2.8E: An Application to Input-Output Economic Models Exercises
- Page ID
- 132807
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Exercises for 1
solutions
2
Find the possible equilibrium price structures when the input-output matrices are:
\(\left[ \begin{array}{rrr} 0.1 & 0.2 & 0.3 \\ 0.6 & 0.2 & 0.3 \\ 0.3 & 0.6 & 0.4 \end{array} \right]\) \(\left[ \begin{array}{rrr} 0.5 & 0 & 0.5 \\ 0.1 & 0.9 & 0.2 \\ 0.4 & 0.1 & 0.3 \end{array} \right]\) \(\left[ \begin{array}{rrrr} 0.3 & 0.1 & 0.1 & 0.2 \\ 0.2 & 0.3 & 0.1 & 0 \\ 0.3 & 0.3 & 0.2 & 0.3 \\ 0.2 & 0.3 & 0.6 & 0.5 \end{array} \right]\) \(\left[ \begin{array}{rrrr} 0.5 & 0 & 0.1 & 0.1 \\ 0.2 & 0.7 & 0 & 0.1 \\ 0.1 & 0.2 & 0.8 & 0.2 \\ 0.2 & 0.1 & 0.1 & 0.6 \end{array} \right]\)
- \(\left[ \begin{array}{r} t \\ 3t \\ t \end{array} \right]\)
- \(\left[ \begin{array}{r} 14t \\ 17t \\ 47t \\ 23t \end{array} \right]\)
Three industries \(A\), \(B\), and \(C\) are such that all the output of \(A\) is used by \(B\), all the output of \(B\) is used by \(C\), and all the output of \(C\) is used by \(A\). Find the possible equilibrium price structures.
\(\left[ \begin{array}{r} t \\ t \\ t \end{array} \right]\)
Find the possible equilibrium price structures for three industries where the input-output matrix is \(\left[ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right]\). Discuss why there are two parameters here.
Prove Theorem [thm:007013] for a \(2 \times 2\) stochastic matrix \(E\) by first writing it in the form \(E = \left[ \begin{array}{cc} a & b \\ 1 - a & 1 - b \end{array} \right]\), where \(0 \leq a \leq 1\) and \(0 \leq b \leq 1\).
\(P = \left[ \begin{array}{c} bt \\ (1 - a)t \end{array} \right]\) is nonzero (for some \(t\)) unless \(b = 0\) and \(a = 1\). In that case, \(\left[ \begin{array}{r} 1 \\ 1 \end{array} \right]\) is a solution. If the entries of \(E\) are positive, then \(P = \left[ \begin{array}{c} b \\ 1 - a \end{array} \right]\) has positive entries.
If \(E\) is an \(n \times n\) stochastic matrix and \(\mathbf{c}\) is an \(n \times 1\) matrix, show that the sum of the entries of \(\mathbf{c}\) equals the sum of the entries of the \(n \times 1\) matrix \(E\mathbf{c}\).
Let \(W = \left[ \begin{array}{ccccc} 1 & 1 & 1 & \cdots & 1 \end{array} \right]\). Let \(E\) and \(F\) denote \(n \times n\) matrices with nonnegative entries.
- Show that \(E\) is a stochastic matrix if and only if \(WE = W\).
- Use part (a.) to deduce that, if \(E\) and \(F\) are both stochastic matrices, then \(EF\) is also stochastic.
Find a \(2 \times 2\) matrix \(E\) with entries between \(0\) and \(1\) such that:
- \(I - E\) has no inverse.
- \(I - E\) has an inverse but not all entries of \((I - E)^{-1}\) are nonnegative.
- \(\left[ \begin{array}{rr} 0.4 & 0.8 \\ 0.7 & 0.2 \end{array} \right]\)
If \(E\) is a \(2 \times 2\) matrix with entries between \(0\) and \(1\), show that \(I - E\) is invertible and \((I - E)^{-1} \geq 0\) if and only if \(tr \;E < 1 + \det E\). Here, if \(E = \left[ \begin{array}{rr} a & b \\ c & d \end{array} \right]\), then \(tr \;E = a + d\) and \(\det E = ad - bc\).
If \(E = \left[ \begin{array}{rr} a & b \\ c & d \end{array} \right]\), then \(I - E = \left[ \begin{array}{cc} 1 - a & -b \\ -c & 1 - d \end{array} \right]\), so \(\det (I - E) = (1 - a)(1 - d) - bc = 1 - tr \;E + \det E\). If \(\det (I - E) \neq 0\), then \((I - E)^{-1} = \frac{1}{\det (I - E)} \left[ \begin{array}{cc} 1 - d & b \\ c & 1 - a \end{array} \right]\), so \((I - E)^{-1} \geq 0\) if \(\det (I - E) > 0\), that is, \(tr \;E < 1 + \det E\). The converse is now clear.
In each case show that \(I - E\) is invertible and \((I - E)^{-1} \geq 0\).
\(\left[ \begin{array}{rrr} 0.6 & 0.5 & 0.1 \\ 0.1 & 0.3 & 0.3 \\ 0.2 & 0.1 & 0.4 \end{array} \right]\) \(\left[ \begin{array}{rrr} 0.7 & 0.1 & 0.3 \\ 0.2 & 0.5 & 0.2 \\ 0.1 & 0.1 & 0.4 \end{array} \right]\) \(\left[ \begin{array}{rrr} 0.6 & 0.2 & 0.1 \\ 0.3 & 0.4 & 0.2 \\ 0.2 & 0.5 & 0.1 \end{array} \right]\) \(\left[ \begin{array}{rrr} 0.8 & 0.1 & 0.1 \\ 0.3 & 0.1 & 0.2 \\ 0.3 & 0.3 & 0.2 \end{array} \right]\)
- Use \(\mathbf{p} = \left[ \begin{array}{r} 3 \\ 2 \\ 1 \end{array} \right]\) in Theorem [thm:007060].
- \(\mathbf{p} = \left[ \begin{array}{r} 3 \\ 2 \\ 2 \end{array} \right]\) in Theorem [thm:007060].
[ex:ex2_8_10] Prove that (1) implies (2) in the Corollary to Theorem [thm:007060].
[ex:ex2_8_11] If \((I - E)^{-1} \geq 0\), find \(\mathbf{p} > 0\) such that \(\mathbf{p} > E\mathbf{p}\).
[ex:ex2_8_12] If \(E\mathbf{p} < \mathbf{p}\) where \(E \geq 0\) and \(\mathbf{p} > 0\), find a number \(\mu\) such that \(E\mathbf{p} < \mu\mathbf{p}\) and \(0 < \mu < 1\).
[Hint: If \(E\mathbf{p} = (q_{1}, \dots, q_{n})^{T}\) and \(\mathbf{p} = (p_{1}, \dots, p_{n})^{T}\), take any number \(\mu\) where \(\func{max}\left\lbrace \frac{q_{1}}{p_{1}}, \dots, \frac{q_{n}}{p_{n}} \right\rbrace < \mu < 1\).]