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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’s 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’s 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’s 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 \(A\) is \(m \times n\)and \(b\) is in \({\mathbb{R}^m}\). What has to be true about the two numbers rank \(\left[ {A\,\,\,{\rm{b}}} \right]\) and \({\rm{rank}}\,A\) in order for the equation \(Ax = b\) to be consistent?

(M) Let \({{\mathop{\rm a}\nolimits} _1},...,{{\mathop{\rm a}\nolimits} _5}\) denote the columns of the matrix \(A\), where \(A = \left( {\begin{array}{*{20}{c}}5&1&2&2&0\\3&3&2&{ - 1}&{ - 12}\\8&4&4&{ - 5}&{12}\\2&1&1&0&{ - 2}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}{{{\mathop{\rm a}\nolimits} _1}}&{{{\mathop{\rm a}\nolimits} _2}}&{{{\mathop{\rm a}\nolimits} _4}}\end{array}} \right)\)

  1. Explain why \({{\mathop{\rm a}\nolimits} _3}\) and \({{\mathop{\rm a}\nolimits} _5}\) are in the column space of \(B\).
  2. Find a set of vectors that spans \({\mathop{\rm Nul}\nolimits} A\).
  3. Let \(T:{\mathbb{R}^5} \to {\mathbb{R}^4}\) be defined by \(T\left( x \right) = A{\mathop{\rm x}\nolimits} \). Explain why \(T\) is neither one-to-one nor onto.

Suppose a nonhomogeneous system of six linear equations in eight unknowns has a solution, with two free variables. Is it possible to change some constants on the equations’ right sides to make the new system inconsistent? Explain.

Let \({M_{2 \times 2}}\) be the vector space of all \(2 \times 2\) matrices, and define \(T:{M_{2 \times 2}} \to {M_{2 \times 2}}\) by \(T\left( A \right) = A + {A^T}\), where \(A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\).

  1. Show that \(T\)is a linear transformation.
  2. Let \(B\) be any element of \({M_{2 \times 2}}\) such that \({B^T} = B\). Find an \(A\) in \({M_{2 \times 2}}\) such that \(T\left( A \right) = B\).
  3. Show that the range of \(T\) is the set of \(B\) in \({M_{2 \times 2}}\) with the property that \({B^T} = B\).
  4. Describe the kernel of \(T\).

(M) Let \(H = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2}} \right\}\) and \(K = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\), where

\({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}5\\3\\8\end{array}} \right),{{\mathop{\rm v}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}1\\3\\4\end{array}} \right),{{\mathop{\rm v}\nolimits} _3} = \left( {\begin{array}{*{20}{c}}2\\{ - 1}\\5\end{array}} \right),{{\mathop{\rm v}\nolimits} _4} = \left( {\begin{array}{*{20}{c}}0\\{ - 12}\\{ - 28}\end{array}} \right)\)

Then \(H\) and \(K\) are subspaces of \({\mathbb{R}^3}\). In fact, \(H\) and \(K\) are planes in \({\mathbb{R}^3}\) through the origin, and they intersect in a line through 0. Find a nonzero vector w that generates that line. (Hint: w can be written as \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2}\) and also as \({c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\). To build w, solve the equation \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2} = {c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\) for the unknown \({c_j}'{\mathop{\rm s}\nolimits} \).)
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