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Chapter 4: Q2E (page 191)

Question 2: A laboratory animal may eat any one of three foods each day. Laboratory records show that if the animal chooses one food on one trial, it will chose the same food on the next trial with a probability of 50%, and it will choose the other foods on the next trial with equal probabilities of 25%.

a. What is the stochastic matrix for this situation?

b. If the animal chooses food #1 on an initial trial, what is the probability that it will choose food #2 on the second trial after the initial trial?

Short Answer

Expert verified
  1. The stochastic matrix is \(P = \left( {\begin{array}{*{20}{c}}{.5}&{.25}&{.25}\\{.25}&{.5}&{.25}\\{.25}&{.25}&{.5}\end{array}} \right)\).
  2. The probability that the animal will choose food #2 after the initial trial is 0.3125.

Step by step solution

01

Determine the stochastic matrix

a)

Let the food be labeled 1, 2, and 3.

The following table represents the animal's behavior.

From:

To:

1

2

3

.5

.25

.25

.25

.5

.25

.25

.25

.5

1

2

3

From this, the stochastic matrix is \(P = \left( {\begin{array}{*{20}{c}}{.5}&{.25}&{.25}\\{.25}&{.5}&{.25}\\{.25}&{.25}&{.5}\end{array}} \right)\).

02

Determine the probability that will choose food #2 after the initial trial

b)

After the initial trial, there are two trials. The initial state vector is \({{\mathop{\rm x}\nolimits} _0} = \left( {\begin{array}{*{20}{c}}1\\0\\0\end{array}} \right)\).

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

\(\begin{array}{c}{{\mathop{\rm x}\nolimits} _1} = P{{\mathop{\rm x}\nolimits} _0}\\ = \left( {\begin{array}{*{20}{c}}{.5}&{.25}&{.25}\\{.25}&{.5}&{.25}\\{.25}&{.25}&{.5}\end{array}} \right)\left( {\begin{array}{*{20}{c}}1\\0\\0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{0.5 + 0 + 0}\\{0.25 + 0 + 0}\\{0.25 + 0 + 0}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{0.5}\\{0.25}\\{0.25}\end{array}} \right)\end{array}\)

\(\begin{array}{c}{{\mathop{\rm x}\nolimits} _2} = P{{\mathop{\rm x}\nolimits} _1}\\ = \left( {\begin{array}{*{20}{c}}{.5}&{.25}&{.25}\\{.25}&{.5}&{.25}\\{.25}&{.25}&{.5}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{.5}\\{.25}\\{.25}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{0.25 + 0.0625 + 0.0625}\\{0.25 + 0.125 + 0.0625}\\{0.125 + 0.0625 + 0.125}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{0.375}\\{0.3125}\\{0.3125}\end{array}} \right)\end{array}\)

Thus, the probability that the animal will choose food #2 after the initial trial is 0.3125.

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Most popular questions from this chapter

Let \(A\) be any \(2 \times 3\) matrix such that \({\mathop{\rm rank}\nolimits} A = 1\), let u be the first column of \(A\), and suppose \({\mathop{\rm u}\nolimits} \ne 0\). Explain why there is a vector v in \({\mathbb{R}^3}\) such that \(A = {{\mathop{\rm uv}\nolimits} ^T}\). How could this construction be modified if the first column of \(A\) were zero?

Is it possible for a nonhomogeneous system of seven equations in six unknowns to have a unique solution for some right-hand side of constants? Is it possible for such a system to have a unique solution for every right-hand side? Explain.

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

13. Show that if \(P\) is an invertible \(m \times m\) matrix, then rank\(PA\)=rank\(A\).(Hint: Apply Exercise12 to \(PA\) and \({P^{ - 1}}\left( {PA} \right)\).)

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

16. If \(A\) is an \(m \times n\) matrix of rank\(r\), then a rank factorization of \(A\) is an equation of the form \(A = CR\), where \(C\) is an \(m \times r\) matrix of rank\(r\) and \(R\) is an \(r \times n\) matrix of rank \(r\). Such a factorization always exists (Exercise 38 in Section 4.6). Given any two \(m \times n\) matrices \(A\) and \(B\), use rank factorizations of \(A\) and \(B\) to prove that rank\(\left( {A + B} \right) \le {\mathop{\rm rank}\nolimits} A + {\mathop{\rm rank}\nolimits} B\).

(Hint: Write \(A + B\) as the product of two partitioned matrices.)

(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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