Discrete Mathematics and Probability Theory Fall 2016 Seshia and Walrand Final Solutions

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1 CS 70 Discrete Mathematics ad Probability Theory Fall 2016 Seshia ad Walrad Fial Solutios CS 70, Fall 2016, Fial Solutios 1

2 1 TRUE or FALSE?: 2x8=16 poits Clearly put your aswers i the aswer box o the bottom of this page! This is what is to be graded No eed to justify! Aswer: Note that the aswers provide explaatios for your uderstadig, eve though o such justificatio was required 1 A fiite irreducible Markov chai that ca go from state 1 to itself i oe step is such that P[X = 1] must coverge as Aswer: True, because it is aperiodic 2 We are give a probability space with sample space Ω ad probability P( ) Let {A 1,A 2,,A 5 } be pairwise disjoit evets whose uio is Ω ad let B be aother evet If P[B A 1 ] = 1, the P[A 1 B] P[A k B] for k = 1,,5 Aswer: False For istace, it could be that P(A 1 ) 1 3 You throw 15 balls idepedetly ad uiformly at radom ito 7 bis The, the umber of balls i bis 3 ad 5 are positively correlated Aswer: False They are egatively correlated 4 The umber of ways of distributig 10 idetical smartphoes to Steve, Bill, ad Larry is ( 10 3 ) Aswer: False Stars ad bars, it should be ( 12 2 ) 5 You draw five cards from a perfectly shuffled stadard 52-card deck The probability that the third card is a ace is strictly smaller tha the probability that the first card is a ace Aswer: False These probabilities are equal, by symmetry 6 If Q[Y X] = 3 + 2X, the L[Y X] = 3 + 2X Aswer: True The best liear estimate of Y caot be better tha the best quadratic estimate 7 Assume that E[(Y 3cos(5X) 13cos(7X))cos(X))] = 0 for = 0,1,,9 The the values of a 0,a 1,,a 9 that miimize E((Y a 0 a 1 cos(x) a 2 cos(2x) a 9 cos(9x)) 2 ) are such that a k = 0 for k / {5,7} ad a 5 = 3,a 7 = 13 Aswer: True This is the projectio property 8 Let p be a prime umber The set of all fuctios from {0,1,2,, p 1} to Q is ucoutably ifiite Aswer: False It is coutably ifiite You ca thik of this set as isomorphic to the p-wise cartesia product Q Q Q, ad oe ca eumerate this set i a maer aalogous to Q as covered i class Aswer Box (Aswer TRUE or FALSE i the correspodig box below) Short Aswers: 4x5=20 poits Provide a clear ad cocise justificatio of your aswer 2

3 1 You roll a six-sided balaced die util you get the first 6 How may dots do you accumulate, o average, icludig the 6 o the last roll? Aswer: The average umber of dots o a roll that does ot yield a 6 is ( )/5 = 3 O average, you have to roll the die 6 times to get the first 6, thus five times before the first 6 You the accumulate, o average, = 21 dots (You may have oticed the use of Wald s idetity) 2 Alice, Bob, ad Charles wat to flip a coi to decide who pays for luch, ad wat each to have the same probability of payig However, they oly have a biased coi ad they do t eve kow its bias Explai a mechaism that achieves the objective Aswer: You flip the coi three times If the outcome has a sigle heads, the Alice pays if it is HTT, Bob pays if it is THT, ad Charles pays if it is TTH Otherwise, you igore the outcome ad you repeat the experimet 3 You have to pass two courses i sequece, A the B, to be able to declare CS as your major Whe you take a course for the first time, you pass it with probability 1/2 If you take it for the secod time, you pass it with probability 2/3 If you fail a course twice, you caot declare CS as your major What is the probability that you ca declare CS as your major? Aswer: The probability that you fail course A twice i a row is (1/2)(1/3) = 1/6 Thus, the probability that you pass course A is 5/6 Similarly, the probability that you pass course B is 5/6 Thus, the probability that you ca declare CS as your major is (5/6) 2 = 25/36 70% 4 You pick poits idepedetly ad uiformly at radom iside a uit circle Let Z be the distace of the furthest poit to the ceter What is E(Z)? Aswer: We see that P(Z < z) = P(Z 1 < z) where Z 1 is the distace from the first poit to the ceter of the circle But P(Z 1 < z) = z 2 Thus, P(Z < z) = z 2, so that the pdf of Z is 2z 2 1 1{0 < z < 1} Hece, E(Z) = 1 0 z(2z2 1 )dz = Suppose we draw cards from a stadard 52-card deck without replacemet What is the expected umber of cards we must draw before we get all 13 spades (icludig the last spade)? Aswer: Defie X to be the umber of cards we draw before we get all 13 spades, icludig the last spade Note that X = 52 Y, where Y is the umber of cards that come after all the spades Arbitrarily label all the o-spade cards from 1 to 39 Defie Y i as the idicator variable that deotes whether the o-spade card with label i comes after all the spades The Y = 39 i=1 Y i Note that E(Y i ) = 1 14 This is because if we cosider oly the relative orders of the 13 spades ad the card labelled i, there is a 1 14 chace that the card labelled i comes last Applyig liearity of expectatio, we get E(X) = E(52 Y ) = 52 E(Y ) = = Rollig Dice I: =12 poits Provide a clear ad cocise justificatio of your aswer I this problem, you roll two balaced six-sided dice 1 What is the probability that the umber of dots o the first die is at least twice the umber o the secod die? Aswer: If the secod die yields 1, the there are 5 possibilities for the first die, if it yields 2, there are 3, ad if it yields 3, there is oe Thus, the probability is (1/36)[ ] = 1/4 2 What is the probability that the first die yields 4 if the total umber of dots o the two dice is 8? Aswer: It is P[(4,4) (2,6),(3,5),(4,4),(5,3),(6,2)] = 1/5 3

4 3 What is the probability that the sum is 8 give that the secod die yields 4? Aswer: It is the probability that the first die yields 4, ie, 1/6 4 What is the probability that the umber of dots o the first die is at least as large as o the secod? Aswer: Let X,Y be the umber of dots o the two dice It s easy to see that P(X = Y ) = 1/6 so by symmetry P(X > Y ) = (1 1/6)/2 = 5/12 Thus, P(X Y ) = P(X > Y )+P(X = Y ) = 5/12+1/6 = 7/12 4 Tossig Cois: 6+6= 12 poits Provide a clear ad cocise justificatio of your aswer 1 You are give two cois A ad B Oe is biased with P(H) = 06 ad the secod is fair You do t kow which is biased; it is equally likely to be A or B You flip the pair of cois repeatedly (both each time) util at least oe of the two cois yields H It turs out that you have to flip them 27 times ad that coi A the yields H while coi B yields T What is the probability that A is the biased coi? Aswer: Call F the evet that we described The, Thus, by Bayes Rule, P[F A is biased] = ad P[F A is fair] = P[A is biased F] = = (1/2)P[F A is biased] (1/2)P[F A is biased] + (1/2)P[F A is fair] = = 06 2 You flip a fair coi 10 times Let X be the total umber of heads i the first 5 flips ad Y the total umber of heads i the odd flips, ie, i flips 1,3,5,7, ad 9 (a) Fid E[X Y ] (b) Fid E[Y X] Aswer: Let Z = 1 if coi yields H The, X = Z 1 + Z 2 + Z 3 + Z 4 + Z 5 ad Y = Z 1 + Z 3 + Z 5 + Z 7 + Z 9 (a) Now, E[Z 1 Y ] = Y /5 by symmetry ad E[Z 2 Y ] = 1/2 by idepedece Thus, E[X Y ] = (3/5)Y + 2(1/2) = 1 + (3/5)Y (b) Similarly, E[Z 1 X] = (1/5)X ad E[Z 9 X] = 1/2 Thus, E[Y X] = (3/5)X + 2(1/2) = 1 + (3/5)X 5 Rollig Dice II: 8+6=14 poits Provide a clear ad cocise justificatio of your aswer 1 You roll a six-sided balaced die util either you get a 6 or the umber of dots o a roll exceeds the umber of dots o the previous roll How may rolls do you eed, o average? Write the equatios that you have to solve to fid the aswer Do ot solve the equatios Aswer: Let β() be the average umber of rolls you still eed, give that the last roll yielded dots ad let β be the total average umber of rolls Thus β(6) = 0 ad the first step equatios are β = 1 + (1/6) β() = 1 + (1/6) 5 =1 m=1 β() β(), = 1,,5 4

5 2 You have a die ad wat to check if it is balaced To do this, you decide to test the mea You roll the die 10 4 times ad fid a average umber of dots per roll equal to 32 (a) Usig Chebyshev s iequality ad a simple upper boud o the variace, fid a 95% cofidece iterval for the mea (Hit: It ca be show that variace is bouded by 36) (b) Are you 95% sure that the die is ot balaced? Aswer: (a) We kow that this iterval is 32 ± 45 σ = 32 ± σ It remais to fid a upper boud o σ Sice X E(X) 6, we coclude that var(x) 36, ie, σ 6 Hece, a 95% cofidece iterval is 32 ± , ie, 32 ± 027 (b) Yes, because 35 is ot i this iterval 6 Graphs ad Chais: 8+6=14 poits Provide a clear ad cocise justificatio of your aswer 1 Imagie you are traversig a 3-dimesioal hypercube Begiig at the vertex 000, you repeatedly radomly pick ay outgoig edge from your curret vertex ad follow it to aother vertex What is the expected umber of edges you must traverse before reachig the vertex 111 for the first time? Aswer: Our jourey ca be modeled as a Markov Chai with odes 0, 1, 2, ad 3, where ode i is the collectio of odes with i oes i it That is, ode 0 cosists of vertex 000; ode 1 cosists of vertices 001, 010, ad 100; ode 2 cosists of vertices 110, 101, ad 011; ad ode 3 cosists of vertex 111 We wat to fid E(X 0 ), where X i deotes the umber of edges we must traverse from ode i to ode 3 We ca costruct the followig equatios: E(X 0 ) = 1 + E(X 1 ) E(X 1 ) = E(X 2) E(X 0) E(X 2 ) = E(X 1) E(X 3) E(X 3 ) = 0 Solvig this liear system of equatios, we get E(X 0 ) = 10 2 Cosider a Markov chai o {0,1} with P(0,1) = α = P(1,0) for some α [0,1] Let X 0,X 1,X 2,,X be the successive values of the Markov chai (i) Propose a method to estimate α (ii) Usig Chebyshev s iequality, what is a 95% cofidece iterval for α Aswer: (i) We ote that the Markov chai flips with probability α at each step, idepedetly Thus, we let Y m = 1{X m 1 X m } = X m 1 X m ad we see that these are iid ad equal to oe with probability α ad to zero otherwise Hece, our estimate is A = Y 1 + +Y (ii) As we kow by lookig at coi flips, a 95% cofidece iterval for α is A ± 225, usig Chebyshev (We ca replace 225 by 1 for large by the CLT, but you are ot expected to remember this) 5

6 7 Erdos-Reyi Radom Graph: =12 poits Provide a clear ad cocise justificatio of your aswer Radom graph model is a key idea i etwork aalysis Erdos-Reyi Graph Model is the most popular ad simplest etwork model Suppose you have vertices ad you leave them ucoected at the begiig The for every two vertices, you draw a udirected edge with probability p (p is a fixed umber for all edges) Suppose we label all the vertices i a order 1, 2,, 1 Prove that i ay E-R graph, if u is a vertex of odd degree i, the there exists a path from u to aother vertex v of the graph where v also has odd degree A path is a sequece of edges (v 1,v 2 ),(v 2,v 3 ),, (v 2,v 1 ), (v 1,v ), where o vertex is repeated Aswer: We will build a path that does ot reuse ay edges As we build the path, imagie erasig the edge we used to leave so that we will ot use it agai Begi at vertex u ad select a arbitrary path away from it This will be the first compoet of the path If, at ay poit, the path reaches a vertex of odd degree, we will be doe, but each time we arrive at a vertex of eve degree, we are guarateed that there is aother vertex out, ad, havig left, we effectively erase two edges from that meet at the vertex Sice the vertex origially was of eve degree, comig i ad goig out reduces its degree by two, so it remais eve I this way, there is always a way to cotiue whe we arrive at a vertex of eve degree Sice there are oly a fiite umber of edges, the tour must ed evetually, ad the oly way it ca ed is if we arrive at a vertex of odd degree 2 What s the probability that there are 0 paths of legth 2 coectig vertices 1 ad 2? Aswer: Let I(v,w) be the idicator that there is a edge coectig the vertices v ad w Usig idepedece of the edges we compute: v=3 (1 P[(I(1,v) = 1)I(v,2) = 1)]) = (1 p 2 ) ( 2) 3 Let X l be the umber of paths of legth l coectig vertices 1 ad 2 What s the expectatio ad variace of X 2? Aswer: We say that there exists a path of legth l coectig two give vertices v w if there exist vertices v 1,,v l 1 such that v v 1 v 2 v l 1 w, where v w meas that there is a edge betwee v ad w By defiitios we have Takig expectatio yields Now we compute E(X 2 2 ): E(X 2 2 ) = = v,w=3 v=3 E(X 2 ) = X 2 = v=3 v=3 E[1(1 v 2)1(1 v 2)] P(1 v)p(v 2) + 1(1 v,v 2) P(1 v)p(v 2) = ( 2)p 2 v,w=3:v w = ( 2)p 2 + ( 2)( 3)p 4 P(1 v)p(v 2)P(1 w)p(w 2) 6

7 So the variace of X 2 becomes: Var(X 2 ) = E(X2 2 ) E(X 2 ) 2 = ( 2)p 2 + ( 2)( 3)p 4 [( 2)p 2 ] 2 = ( 2)(1 p 2 )p 2 4 What s the expectatio of X l? (You should give a expressio i terms of, p ad l ad you could leave a summatio, itegral or product if that s easier to write the fial aswer) Aswer: By defiitios we have X l = 1(1 v 1,v 1 v 2,,v l 1 2) where the sum is over ordered but distict vertices v i take from the set {3,,} Let s first thik about how may ways to choose l + 1 vertices (to form a path of legth l) from : ( l+1) = ( 1) ( l) The situatio i this problem is slightly differet where you eed to fix the first two vertices to be 1 ad 2 respectively So the total umber of ways to choose vertices from becomes ( 2)( 3) ( l) Takig expectatio yields E(X l ) = P(1 v 1,v 1 v 2,,v l 1 2) = P(1 v 1 )P(v 1 v 2 ) P(v l 1 2) = ( 2)( 3)( l)p l 8 Pollig for 2020: 10 poits Provide a clear ad cocise justificatio of your aswer We poll 100 me ad 200 wome Amog those, 30 me ad 160 wome declare they would vote for Michelle Obama for Presidet i 2020 Assume that, amog the voters i the geeral electio, there will be a equal umber of me ad wome ad that your poll is represetative of the votig i the geeral electio What is your 95% cofidece iterval for the fractio of people who will vote for Michelle? Would you advise her to ru? Justify your aswer To solve this problem, we suggest that you follow the followig steps: (1) What is your estimate of the fractio of the people who will vote for Michelle i the geeral electio? (2) What is its variace?; (3) What is a upper boud o that variace?; (4) Write dow Chebyshev ad use the upper boud; (5) Choose the distace from the mea so that the Chebyshev boud is 5%; (6) What is the resultig cofidece iterval? Aswer: Let X be the idicators that me vote for Michelle ad Y for wome Thus, we kow that V := 100 =1 X = 30 ad W := 200 =1 Y = 160 (1) The fractio of people votig for Michelle is p := E(X 1 +Y 1 )/2 Thus, a good estimate is Z := V /200 + W/400 sice E(Z) = p Now, (2) Oe has σ 2 := var(z) = 1 (200) 2 var(v ) + 1 (400) 2 var(w) = var(x 1) var(y 1) 7

8 (3) Sice var(x 1 ) 1/4 ad var(y 1 ) 1/4, oe has (4) Accordig to Chebyshev, (5) If we wat this probability to be 5%, we eed σ 2 ( )1 4 = P( Z E(Z) > ε) σ 2 ε ε ε 2 = 5% = 1 20, so that ε 2 = = = 2% (6) Thus, a 95% cofidece iterval is Z ± 2% = ( )/2 ± 2% = 055 ± 002 = [053,057] Yes, I would advise Michelle to ru The probability she would wi is larger tha 95% 9 Radom Variables: =30 poits Provide a clear ad cocise justificatio of your aswer 1 Defie radom variable i oe setece (No eed for a justificatio) Aswer: A radom variable is a real-valued fuctio of the outcome of a radom experimet 2 Let X, 1 be iid with mea µ ad variace σ 2 Let also A = (X X )/ (a) Usig Chebyshev, fid a 90% cofidece iterval for µ (b) Apply this result to the situatio where the X are 1 whe a coi yields H ad zero otherwise Aswer: (a) We have P( A µ > ε) σ 2 ε 2 We choose ε so that the right-had side is 01 This gives σ 2 ε 2 = 01, ie, ε2 = 10 σ 2 Thus, ε = 10σ/ The 90% cofidece iterval is the A ± 10 σ (b) For coi flips, µ = p ad σ 1/2 Hece, the 90% cofidece iterval for p is A ± 2 A ± 16 3 You choose a poit (X,Y ) uiformly at radom i the triagle with vertices ( 1,0),(0,1),(1,0) (a) Fid E[Y X] (b) Fid L[Y X] (c) Are X ad Y positively-, egatively-, or u-correlated? Aswer: (a) Give that X = x, we kow that Y = U[0,1 x ] Hece, E[Y X] = (1/2)(1 X ) (b) By symmetry, L[Y X] is costat Therefore, it must be equal to E[Y ] Now, P(Y > y) = (1 y) 2, sice this happes whe (X,Y ) falls i a triagle with base 2(1 y) ad height (1 y) Thus, f Y (y) = 2(1 y), so that E[Y ] = 1 0 2y(1 y)dy = 1/3, as we might have guessed Hece, L[Y X] = 1/3 (c) Sice LLSE is a costat, clearly X ad Y are ucorrelated Or we ca do a computatio We kow that E((Y L[Y X])X) = 0, so that E(XY ) = E(XL[Y X]) = (1/3)E[X] = 0 Thus, E(XY ) = 0 = E(X)E(Y ) ad X,Y are ucorrelated 8 10

9 4 Let Y ad Z be idepedet radom variables where Y is Poisso with mea 60 ad Z is Poisso with mea 10 (a) Fid L[Y X] where X = Y + Z (b) Fid E[Y X] (Hit: Recall that if V is Poisso with mea λ, the E(V ) = λ ad var(v ) = λ Also, recall that the sum of Poisso idepedet radom variables is Poisso) Aswer: (a) We kow that L[Y X] = E(Y ) + cov(x,y ) (X E(X)) var(x) Now, E(Y ) = 60 ad E(X) = E(Y ) + E(Z) = 70 Also, var(x) = var(y ) + var(z) = 70 Fially, Hece, cov(x,y ) = cov(y + Z,Y ) = var(y ) = 60 L[Y X] = (X 70) = 6 7 X (b) Let Z,Z 1,,Z 6 be idepedet ad Poisso with mea 10 We ca write Y = Z 1 + +Z 6 because this sum is Poisso with mea 60 ad idepedet of Z The, X = Z Z 6 + Z ad oe has by symmetry E[Y X] = E[Z Z 6 Z Z 6 + Z] = 6 7 (Z Z 6 + Z) = 6 7 X, 5 Let X be a discrete radom variable that takes values i [0,1] Show that its variace is at most 1/4 Hit: Show that the radom variable that takes at most differet values i (0,1) ad that has the largest variace is Beroulli 1/2 To do this, argue by cotradictio ad assume that X takes a value x i [05,1) with probability p Replace that value by 1 with probability 1/2 ad 2x 1 with probability 1/2 Show that the variace icreases Aswer: The mea does ot chage It suffices to show that the mea value of the square icreases This is the case because i the calculatio of E(X 2 ) the term x 2 p gets replaced by (p/2)(1) 2 + (p/2)(2x 1) 2 But x 2 < 1/2 + (1/2)(2x 1) 2 (Scratch space) 9

Problem Set 2 Solutions

Problem Set 2 Solutions CS271 Radomess & Computatio, Sprig 2018 Problem Set 2 Solutios Poit totals are i the margi; the maximum total umber of poits was 52. 1. Probabilistic method for domiatig sets 6pts Pick a radom subset S