Stat 330 (Spring 2015): Homework 8
Due: March 27, 2015
Show all of your work, and please staple your assignment if you use more than one sheet. Write your name,
the course number and the section on every sheet. Problems marked with * will be graded and one additional
randomly chosen problem will be graded. Show all work to be considered for partial credit.
1. An auto insurance company classifies its customers in three categories: poor, satisfactory and preferred.
No one moves from poor to preferred or from preferred to poor in one year. 40% of the customers in the
poor category become satisfactory, 30% of those in the satisfactory category moves to preferred, while
10% become poor; 20% of those in the preferred category are downgraded to satisfactory. We will model
this process with a Markov chain with three states corresponding to the three categories.
(a) Write the 1-step transition matrix for the model.
(b) What is the probability that a customer in the satisfactory category remain so next year?
(c) Customers in which category are more likely to be downgraded next year? Why?
(d) What is the probability that a customer in the preferred category still remains in the same category
after 2 years?
(e) What is the probability that a customer in the preferred category remains in the same category for
2 years in a row, i.e. he was preferred in both years? (This probability is different from the one
calculated in part (d).)
(f) What is the limiting fraction of customers in each of these categories? (Find the steady-state
distribution of this Markov chain.)
Answer:
(a)
P oor
P = Satisf actory
P ref erred
P oor
0.6
0.1
0
Satisf actory
0.4
0.6
0.2
P ref erred
!
0
0.3
0.8
(b) 0.6
(c) Only those in preferred and satisfactory categories may be downgraded. Since .2 > .1, those in
preferred category are more likely to be downgraded to satisfactory.
(d) This is the 2-step transition probability starting from Preferred and ending at Preferred, i.e. the
bottom right element. The 2-step probabilty matrix is obtained using (P (2) = P 2 = P · P ) as:
P oor
P 2 = Satisf actory
P ref erred
P oor
0.40
0.12
0.02
Satisf actory
0.48
0.46
0.28
P ref erred
0.12 !
,
0.42
0.70
Thus the required probability is 0.70.
(e) In order for the customer to be Preferred, the Markov chain must go from Preferred to Preferred to
Preferred Since each of these transitions has probability 0.8 and they are independent of each other
(by Markov property), then the probability of the set of transitions is 0.82 = 0.64.
(f) The steady state distribution of the chain can be found by solving π = πP or, by multiplying P by itself until all rows are the same. Solving the equations we get that the steady-state
is π = (1/11, 4/11, 6/11). Using the second strategy, after many multiplications, we get π =
(0.0909091, 0.3636364, 0.5454544).
2. * Jobs are sent to a server at a rate of 2 jobs per minute. We will model job arrivals using a homogenous
Poisson process.
(a) What is the probability of receiving more than 3 jobs in a period of one minute?
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Stat 330 (Spring 2015): Homework 8
Due: March 27, 2015
(b) What is the probability of receiving more than 30 jobs in a period of 10 minutes?
(c) What is the expected value and the variance of interarrival times?
(d) Compute the probability that the next job does not arrive during the next 30 seconds?
(e) Compute the probability that the time till the fourth job arrives exceeds 40 seconds?
Answer:
(a) Let X(t) be the number of calls in t minutes, then X(t) ∼ P o(λt) where λ = 2 jobs per minute.
P [X(1) > 3] = 1 − P [X(1) ≤ 3] = 1 − 0.8571 = 0.1429
(b)
P [X(10) > 30] = 1 − P [X(10) ≤ 30] = 1 − 0.9865 = 0.0135
iid
(c) Let Ij = be the inter-arrival time for the jth job. Then Ij ∼ Exp(2) and
E(Ij ) =
1
= 0.5 min
λ
V ar(Ij ) =
1
= 0.25 min2
λ2
(d) Since 30 seconds equals 0.5 minutes, we have
P [I1 > 0.5 min] = e−λ(0.5) = e−1 = 0.3679
(e) Let O4 = (I1 + I2 + I3 + I4 ) be the time until the fourth job, then O4 ∼ Gamma(4, 2). Now
40 seconds is 2/3 of a minute. We could get this answer directly by using the gamma cumulative
distribution function to find P (O4 > 2/3) = 1 − P (O4 ≤ 2/3) = 0.9535.
Alternatively, we could use the Poisson-Gamma formula by recognizing the only way the 4th job
can take longer than 40 seconds is if, only 0, 1, 2, or 3 jobs arrive in the first 40 seconds (2/3 of a
minute). So remembering that X(2/3) ∼ P o(2 × 2/3), we can calculate
P (O4 > 40/60) = P (X(2/3) ≤ 3)
= P (X(2/3) = 0) + P (X(2/3) = 1) + P (X(2/3) = 2) + P (X(2/3) = 3)
= 0.9539
3. Suppose that the telephone calls arriving at the switchboard of a small corporation follows a Poisson
process. Let X(t) be the number of calls received by time t that day, with X(0) = 0. Suppose that the
intensity parameter is λ = 0.5 calls per minute. Assume that the number of calls arriving during any
period has a Poisson distribution.
(a) What is the probability that no calls will arrive in a 5-minute period?
(b) What is the expected amount of time between the 10th and 11th callers?
(c) On average, how much time is needed to get their 100th caller of the day?
(d) What is the probability that the 100th caller arrives by their lunch break at t = 240?
Answer:
(a) As X(t) values follows a Poisson(λ t) probability distribution, then X(5) ∼ P oisson(λ · 5 = 2.5), so
P (X(5) = 0) = e−2.5 = .0821.
(b) Let I be the interarrival time between any two calls, and I ∼ Exp(λ), so E(I) =
1
λ
= 2 minutes.
(c) Let T be the time needed until the 100th caller. Then T ∼ Gamma(100, 0.5) so E(T ) =
minutes.
100
0.5
= 200
(d) Again let T be the time until the 100th call. As T ∼ Gamma(100, 0.5), P (T < 240) = 0.9721.
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Stat 330 (Spring 2015): Homework 8
Due: March 27, 2015
4. (Baron’s book): 6.3
Answer:
5. * (Baron’s book): 6.4
Answer:
6. (Baron’s book): 6.7
Answer:
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Stat 330 (Spring 2015): Homework 8
Due: March 27, 2015
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