr(A = /2. Alteratively, if the probability that Bob has more heads tha Alice i the first tosses is p, the the probability that Bob has feer heads tha Alice i the first tosses is also p, ad the probability that they are tied after tosses is 2p. So, the probability that Bob is is p + (/2( 2p = /2 sice Bob ca either i by havig more heads tha Alice i the first tosses, or by havig the same umber of heads as Alice i the first tosses ad the flippig heads o the last toss.

2 Problem 3. Each of k jars cotais hite ad b black balls. A ball is radomly chose from jar ad trasferred to jar 2, the a ball is radomly chose from jar 2 ad trasferred to jar 3, etc. Fially, a ball is radomly chose from jar k. Sho that the probability that the last ball is hite is the same as the probability that the first ball is hite, i.e., it is /( + b. Solutio 3. We derive a recursio for the probability p i that a hite ball is chose from the ith jar. We have, usig the total probability theorem, p i+ = + + b + p i + + b + ( p i = + b + p i + startig ith the iitial coditio p = /( + b. Thus, e have p 2 = + b + + b + + b + = + b. + b +, More geerally, this calculatio shos that if p i = /( +b, the p i = /( +b. Thus, e obtai p i = /( + b for all i. Problem 4. There are passegers i a plae ith assiged seats, but after boardig, the passegers take the seats radomly. Assumig all seatig arragemets are equally likely, hat is the probability that o passeger is i their assiged seat? Compute the probability he. Hit: Use the iclusio-exclusio priciple. Solutio 4. First, let us calculate the probability that at least oe passeger sits i his or her assiged seat usig iclusio-exclusio. Let A i, i =,...,, be the evet that passeger i sits i his or her assiged seat. We first add the probabilities of the sigle evets (of hich there are, ad the probability of each evet is (!/! (ideed there are (! ays to permute the remaiig passegers oce a specific passeger is fixed, ad! total permutatios, so the probability is (!/!; ext, e subtract the probabilities of the pairise itersectios of evets (of hich there are ( 2, ad the probability of each evet is ( 2!/! (there are ( 2! ays to permute the passegers other tha the fixed to; cotiuig o, e see that ( Pr A i = ( ( j! ( j+ = j! ( j+ j!. o, the evet that o passeger sits i his or her assiged seat is the complemet of the evet just discussed: ( Pr A i = ( j+ j! = ( j. j! Takig the limit as, the expressio coverges to j=0 ( j /j!, ad usig the expressio for the poer series of the expoetial fuctio, e coclude that the probability coverges to exp( j=0 2

3 Problem 5. (i Let Z >0 ad A,..., A be ay evets. Prove the uio boud: Pr( A i Pr(A i (Hit: Use Iductio. (ii Let A A 2 be a sequece of icreasig evets. Prove that lim Pr(A = Pr( A i. [This ca be vieed as a cotiuity property for Probability.] (iii Let A, A 2,... be a sequece of evets. Prove that the uio boud holds for the folloig settig: Pr( A i Pr(A i. Solutio 5.. From iclusio-exclusio, Pr(A B = Pr(A + Pr(B Pr(A B Pr(A + Pr(B. The result o follos from iductio. Formally, the case of = is trivial ad the case of = 2 as prove above; let 3. Pr( A i = Pr(A Pr(A + A i Pr(A + Pr( A i Pr(A i = Pr(A i. 2. Write A = A ad A i = A i \ i A j for i, i 2. o, the A i for i Z >0 are disjoit, ad A i = A i = A, so Pr(A = Pr( A i = Pr(A i. Hece, lim Pr(A = lim by coutable additivity. Pr(A i = Pr(A i = Pr( A i = Pr( A i, ote: A fuctio f : R R is called cotiuous if for every x R ad every sequece {x } covergig to x, e have lim f(x = f(lim x = f(x. If e vie A i as lim i A i, the the cotiuity property of probability says that lim Pr(A = Pr(lim A, hich explais the ame. ote 2: Some studets oticed that by iclusio-exclusio, Pr( A i = Pr(A + Pr( A i Pr(A ( A i, but A ( A i = A, so Pr( A i = Pr( A i; ideed, oe ca prove by iductio that for ay Z >0, Pr( A i = Pr( i= A i. This is ot sufficiet to prove the problem, but it ca be explaied by oticig that the LHS, lim Pr(A, does ot deped o fiitely may evets (sice e are takig a limit, so throig aay the evets A,..., A does ot chage the probability. 3. As i the previous part, defie A = A ad A i = A i \ i A i for i, i 2. o, Pr( A i = Pr( A i = Pr(A i, ad for all i Z >0 e have Pr(A i Pr(A i sice A i A i, so Pr( A i Pr(A i. ote: The fact e used above is that if B A, the Pr(B Pr(A; this follos because A = B (A\B is a disjoit uio, so Pr(A = Pr(B+Pr(A\ B Pr(B. 3

4 Problem 6. (i Superma ad Captai America are playig a game of basketball. At the ed of the game, Captai America scored poits ad Superma scored m poits, here > m are positive itegers. Supposig that each basket couts for exactly oe poit, hat is the probability that after the start of the game (he they are iitially tied, Captai America as alays strictly ahead of Superma? (Assume that all sequeces of baskets hich result i the fial score of baskets for Captai America ad m baskets for Superma are equally likely.(hit: Thik about symmetry. First, try to figure out hich is more likely: there as a tie ad Superma scored the first poit, or there as a tie ad Captai America scored the first poit? Solutio 6. Let T be the evet that Captai America ad Superma ere tied at least oce after the first poit. Let C be the evet that Captai America scores the first poit, ad S be the evet that Superma scores the first poit. I fact, Pr(C T = Pr(S T for the folloig reaso: give ay sequece of baskets here the first poit is scored by Captai America ad there is a tie, flip all of the baskets util the first tie. This yields a sequece of baskets here the first poit is scored by Superma ad there is a tie. Thus, the outcomes i C T are i oe-to-oe correspodece ith the outcomes i S T. Hoever, sice Captai America o the game, Pr(S T = Pr(S (if Superma scored the first poit, the there must have bee a poit he Captai America caught up. So far, e have Pr(T = Pr(C T + Pr(S T = 2 Pr(S T = 2 Pr(S. We ca calculate Pr(S = m/(m +. Thus, Pr(T = 2m/(m +. Fially, the questio is askig for Pr(T c = 2m/(m + = ( m/(m +. ote: This is yet aother famous problem ko as the ballot problem. Problem 7 (Bous. I a touramet ith players (here is a positive iteger, each player plays agaist every other player for a total of ( 2 games (assume that there are o ties. Let k be a positive iteger. Is it alays possible to fid a touramet such that for ay subset A of k players, there is a player ho has beate everyoe i A? For such a touramet, let us say that every k-subset is domiated. For example, Figure depicts the smallest touramet i hich every 2-subset is domiated. I fact, as log as ( k ( 2 k k <, it is possible to fid a touramet of players such that every k-subset is domiated. Prove this fact, ad explai hy it implies that for ay positive iteger k there exists a touramet such that every k-subset is domiated. ote: The bous questio is just for fu. You are ot required to submit the bous questio, but do give it a try ad rite do your progress. Solutio 7. Let us build a touramet ith players radomly, by decidig each game ith a idepedet fair coi flip. Let A j, for j =,..., ( k, be the evet that the jth subset of k players is ot domiated by ayoe. What is Pr(A j? There are k other players, ad the probability that a player fails to domiate the jth subset is 2 k, so the probability 4

5 Figure : A touramet ith 7 vertices such that every pair of players is beate by a third player. that oe of the k other players domiates the jth subset is Pr(A j = ( 2 k k. Hece: (( k Pr ( k A j Pr(A j = ( ( 2 k k. k Thus, if ( k ( 2 k k <, the Pr( ( k Ac j > 0, i.e., there is a positive probability that every subset of k players is domiated, but this meas there exists some touramet ith players such that every k-subset is domiated. o, the questio is hether e ca pick so that ( ( k ( 2 k k <. The term k gros roughly at the rate k, ad the term ( 2 k k gros roughly at the rate c for some c (0,, ad sice expoetial decay outpaces polyomial groth, e ca ideed alays pick sufficietly large so that ( k ( 2 k k < holds. Ca e figure out the order of magitude of? Usig the simpler bouds ( k k ad ( 2 k k exp( ( k2 k, the it suffices to have k exp( ( k2 k <, ad upo takig logarithms ad rearragig e get the coditio k2 k l k > 0. From this e at least eed > 2 k so l > k but the k2 k l > k 2 2 k, hich meas e eed to take > k 2 2 k. If e let = Ck 2 2 k for some costat C >, the k2 k l k (C k 2 2 k > 0 for large k, so roughly, a costat multiple of k 2 2 k players suffices. 5