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[学科前沿] [QuantEcon]MATLAB混编FORTRAN语言 [推广有奖]

131
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-18 22:12:03
$$
{{C}}_{t}+{{\left({\frac{P^E}{P}}\right)}}_{t}+\frac{1}{{{R^R}}_{t}}\, {\nu}\, {K}_{t}={{N}}_{t}\, {{\left({\frac{W}{P}}\right)}}_{t}+{{\left({\frac{P^E}{P}}\right)}}_{t}+{{\left({\frac{D^E}{P}}\right)}}_{t}+{\nu}\, {K}_{t-1}
$$

132
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-19 09:14:25
$$
y=\gamma \left( \delta x_1^{-\rho} + \left(1-\delta \right) x_2^{-\rho} \right)^{-\frac{\nu}{\rho}},
$$
where $y$ is the output quantity,
$x_1$ and $x_2$ are the input quantities,
and $\gamma$, $\delta$, $\rho$, and $\nu$ are parameters.
Parameter $\gamma \in [0,\infty)$ determines the productivity,
$\delta \in [ 0,1 ]$ determines the optimal distribution of the inputs,
$\rho \in [-1,0) \cup (0,\infty)$ determines the (constant) elasticity of substitution,
which is $\sigma = 1 \left/ \left( 1 + \rho \right) \right.$,
and $\nu \in [0,\infty)$ is equal to the elasticity of scale.

133
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-19 09:15:25
The CES function includes three special cases: for $\rho \rightarrow 0$,
$\sigma$ approaches $1$
and the CES turns to the Cobb-Douglas form;
for $\rho \rightarrow \infty$, $\sigma$ approaches $0$ and the CES turns
to the Leontief production function;  
and for $\rho \rightarrow -1$, $\sigma$ approaches infinity
and the CES turns to a linear function
if $\nu$ is equal to 1.
$$
\begin{align}
y = \; & \gamma \left( \sum_{i=1}^n \delta_i x_i^{-\rho} \right)^{-\frac{\nu}{\rho}}
\label{eq:n-ces-plain}\\
& \text{with } \sum_{i=1}^n \delta_i = 1,
\end{align}
$$

134
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-19 09:16:45
$$
y = \gamma \left[ \delta \left( \delta_1 x_1^{-\rho_1} + ( 1 - \delta_1 ) x_2^{-\rho_1} \right)^{\rho/\rho_1}
    +  ( 1 - \delta ) \left( \delta_2 x_3^{-\rho_2} + ( 1 - \delta_2 ) x_4^{-\rho_2} \right)^{\rho/\rho_2} \right]^{-\nu/\rho}.
$$
If $\rho_1 = \rho_2 = \rho$,
the four-input nested CES function defined in Equation~\ref{eq:4-ces-nested}
reduces to the plain four-input CES function defined in Equation

135
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-19 09:18:18
In this case,
the parameters of
the four-input nested CES function defined in Equation~\ref{eq:4-ces-nested}
(indicated by the superscript $n$)
and the parameters of
the plain four-input CES function defined in Equation~\ref{eq:n-ces-plain}
(indicated by the superscript $p$)
correspond in the following way:
where $\rho^p = \rho_1^n = \rho_2^n = \rho^n$,
$\delta_1^p = \delta_1^n \; \delta^n$,
$\delta_2^p = ( 1 - \delta_1^n ) \; \delta^n$,
$\delta_3^p = \delta_2^n \; ( 1 - \delta^n )$,
$\delta_4^p = ( 1 - \delta_2^n ) \; ( 1 - \delta^n )$,
$\gamma^p = \gamma^n$,
$\delta_1^n = \delta_1^p / ( \delta_1^p + \delta_2^p )$,
$\delta_2^n = \delta_3^p / ( \delta_3^p + \delta_4^p )$, and
$\delta^n = \delta_1^p + \delta_2^p$.

$$
y = \gamma \left[ \delta \gamma_1^{-\rho} \left( \delta_1 x_1^{-\rho_1} + (1-\delta_1) x_2^{-\rho_1}\right)^{\rho/\rho_1}
+ ( 1 - \delta ) x_3^{-\rho} \right]^{-\nu/\rho}.
$$
However, adding the term $\gamma_1^{-\rho}$ does not increase the flexibility
of this function
as $\gamma_1$ can be arbitrarily normalised to one;
normalising $\gamma_1$ to one
changes $\gamma$ to
$\gamma \left( \delta \gamma_1^{-\rho} + ( 1 - \delta ) \right)^{-( \nu / \rho )}$
and changes $\delta$ to
$\left. \left( \delta \gamma_1^{-\rho} \right) \right/
\left( \delta \gamma_1^{-\rho} + ( 1 - \delta ) \right)$,
but has no effect on the functional form.
Hence, the parameters $\gamma$, $\gamma_1$, and $\delta$
cannot be (jointly) identified in econometric estimations
(see also explanation for the four-input nested CES function
above Equation

136
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-19 09:19:49
approximation of the classical two-input CES production function
that could be estimated by ordinary least-squares techniques.
$$
\begin{align}
\ln y = & \ln \gamma + \nu \; \delta \ln x_1 + \nu
    \left( 1 - \delta \right) \ln x_2
   \label{eq:kmenta}\\
  &- \frac{\rho \, \nu}{2} \;
      \delta \left( 1 - \delta \right) \left( \ln x_1 - \ln x_2 \right)^2
   \nonumber
\end{align}
$$
While \citet{kmenta67} obtained this formula by logarithmising the CES function
and applying a second-order Taylor series expansion to
$\ln \left( \delta x_1^{-\rho} + ( 1 - \delta ) x_2^{-\rho} \right)$
at the point $\rho = 0$,
the same formula can be obtained
by applying a first-order Taylor series expansion
to the entire logarithmised CES function

137
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-19 09:21:18
$$
\begin{align}
\ln y =& \alpha_0 + \alpha_1 \ln x_1 + \alpha_2 \ln x_2
   \label{eq:kmentaTranslog}\\
   & + \frac{1}{2} \; \beta_{11} \left( \ln x_1 \right)^2
   + \frac{1}{2} \; \beta_{22} \left( \ln x_2 \right)^2
   + \beta_{12} \, \ln x_1 \ln x_2,
   \nonumber
\end{align}
$$
where the two restrictions are
$$
\beta_{12} = - \beta_{11} = - \beta_{22}.
$$
If this is the case,
a simple $t$-test for the coefficient $\beta_{12} = - \beta_{11} = - \beta_{22}$
can be used to check if the Cobb-Douglas functional form is an acceptable
simplification of the Kmenta approximation of the CES function.
$$
\begin{align}
\gamma &= \exp( \alpha_0 )
   \label{eq:kmentaTranslogGamma}\\
\nu &= \alpha_1 + \alpha_2
   \label{eq:kmentaTranslogNu}\\
\delta &= \frac{ \alpha_1 }{ \alpha_1 + \alpha_2 }
   \label{eq:kmentaTranslogDelta}\\
\rho &= \frac{ \beta_{12} \left( \alpha_1 + \alpha_2 \right) }{
   \alpha_1 \cdot \alpha_2 }
   \label{eq:kmentaTranslogRho}
\end{align}
$$

138
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-19 09:23:23
For the traditional two-input CES function, these partial derivatives are:$$\begin{align}
\frac{\partial y}{\partial \gamma } =\;&
   e^{\lambda \, t} \, \left( \delta x_1^{-\rho} + ( 1 - \delta ) x_2^{-\rho} \right)^{-\frac{ \nu }{\rho}}
   \label{eq:derivYGamma}\\
\frac{\partial y}{\partial \lambda } =\;&
    \gamma \; t \; \frac{\partial y}{\partial \gamma }
   \label{eq:derivYLambda}\\
\frac{\partial y}{\partial \delta } =\;&
   -  \gamma \, e^{\lambda \, t} \, \frac{\nu }{ \rho } \left( x_1^{-\rho} - x_2^{-\rho} \right)
   \left( \delta x_1^{-\rho} + ( 1 - \delta ) x_2^{-\rho} \right)^{-\frac{\nu}{\rho} - 1}
   \label{eq:derivYDelta}\\
\frac{\partial y}{\partial \rho } =\;&
   \gamma \, e^{\lambda \, t} \, \frac{ \nu }{ \rho^2 } \;
   \ln \left( \delta x_1^{-\rho} + ( 1 - \delta ) x_2^{-\rho} \right)
   \left( \delta x_1^{-\rho} + ( 1 - \delta ) x_2^{-\rho} \right)^{-\frac{ \nu }{\rho}}
   \label{eq:derivYRho}\\
   & + \gamma \, e^{\lambda \, t} \, \frac{ \nu }{ \rho }
   \left( \delta \ln( x_1 ) x_1^{-\rho} + ( 1 - \delta ) \ln( x_2 ) x_2^{-\rho} \right)
   \left( \delta x_1^{-\rho} + ( 1 - \delta ) x_2^{-\rho} \right)^{-\frac{\nu}{\rho} -1}
   \nonumber\\
\frac{\partial y}{\partial \nu } =\;&
   - \gamma \, e^{\lambda \, t} \, \frac{ 1 }{ \rho } \;
   \ln \left( \delta x_1^{-\rho} + ( 1 - \delta ) x_2^{-\rho} \right)
   \left( \delta x_1^{-\rho} + ( 1 - \delta ) x_2^{-\rho} \right)^{-\frac{ \nu }{\rho }}
   \label{eq:derivYNu}
\end{align}
$$

139
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-19 09:27:16
$$
\begin{align}
\frac{\partial y}{\partial \gamma } =\;&
   e^{\lambda \, t} \, x_1^{\nu \, \delta} \; x_2^{\nu \, ( 1 - \delta )}
   \exp \left( - \frac{\rho}{2} \, \nu \, \delta \, ( 1 - \delta )
      \left( \ln x_1 - \ln x_2 \right)^2
   \right)
   \label{eq:derivYGammaApprox}\\
\frac{\partial y}{\partial \delta } =\;&
   \gamma \, e^{\lambda \, t} \, \nu \, \left( \ln x_1 - \ln x_2 \right)
   x_1^{ \nu \, \delta } \; x_2^{ \nu ( 1 - \delta ) }
   \label{eq:derivYDeltaApprox}\\
   & \left( 1 - \frac{ \rho }{ 2 }
      \big[ 1 - 2 \, \delta + \nu \, \delta ( 1 - \delta )
         \left( \ln x_1 - \ln x_2 \right) \big]
      \left( \ln x_1 - \ln x_2 \right)
   \right)
   \nonumber\\
\frac{\partial y}{\partial \rho } =\;&
   \gamma \, e^{\lambda \, t} \, \nu \, \delta \, ( 1 - \delta ) \,
   x_1^{ \nu \, \delta } \; x_2^{ \nu ( 1 - \delta ) }
   \bigg( - \frac{1}{2} \left( \ln x_1 - \ln x_2 \right)^2
   \label{eq:derivYRhoApprox}\\
      & + \frac{\rho}{3} ( 1 - 2 \, \delta ) \left( \ln x_1 - \ln x_2 \right)^3
      + \frac{\rho}{4} \, \nu \, \delta ( 1 - \delta )
         \left( \ln x_1 - \ln x_2 \right)^4
   \bigg)
   \nonumber\\
\frac{\partial y}{\partial \nu } =\;&
   \gamma \, e^{\lambda \, t} \, x_1^{ \nu \, \delta } \; x_2^{ \nu ( 1 - \delta ) }
   \bigg( \delta \ln x_1 + ( 1 - \delta ) \ln x_2
      \label{eq:derivYNuApprox}\\
      & - \frac{\rho}{2} \, \delta ( 1 - \delta )
      \left( \ln x_1 - \ln x_2 \right)^2
      \left[ 1 + \nu
         \left( \delta \ln x_1 + ( 1 - \delta ) \ln x_2 \right)
      \right]
   \bigg)
   \nonumber
\end{align}
$$
If $\rho$ is zero or close to zero,
the partial derivatives with respect to $\lambda$ are calculated
also with Equation~\ref{eq:derivYLambda},
but now $\partial y / \partial \gamma$ is calculated
with Equation

140
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-19 09:32:09
Allen-Uzawa elasticity of substitution
$$
\sigma_{i,j} = \begin{cases}
        (1 - \rho )^{-1}
        & \text{for } i = 1,2; \; j=3 \\
        & \\
        \dfrac{(1-\rho_1)^{-1} - (1-\rho)^{-1}}
        {\delta \left( \dfrac{y}{
        B_1^{-\frac{1}{\rho_1}}}\right)^{1+\rho}}+(1-\rho)^{-1}
        & \text{for } i=1; \; j=2
   \end{cases}
$$

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