《Mathematics for Computer》Notes for recitation 22

6.042/18.] Mathematics for Computer Science May6,2005 Srini devadas and Eric Lehman Notes for recitation 22 1 Conditional Expectation and Total Expectation There are conditional expectations, just as there are conditional probabilities. If R is a random variable and e is an event, then the conditional expectation Ex(r e)is defined bv x(R|E)=∑R(),Pr(m|E) For example, let R be the number that comes up on a roll of a fair die, and let E be the event that the number is even. Let's compute Ex(RI E), the expected value of a die roll, given that the result is even Ex(R|E)=∑R(o)P(E) =1.0+2.-+3·0+4.-+5·0+6 It helps to note that the conditional expectation, Ex(R I E)is simply the expectation of R with respect to the probability measure Preo defined in PSet 10. So it's linear Ex(R1+R2 E)=Ex(R1 E)+Ex(r2 E) Conditional expectation is really useful for breaking down the calculation of an ex- pectation into cases. The breakdown is justified by an analogue to the Total Probability The Theorem 1(Total Expectation). Let E1,..., En be events that partition the sample space and all have nonzero probabilities. If R is a random variable, then Ex(B)=Ex(B|E1)Pr(E1)+…+Ex(B|En)·Pr(En) For example, let R be the number that comes up on a fair die and e be the event that result is even, as before. Then E is the event that the result is odd. So the Total Expectation heorem says Pr(E)+Ex(R E). Pr(E)

� � � 6.042/18.062J Mathematics for Computer Science May 6, 2005 Srini Devadas and Eric Lehman Notes for Recitation 22 1 Conditional Expectation and Total Expectation There are conditional expectations, just as there are conditional probabilities. If R is a random variable and E is an event, then the conditional expectation Ex (R | E) is defined by: � Ex (R | E) = R(w) · Pr (w | E) w∈S For example, let R be the number that comes up on a roll of a fair die, and let E be the event that the number is even. Let’s compute Ex (R | E), the expected value of a die roll, given that the result is even. Ex (R | E) = R(w) · Pr (w | E) w∈{1,...,6} 1 1 1 = 1 · 0 + 2 · + 3 · 0 + 4 · + 5 · 0 + 6 · 3 3 3 = 4 It helps to note that the conditional expectation, Ex (R | E) is simply the expectation of R with respect to the probability measure PrE () defined in PSet 10. So it’s linear: Ex (R1 + R2 | E) = Ex (R1 | E) + Ex (R2 | E). Conditional expectation is really useful for breaking down the calculation of an ex￾pectation into cases. The breakdown is justified by an analogue to the Total Probability Theorem: Theorem 1 (Total Expectation). Let E1, . . . , En be events that partition the sample space and all have nonzero probabilities. If R is a random variable, then: Ex (R) = Ex (R | E1) · Pr (E1) + · · · + Ex (R E| n) · Pr (En) For example, let R be the number that comes up on a fair die and E be the event that result is even, as before. Then E is the event that the result is odd. So the Total Expectation theorem says: Ex (R) = Ex (R E)·Pr (E) + Ex R E ·Pr (E) � �� � � ��| � � �� � � ��| � � �� � = 7/2 = 4 = 1/2 = ? = 1/2

Recitation 22 The only quantity here that we don' t already know is Ex(RIE), which is the expected die roll, given that the result is odd. Solving this equation for this unknown, we conclude that Ex(rie=3 To prove the Total Expectation Theorem we begin with a lemma Lemma. Let R be a random variable, e be an event with positive probability, and Ie be the indicator variable for e. then Ex(R·IE) Pr(e Proof. Note that for any outcome, s, in the sample space, Pr({s}∩E)= ∫oi(s)=0 Pr(s)if Ie(s)=1, d Pr(s)nE)=Ie(s Pr(s) Ex(R|E)=∑R(s)P({s}E (ef of Ex(E)) ∑R(s) Pr({s}∩E ( Def of pr(·|E) R(s IE(s)·Pr(s) Pr(e) (by(2) ∑ses(F(s)·Is(s)·Pr(s) Pr(e) Ex(R·IE ( Def of ex(R·IE) Now we prove the Total Expectation Theorem: Proof. Since the Eis partition the sample space, R R·I

� � � � � � Recitation 22 2 The only quantity here that we don’t already know is Ex R | E , which is the expected die roll, given that the result is odd. Solving this equation for this unknown, we conclude that Ex R | E = 3. To prove the Total Expectation Theorem, we begin with a Lemma. Lemma. Let R be a random variable, E be an event with positive probability, and IE be the indicator variable for E. Then Ex (R IE) Ex (R | E) = · (1) Pr (E) Proof. Note that for any outcome, s, in the sample space, 0 if IE(s) = 0, Pr ({s} ∩ E) = Pr (s) if IE(s) = 1, and so Pr ({s} ∩ E) = IE(s) · Pr (s). (2) Now, � Ex (R | E) = s∈S R(s) · Pr ({s} | E) = � s∈S R(s) · Pr ({s} ∩ E) Pr (E) = � s∈S R(s) · IE(s) · Pr (s) Pr (E) � (Def of Ex (· | E)) (Def of Pr (· | E)) (by (2)) = s∈S(R(s) · IE(s)) · Pr (s) Pr (E) = Ex (R · IE) Pr (E) (Def of Ex (R · IE)) Now we prove the Total Expectation Theorem: Proof. Since the Ei’s partition the sample space, R = R IEi · (3) i

Recitation 22 for any random variable, R. So Ex(R)=Ex R·I ∑Ex(R·IB) (linearity of Ex o) ∑Bx(R|E)Pr(E) by(1)

Recitation 22 3 for any random variable, R. So � � � Ex (R) = Ex R · IEi (by (3)) � i = Ex (R · IEi ) (linearity of Ex ()) � i = i Ex (R | Ei) · Pr (Ei) (by (1))

Recitation 22 Problem 1. Final exams in 6.042 are graded according to a rigorous procedure With probability the exam is graded by a recitation instructor with probability, it is graded by a lecturer, and with probability ,, it is accidentally dropped behind the radiator and arbitrarily given a score of 84 Recitation instructors score an exam by scoring each problem individually and then taking th ne sum There are ten true/false questions worth 2 points each. For each, full credit given with probability ,, and no credit is given with probability There are four questions worth 15 points each. For each, the score is deter- mined by rolling two fair dice, summing the results, and adding 3 The single 20 point question is awarded either 12 or 18 points with equal prob- Lecturers score an exam by rolling a fair die twice, multiplying the results, and then adding a general impression"score With probability io, the general impression score is 40 With probability ao, the general impression score With probability io, the general impression score Assume all random choices during the grading process are mutually independent (a)What is the expected score on an exam graded by a recitation instructor? Solution. Let X equal the exam score and C be the event that the exam is graded by a recitation instructor. We want to calculate Ex(X C) By linearity of (conditional) expectation, the expected sum of the problem scores is the sum of the expected problem scores. Therefore, we have Ex(X I C)=10. Ex(T/F score C)+4. Ex(15pt prob score |C)+Ex(20pt prob score I C 7 1 +-0+4·2·+3)+ 12+-·18 =102+4:10+15=70 (b) What is the expected score on an exam graded by a lecturer?

Recitation 22 4 Problem 1. Final exams in 6.042 are graded according to a rigorous procedure: • With probability 4 7 the exam is graded by a recitation instructor, with probability 2 it 7 is graded by a lecturer, and with probability 1 7 , it is accidentally dropped behind the radiator and arbitrarily given a score of 84. • Recitation instructors score an exam by scoring each problem individually and then taking the sum. – There are ten true/false questions worth 2 points each. For each, full credit is given with probability 3 4 , and no credit is given with probability 1 . 4 – There are four questions worth 15 points each. For each, the score is deter￾mined by rolling two fair dice, summing the results, and adding 3. – The single 20 point question is awarded either 12 or 18 points with equal prob￾ability. • Lecturers score an exam by rolling a fair die twice, multiplying the results, and then adding a “general impression” score. 4 – With probability 10 , the general impression score is 40. 3 – With probability 10 , the general impression score is 50. 3 – With probability 10 , the general impression score is 60. Assume all random choices during the grading process are mutually independent. (a) What is the expected score on an exam graded by a recitation instructor? Solution. Let X equal the exam score and C be the event that the exam is graded by a recitation instructor. We want to calculate Ex (X | C). By linearity of (conditional) expectation, the expected sum of the problem scores is the sum of the expected problem scores. Therefore, we have: Ex (X C| ) = 10 · Ex (T/F score C) + 4 · Ex (15pt prob score | C) + Ex (20pt prob score C) � � | � � � � | 3 1 7 1 1 = 10 · 2 + 0 + 4 · 2 · + 3 + 12 + 18 4 · 4 · 2 2 · 2 · 3 = 10 · + 4 · 10 + 15 = 70 2 (b) What is the expected score on an exam graded by a lecturer?

Recitation 22 Solution. Now we want Ex( C), the expected score a lecturer would give Employing linearity again, we have Ex(x iC)=Ex(product of dice IC) +Ex(general impression C) (because the dice are independent) 50+ 60 (c) What is the expected score on a 6.042 exam? Solution. Let X equal the true exam score. The Total Expectation Theorem implies Ex(X)=Ex(X | C)Pr(C)+Ex(X I C)Pr(c) 4 2 +49 +84 4 40+(-+14)+12=6

� � � � � � | � ¯ � � � � � � � | � � � � Recitation 22 5 Solution. Now we want Ex X ¯ | C , the expected score a lecturer would give. Employing linearity again, we have: Ex X C ¯ = Ex product of dice ¯ | C + Ex general impression | C � �2 7 = (because the dice are independent) 2 4 3 3 + 40 + 50 + 60 10 · 10 · 10 · 49 1 = + 49 = 61 4 4 (c) What is the expected score on a 6.042 exam? Solution. Let X equal the true exam score. The Total Expectation Theorem implies: Ex (X ¯ ¯ ) = Ex (X C) Pr (C) + Ex X | C Pr C 4 49 2 1 = 70 · + + 49 + 84 7 4 · 7 · 7 7 1 = 40 + + 14 + 12 = 69 2 2

Recitation 22 Problem 2. Here's yet another fun 6.042 game! You pick a number between l and 6. Then you roll three fair, independent dice If your number never comes up, then you lose a dollar If your number comes up once, then you win a dollar. If your number comes up twice, then you win two dollars If your number comes up three times, you win four dollars! What is your expected payoff? Is playing this game likely to be profitable for you or not? Solution. Let the random variable r be the amount of money won or lost by the player in a round. We can compute the expected value of R as follows Ex(r)=-1 Pr(0 matches)+1. Pr(1 match)+2. Pr(2 matches)+ 4. Pr(3 matches +1·3 +2.3 6/)(6/+4./1)3 -125+75+30+4 216 216 game!pect to lose 16/216 of a dollar (about 7. 4 cents)in every round. This is a horrible n e

Recitation 22 6 Problem 2. Here’s yet another fun 6.042 game! You pick a number between 1 and 6. Then you roll three fair, independent dice. • If your number never comes up, then you lose a dollar. • If your number comes up once, then you win a dollar. • If your number comes up twice, then you win two dollars. • If your number comes up three times, you win four dollars! What is your expected payoff? Is playing this game likely to be profitable for you or not? Solution. Let the random variable R be the amount of money won or lost by the player in a round. We can compute the expected value of R as follows: Ex (R) = −1 · Pr (0 matches) + 1 · Pr (1 match) + 2 · Pr (2 matches) + 4 · Pr (3 matches) � �3 � � � �2 � �2 � � � �3 5 1 5 1 5 1 = −1 · + 1 · 3 + 2 · 3 + 4 · 6 6 6 6 6 6 −125 + 75 + 30 + 4 = 216 −16 = 216 You can expect to lose 16/216 of a dollar (about 7.4 cents) in every round. This is a horrible game!

Recitation 22 Problem 3. The number of squares that a piece advances in one turn of the game Monopoly is determined as follows: Roll two dice, take the sum of the numbers that come up, and advance that number lares If you roll doubles(that is, the same number comes up on both dice), then you roll a second time, take the sum, and advance that number of additional squares If you roll doubles a second time, then you roll a third time, take the sum, and advance that number of additional squares (a)What is the expected sum of two dice, given that the same number comes up on Solution. There are six equally-probable sums: 2, 4, 6, 8, 10, and 12. Therefore, the expected sum is 1 6 6 (b)What is the expected sum of two dice, given that different numbers come up? (Use your previous answer and the Total Expectation Theorem Solution. Let the random variables D1 and D2 be the numbers that come up on the two dice. Let e be the event that they are equal. The Total Expectation Theorem Ex(D1+D2)=Ex(D1+ D2 I E). Pr(E)+Ex(D2+ D2 E). Pr(E Two dice are equal with probability Pr(e)=1/6, the expected sum of two indepen dent dice is 7, and we just showed that Ex(D1+ D2 E)=7. Substituting in these quantities and solving the equation, we find 7=7·+Ex(D2+D2|E) Ex(D2+D2|E)=7 (c) To simplify the analysis, suppose that we always roll the dice three times, but may ignore the second or third rolls if we didnt previously get doubles. Let the random variable X i be the sum of the dice on the i-th roll, and let ei be the event that the i-th roll is doubles. Write the expected number of squares a piece advances in these terms

� � � � � � Recitation 22 7 Problem 3. The number of squares that a piece advances in one turn of the game Monopoly is determined as follows: • Roll two dice, take the sum of the numbers that come up, and advance that number of squares. • If you roll doubles (that is, the same number comes up on both dice), then you roll a second time, take the sum, and advance that number of additional squares. • If you roll doubles a second time, then you roll a third time, take the sum, and advance that number of additional squares. • However, as a special case, if you roll doubles a third time, then you go to jail. Regard this as advancing zero squares overall for the turn. (a) What is the expected sum of two dice, given that the same number comes up on both? Solution. There are six equally­probable sums: 2, 4, 6, 8, 10, and 12. Therefore, the expected sum is: 1 1 1 2 + 4 + . . . + 12 = 7 6 · 6 · 6 · (b) What is the expected sum of two dice, given that different numbers come up? (Use your previous answer and the Total Expectation Theorem.) Solution. Let the random variables D1 and D2 be the numbers that come up on the two dice. Let E be the event that they are equal. The Total Expectation Theorem says: Ex (D1 + D2) = Ex (D1 + D2 | E) · Pr (E) + Ex D2 + D2 | E · Pr E Two dice are equal with probability Pr (E) = 1/6, the expected sum of two indepen￾dent dice is 7, and we just showed that Ex (D1 + D2 | E) = 7. Substituting in these quantities and solving the equation, we find: 1 � � 5 7 = 7 · + Ex D2 + D2 | E 6 · 6 Ex D2 + D2 | E = 7 (c) To simplify the analysis, suppose that we always roll the dice three times, but may ignore the second or third rolls if we didn’t previously get doubles. Let the random variable Xi be the sum of the dice on the i­th roll, and let Ei be the event that the i­th roll is doubles. Write the expected number of squares a piece advances in these terms

Recitation 22 Solution. From the total expectation formula, we get Ex(advance)=Ex(X1| E1). Pr(E1 +Ex(X1+X2|E1∩E)P(E1∩E) +Ex(X1+X2+X3E1∩E2∩E3)Pr(E1∩E2∩E3) (0|E1∩E2∩E3)·Pr(E1∩E2∩E3) Then using linearity of (conditional) expectation, we refine this to Ex(advance) Ex(X1|E1. Pr(E1) +(Ex(X1E1∩E2)+Ex(X2|E1∩E2)·Pr(E1∩E) +(Ex(X1E1∩E2∩E3)+Ex(X2|E1∩E2∩E)+Ex(X8|E1∩E2∩E3) (E1∩E2∩E3) +0. Using mutual independence of the rolls, we simplify this to (X1l E1) Pr(E1) +(Ex(X1 l E1)+Ex(X2 1 E2). Pr (E1).Pr(E2 +(Ex(X1 E1)+Ex(X2 E2)+Ex(X3 E3).Pr(E1). Pr(E2). Pr(E3 (d)What is the expected number of squares that a piece advances in Monopoly? Solution. We plug the values from parts(a)and(b) into equation(4) 5 )=7·2+(7+7) 15 115 66 (7+7+7) 66

� � � � � � � � � � | | � � � � Recitation 22 8 Solution. From the total expectation formula, we get: Ex (advance) = Ex X1 | E1 Pr E1 � · � � � + Ex X1 + X2 | E1 ∩ E2 Pr E1 ∩ E2 � · � � � + Ex X1 + X2 + X3 | E1 ∩ E2 ∩ E3 · Pr E1 ∩ E2 ∩ E3 + Ex (0 | E1 ∩ E2 ∩ E3) · Pr (E1 ∩ E2 ∩ E3) Then using linearity of (conditional) expectation, we refine this to Ex (advance) = Ex X1 E1 Pr E1 � � | · � � �� � � + Ex X1 | E1 ∩ E2 + Ex X2 E1 ∩ E2 Pr E1 ∩ E2 � � � | � · � � �� + Ex X1 | E1 ∩ E2 ∩ E3 + Ex X2 E1 ∩ E2 ∩ E3 + Ex X3 E1 ∩ E2 ∩ E3 · Pr E1 ∩ E2 ∩ E3 + 0. Using mutual independence of the rolls, we simplify this to Ex (advance) = Ex X1 | E1 Pr E1 (4) � · � �� � � + Ex (X1 | E1) + Ex X2 E2 · Pr (E1) Pr E2 � | � · �� � � + Ex (X1 | E1) + Ex (X2 E2) + Ex X3 E3 | | · Pr (E1) · Pr (E2) Pr · E3 (d) What is the expected number of squares that a piece advances in Monopoly? Solution. We plug the values from parts (a) and (b) into equation (4): 5 1 5 1 1 5 Ex (advance) = 7 + (7 + 7) + (7 + 7 + 7) · 6 · 6 · 6 · 6 · 6 · 6 19 = 8 72