CS 171 Lecture Outline October 09, 2008

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1 CS 171 Lecture Outlie October 09, 2008 The followig theorem comes very hady whe calculatig the expectatio of a radom variable that takes o o-egative iteger values. Theorem: Let Y be a radom variable that takes o o-egative iteger values. The E[Y ] Proof. Pr[Y i] E[Y ] Pr[Y ] 1 1 i Pr[Y ] Pr[Y ] Pr[Y i] Liearity of Expectatio Oe of the most importat properties of expectatio that simplifies its computatio is the liearity of expectatio. By this property, the expectatio of the sum of radom variables equals the sum of their expectatios. This is give formally i the followig theorem. I did t cover the proof i the class but I am icludig it here for ayoe who is iterested. Theorem. For ay fiite collectio of radom variables X 1,X 2,...,X, [ ] E X i Proof. We will prove the statemet for two radom variables X ad Y. The geeral claim ca be prove usig iductio.

2 2 Lecture Outlie October 09, 2008 E[X + Y ] i i i i (i + )Pr[X i Y ] (ipr[x i Y ] + Pr[X i Y ]) ipr[x i Y ] + i Pr[X i Y ] i Pr[X i Y ] + Pr[X i Y ] i i ipr[x i] + Pr[Y ] E[X] + E[Y ] It is importat to ote that o assumptios have bee made about the radom variables while provig the above theorem. For example, the radom variables do ot have to be idepedet for liearity of expectatio to be true. Lemma. For ay costat c ad discrete radom variable X, E[cX] ce[x] Proof. The lemma clearly holds for c 0. For c 0 E[cX] c c k Pr[cX ] (/c)pr[x /c] k Pr[X k] ce[x] Example 1. Usig liearity of expectatio calculate the expected value of the sum of the umbers obtaied whe two dice are rolled. Solutio. Let X 1 ad X 2 deote the radom variables that deote the result whe die 1 ad die 2 are rolled respectively. We wat to calculate E[X 1 + X 2 ]. By liearity of expectatio E[X 1 + X 2 ] E[X 1 ] + E[X 2 ] 1 6 ( ) + 1 ( )

3 October 09, 2008 Lecture Outlie 3 Example 2. Suppose that people leave their hats at the hat check. If the hats are radomly retured what is the expected umber of people that get their ow hat back? Solutio. Let X be the radom variable that deotes the umber of people who get their ow hat back. Let X i,1 i, be the radom variable that is 1 if the ith perso gets his/her ow hat back ad 0 otherwise. Clearly, By liearity of expectatio we get X X 1 + X 2 + X X E[X] ( 1)!! 1 1 Example 3. Suppose we throw balls ito bis with the probability of a ball ladig i each of the bis beig equal. What is the expected umber of empty bis? Solutio. Let X be the radom variable deotig the umber of empty bis. Let X i be a radom variable that is 1 if the ith bi is empty ad is 0 otherwise. Clearly X X i By liearity of expectatio, we have E[X] Pr[X i 1] ( 1 ( 1 1 ) ) As, (1 1 ) 1 e. Hece, for large eough values of we have E[X] e Example 4. The followig pseudo-code computes the miimum of distict umbers that are stored i a array A. What is the expected umber of times that the variable mi is assiged a value if the array A is a radom permutatio of the elemets.

4 4 Lecture Outlie October 09, 2008 FidMi(A, ) 1 mi A[1] 2 for i 2 to do 3 if (A[i] < mi) the 4 mi A[i] 5 retur mi Solutio. Let X be the radom variable deotig the umber of times that mi is assiged a value. We wat to calculate E[X]. Let X i be the radom variable that is 1 if mi is assiged A[i] ad 0 otherwise. Clearly, Usig the liearity of expectatio we get X X 1 + X 2 + X X E[X] Pr[X i 1] (1) Note that Pr[X i 1] is the probability that A[i] cotais the smallest elemet amog the elemets A[1],A[2],...,A[i]. Sice the smallest of these elemets is equally likely to be i ay of the first i locatios, we have Pr[X i 1] 1 i. Thus equatio (1) becomes E[X] where c is a costat less tha 1. 1 i H() l + c Example 5. (Birthday Paradox) Suppose there are k people i a room ad days i a year. We are iterested i the probability that there are at least two people i the room with the same birthday. What is the smallest value of k for which this probability is at least 1/2? Assume that it is equally likely for a perso to be bor o ay of the days of the year. Solutio. Let B be the evet that at least two people i the room have the same birthday. We are iterested i Pr[B]. Pr[B] 1 Pr[B] 1 P(,k) k For 365, the smallest value of k for which the RHS is at least 1/2 is 23. If m 40 the Pr[E] 0.89, ad if m 60 the Pr[E] This meas that if there are 60 people the it is almost certai that there exists two amog them sharig the same birthday. To

5 October 09, 2008 Lecture Outlie 5 illustrate how good our model is, cosider the set of presidets of the Uited States of America. Through Bill Clito, 41 people belog to this set. The chaces of two of them sharig the same birthday is at least 89%. Ideed, James Polk (11th presidet) ad Warre Hardig (29th presidet) are both bor o Nov. 2. Example 6. I the above example, o average how may pairs of people share the same birthday? Solutio. Let X be the radom variable deotig the umber of pairs of people sharig the same birthday. For ay two people i ad, let X i be a idicator radom variable that is 1 if i ad have the same birthday ad is 0 otherwise. Clearly X i, X i. Usig the liearity of expectatio we get E[X] i, i, i, Pr[X i 1] Pr[X i 1] 1 i, ( k 2) k(k 1) 2 Assumig 365, the smallest value of k for which the RHS is at least 1 is 28.

Randomized Algorithms I, Spring 2018, Department of Computer Science, University of Helsinki Homework 1: Solutions (Discussed January 25, 2018)

Randomized Algorithms I, Spring 2018, Department of Computer Science, University of Helsinki Homework 1: Solutions (Discussed January 25, 2018) Radomized Algorithms I, Sprig 08, Departmet of Computer Sciece, Uiversity of Helsiki Homework : Solutios Discussed Jauary 5, 08). Exercise.: Cosider the followig balls-ad-bi game. We start with oe black