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Chapter 4: Q3E (page 191)

Question 3: On any given day, a student is either healthy or ill. Of the students who are healthy today, 95% will be healthy tomorrow. Of the students who are ill today, 55% will still be ill tomorrow.

a. What is the stochastic matrix for this situation?

b. Suppose 20% of the students are ill on Monday. What fraction or percentage of the students are likely to be ill on Tuesday? On Wednesday?

c. If a student is well today, what is the probability that he or she will be well two days from now?

Short Answer

Expert verified
  1. The stochastic matrix is \(P = \left( {\begin{array}{*{20}{c}}{.95}&{.45}\\{.05}&{.55}\end{array}} \right)\).
  2. 15% of the students are ill on Tuesday, and 12.5% of the students are ill on Wednesday.
  1. The probability that the students will be well two days from now is 0.925.

Step by step solution

01

Determine the stochastic matrix

a)

Let \(H\) represent a healthy student and \(I\) represent an ill student.

The following table represents the student鈥檚 conditions.

From:

To:

H

I

.95

.45

H

.05

.55

I

Therefore, the stochastic matrix is \(P = \left( {\begin{array}{*{20}{c}}{.95}&{.45}\\{.05}&{.55}\end{array}} \right)\).

02

Determine the percentage of students who are likely to be ill on Tuesday

b)

Theorem 18states that if \(P\) is a \(n \times n\) regular stochastic matrix,then \(P\) has a unique steady-state vector \({\mathop{\rm q}\nolimits} \).Further, if \({{\mathop{\rm x}\nolimits} _0}\) is any initial state and \({{\mathop{\rm x}\nolimits} _{k + 1}} = P{{\mathop{\rm x}\nolimits} _k}\) for \(k = 0,1,2,....\) then the Markov chain \(\left\{ {{{\mathop{\rm x}\nolimits} _k}} \right\}\) converges to \({\mathop{\rm q}\nolimits} \)as \(k \to \infty \).

The initial state vector is \({{\mathop{\rm x}\nolimits} _0} = \left( {\begin{array}{*{20}{c}}{.8}\\{.2}\end{array}} \right)\) because 20% of the students are ill on Monday.

Compute \({{\mathop{\rm x}\nolimits} _1}\) to find Tuesday鈥檚 percentage.

\(\begin{array}{c}{{\mathop{\rm x}\nolimits} _1} = P{{\mathop{\rm x}\nolimits} _0}\\ = \left( {\begin{array}{*{20}{c}}{.95}&{.45}\\{.05}&{.55}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{.8}\\{.2}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{0.76 + 0.09}\\{0.04 + 0.11}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{0.85}\\{0.15}\end{array}} \right)\end{array}\)

Thus, 15% of the students are ill on Tuesday.

03

Determine the percentage of students likely to be ill on Wednesday

Compute \({{\mathop{\rm x}\nolimits} _1}\) to find Tuesday鈥檚 percentage.

\(\begin{array}{c}{{\mathop{\rm x}\nolimits} _2} = P{{\mathop{\rm x}\nolimits} _1}\\ = \left( {\begin{array}{*{20}{c}}{.95}&{.45}\\{.05}&{.55}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{0.85}\\{0.15}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{0.8075 + 0.0675}\\{0.0425 + 0.0825}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{0.875}\\{0.125}\end{array}} \right)\end{array}\)

Hence, 12.5% of the students are ill on Wednesday.

04

Determine the probability that the student will be well in two days

c)

Suppose the student is well today, then the initial state vector is \({{\mathop{\rm x}\nolimits} _0} = \left( {\begin{array}{*{20}{c}}1\\0\end{array}} \right)\).

Compute \({{\mathop{\rm x}\nolimits} _2}\) as shown below:

\(\begin{array}{c}{{\mathop{\rm x}\nolimits} _1} = P{{\mathop{\rm x}\nolimits} _0}\\ = \left( {\begin{array}{*{20}{c}}{.95}&{.45}\\{.05}&{.55}\end{array}} \right)\left( {\begin{array}{*{20}{c}}1\\0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{0.95 + 0}\\{0.05 + 0}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{0.95}\\{0.05}\end{array}} \right)\end{array}\)

\(\begin{array}{c}{{\mathop{\rm x}\nolimits} _2} = P{{\mathop{\rm x}\nolimits} _1}\\ = \left( {\begin{array}{*{20}{c}}{.95}&{.45}\\{.05}&{.55}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{0.95}\\{0.05}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{0.9025 + 0.0225}\\{0.0475 + 0.0275}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{0.925}\\{0.075}\end{array}} \right)\end{array}\)

Thus, the probability that the students will be well two days from now is 0.925.

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Most popular questions from this chapter

Suppose a \({\bf{4}} \times {\bf{7}}\) matrix A has four pivot columns. Is \({\bf{Col}}\,A = {\mathbb{R}^{\bf{4}}}\)? Is \({\bf{Nul}}\,A = {\mathbb{R}^{\bf{3}}}\)? Explain your answers.

Question 18: Suppose A is a \(4 \times 4\) matrix and B is a \(4 \times 2\) matrix, and let \({{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}\) represent a sequence of input vectors in \({\mathbb{R}^2}\).

  1. Set \({{\mathop{\rm x}\nolimits} _0} = 0\), compute \({{\mathop{\rm x}\nolimits} _1},...,{{\mathop{\rm x}\nolimits} _4}\) from equation (1), and write a formula for \({{\mathop{\rm x}\nolimits} _4}\) involving the controllability matrix \(M\) appearing in equation (2). (Note: The matrix \(M\) is constructed as a partitioned matrix. Its overall size here is \(4 \times 8\).)
  2. Suppose \(\left( {A,B} \right)\) is controllable and v is any vector in \({\mathbb{R}^4}\). Explain why there exists a control sequence \({{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}\) in \({\mathbb{R}^2}\) such that \({{\mathop{\rm x}\nolimits} _4} = {\mathop{\rm v}\nolimits} \).

Exercises 23-26 concern a vector space V, a basis \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\) and the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\).

Show that a subset \(\left\{ {{{\bf{u}}_1},...,{{\bf{u}}_p}} \right\}\) in V is linearly independent if and only if the set of coordinate vectors \(\left\{ {{{\left( {{{\bf{u}}_{\bf{1}}}} \right)}_B},.....,{{\left( {{{\bf{u}}_p}} \right)}_B}} \right\}\) is linearly independent in \({\mathbb{R}^n}\)(Hint: Since the coordinate mapping is one-to-one, the following equations have the same solutions, \({c_{\bf{1}}}\),鈥�.,\({c_p}\))

\({c_{\bf{1}}}{{\bf{u}}_{\bf{1}}} + ..... + {c_p}{{\bf{u}}_p} = {\bf{0}}\) The zero vector V

\({\left( {{c_{\bf{1}}}{{\bf{u}}_{\bf{1}}} + ..... + {c_p}{{\bf{u}}_p}} \right)_B} = {\left( {\bf{0}} \right)_B}\) The zero vector in \({\mathbb{R}^n}\)a

Exercises 23-26 concern a vector space V, a basis \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\)and the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\).

Given vectors, \({u_{\bf{1}}}\),鈥�.,\({u_p}\) and w in V, show that w is a linear combination of \({u_{\bf{1}}}\),鈥�.,\({u_p}\) if and only if \({\left( w \right)_B}\) is a linear combination of vectors \({\left( {{{\bf{u}}_{\bf{1}}}} \right)_B}\),鈥�.,\({\left( {{{\bf{u}}_p}} \right)_B}\).

Question: Exercises 12-17 develop properties of rank that are sometimes needed in applications. Assume the matrix \(A\) is \(m \times n\).

17. A submatrix of a matrix A is any matrix that results from deleting some (or no) rows and/or columns of A. It can be shown that A has rank \(r\) if and only if A contains an invertible \(r \times r\) submatrix and no longer square submatrix is invertible. Demonstrate part of this statement by explaining (a) why an \(m \times n\) matrix A of rank \(r\) has an \(m \times r\) submatrix \({A_1}\) of rank \(r\), and (b) why \({A_1}\) has an invertible \(r \times r\) submatrix \({A_2}\).

The concept of rank plays an important role in the design of engineering control systems, such as the space shuttle system mentioned in this chapter鈥檚 introductory example. A state-space model of a control system includes a difference equation of the form

\({{\mathop{\rm x}\nolimits} _{k + 1}} = A{{\mathop{\rm x}\nolimits} _k} + B{{\mathop{\rm u}\nolimits} _k}\)for \(k = 0,1,....\) (1)

Where \(A\) is \(n \times n\), \(B\) is \(n \times m\), \(\left\{ {{{\mathop{\rm x}\nolimits} _k}} \right\}\) is a sequence of 鈥渟tate vectors鈥� in \({\mathbb{R}^n}\) that describe the state of the system at discrete times, and \(\left\{ {{{\mathop{\rm u}\nolimits} _k}} \right\}\) is a control, or input, sequence. The pair \(\left( {A,B} \right)\) is said to be controllable if

\({\mathop{\rm rank}\nolimits} \left( {\begin{array}{*{20}{c}}B&{AB}&{{A^2}B}& \cdots &{{A^{n - 1}}B}\end{array}} \right) = n\) (2)

The matrix that appears in (2) is called the controllability matrix for the system. If \(\left( {A,B} \right)\) is controllable, then the system can be controlled, or driven from the state 0 to any specified state \({\mathop{\rm v}\nolimits} \) (in \({\mathbb{R}^n}\)) in at most \(n\) steps, simply by choosing an appropriate control sequence in \({\mathbb{R}^m}\). This fact is illustrated in Exercise 18 for \(n = 4\) and \(m = 2\). For a further discussion of controllability, see this text鈥檚 website (Case study for Chapter 4).
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