(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 4.2' Mathematica-Compatible Notebook This notebook can be used with any Mathematica-compatible application, such as Mathematica, MathReader or Publicon. The data for the notebook starts with the line containing stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). NOTE: If you modify the data for this notebook not in a Mathematica- compatible application, you must delete the line below containing the word CacheID, otherwise Mathematica-compatible applications may try to use invalid cache data. For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. *******************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 43882, 1166]*) (*NotebookOutlinePosition[ 44637, 1192]*) (* CellTagsIndexPosition[ 44593, 1188]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["The Quadratic-Linear Dynamic Programming Problem", "Title"], Cell["by Duncan K. Foley", "Subtitle"], Cell[CellGroupData[{ Cell["1. Computational limits to dynamic programming", "Section"], Cell["\<\ \"New Classical\" macro-economics founds itself on the project of \ rigorously deriving macroeconomic predictions from the full-information \ general equilibrium model. This research program is regarded as \"providing \ micro-foundations for macro-economics\". Since the full information general \ equilibrium model without uncertainty is completely implausible as a basis \ for macroeconomic predictions, the New Classical research program naturally \ focuses on general equilibrium models with uncertainty. The program runs into \ two practical problems. First, even without uncertainty it is very difficult \ to calculate general equilibrium prices even in static models when the \ conditions for aggregating the economy into a representative agent are not \ satisfied. Thus the New Classical program leads very rapidly to a focus on \ representative agent models. But even this drastic simplification is not \ enough to make the problem of computing equilibria (or what is the same \ thing, optima) tractable, even with very powerful computational resources. In \ fact, as we shall see in more detail, there is really only one setting in \ which computation of equilibria/optima in stochastic intertemporal models is \ tractable, which is when the utility function of the representative agent is \ quadratic in the consumption vector, and the technology is linear. This model \ is a workhorse of optimal control theory, and a great deal is known about the \ solutions to it. Thus despite the apparent generality of the New Classical \ research program, practical applications are, at least at the present stage \ of computational technique, limited to quadratic-linear models. The quadratic-linear model arises naturally as an approximation to a general \ stochastic control problem when the stochastic shocks to the economy are \ small, so that what is at issue is small deviations from an equilibrium \ without stochastic shocks. A quadratic objective function can approximate any \ utility function locally, and a linear technology can be regarded as a first \ approximation the local behavior of any technology. (It is striking that New \ Classical macro-economics has in this way returned to the preoccupation of \ the Classical/Marxian tradition with linear production models.) Thus the most \ convincing argument for the relevance of the computation of equilibria/optima \ in the linear-quadratic regulator to real economies presumably rests on the \ idea that the shocks being modeled are both small and stationary. Whether \ this point of view is acceptable depends on a researcher's judgment as to its \ relevance to real economic data. Since the notation and generality (in some respects) of the quadratic-linear \ model are rather confusing and hard to see through, it is useful to approach \ the problem first in a drastically simplified setting.\ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["2. Two-period consumption equilibrium", "Section"], Cell[TextData[{ "The simplest setting in which the economic issues addressed by the \ quadratic-linear regulator appear is the two-period Ramsey problem with a \ linear technology subject to additive shocks. The consumer maximizes ", Cell[BoxData[ \(TraditionalForm\`u[c\_1] + \[Beta]\ u[c\_2]\)]], " subject to a budget (or technological constraint, since they are the same \ in the linear case) ", Cell[BoxData[ \(TraditionalForm\`c\_2 = R(k - c\_1)\)]], ", or ", Cell[BoxData[ \(TraditionalForm\`c\_1 + c\_2\/R = k\)]], ", Here ", Cell[BoxData[ \(TraditionalForm\`c\_1\)]], "and ", Cell[BoxData[ \(TraditionalForm\`c\_2\)]], "are consumption in periods 1 and 2 respectively, ", Cell[BoxData[ \(TraditionalForm\`k\)]], " is the endowment of the single good in period 1, and ", Cell[BoxData[ \(TraditionalForm\`R\)]], " is the productivity of capital. The Lagrangian for this problem leads to \ the first-order conditions:" }], "Text"], Cell[BoxData[{\(\[PartialD]L\/\[PartialD]c\_1 = \[PartialD]u[c\_1]\/\ \[PartialD]c\_1 - \[Lambda]\ \[LessEqual] \ 0\ \((\(=\)\(\ \)\(if\ c\_1 > 0\))\)\), "\[IndentingNewLine]", \(\[PartialD]L\/\[PartialD]c\ \_2 = \[Beta] \[PartialD]u[c\_2]\/\[PartialD]c\_2 - \[Lambda]\/R\ \ \[LessEqual] \ 0\ \((\(=\)\(\ \)\(if\ c\_2 > 0\))\)\), "\[IndentingNewLine]", RowBox[{\(\[PartialD]L\/\[PartialD]\[Lambda]\), "=", RowBox[{ RowBox[{"-", RowBox[{"(", FormBox[\(c\_1 + c\_2\/R - k\), "TraditionalForm"], ")"}]}], "\[GreaterEqual]", \(0\ \((\(=\)\(\ \)\(if\ \[Lambda]\ > 0\))\)\)}]}]}], "DisplayFormula"], Cell[TextData[{ "The solution to this problem is much simplified if the marginal utilities \ of consumption are linear functions of the consumption levels, so that, for \ example, ", Cell[BoxData[ \(TraditionalForm\`\[PartialD]u[c]\/\[PartialD]c = u\_1 - \(u\_11\) c\)]], ", where ", Cell[BoxData[ FormBox[Cell[TextData[Cell[BoxData[ \(TraditionalForm\`u\_1, u\_11\)]]]], TraditionalForm]]], "are given constants. In order to have diminishing marginal utility, we \ have to assume ", Cell[BoxData[ \(TraditionalForm\`u\_11 > 0\)]], ", so that marginal utility itself will be positive only if we assume ", Cell[BoxData[ \(TraditionalForm\`u\_1 > 0\)]], ", and restrict ", Cell[BoxData[ \(TraditionalForm\`c\)]], " to the interval ", Cell[BoxData[ \(TraditionalForm\`0 \[LessEqual] c \[LessEqual] u\_1\/u\_11 = c\&_\)]], ", where ", Cell[BoxData[ \(TraditionalForm\`c\&_\)]], " represents a satiation level of consumption. Integration gives a \ quadratic felicity function, with an arbitrary constant of integration, ", Cell[BoxData[ \(TraditionalForm\`u\)]], ", which can be set so that : " }], "Text"], Cell[BoxData[{ \(TraditionalForm\`u[c] = u\_1\^2\/\(2\ u\_11\) + \(u\_1\) c - \(u\_11\/2\) c\^2\), "\[IndentingNewLine]", \(TraditionalForm\`\(\(=\)\(\(-\(u\_11\/2\)\) \((c - c\&_)\)\^2 \ \[Congruent] \(-\(u\_11\/2\)\) c\&^\^2\)\)\)}], "DisplayFormula"], Cell[TextData[{ "Satiation utility is 0 when ", Cell[BoxData[ \(TraditionalForm\`c = c\&_\)]], " and ", Cell[BoxData[ \(TraditionalForm\`c\&^ = \(c - c\&_ = 0\)\)]], ". The quadratic utility function has diminishing marginal utility only for \ consumption levels below the satiation point.\nWith a quadratic utility \ function, the consumer's optimal consumption bundle is the solution of the \ equations:" }], "Text"], Cell[BoxData[{ \(\(-u\_11\) c\&^\_1 = \[Lambda]\), "\n", \(\(-u\_11\) c\&^\_2 = \[Lambda]/\[Beta]\ R\), "\[IndentingNewLine]", \(c\&^\_1 + c\&^\_2\/R = k - \((1 + 1\/R)\) c\&_ \[Congruent] \(k\&^\)\)}], "DisplayFormula"], Cell["These equations can be written in matrix form:", "Text"], Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{"(", GridBox[{ {\(u\_11\), "0", "1"}, {"0", \(u\_11\), \(1\/\(\[Beta]\ R\)\)}, {"1", \(1\/R\), "0"} }], ")"}], RowBox[{"(", GridBox[{ {\(c\&^\_1\)}, {\(c\&^\_2\)}, {"\[Lambda]"} }], ")"}]}], "=", RowBox[{"(", GridBox[{ {"0"}, {"0"}, {\(k\&^\)} }], ")"}]}]], "DisplayFormula"], Cell["\<\ This system of linear equations, is easy to solve it (either by \ substitution or by inverting the matrix).\ \>", "Text"], Cell[TextData[{ StyleBox["Problems:", FontWeight->"Bold"], "\n2.1 Solve the quadratic-linear one-good two-period model explicitly for \ the demands for ", Cell[BoxData[ \(TraditionalForm\`c\_1\)]], "and ", Cell[BoxData[ \(TraditionalForm\`c\_2\)]], "in terms of the parameters ", Cell[BoxData[ \(TraditionalForm\`R, u\_1, u\_11, k\)]], ".\n2.2 Find the comparative static effects of a change in ", Cell[BoxData[ \(TraditionalForm\`R\)]], " (or ", Cell[BoxData[ \(TraditionalForm\`1\/R\)]], ") and ", Cell[BoxData[ \(TraditionalForm\`k\)]], " on the optimal consumption in the first period in the one-good, \ two-period quadratic-linear model.\n2.3 Write a Mathematica program to solve \ the quadratic-linear two-period model. How difficult is it to generalize this \ to more than one good in each period, and more than one period?" }], "ItemizedText"] }, Open ]], Cell[CellGroupData[{ Cell["\<\ 3. Stochastic shocks in the quadratic-linear two-period model\ \>", \ "Section"], Cell[TextData[{ "The quadratic-linear model can be thought of as a first approximation to \ more general preferences and technology. It also has the remarkable property \ of ", StyleBox["certainty equivalence", FontSlant->"Italic"], " when we add a random shock to the endowment. To see this, consider the \ modified model:" }], "Text"], Cell[BoxData[{ \(max\+{c\_1, c\_2}E[\(-\(u\_11\/2\)\) c\&^\_1\^2 - \[Beta]\ \(u\_11\/2\) c\&^\_2\^2]\), "\[IndentingNewLine]", \(subject\ to\ c\&^\_1 + c\&^\_2\/R \[LessEqual] \(k\&^\) + e\)}], "DisplayFormula"], Cell[TextData[{ "where ", Cell[BoxData[ \(TraditionalForm\`e\)]], " is a random variable with ", Cell[BoxData[ \(TraditionalForm\`E[e] = 0, E[e\^2] = \[Sigma]\_e\^2\)]], ". We can substitute out the constraint (assuming that the solution is \ below the satiation point of the utility function) by writing ", Cell[BoxData[ \(TraditionalForm\`c\&^\_2 = R(k\&^ - c\&^\_1 + e)\)]], ". Then we see that the problem is:" }], "Text"], Cell[BoxData[{ RowBox[{ RowBox[{\(max\+\(c\&^\_1\)\), RowBox[{"(", RowBox[{\(\(-\(u\_11\/2\)\) c\&^\_1\^2\), "-", RowBox[{\(\(\(\[Beta]\)\(\ \)\)\/2\), RowBox[{"E", "[", SuperscriptBox[ RowBox[{"(", FormBox[\(R(k\&^ - c\&^\_1 + e)\), "TraditionalForm"], ")"}], "2"], "]"}]}]}], ")"}]}], "=", "\[IndentingNewLine]", \(max\+\(c\&^\_1\)\((\(-\(u\_11\/2\)\) c\&^\_1\^2 - \[Beta]\ \(u\_11\/2\) \(R\^2\) \((E[\((\(k\&^\) - \ c\&^\_1)\)\^2] + 2 \((\(k\&^\) - c\&^\_1)\) E[e] + E[e\^2])\))\)\)}], "\[IndentingNewLine]", \(\(=\)\(max\+\(c\ \&^\_1\)\((\(-\(u\_11\/2\)\) c\&^\_1 - \[Beta]\ \(u\_11\/2\) \(R\^2\) \((\(k\&^\) - \ c\&^\_1)\)\^2)\) - \[Beta]\ \(u\_11\/2\) \(R\^2\) \[Sigma]\_e\^2\)\)}], \ "DisplayFormula"], Cell[TextData[{ "The remarkable thing about this expression is that terms in the objective \ function involving ", Cell[BoxData[ \(TraditionalForm\`c\&^\_1\)]], " do not involve ", Cell[BoxData[ \(TraditionalForm\`\[Sigma]\_e\^2\)]], ", which implies that the solution to the problem is exactly the same \ whatever the size of ", Cell[BoxData[ \(TraditionalForm\`\[Sigma]\_e\^2\)]], ". This is the ", StyleBox["certainty equivalence property", FontSlant->"Italic"], ". The optimal level of first period consumption in the quadratic-linear \ problem is unaffected by a stochastic shock to the endowment. The maximal \ utility the consumer can reach does depend on ", Cell[BoxData[ \(TraditionalForm\`\[Sigma]\_e\^2\)]], "(negatively, since ", Cell[BoxData[ \(TraditionalForm\`u\_11 > 0\)]], "), but not her optimal consumption in the first period.\nCertainty \ equivalence does not work if the utility function is not quadratic. (Since an \ arbitrary differentiable function can be approximated by its polynomial \ Taylor's Series, it is easy to test this just by introducing a cubic or \ quartic term in the utility function.) \nFurthermore, certainty equivalence \ does not work for a multiplicative shock to productivity of capital. To see \ this, modify the problem to:" }], "Text"], Cell[BoxData[{ RowBox[{ RowBox[{\(max\+\(c\&^\_1\)\), RowBox[{"(", RowBox[{\(\(-\(u\_11\/2\)\) c\&^\_1\^2\), "-", RowBox[{"\[Beta]", " ", \(u\_11\/2\), RowBox[{"E", "[", SuperscriptBox[ RowBox[{"(", FormBox[\(\((1 + e)\)\ \(R(k\&^ - c\&^\_1)\)\), "TraditionalForm"], ")"}], "2"], "]"}]}]}], ")"}]}], "=", \(max\+\(c\&^\_1\)\((\(-\(u\_11\/2\)\) c\&^\_1\^2 - \[Beta]\ \(u\_11\/2\) \(R\^2\) \(\((k\&~ - c\&^\_1)\)\^2\) \((E[ 1 + 2 e + e\^2])\))\)\)}], "\[IndentingNewLine]", \(\(=\)\(max\+\(c\ \&^\_1\)\((\(-\(u\_11\/2\)\) c\&^\_1\^2 - \[Beta]\ \(u\_11\/2\) \(R\^2\) \(\((\[Omega]\&~ - c\&^\_1)\)\^2\) \((1 + \[Sigma]\_e\^2)\))\)\)\)}], \ "DisplayFormula"], Cell[TextData[{ "Clearly the solution to this problem does depend on ", Cell[BoxData[ \(TraditionalForm\`\[Sigma]\_e\^2\)]], ", so that certainty-equivalence fails." }], "Text"], Cell[TextData[{ StyleBox["Problems:", FontWeight->"Bold"], "\n3.1 Show that certainty equivalence fails for a concave utility function \ with a term in ", Cell[BoxData[ \(TraditionalForm\`c\&^\^4\)]], ".\n3.2 Use Mathematica to find the dependence of the optimal ", Cell[BoxData[ \(TraditionalForm\`c\&^\_1\)]], "on ", Cell[BoxData[ \(TraditionalForm\`\[Sigma]\_e\^2\)]], "in the model with a multiplicative shock to capital productivity. Is the \ effect of the multiplicative shock equivalent to any parametric change in the \ certainty model?" }], "ItemizedText"] }, Open ]], Cell[CellGroupData[{ Cell["\<\ 4. The finite horizon one-good quadratic-linear consumption problem\ \ \>", "Section"], Cell[TextData[{ "The next simplest extension of the quadratic-linear regulator is to allow \ more than two (but still a finite number of) periods. With a ", Cell[BoxData[ \(TraditionalForm\`T\)]], "period horizon, assuming no discount, we can write the quadratic-linear \ regulator problem for the certainty case as:" }], "Text"], Cell[BoxData[{ \(V[k, T] = \(max\+\({c\_t, k\_\(t + 1\)}\_\(t = 1\)\^T\)\(-\(\(\[Sum]\+\(t \ = 1\)\)\&T\( u\_11\/2\) \((c\_t - c\&_)\)\)\) = max\+\({c\&^\_t, k\_\(t + 1\)}\_\(t = 1\)\^T\)\(-\(\(\[Sum]\+\(t = \ 1\)\)\&T\( u\_11\/2\) c\&^\_t\^2\)\)\)\), "\[IndentingNewLine]", \(subject\ to\ k\_\(t + 1\) \[LessEqual] R\ \((k\_t - \((c\&^\_t + c\&_)\))\), \ t = 1, 2, \[Ellipsis], T\), "\[IndentingNewLine]", \(R, \ c\&_, k\_1\ given\)}], "DisplayFormula"], Cell["\<\ If we can be sure that the sequence of consumption plans can always \ be kept below satiation, the inequality in the budget constraint will be an \ equality, and the complementary slackness conditions for the problem are \ eliminated. Some care has to be taken in setting up problems of this type to \ ensure that this condition is met. When the constraint is an equality, this \ problem can be written in a recursive form. In the last period the \ consumption will be equal to wealth, assuming that wealth is smaller than \ satiation:\ \>", "Text"], Cell[BoxData[{ \(V[k, 1] = \(\(-\ u\_11\)\/2\) \((\ k - c\&_)\)\^2\), "\[IndentingNewLine]", \(V[k, T] = max\+\(c\&^\)\((\(\(-\ u\_11\)\/2\) \(c\&^\)\^2 + \ V[R \((\ k - \((\(c\&^\) + c\&_)\))\), T - 1])\)\)}], "DisplayFormula"], Cell[TextData[{ "We can use ", StyleBox["Mathematica", FontSlant->"Italic"], " to calculate the resulting series of valuation functions." }], "Text"], Cell[BoxData[{ \(\(Clear[QLV, cmax, chatopt];\)\), "\[IndentingNewLine]", \(cmax[U_, V_] := c /. Solve[D[U[c] + \ V[c], c] \[Equal] 0, c]\), "\[IndentingNewLine]", \(\(chatopt[params : {u11_, \[Beta]_, R_, cbar_}, 1]\)[k_] := k - cbar\), "\[IndentingNewLine]", \(\(chatopt[params : {u11_, \[Beta]_, R_, cbar_}, T_]\)[k_] := cmax[\((\(\(-u11\)\/2\) \((#)\)\^2)\) &, \[Beta]\ \((\(QLV[params, T - 1]\)[ R \((k - # - cbar)\)])\) &]\), "\[IndentingNewLine]", \(\(QLV[params : {u11_, \[Beta]_, R_, cbar_}, 1]\)[ k_] := \(\(-u11\)\/2\) \((k - cbar)\)\^2\), "\[IndentingNewLine]", \(\(QLV[params : {u11_, \[Beta]_, R_, cbar_}, T_]\)[ k_] := \(\(QLV[params, T]\)[ k] = \(\(-u11\)\/2\) \(chatopt[params, T]\)[k]\^2 + \[Beta]\ \(QLV[ params, T - 1]\)[ R \((k - \(chatopt[params, T]\)[k] - cbar)\)] // Simplify\)\)}], "Input", CellLabel->"In[13]:="], Cell[CellGroupData[{ Cell[BoxData[ \(param1 = {u\_11, 1, R, cbar}\)], "Input", CellLabel->"In[7]:="], Cell[BoxData[ \({u\_11, 1, R, cbar}\)], "Output", CellLabel->"Out[7]="] }, Open ]], Cell[TextData[{ "This follows the general pattern of ", StyleBox["Mathematica", FontSlant->"Italic"], " programs for solving recursive systems. The separate definition of the \ function cmax[] ensures that ", StyleBox["Mathematica", FontSlant->"Italic"], " will evaluate the functions in the correct sequence. For example, for ", Cell[BoxData[ \(TraditionalForm\`T = 2\)]], ", we have " }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\(chatopt[param1, 2]\)[k] // Simplify\)], "Input", CellLabel->"In[8]:="], Cell[BoxData[ \({\(-\(\(R\ \((cbar + cbar\ R - k\ R)\)\)\/\(1 + R\^2\)\)\)}\)], "Output",\ CellLabel->"Out[8]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\(QLV[param1, 2]\)[k] // FullSimplify\)], "Input", CellLabel->"In[11]:="], Cell[BoxData[ \({\(-\(\(\((cbar + cbar\ R - k\ R)\)\^2\ u\_11\)\/\(2\ \((1 + R\^2)\)\)\)\)}\)], "Output", CellLabel->"Out[11]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(Table[{T, \(QLV[param1, T]\)[k] // Simplify}, {T, 1, 3}] // TableForm\)], "Input", CellLabel->"In[25]:="], Cell[BoxData[ InterpretationBox[GridBox[{ {"1", \(\(-\(1\/2\)\)\ \((cbar - k)\)\^2\ u\_11\)}, {"2", GridBox[{ {\(-\(\(\((cbar + cbar\ R - k\ R)\)\^2\ u\_11\)\/\(2\ \((1 + R\^2)\)\)\)\)} }, RowSpacings->0.25, ColumnSpacings->1, RowAlignments->Baseline, ColumnAlignments->{Left}]}, {"3", GridBox[{ {\(-\(\(\((1 + 2\ R\^2 + 4\ R\^3 + 4\ R\^4 + 4\ R\^5 + 9\ R\^6 + 8\ R\^7 + 5\ R\^8 + 8\ R\^9 + 9\ R\^10 + 4\ R\^11 + 3\ R\^12 + 4\ R\^13 + 2\ R\^14)\)\ \((k\ R\^2 - cbar\ \((1 + R + \ R\^2)\))\)\^2\ u\_11\)\/\(2\ \((1 + R\^2)\)\ \((1 + R\^4)\)\^2\ \((1 + R\^2 + 2\ \ R\^3 + R\^4)\)\^2\)\)\)} }, RowSpacings->0.25, ColumnSpacings->1, RowAlignments->Baseline, ColumnAlignments->{Left}]} }, RowSpacings->1, ColumnSpacings->3, RowAlignments->Baseline, ColumnAlignments->{Left}], TableForm[ {{1, Times[ Rational[ -1, 2], Power[ Plus[ cbar, Times[ -1, k]], 2], Subscript[ u, 11]]}, {2, { Times[ Rational[ -1, 2], Power[ Plus[ cbar, Times[ cbar, R], Times[ -1, k, R]], 2], Power[ Plus[ 1, Power[ R, 2]], -1], Subscript[ u, 11]]}}, {3, { Times[ Rational[ -1, 2], Power[ Plus[ 1, Power[ R, 2]], -1], Power[ Plus[ 1, Power[ R, 4]], -2], Power[ Plus[ 1, Power[ R, 2], Times[ 2, Power[ R, 3]], Power[ R, 4]], -2], Plus[ 1, Times[ 2, Power[ R, 2]], Times[ 4, Power[ R, 3]], Times[ 4, Power[ R, 4]], Times[ 4, Power[ R, 5]], Times[ 9, Power[ R, 6]], Times[ 8, Power[ R, 7]], Times[ 5, Power[ R, 8]], Times[ 8, Power[ R, 9]], Times[ 9, Power[ R, 10]], Times[ 4, Power[ R, 11]], Times[ 3, Power[ R, 12]], Times[ 4, Power[ R, 13]], Times[ 2, Power[ R, 14]]], Power[ Plus[ Times[ k, Power[ R, 2]], Times[ -1, cbar, Plus[ 1, R, Power[ R, 2]]]], 2], Subscript[ u, 11]]}}}]]], "Output", CellLabel->"Out[25]//TableForm="] }, Open ]], Cell[TextData[{ "Thus if ", Cell[BoxData[ \(TraditionalForm\`V[k, T - 1]\)]], " is quadratic in the state variable ", Cell[BoxData[ \(TraditionalForm\`k\)]], ", then ", Cell[BoxData[ \(TraditionalForm\`V[k, T]\)]], " will be quadratic in the state variable as well, so the series of \ valuation functions is quadratic, which is a considerable simplification. The \ actual series of valuation functions is not so simple, however, at least \ written out in this way. It turns out to be more straightforward to write the \ quadratic part of these expressions in matrix form. Any quadratic form can be \ written as ", Cell[BoxData[ \(TraditionalForm\`x . P . x\)]], ", where ", Cell[BoxData[ \(TraditionalForm\`P\)]], " is a symmetric matrix. For example, in two dimensions:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Collect[{x1, x2} . {{a, b/2}, {b/2, c}} . {x1, x2}, {x1, x2}]\)], "Input", CellLabel->"In[24]:="], Cell[BoxData[ \(a\ x1\^2 + b\ x1\ x2 + c\ x2\^2\)], "Output", CellLabel->"Out[24]="] }, Open ]], Cell[TextData[{ "This suggests that the problem is actually simpler to write if we add the \ constant 1 as a state variable, so that the vector of state variables is ", Cell[BoxData[ \(TraditionalForm\`{k, 1}\)]], ", so that we can express the valuation function as ", Cell[BoxData[ \(TraditionalForm\`V[k, T] = {k, 1} . P\_T . {k, 1}\)]], " for some sequence of symmetric matrices ", Cell[BoxData[ \(TraditionalForm\`{P\_T}\)]], ". We could re-write the initial problem in the following way (cf Ljunqvist \ and Sargent, pp 643 ff):" }], "Text"], Cell[BoxData[{ RowBox[{ RowBox[{\(max\+\({\(c\&^\)\_t}\_\(t = 1\)\^T\)\), RowBox[{"\[Sum]", "\[NegativeThinSpace]", RowBox[{"2", \((k\_t\ 1)\), RowBox[{"(", "\[NegativeThinSpace]", GridBox[{ {"0"}, {\(u\_1\/2\)} }], "\[NegativeThinSpace]", ")"}], \(c\_t\)}]}]}], "+", \(\(c\_t\) \((\(-u\_11\)\/2)\) c\_t\)}], "\[IndentingNewLine]", RowBox[{ RowBox[{"subject", " ", "to", " ", RowBox[{"(", "\[NegativeThinSpace]", GridBox[{ {\(k\_\(t + 1\)\)}, {"1"} }], "\[NegativeThinSpace]", ")"}]}], "=", RowBox[{ RowBox[{ RowBox[{"(", "\[NegativeThinSpace]", GridBox[{ {"R", "0"}, {"0", "1"} }], "\[NegativeThinSpace]", ")"}], RowBox[{"(", "\[NegativeThinSpace]", GridBox[{ {\(k\_t\)}, {"1"} }], "\[NegativeThinSpace]", ")"}]}], "+", RowBox[{ RowBox[{"(", "\[NegativeThinSpace]", GridBox[{ {\(-R\)}, {"0"} }], "\[NegativeThinSpace]", ")"}], \(c\_t\)}]}]}]}], "DisplayFormula"], Cell[TextData[{ "Problems:\n4.1 Verify that a general quadratic form in three dimensions \ can be written ", Cell[BoxData[ \(TraditionalForm\`x . P . x\)]], " where ", Cell[BoxData[ \(TraditionalForm\`x \[Element] R\^3\)]], ", and ", Cell[BoxData[ \(TraditionalForm\`P\)]], " is a symmetric ", Cell[BoxData[ \(TraditionalForm\`3\[Times]3\)]], " matrix.\n4.2 Verify that the matrix representation of the ", Cell[BoxData[ FormBox[ RowBox[{"T", Cell[""]}], TraditionalForm]]], "period quadratic-linear consumption problem is the same as the original \ statement of the problem." }], "ItemizedText"] }, Open ]], Cell[CellGroupData[{ Cell["5. The general quadratic-linear regulator", "Section"], Cell[TextData[{ "The linearity of the solutions to the quadratic-linear regulator reflects \ the assumption that marginal utilities are linear functions of consumption, \ and that the production set is linear. The assumption of linearity makes it \ \"cheap\" to generalize the problem to arbitrary finite-dimensional state and \ control spaces. The general problem can be written (using the notation ", Cell[BoxData[ \(TraditionalForm\`A'\)]], " for the transpose of any matrix ", Cell[BoxData[ \(TraditionalForm\`A\)]], "):" }], "Text"], Cell[BoxData[{ RowBox[{ RowBox[{\(max\+\({u\_t}\_\(t = 1\)\^T\)\), RowBox[{\(\(\[Sum]\+\(t = 1\)\)\&T\), RowBox[{\((x\_t'\ u\_t')\), RowBox[{"(", "\[NegativeThinSpace]", GridBox[{ {\(R\_t\), \(W\_t\)}, {\(W\_t'\), \(Q\_t\)} }], "\[NegativeThinSpace]", ")"}], RowBox[{"(", "\[NegativeThinSpace]", GridBox[{ {\(x\_t\)}, {\(u\_t\)} }], "\[NegativeThinSpace]", ")"}]}]}]}], "+", \(x\_T' \(P\_0\) x\_T\)}], "\[IndentingNewLine]", \(subject\ to\ x\_\(t + 1\) = \(A\_t\ \) x\_t + \(B\_t\) u\_t\), "\[IndentingNewLine]", \(x\_1\ given\)}], "DisplayFormula"], Cell[TextData[{ "The state vector ", Cell[BoxData[ \(TraditionalForm\`x\_t \[Element] R\^n\)]], ", the control vector ", Cell[BoxData[ \(TraditionalForm\`u\_t \[Element] R\^k\)]], ", so that the matrices ", Cell[BoxData[ \(TraditionalForm\`R\_t, A\_t, P\_T\)]], " are ", Cell[BoxData[ \(TraditionalForm\`n\[Times]n\)]], ", ", Cell[BoxData[ \(TraditionalForm\`W\_t, B\_t\)]], " are ", Cell[BoxData[ \(TraditionalForm\`n\[Times]k\)]], ", and ", Cell[BoxData[ \(TraditionalForm\`Q\_t\)]], " are ", Cell[BoxData[ \(TraditionalForm\`k\[Cross]k\)]], ". The idea is to control the system by choosing the values of the vectors \ ", Cell[BoxData[ \(TraditionalForm\`u\_t\)]], ", given its basic law of motion defined by ", Cell[BoxData[ \(TraditionalForm\`A\_t\)]], ". Several points about this formulation are worth noting. \nFirst, \ considerable care has to be taken to make sure that the problem makes sense. \ For example, there has to be some appropriate relationship between the \ control matrices ", Cell[BoxData[ \(TraditionalForm\`B\_t\)]], "and the state-evolution matrices ", Cell[BoxData[ \(TraditionalForm\`A\_t\)]], " to ensure that the system is in some sense controllable at all. \nSecond, \ this formulation allows for a direct impact of the state variables on the \ objective function through the matrices ", Cell[BoxData[ \(TraditionalForm\`R\_t\)]], ". In economic terms this would represent the possibility of the consumer \ getting utility from some of the capital stock (such as household durable \ goods). \nThird, this formulation allows for an interaction of the state \ variables and the control variables in the objective function through the \ matrices ", Cell[BoxData[ \(TraditionalForm\`W\_t\)]], ". As we have seen in the example in the last section, this is convenient \ to express target levels of consumption, and in the current very general \ formulation, to allow for changes over time in the target levels of \ consumption, which we called ", Cell[BoxData[ \(TraditionalForm\`c\&_\)]], ". This feature of the general setup also allows us to transform discounted \ quadratic-linear problems into undiscounted ones through a transformation of \ variables, as Ljunqvist and Sargent show.\nFourth, the fact that the \ dimensions of the state and control vectors are arbitrary allows in principle \ for tremendous flexibility in the representation of an economic system, \ within the limits posed by linearity. For example, arbitrary time patterns of \ production and depreciation can in principle be incorporated into this type \ of model by expanding the state space to include machines of different ages. \ The appearance of the controls in the objective function also allows for a \ limited modeling of some non-constant inputs, such as labor supply, which can \ be treated as the negative consumption of an endowment of leisure, and \ included among the controls.\nFifth, this formulation assumes that the \ constraints are always satisfied with equality, so that the complementary \ slackness issues that arise in general Lagrangian problems are eliminated. \ Considerable care has to be taken in the setting up of these problems to \ ensure that this assumption is valid, and that, for example, no negative \ stocks are implied along the optimal path. \nSixth, this formulation allows \ for arbitrary changes in technology and tastes, since the matrices are all \ subscripted by time. A special case of the model would assume stationary \ technology and tastes.\nFinally, it is not difficult to prove that certainty \ equivalence holds in this model for ", StyleBox["additive", FontSlant->"Italic"], " shocks to the resource contraint. (The demonstration of this fact follows \ exactly the same pattern as in the simpler model we looked at above.) As the \ simpler example showed, certainty equivalence will not hold for \ multiplicative shocks. Thus although underlying technology and tastes can be \ changing in a pattern known to the consumer (following seasonal, or \ life-cycle patterns, for example) the random shocks have to be additive in \ order to preserve certainty equivalence, which is the property that \ ultimately makes the solution of these systems tractable.\nA further general \ simplification of this model can be achieved by a linear transformation of \ variables that eliminates the explicit dependence of the objective function \ on interaction terms between the state variables and the conrols. (Ljundqvist \ and Sargent describe this transformation in detail.) Thus there is no loss of \ generality in assuming that ", Cell[BoxData[ \(TraditionalForm\`W = 0\)]], ". Of course in practice, it is necessary to keep track of this \ transformation in order to interpret the solutions of the model in terms of \ the primitive economic concepts that motivate it. This is basically a \ book-keeping issue, but can be confusing in actually working with this \ technique. We can therefore understand the mathematical properties of the \ general quadratic-linear regulator by considering the problem:" }], "Text"], Cell[BoxData[{ \(V\_T[x] = max\+\({u\_t}\_\(t = 1\)\^T\)\(\(\[Sum]\+\(t = 1\)\)\&T x\_t' \(R\_t\) x\_t\) + u\_t' Q\ u\_t\ + \ x\_T' \(P\_0\) x\_T\), "\[IndentingNewLine]", \(subject\ to\ x\_\(t + 1\) = A\ x\_t + \ B\ u\_t\), "\[IndentingNewLine]", \(x\_1 = \ x\ given\)}], "DisplayFormula"], Cell[TextData[{ "This problem can also be stated in recursive form. To simplify the \ expressions, consider the stationary case where the matrices are the same \ from period to period. Suppose (as is the case) that the valuation function \ is quadratic in the state variables, so that ", Cell[BoxData[ \(TraditionalForm\`V\_T[x] = \(x . P\_T . x = x' P\_T\ x\)\)]], "." }], "Text"], Cell[BoxData[{ \(V\_1[x] = x' \(P\_0\) x\), "\[IndentingNewLine]", \(V\_T[ x] = \(max\+u\((x' R\ x\ + u' Q\ u + V\_\(T - 1\)[A\ x + B\ u])\) = max\+u\((x' R\ x\ + u' Q\ u + \((A\ x\ + B\ u)\)' \(P\_\(T - 1\)\) \((A\ x + B\ u)\))\)\)\)}], "DisplayFormula"], Cell[TextData[{ "We can solve this maximum problem by finding the critical points where the \ partial derivatives are all equal to zero. (The rules for differentiating \ bilinear and quadratic forms are ", Cell[BoxData[ FormBox[ RowBox[{\(\(\[PartialD]x' A\ x\)\/\[PartialD]x\), "=", RowBox[{\((A + A')\), "x", Cell[""]}]}], TraditionalForm]]], "or ", Cell[BoxData[ \(TraditionalForm\`2\ A\ x\)]], " when ", Cell[BoxData[ \(TraditionalForm\`A\)]], " is symmetric, ", Cell[BoxData[ \(TraditionalForm\`\(\[PartialD]y'\ B\ z\)\/\[PartialD]y = B\ z\)]], ", and ", Cell[BoxData[ \(TraditionalForm\`\(\[PartialD]y'\ B\ z\)\/\[PartialD]z = B'\ y\)]], ".)" }], "Text"], Cell[BoxData[{ \(Q\ u\ + \ B' \(P\_\(T - 1\)\) \((A\ x + B\ u)\) = 0\ so\ that\), "\[IndentingNewLine]", \(\((Q + B' \(P\_\(T - 1\)\) B)\) u = \(-B'\) \(P\_\(T - 1\)\) A\ x\), "\[IndentingNewLine]", \(u = \(\(-\((Q + B' \(P\_\(T - 1\)\) B)\)\^\(-1\)\) B' \(P\_\(T - 1\)\) A\ x = \(-F\_T\) x\)\)}], "DisplayFormula"], Cell["\<\ Plugging this optimal control back into the expression recursively \ defining the value function, we see that:\ \>", "Text"], Cell[BoxData[{ \(V\_T[x] = x' R\ x\ + x' F\_T' Q\ \(F\_T\) x + x' \((A\ - B\ F\_T)\)' \(P\_\(T - 1\)\) \((A\ - B\ F\_T)\) x\), "\[IndentingNewLine]", \(\(\(=\)\(x' \((R + F\_T' Q\ F\_T + \((A\ - B\ F\_T)\)' \(P\_\(T - 1\)\) \((A\ - B\ F\_T)\))\) x = x' \(P\_T\) x\)\)\)}], "DisplayFormula"], Cell["Unpacking the optimal control:", "Text"], Cell[BoxData[{ \(P\_T = \(R + A' \(P\_\(T - 1\)\) A + F\_T' \((Q + B' \(P\_\(T - 1\)\) B)\) F\_T = R + A' \(P\_\(T - 1\)\) A + \((\(\((Q + B' \(P\_\(T - 1\)\) B)\)\^\(-1\)\) B' \(P\_\(T - 1\)\) A\ )\)' \((Q + B' \(P\_\(T - 1\)\) B)\) \(\((Q + B' \(P\_\(T - 1\)\) B)\)\^\(-1\)\) B' \(P\_\(T - 1\)\) A\)\ \), "\[IndentingNewLine]", \(\(\(=\)\(A' \(P\_\(T - 1\)\) A + \(\(\((\(\((Q + B' \(P\_\(T - 1\)\) B)\)\^\(-1\)\) B' \(P\_\(T - 1\)\) A\ )\)'\) \(B'\) \(P\_\(T - 1\)\) \(A\)\(\ \)\)\)\)\)}], "DisplayFormula"], Cell[TextData[{ "This is the ", StyleBox["matrix Riccati equation", FontSlant->"Italic"], ", which recursively solves the quadratic-linear regulator, given ", Cell[BoxData[ \(TraditionalForm\`P\_0\)]], ". If this sequence of matrices converges, the limit is the solution to the \ infinite-horizon quadratic-linear regulator problem." }], "Text"], Cell[TextData[{ "Problems:\n5.1 Show that if ", Cell[BoxData[ \(TraditionalForm\`P\_\(T - 1\)\)]], "is symmetric, then ", Cell[BoxData[ \(TraditionalForm\`P\_T\)]], "is also symmetric.\n5.2 For a ", Cell[BoxData[ \(TraditionalForm\`2\[Times]2\)]], "matrix ", Cell[BoxData[ \(TraditionalForm\`A\)]], ", show that ", Cell[BoxData[ \(TraditionalForm\`\(\[PartialD]x' A\ x\)\/\[PartialD]x = \((A + A')\) x\)]], "." }], "ItemizedText"] }, Open ]], Cell[CellGroupData[{ Cell["6. The Lagrangian Approach to the Quadratic-Linear Regulator", "Section"], Cell["\<\ Since the quadratic-linear regulator is a constrained maximization \ problem, it is possible to set it up in the Lagrangian form. The Lagrangian \ (for the stationary problem, to keep things simple) is:\ \>", "Text"], Cell[BoxData[ \(\[ScriptCapitalL][{u\_t, x\_\(t + 1\)}\_\(t = 1\)\^T, 2 {\[Mu]\_\(t + 1\)}\_\(t = 1\)\^T] = \(\[Sum]\+\(t = \ 1\)\)\&T\((x\_t' R\ x\_t + u\_t' Q\ u\_t - 2 \[Mu]\_\(t + 1\)' \((x\_\(t + 1\) - A\ x\_t - B\ u\_t)\))\)\)], "DisplayFormula"], Cell["The first-order conditions are:", "Text"], Cell[BoxData[{ \(\[PartialD]\[ScriptCapitalL]\/\[PartialD]u\_t = \(\(Q\_t\) u\_t + B' \[Mu]\_\(t + 1\) = 0\)\), "\[IndentingNewLine]", \(\[PartialD]\[ScriptCapitalL]\/\[PartialD]x\_\(t + 1\) = \(R\ x\_\(t + 1\ \) - \[Mu]\_\(t + 1\) + A' \[Mu]\_\(t + 2\) = 0\)\), "\[IndentingNewLine]", \(\[PartialD]\[ScriptCapitalL]\/\[PartialD]\[Mu]\_t = \(\(-\((x\_\(t + \ 1\) - A\ x\_t - B\ u\_t)\)\) = 0\)\)}], "DisplayFormula"], Cell[TextData[{ "Since we are maintaining the assumption that the optimal solution is \ interior, so that the first-order conditions always hold with equality, it is \ possible to use the first-order condition with respect to the controls to \ eliminate the controls from the system, since ", Cell[BoxData[ \(TraditionalForm\`u\_t = \(-Q\^\(-1\)\) B' \[Mu]\_\(t + 1\)\)]], ". ", Cell[BoxData[ \(TraditionalForm\`x\_1\)]], "is given by the initial conditions of the problem, and ", Cell[BoxData[ \(TraditionalForm\`\[Mu]\_\(T + 2\) = 0\)]], ", since there is no value to the state vector after the end of the \ problem." }], "Text"], Cell[BoxData[{ \(x\_\(t + 1\) = \(A\ x\_t - B\ \(Q\^\(-1\)\) B'\ \[Mu]\_\(t + 1\)\ t = 1\), \[Ellipsis], T\), "\[IndentingNewLine]", \(\[Mu]\_\(t + 1\) = \(R\ x\_\(t + 1\) + A' \[Mu]\_\(t + 2\)\ t = 1\), \[Ellipsis], T\), "\[IndentingNewLine]", \(x\_1, \[Mu]\_\(T + 2\) = 0\ given\)}], "DisplayFormula"], Cell[TextData[{ "This reveals the characteristic mathematical structure of recursive models \ of this kind. The evolution of the state vector, ", Cell[BoxData[ \(TraditionalForm\`x\)]], ", moves forward in time from the present to the future, while the \ evolution of the co-state vector ", Cell[BoxData[ \(TraditionalForm\`\[Mu]\)]], " moves backward in time, from the future to the present. In the \ finite-horizon case, these are just a system of ", Cell[BoxData[ \(TraditionalForm\`2 T\)]], " linear equations in the ", Cell[BoxData[ \(TraditionalForm\`2\ T\)]], " unknowns ", Cell[BoxData[ \(TraditionalForm\`{x\_\(t + 1\), \[Mu]\_\(t + 1\)}\_\(t = 1\)\^T\)]], ". To get a feel for the drama of this problem, suppose that ", Cell[BoxData[ \(TraditionalForm\`A\)]], " is invertible. Then the system could be written:" }], "Text"], Cell[BoxData[{ \(x\_\(t + 1\) = \(A\ x\_t - B\ \(Q\^\(-1\)\) B'\ \[Mu]\_\(t + 1\)\ t = 1\), \[Ellipsis], T\), "\[IndentingNewLine]", \(\[Mu]\_\(t + 2\) = \(\(-A'\^\(-1\)\) R\ x\_\(t + 1\) + \(A'\^\(-1\)\) \[Mu]\_\(t + 1\)\ t = 1\), \[Ellipsis], T\), "\[IndentingNewLine]", \(\(x\_1\) given\)}], "DisplayFormula"], Cell[TextData[{ "Given an initial ", Cell[BoxData[ \(TraditionalForm\`\[Mu]\_2\)]], ",this system determines a path ", Cell[BoxData[ \(TraditionalForm\`{x\_\(t + 1\), \[Mu]\_\(t + 2\)}\_\(t = 1\)\^T\)]], ", with a terminal value ", Cell[BoxData[ \(TraditionalForm\`\[Mu]\_\(T + 2\)\)]], ". If the terminal value is indeed ", Cell[BoxData[ \(TraditionalForm\`\[Mu]\_\(T + 2\) = 0\)]], ", then this path is the solution, and the optimal path for the initial \ problem. If, however, ", Cell[BoxData[ \(TraditionalForm\`\[Mu]\_\(T + 2\) \[NotEqual] 0\)]], " on this path, it is not the solution, and it is necessary to choose a \ different initial ", Cell[BoxData[ \(TraditionalForm\`\[Mu]\_2\)]], ".The condition ", Cell[BoxData[ \(TraditionalForm\`\[Mu]\_\(T + 2\) = 0\)]], " is a \"transversality\" condition that characterizes the optimal \ solution. The process of finding the appropriate ", Cell[BoxData[ \(TraditionalForm\`\[Mu]\_2\)]], "that starts the optimal path is sometimes called the \"shooting method\", \ since the idea is to \"aim\" at the terminal condition ", Cell[BoxData[ \(TraditionalForm\`\[Mu]\_\(T + 2\) = 0\)]], ".\nWe can also calculate shadow prices by treating the valuation \ functional equation as a Lagrangian, keeping the next period state vector, ", Cell[BoxData[ \(TraditionalForm\`x\_\(+1\)\)]], "in the problem explicitly:" }], "Text"], Cell[BoxData[{ \(V\_T[ x] = \(max\+u\((x' R\ x\ + u' Q\ u + V\_\(T - 1\)[A\ x + B\ u])\) = max\+u\((x' R\ x\ + u' Q\ u + \((A\ x\ + B\ u)\)' \(P\_\(T - 1\)\) \((A\ x + B\ u)\))\)\)\), "\[IndentingNewLine]", \(\(\(=\)\(\ \)\(\(\(\(max\)\(\ \)\)\+u\) \(min\+\[Mu] \[ScriptCapitalL][ u, x\_\(+1\), \[Mu], x]\) = \ \(\(\(max\)\(\ \)\)\+u\) \(min\+\[Mu]\((x' R\ x\ + u' Q\ u + x\_\(+1\)'\ \(P\_\(T - 1\)\) x\_\(+1\) - 2 \[Mu] \((x\_\(+1\) - A\ x - B\ u)\))\)\)\)\)\)}], "DisplayFormula"], Cell[TextData[{ "Problems:\n6.1 Use the Lagrangian form of the valuation functional \ equation to show that ", Cell[BoxData[ \(TraditionalForm\`\[Mu] = \(P\_\(T - 1\)\) x\)]], ". (Hint: find the first-order condition corresponding to ", Cell[BoxData[ \(TraditionalForm\`x\_\(+1\)\)]], ".)\n6.2 Write down all of the first-order conditions for Lagrangian form \ of the valuation functional equation.\n6.3 Solve the one-good Ramsey problem \ we studied above for ", Cell[BoxData[ \(TraditionalForm\`T = 2\)]], "using the matrix form of the valuation functional equation. (You may use \ Mathematica for this problem, but you do not have to.)" }], "ItemizedText"] }, Open ]], Cell[CellGroupData[{ Cell["References", "Section"], Cell["\<\ Lars Ljunqvist and Thomas J. Sargent. 2000. Recursive Macroeconomic \ Theory. Cambridge: M.I.T. Press.\ \>", "ItemizedText"] }, Open ]] }, Open ]] }, FrontEndVersion->"4.2 for Macintosh", ScreenRectangle->{{4, 1152}, {0, 746}}, WindowToolbars->"EditBar", WindowSize->{725, 624}, WindowMargins->{{4, Automatic}, {Automatic, 4}}, PrintingCopies->1, PrintingPageRange->{1, Automatic}, StyleDefinitions -> "TutorialBook.nb" ] (******************************************************************* Cached data follows. If you edit this Notebook file directly, not using Mathematica, you must remove the line containing CacheID at the top of the file. The cache data will then be recreated when you save this file from within Mathematica. *******************************************************************) (*CellTagsOutline CellTagsIndex->{} *) (*CellTagsIndex CellTagsIndex->{} *) (*NotebookFileOutline Notebook[{ Cell[CellGroupData[{ Cell[1776, 53, 65, 0, 118, "Title"], Cell[1844, 55, 38, 0, 34, "Subtitle"], Cell[CellGroupData[{ Cell[1907, 59, 65, 0, 92, "Section"], Cell[1975, 61, 2887, 39, 512, "Text"] }, Open ]], Cell[CellGroupData[{ Cell[4899, 105, 56, 0, 92, "Section"], Cell[4958, 107, 1008, 27, 110, "Text"], Cell[5969, 136, 708, 14, 124, "DisplayFormula"], Cell[6680, 152, 1215, 33, 114, "Text"], Cell[7898, 187, 280, 5, 84, "DisplayFormula"], Cell[8181, 194, 440, 11, 80, "Text"], Cell[8624, 207, 241, 4, 82, "DisplayFormula"], Cell[8868, 213, 62, 0, 28, "Text"], Cell[8933, 215, 509, 17, 79, "DisplayFormula"], Cell[9445, 234, 131, 3, 28, "Text"], Cell[9579, 239, 919, 26, 174, "ItemizedText"] }, Open ]], Cell[CellGroupData[{ Cell[10535, 270, 90, 3, 92, "Section"], Cell[10628, 275, 346, 8, 68, "Text"], Cell[10977, 285, 242, 4, 81, "DisplayFormula"], Cell[11222, 291, 458, 12, 48, "Text"], Cell[11683, 305, 902, 18, 122, "DisplayFormula"], Cell[12588, 325, 1345, 30, 232, "Text"], Cell[13936, 357, 912, 19, 120, "DisplayFormula"], Cell[14851, 378, 189, 5, 28, "Text"], Cell[15043, 385, 603, 16, 104, "ItemizedText"] }, Open ]], Cell[CellGroupData[{ Cell[15683, 406, 96, 3, 92, "Section"], Cell[15782, 411, 341, 7, 48, "Text"], Cell[16126, 420, 498, 9, 94, "DisplayFormula"], Cell[16627, 431, 559, 9, 108, "Text"], Cell[17189, 442, 296, 7, 76, "DisplayFormula"], Cell[17488, 451, 159, 5, 28, "Text"], Cell[17650, 458, 1007, 19, 313, "Input"], Cell[CellGroupData[{ Cell[18682, 481, 85, 2, 29, "Input"], Cell[18770, 485, 77, 2, 27, "Output"] }, Open ]], Cell[18862, 490, 426, 12, 68, "Text"], Cell[CellGroupData[{ Cell[19313, 506, 94, 2, 29, "Input"], Cell[19410, 510, 119, 3, 46, "Output"] }, Open ]], Cell[CellGroupData[{ Cell[19566, 518, 95, 2, 29, "Input"], Cell[19664, 522, 153, 3, 51, "Output"] }, Open ]], Cell[CellGroupData[{ Cell[19854, 530, 134, 3, 49, "Input"], Cell[19991, 535, 2979, 95, 118, "Output"] }, Open ]], Cell[22985, 633, 835, 22, 108, "Text"], Cell[CellGroupData[{ Cell[23845, 659, 128, 3, 29, "Input"], Cell[23976, 664, 90, 2, 30, "Output"] }, Open ]], Cell[24081, 669, 577, 13, 68, "Text"], Cell[24661, 684, 1241, 31, 89, "DisplayFormula"], Cell[25905, 717, 655, 20, 124, "ItemizedText"] }, Open ]], Cell[CellGroupData[{ Cell[26597, 742, 60, 0, 92, "Section"], Cell[26660, 744, 559, 12, 88, "Text"], Cell[27222, 758, 740, 16, 94, "DisplayFormula"], Cell[27965, 776, 5230, 103, 884, "Text"], Cell[33198, 881, 343, 7, 94, "DisplayFormula"], Cell[33544, 890, 393, 8, 68, "Text"], Cell[33940, 900, 320, 6, 77, "DisplayFormula"], Cell[34263, 908, 732, 21, 75, "Text"], Cell[34998, 931, 357, 6, 62, "DisplayFormula"], Cell[35358, 939, 134, 3, 28, "Text"], Cell[35495, 944, 367, 7, 42, "DisplayFormula"], Cell[35865, 953, 46, 0, 28, "Text"], Cell[35914, 955, 665, 12, 69, "DisplayFormula"], Cell[36582, 969, 365, 9, 48, "Text"], Cell[36950, 980, 493, 18, 86, "ItemizedText"] }, Open ]], Cell[CellGroupData[{ Cell[37480, 1003, 79, 0, 92, "Section"], Cell[37562, 1005, 226, 4, 48, "Text"], Cell[37791, 1011, 298, 5, 55, "DisplayFormula"], Cell[38092, 1018, 47, 0, 28, "Text"], Cell[38142, 1020, 441, 6, 126, "DisplayFormula"], Cell[38586, 1028, 663, 15, 88, "Text"], Cell[39252, 1045, 335, 5, 63, "DisplayFormula"], Cell[39590, 1052, 896, 23, 88, "Text"], Cell[40489, 1077, 357, 6, 65, "DisplayFormula"], Cell[40849, 1085, 1470, 37, 160, "Text"], Cell[42322, 1124, 635, 11, 119, "DisplayFormula"], Cell[42960, 1137, 691, 15, 152, "ItemizedText"] }, Open ]], Cell[CellGroupData[{ Cell[43688, 1157, 29, 0, 92, "Section"], Cell[43720, 1159, 134, 3, 28, "ItemizedText"] }, Open ]] }, Open ]] } ] *) (******************************************************************* End of Mathematica Notebook file. *******************************************************************)