statistics

STAT 4382: Homework 8

Due by November 2, 2020

Yuly Koshevnik

First Name Last Name

Problem 1 2 3 4 5 6 7 Total

Points

Maximum 15 15 15 15 10 10 20 100

Please submit the title page with your homework!

Topics included into this assignment are:

• Finite-State Continuous Time Markov Chains

• Renewal Phenomena and related distributions

Please, read Sections 6.6, 7.1 and 7.3 from textbook. Alternatively, look at Notes, Part 7.

1

General Techniques and Notation

Recall that a notion of the Renewal applies to both the counting process,

N = {N(t) : t ≥ 0} ,

and to the sequence of waiting or arrival times,

Wk = min{t ≥ 0 : N(t) = k} , for k = 1, 2, . . . .

Our attention is focused on N being a Poisson process, for which we have already accrued necessary knowledge. The arrival times, {Wk : k ≥ 1}, are expressed as partial sums of sojourn or inter-arrival times,

Wk = k∑

j=1

Sj, for k = 1, 2, . . . ,

which are assumed to be independent and identically distributed, or i.i.d. with a common cumulative distribution function,

F(x) = P [Sj ≤ x] = 1 − exp(−λ ·x) for x ≥ 0.

Since N is a Poisson process with the rate λ > 0, inter-arrival times share the common exponential density

f(x) = λ · exp(−λ ·x), for x > 0,

and

E [Sj] = 1

λ for all j ≥ 1

You need to recognize that the kth arrival time, Wk, has Gamma distribution with density:

f(k)(w) = λk

(k − 1)! ·wk−1 · exp(−λw) for w > 0.

Another important detail is that event N(t) = k occurs when [Wk ≤ t < Wk+1], hence related probabilities can be expressed using Poisson distribution for N(t).

2

Problem 1: [15 points = 5 + 5 + 5]

Assume that customers’ inter-arrival times, {Sj : j ≥ 1}, at a car service station are independent expo- nentially distributed with the average, E [Sj] = 10 minutes. Time variable for N(t) is measured in hours, and t = 0 corresponds to the time when service station opens. A variable Wk is the exact time when the k

th customer arrived.

1. Derive conditional expectation of the arrival time for a third customer, given that W7 ≤ 2 < W8.

2. Evaluate conditional expectation of the arrival time for a third customer, given that the seventh arrival occurs at W7 = 2.

3. Find expected value of the ratio,

T = W7 −W3 W3

Solution

3

Problem 2 [15 points = 5 + 5 + 5]

Assume that customers’ inter-arrival times at a small car service are independent exponentially distributed random variables, {Sj : j ≥ 1} with expectation E [Sj] = 30 minutes. Measure a time variable, t, in hours and consider arrival times,

Wk = k∑

j=1

Sj for k ≥ 1.

Then introduce the counting process, N = {N(t) : t ≥ 0}, so that

[N(t) = k] ⇐⇒ [Wk ≤ t < Wk+1]

1. Find conditional expectation of the 6th customer’s arrival, given that the second customer arrived exactly two hours after the service was open.

2. Evaluate conditional expectation of the second arrival time, given that the sixth customer arrived exactly two hours after the service was open.

3. Find expected time between the second and sixth arrivals.

Solution

4

Problem 3 [15 points = 5 + 5 + 5]

Customers inter-arrival times, {Sj : j ≥ 1}, at a small car service center are independent exponentially distributed random variables with common expectation, E [Sj] = 15 minutes. As before, Wk denotes the arrival time of a kth customer.

1. Find expectation of a ratio, Q = W3/W4

2. Determine expected value of a ratio, (W4/W3)

3. Find expected value of the ratio, (S4/W3).

Solution

5

Problem 4 [15 points = 5 + 5 + 5]

Customers inter-arrival times at a small car service center are independent exponentially distributed ran- dom variables with common expectation, E [Sj] = 15 minutes. As before, Wk denotes the arrival time of a kth customer. Introduce an event B = [W5 ≤ 2 < W6] and random variable,

Y = 5∑

j=1

Wj

1. Evaluate probability of event B.

2. Derive conditional expectation, E [Y |B]

3. Find marginal expected value of Y

Solution

6

Problem 5 [10 points = 5 + 5]

Consider a process, N = {N(t) : t ≥ 0} defined using the independent exponentially distributed sojourn times, {Sj : j ≥ 1} with E [Sj] = 30 minutes. Measure time (t) in hours and consider arrival times for process N(t) as follows:

Wk = k∑

j=1

Sj

Introduce two events as follows:

A = [W2 ≤ 1 < W3] and B = [W5 ≤ 3 < W6]

1. Evaluate P [A |B]

2. Find P [B |A]

Solution

7

Problem 6 [10 points = 5 + 5]

Assume that inter-arrival times for a process, N = {N(t) : t ≥ 0} are independent exponentially distributed variables with E [Sj] = 15 minutes. Time variable, t is measured in hours. Consider sums,

Wk = k∑

j=1

Sj for k ≥ 1

such that [N(t) = k] ⇐⇒ [Wk ≤ t < Wk+1]. Find conditional expectations listed below.

1. E [N(1) |W3 ≤ 2 < W4]

2. E [W1 |W3 ≤ 2 < W4]

Solution

8

Problem 7 [20 points = 10 + 10]

A four-state continuous time Markov chain describes a telecommunications network health. A process, X = {X(t) : t ≥ 0} has states interpreted as follows. When X(t) = 0, it means that there is no failure at time t,, while three other states, {1, 2, 3} are failure indicators of three different types. Instant transition probabilities are described below under assumption that the time increment, h ↓ 0.

• P [X(t + h) = j |X(t) = 0] = λj ·h + o(h) for j = 1, 2, or 3.

• States {1, 2, 3} do not instantly communicate with each other, and restoration times are independent exponentially distributed, with parameters {µj : j = 1, 2, 3, hence P [X(t + h) = 0 |X(t) = j] = µj ·h + o(h) and instant transitions (i → j) are impossible when i 6= j, where i ∈{1, 2, 3} and j ∈{1, 2, 3}

• Failure and repair rates are given as: λ1 = 2, λ2 = 4, and λ3 = 10 per month, while repair rates are: µ1 = 40, µ2 = 40 and µ3 = 100 per month.

Network reliability is measured by the stationary probability that the process is at state 0.

1. Evaluate network reliability, π0, under stationary distribution

2. Technicians decided to combine all failures into one type, setting the failure rate as λ = λ1 + λ2 + λ3 and similarly aggregate restoration rates into a single one, µ = µ1 + µ2 + µ3. Then they consider a two-state Markov chain instead of the original one. Evaluate reliability they will obtain.

Solution

9