Tenth Problem Assignment

8/16/2019 Tenth Problem Assignment

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1 Due on April 6, 2007

Tenth Problem Assignment

EECS 401

Due on April 6, 2007

PROBLEM 1 (28 points) Dave is taking a multiple-choice exam. You may assume that

the number of questions is infinite. Simultaneously, but independently, his conscious and

subconscious facilities are generating answers for him, each in a Poisson manner. (His

conscious and subconscious are always working on different questions.)

Average rate at which conscious responses are generated = λc responses/min

Average rate at which subconscious responses are generated = λs responses/min

Each conscious response is an independent Bernoulli trial with probability pc of beingcorrect. Similarly, each subconscious response is an independent Bernoulli trial with

probability ps of being correct.

Dave responds only once to eachquestion, and you can assume that his time for recording

these conscious and subconscious responses is negligible.

(a) Determine pK(k ), the probability mass function for the number of conscious responses

Dave makes in an interval of T minutes.

Solution The number of conscious responses is a Poisson random variable with parameter

λ = λcT . Thus,

pk(k ) = (λcT )k

e−λcT

k !

(b) If we pick any question to which Dave has responded, what is the probability that

his answer to that question:

(a) Represents a conscious response

Solution The probability of a conscious response is

λc

λc + λs

(b) Represents a subconscious response

Solution The probability of a subconscious response is

λs

λc + λs


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Solutions Tenth Problem Assignment


(c) If we pick an interval of T minutes, what is the probability that in that interval Dave

will make exactly r conscious responses and exactly s subconscious responses.

Solution The probability of making r conscious and s subconscious responsed in T

minutes is

(λcT )re−λcT

r! ×

(λsT )se−λsT

s!

(d) Determine the moment generating function for the probability density function for

random variable X, where X is the time from the start of the exam until Dave makes

his first conscious response which is preceded by at least one subconscious response.

Solution Let Y s denote the random variable when the first subconscious response is

generated and Y c denote the random time starting for the first subconscious responsewhen the first conscious response is generated. Then

X = Y s + Y c

Thus,

MX(s) = MY s(s) · MY c(s) = λs

s − λs·

λc

s − λc

(e) Determine the probability mass function for the total number of responses up to and

including his third conscious response.

Solution Consider the arrivals of the merged process. Each arrival belongs to theconscious process with probability λc/(λc + λs). Thus, if we only count the arrivals,

then the arrivals from the conscious process form a Bernoulli process with parameter

p = λc/(λc + λs). Then the number of responses (‘trials’) up to and including his third

conscious response (‘success’) has Pascal distribution with n = 3, that is

pK(k ) =

k − 1

2

λc

λc + λs

3 λs

λc + λs

k−3

(f) The papers are to be collected as soon as Dave has completed exactly N responses.

Determine:

Solution The number of responses are generated by a Poisson process with rate λc+ λs.The correct responses are generated by a Poisson process with rate pcλc + psλs. Thus

each response is correct with probability ( pcλc + psλs)/(λc + λs).

(i) The expected number of questions he will answer correctly


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Solution The expected number of questions answered correctly is the mean of a binomial

random variable and equal to

N pcλc + psλs

λc + λs

(ii) The probability mass function for L, the number of questions he answers correctly

Solution This probability distribution of L, the number of questions answered correctly

is Binomial

N, ( pcλc + psλs)/(λc + λs)

, that is

pL(l) =

N

l

pcλc + psλs

λc + λs

l 1 −

pcλc + psλs

λc + λs

N−l

(g) Repeat part (f) for the case in which the exam papers are to be collected at the end of

a fixed interval T minutes.

Solution The number of correct responses in a fixed interval T is a Poisson process with

parameter ( pcλc + psλs)T . So the PMF of the number of correct responses is

pL(l) =

( pcλc + psλs)T

le−(pcλc+psλs)T

l!

with mean ( pcλc + psλs)T .

PROBLEM 2 (16 points) All ships travel at the same speed through a wide canal. Eastbound

ship arrivals at the canal are a Poisson process with an average arrival rate λE ships per

day. Westbound ships arrive as an independent Poisson process with average arrivalrate λW per day. An indicator at a point in the canal is always pointing in the direction

of travel of the most recent ship to pass it. Each ship takes t days to traverse the length

of the canal.

(a) Given that the pointer is pointing west:

(i) What is the probability that the next ship to pass it will be westbound?

Solution The direction of the next ship does not depend on the previous ships. There-

fore, this is just the probability λW /(λE + λW ) that the next ship is westbound.

(ii) What is the PDF for the remaining time until the pointer changes direction?

Solution The pointer will change directions on the next arrival of an eastbound ship. Thetime until an eastbound ship arrives is an exponential random variable with parameter

λE, and its PDF is

λEe−λEt, t 0


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(b) What is the probability that an eastbound ship will see no westbound ships during

its eastward journey through the canal?

Solution Suppose that an eastbound ship enters the canal at time t0. This ship will meet

any westbound ship that entered the canal between times t0 − t and t0 + t. Thus, the

desired probability is the probability that there are no westbound ship arrivals during an

interval of length 2t, and using the Poisson PMF, it is equal to

e−λW 2t

(c) We begin observing at an arbitrary time. Let V be the time we have to continue

observing until we see the seventh eastbound ship. Determine the PDF for V .

Solution The time until we see the seventh eastbound ship is an Erlang random variable

of order 7, with parameter λE, of the form

λ7Et6e−λEt

6!

PROBLEM 3 (9 points) The number of customers N who shop at a supermarket in a day

is Poisson with parameter λ. The number of items purchased by any customer is Pois-

son with parameter µ , and the number of items purchased by different customers are

independent of each other.

(a) Assume that the number of items purchased by each customer is independent of N.

Find E[S ] and V ar(S), where S is the total number of items sold.

Solution Let Xi be the number of items bought by the ith customer. Then

S =

N

i=0

Xi

which is the random sum of a random number of random variables. Thus,

E[S ] = E[N ]E[X ] = λµ

and

Var(S) = E[N ] Var(X) +

E[X ]2

Var(N) = λµ + λµ 2

(b) The supermarket has two advertising strategies, one can increase λ by 10% and the

other increases µ by 10%. What are the effects of these two strategies on the mean

and variance of S? Which is the better strategy?


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Solution The better strategy will be one which keeps the variance low.

(i) Increase µ by 10%. The variance becomes 1.1λµ + 1.21λµ 2.

(ii) Increase λ by 10%. The variance becomes 1.1λµ + 1.1λµ 2.Thus we have to compare

1.1λµ + 1.21λµ 2 ≷ 1.1λµ + 1.1λµ 2

The R.H.S. is smaller so we choose option (b), i.e., increase λ by 10%.

(c) Because of congestion, when there are more customers around, the amount of time

each customer spends in a store tends to be shorter and hence they will more likely

buy fewer items. To incorporate that, we can revise the above model so that where

there are n customers, µ = c/n, where c is some constant.

(i) Is the number of items bought by a customer independent of N?Solution No. The rate at which each customer buys items depends on the number of

customers in the store. So, the number of items bought is dependent on N.

(ii) Find E[S ] and V ar(S) in this new model.

Solution

E[S ] = E

E[S|N ]

= E

E[

n

i=0

Xi|N ]

= E

N ·c

N

= c.

Further, as conditioned on N, Xis are independent, we have

Var(S|N = n) = V ar ni=0

Xi|N = n

=

n

i=0

Var(Xi|N = n) = n ·c

n = c

Thus,

Var(S) = E

Var(S|N)

+ V ar

E[S|N ]

− = E[c ] + V ar(c) = c

Notice that the mean and variance are same. One would suspect that it likely that S is a

Poisson random variable. In fat it is easy to check this fact by evaluating the transform of

S.

PROBLEM 4 (18 points) The Markov chain with transition probabilities listed below is in

state 3 immediately before the first trial.

p1,1 = p2,2 = 0.4, p1,2 = p2,1 = 0.6, p3,3 = 0.2, p3,4 = 0.3, p4,5 = p5,6 = p6,4 = 1.0


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(a) Draw the state-transition diagram for this Markov Chain. Indicate which states, if

any, are recurrent, transient, and periodic.

Solution

1 2 3 4 5 60.4

. .

0.6

0.6

0.5 0.3 1 1

1

Recurrent States: 1,2,4, 5, 6

Transient States: 3

Periodic States: 4, 5, 6

(b) Find the probability that the process is in state 3 after n trials.

Solution If the process leaves state 3

, it can never return back to it. Thus the probabilitythat the process is in state 3 is the same as the probability that the process remains in state

3 for all times until n. That is, the probability is 0.2n.

(c) Find the expected number of trials up to and including the trial on which the process

leaves state 3.

Solution Let N be the trial on which the process leaves state 3. From the previous part,

we know that N is a geometric random variable with success rate p = 0.8 (because given

that we are in state 3, we will leave with probability 0.8). Thus, we have for the expected

value of N:

E [N ] =

∞

n=1

n0.2n−10.8 = 5 4

(d) Find the probability that the process never enters state 1.

Solution The process cannot stay in state 3 forever. At some finite time, it will make

a transition to either state 2 or 4. If the process jumps to state 2, it cannot stay in state 2

forever and at some finite time it will make a transition to state 1. However, if the process

makes a transition from state 3 to 4, it can never return to state 1. Thus the probability of

never entering state 1 is the same as the probability of jumping from state 3 to 4 (rather

than state 2). That is, we have:

0.30.3 + 0.5

= 38

(e) Find the probability that the process is in state 4 after 10 trials.


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Solution The process will be in state 4 after 10 trials if and only if makes a jump from

state 3 to 4 in trials 1, 4, 7 or 10. The probability of this happening is:

0.3 + (0.2)3 + (0.2)6 + (0.2)9 ∗ (0.3) = 0.31 − (0.2)12

1 − (0.2)3 = 0.3024

(f) Given that the process is in state 4 after 10 trials, find the probability that the process

was in state 4 after the first trial.

Solution Let A be the event that the process is in state 4 after 10 trials and B be the event

that the process was in state 4 after the first trial. Observe that B ⊂ A. Thus,

P(B|A) = P(A ∩ B

P(A) =

P(B)

P(A) =

0.3

0.3024 = 0.992

PROBLEM 5 (20 points)

(a) Buses depart from Ann Arbor to Detroit every hour on the hour. Passengers arrive

according to a Poisson process of rate λ per hour. Find the expected number of

passengers on a bus. (Ignore issues of limited seating.)

Solution The expected number of passengers on a bus is the expected number of arrivals

of a Poisson process of rate λ per hour in an hour, hence equal to λ.

(b) Now suppose that the buses are no longer operating on a deterministic schedule,

but rather their interdeparture times are independent and exponentially distributed

with rate µ per hour. Find the PMF for the number of buses arriving in one hour.

Solution The interdeparture time between the buses is exponential process with rate µ ,

hence the departure process of the buses is a Poisson process with rate µ . Thus the PMF

for the number of buses arriving in one hour is

pK(k ) = µ ke−µ

k !

(c) Let us define an “event” at the bus stop to be either the arrival of a passenger, or the

departure of a bus. With the same assumptions as in part 2 above, find the expected

number of “events” that occur in one hour.

Solution The “event” process is the merged process of two Poisson processes, hence a

Poisson process with rate λ + µ per hour. Thus the expected number of “events” in an

hour is λ + µ .


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(d) If a passenger arrives at thebus stop andsees2λ people waiting, find his/her expected

time to wait until the next bus.


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Solution The interarrival time between the buses is an exponential process, and hence

memoryless. The fact that there are 2λ people waiting gives some information about the

past of the process. But as the arrival process is memoryless, this does not convey anyinformation about the future of the process and hence the waiting time is also exponential

with rate µ . Thus, the expected waiting time is 1/µ .

(e) Find the PMF for the number of people on a bus.

Solution We are interested only in the number of passengers who arrive between the

arrivals of buses. Suppose we concentrate only on arrivals and consider the arrival of

buses as sucesses and the arrival of a passenger as faliure. Thus, we are interested in

the number of failures between two successes of a Bernoulli process. This has a shifted

geometric distribution given by

pN(n) =

λλ + µ

n

µ λ + µ

PROBLEM 6 (9 points) For a series of dependent trials, the probability of success on any

given trial is given by (k + 1)/(k + 3), where k is the number of successes in the previous

three trials. Define a state description and a set of transition probabilities which allow

this process to be described as a Markov chain. Draw the state transition diagram. Try

to use the smallest possible number of states.

Solution Let the outcome of the previous three trials be the “state”. Then

000 001 100 010 101 011 110 1112

3

2

3

1

3

1

2

1

2

1

2

1

2

1

2

3

5

2

5

35

1

2

1

2

3

52

5

1

3

Tenth Problem Assignment

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