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?

a)

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

The following table represents the student’s conditions.

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

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.

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.

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.