Massachusetts Institute of Technology

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1 6.04/6.4: Probabilistic Systems Analysis Fall 00 Quiz Solutions: October, 00 Problem.. 0 points Let R i be the amount of time Stephen spends at the ith red light. R i is a Bernoulli random variable with p /. The PMF for R i is: /, if r 0, P Ri r /, if r, 0, otherwise. The expectation and variance for R i are: E[R i ] p, varr i p p 9 Let T S be the total length of time of Stephen s commute in minutes. Then, T S 8 + R i. T S is a shifted binomial with n trials and p /. The PMF for T S is then: k 8 k, if k {8, 9, 0,,, }, P TS k k 8 0, otherwise. The expectation and variance for T S are: [ ] E[T S ] E 8 + i i R i 9 vart S var points Let N be the number of red lights Stephen encountered on his commute. Given that T S 9, then N 0 or N The unconditional probability of N 0 is PN 0. The unconditional probability of N is PN 4. i R i Page of

2 6.04/6.4: Probabilistic Systems Analysis Fall 00 To find the conditional expectation, the following conditional PDF is calculated:, if n 0, + 4 /7, if n 0, P N TS 9n T S 9 4 /7, if n,, if n, + 4 0, otherwise. 0, otherwise, Therefore, E[N T S 9] 7. 0 points Given that the last red light encountered by Stephen was the fourth light, R 4 and R 0. We are asked to compute varn {R 4 } {R 0}. Therefore, varn {R 4 } {R 0} varr + R + R + R 4 + R {R 4 } {R 0} varr + R + R {R 4 } {R 0} varr + R + R + varr + R + R varr points Under the given condition, the discrete uniform law can be used to compute the probability of interest. There are ways that Stephen can encounter a total of three red lights. There are ways that two out of the first three lights were red. This leaves one additional red light out of the last two lights and there are possible ways that this event can occur. Putting it all together, Ptwo of first three lights were red total of three red lights. points Let T J be the total length of time of Jon s commute in minutes. The PMF of Jon s commute is: {, if l {0,,, }, P TJ l 4 0, otherwise points Let A be the event that Jon arrives at work in 0 minutes and let B be the event that exactly one person arrives in 0 minutes. PA B PA B PB P{T J 0} {T S 0} P{T J 0} {T S 0} + P{T J 0} {T S 0} PT J 0PT S 0 PT J 0PT S 0 + PT J 0PT S 0 Page of

3 6.04/6.4: Probabilistic Systems Analysis Fall 00 Jon arrives at work in 0 minutes or T J 0 if he does not have to wait for the train at the station or X 0. The probability of this event occurring is: PT J 0 PX 0 4 Stephen arrives at work in 0 minutes if he encounters red lights. The probability of this event is a binomial probability: PT S 0 Thus, 4 PA B points The probability of interest is PT S T J. This can be calculated using the total probability theorem by conditioning on the length of Jon s commute or Jon s wait at the station. If Jon s commute is 0 minutes or X 0, then Stephen can encounter up to red lights to satisfy T S T J. Similarly if Jon s commute is minutes or X, Stephen can encounter up to red lights and so on. PT S T J PT S T J X xpx x x0 +x k k 4 k x0 k An alternative approach follows. We first compute the joint PMF of the commute times of Stephen and Jon P TS,T J k, l. Because of independence, P TS,T J k, l P TS kp TJ l. Therefore, PT S T J PT S 8 + PT S 9 + PT S 0 + P{T S } {T J } + P{T S } {T J } + P{T S } {T J } points We express the conditional probability as such: P{X } {T S T J } PX T S T J PT S T J Page of

4 6.04/6.4: Probabilistic Systems Analysis Fall 00 If Jon waited minutes at the train, his commute was minutes and Stephen s commute takes at most as long as Jon s commute since the longest possible commute for Stephen is minutes. Therefore, the numerator in the previous expression is equal to PX 4. The denominator was computed in the previous part. Problem. PX T S T J +x k k k x0 k points Always True. We need to show that PA B c PAPB c. We start with expressing PA as PA B + PA B c. Therefore, which shows that A and B c are independent. PA B c PA PA B PA PAPB PA PB PAPB c,. 0 points Not Always True. Using the diagram below, let C A B and let PA > PC and let PB > PC. The conditional probability PA B C Furthermore, PA C and PB C Since PA B C PA CPB C, A and B are conditionally independent given a third event C. Given C c, A and B are disjoint which means that A and B are not independent. The following is an alternative counterexample. Imagine having coins with the following probability of heads: p /, p / and p /, respectively. Each coin has equal probability of being selected. Let C be the event that you select the coin with p /. Let C c be the event that you choose one of the other two coins. Let A be the event that the first coin toss results in heads. Let B be the event that the second coin toss results in heads. For a given coin, the tosses are independent such that: PB A C PB C. Page 4 of

5 6.04/6.4: Probabilistic Systems Analysis Fall 00 Given C c, A and B are not independent since we can have either the p / coin or the p / coin. Knowing A changes our beliefs of the result of the second coin toss. However, PB A C c B A C c A C c As shown, PB A C c PB C c. PB C c PB C c PC c + 0 points Always True. Using independence of X and Y, varx + Y varx + vary. Since variance is always non-negative, varx + vary varx. Page of

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