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

151
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-19 09:49:09
$$\begin{eqnarray}
f_{\nu}'\left(0\right) & = & \lim_{\rho\rightarrow0}f_{\nu}'\left(\rho\right)\\
& = & \lim_{\rho\rightarrow0}\left(g_{\nu}'\left(\rho\right)\,\exp\left(f\left(\rho\right)\right)+g_{\nu}\left(\rho\right)\,\exp\left(f\left(\rho\right)\right)\, f'\left(\rho\right)\right)\\
& = & \lim_{\rho\rightarrow0}g_{\nu}'\left(\rho\right)\,\exp\left(\lim_{\rho\rightarrow0}f\left(\rho\right)\right)+\lim_{\rho\rightarrow0}g_{\nu}\left(\rho\right)\,\exp\left(\lim_{\rho\rightarrow0}f\left(\rho\right)\right)\lim_{\rho\rightarrow0}f'\left(\rho\right)\\
& = & g_{\nu}'\left(0\right)\,\exp\left(f\left(0\right)\right)+g_{\nu}\left(0\right)\,\exp\left(f\left(0\right)\right)f'\left(0\right)\\
& = & \exp\left(f\left(0\right)\right)\left(g_{\nu}'\left(0\right)+g_{\nu}\left(0\right)\, f'\left(0\right)\right)\\
& = & \exp\left(\nu\left(\delta\,\ln x_{1}+\left(1-\delta\right)\,\ln x_{2}\right)\right)\left(\frac{\delta\left(1-\delta\right)}{2}\left(\ln x_{1}-\ln x_{2}\right)^{2}\right.\\
&  & \left.+\left(-\delta\,\ln x_{1}-\left(1-\delta\right)\,\ln x_{2}\right)\left(-\frac{\nu\delta\left(1-\delta\right)}{2}\left(\ln x_{1}-\ln x_{2}\right)^{2}\right)\right)\nonumber \\
& = & x_{1}^{\nu\delta}x_{2}^{\nu\left(1-\delta\right)}\frac{\delta\left(1-\delta\right)}{2}\left(\ln x_{1}-\ln x_{2}\right)^{2}\left(1+\nu\left(\delta\,\ln x_{1}+\left(1-\delta\right)\,\ln x_{2}\right)\right)\end{eqnarray}$$

152
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-19 09:49:37
and approximate $\partial y/\partial\nu$ by $$\begin{eqnarray}
\frac{\partial y}{\partial\nu} & \approx & -\gamma \, e^{\lambda \, t} \, \left(f_{\nu}\left(0\right)+\rho f_{\nu}'\left(0\right)\right)\\
& = & \gamma \, e^{\lambda \, t} \, \left(\delta\,\ln x_{1}+\left(1-\delta\right)\,\ln x_{2}\right)x_{1}^{\nu\delta}x_{2}^{\nu\left(1-\delta\right)}\\
&  & -\gamma \, e^{\lambda \, t} \, \rho x_{1}^{\nu\delta}x_{2}^{\nu\left(1-\delta\right)}\frac{\delta\left(1-\delta\right)}{2}\left(\ln x_{1}-\ln x_{2}\right)^{2}\nonumber \\
&  & \left(1+\nu\left(\delta\,\ln x_{1}+\left(1-\delta\right)\,\ln x_{2}\right)\right)\nonumber \\
& = & \gamma \, e^{\lambda \, t} \, x_{1}^{\nu\delta}x_{2}^{\nu\left(1-\delta\right)}\biggl(\delta\,\ln x_{1}+\left(1-\delta\right)\,\ln x_{2}\\
&  & -\frac{\rho\delta\left(1-\delta\right)}{2}\left(\ln x_{1}-\ln x_{2}\right)^{2}\left(1+\nu\left(\delta\,\ln x_{1}+\left(1-\delta\right)\,\ln x_{2}\right)\right)\biggr)\nonumber \end{eqnarray}$$

153
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-19 09:50:26
$$
\begin{eqnarray}
\frac{\partial y}{\partial\rho} & = & \gamma \, e^{\lambda \, t} \, \frac{\nu}{\rho^{2}}\left(\delta\, x_{1}^{-\rho}+\left(1-\delta\right)\, x_{2}^{-\rho}\right)^{-\frac{\nu}{\rho}}\ln\left(\delta\, x_{1}^{-\rho}+\left(1-\delta\right)\, x_{2}^{-\rho}\right)\\
&  & +\gamma \, e^{\lambda \, t} \, \frac{\nu}{\rho}\left(\delta\, x_{1}^{-\rho}+\left(1-\delta\right)\, x_{2}^{-\rho}\right)^{-\left(\frac{\nu}{\rho}+1\right)}\left(\delta\, x_{1}^{-\rho}\ln x_{1}+\left(1-\delta\right)\, x_{2}^{-\rho}\ln x_{2}\right)\nonumber \\
& = & \gamma \, e^{\lambda \, t} \, \nu \, \left(\frac{1}{\rho^{2}}\left(\delta\, x_{1}^{-\rho}+\left(1-\delta\right)\, x_{2}^{-\rho}\right)^{-\frac{\nu}{\rho}}\ln\left(\delta\, x_{1}^{-\rho}+\left(1-\delta\right)\, x_{2}^{-\rho}\right)\right.\\
&  & \left.+\frac{1}{\rho}\left(\delta\, x_{1}^{-\rho}+\left(1-\delta\right)\, x_{2}^{-\rho}\right)^{-\left(\frac{\nu}{\rho}+1\right)}\left(\delta\, x_{1}^{-\rho}\ln x_{1}+\left(1-\delta\right)\, x_{2}^{-\rho}\ln x_{2}\right)\right)\nonumber \end{eqnarray}
$$
Now we define the function $f_{\rho}\left(\rho\right)$ $$\begin{eqnarray}
f_{\rho}\left(\rho\right) & = & \frac{1}{\rho^{2}}\left(\delta\, x_{1}^{-\rho}+\left(1-\delta\right)\, x_{2}^{-\rho}\right)^{-\frac{\nu}{\rho}}\ln\left(\delta\, x_{1}^{-\rho}+\left(1-\delta\right)\, x_{2}^{-\rho}\right)\\
&  & +\frac{1}{\rho}\left(\delta\, x_{1}^{-\rho}+\left(1-\delta\right)\, x_{2}^{-\rho}\right)^{-\left(\frac{\nu}{\rho}+1\right)}\left(\delta\, x_{1}^{-\rho}\ln x_{1}+\left(1-\delta\right)\, x_{2}^{-\rho}\ln x_{2}\right)\nonumber \end{eqnarray}$$

154
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-19 09:51:37
so that we can approximate $\partial y/\partial\rho$ by using the
first-order Taylor series approximation of $f_{\rho}\left(\rho\right)$:
$$\begin{eqnarray}
\frac{\partial y}{\partial\rho} & = & \gamma \, e^{\lambda \, t} \, \nu \, f_{\rho}\left(\rho\right)\\
& \approx & \gamma \, e^{\lambda \, t} \, \nu \, \left(f_{\rho}\left(0\right)+\rho f_{\rho}'\left(0\right)\right)\end{eqnarray}$$

We define the helper function $g_{\rho}\left(\rho\right)$
$$
g_{\rho}\left(\rho\right)=\delta\, x_{1}^{-\rho}\ln x_{1}+\left(1-\delta\right)\, x_{2}^{-\rho}\ln x_{2}$$

155
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-19 09:52:11
with first and second derivative$$\begin{eqnarray}
g_{\rho}'\left(\rho\right) & = & -\delta\, x_{1}^{-\rho}\left(\ln x_{1}\right)^{2}-\left(1-\delta\right)\, x_{2}^{-\rho}\left(\ln x_{2}\right)^{2}\\
g_{\rho}''\left(\rho\right) & = & \delta\, x_{1}^{-\rho}\left(\ln x_{1}\right)^{3}+\left(1-\delta\right)\, x_{2}^{-\rho}\left(\ln x_{2}\right)^{3}\end{eqnarray}$$
and use the functions $g\left(\rho\right)$ and $g_{\nu}\left(\rho\right)$

156
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-20 14:21:46
Hence, we can approximate $\partial y/\partial\rho$ by a second-order
Taylor series approximation:$$\begin{eqnarray}
\frac{\partial y}{\partial\rho} & \approx & \gamma \, e^{\lambda \, t} \, \nu \, \left(f_{\rho}\left(0\right)+\rho\, f_{\rho}'\left(0\right)\right)\\
& = & -\frac{1}{2}\gamma \, e^{\lambda \, t} \, \nu \, \delta \, \left(1-\delta\right)\, x_{1}^{\nu\delta}\, x_{2}^{\nu\left(1-\delta\right)}\left(\ln x_{1}-\ln x_{2}\right)^{2}\\
&  & +\gamma \, e^{\lambda \, t} \, \nu \, \rho \, \delta \, \left(1-\delta\right)\, x_{1}^{\nu\delta}x_{2}^{\nu\left(1-\delta\right)}\nonumber \\
&  & \left(\frac{2}{3}\left(1-2\delta\right)\left(\ln x_{1}-\ln x_{2}\right)^{3}+\frac{1}{2}\nu\delta\left(1-\delta\right)\left(\ln x_{1}-\ln x_{2}\right)^{4}\right)\nonumber \\
& = & \gamma \, e^{\lambda \, t} \, \nu \, \delta \, \left(1-\delta\right)\, x_{1}^{\nu\delta}\, x_{2}^{\nu\left(1-\delta\right)}\Biggl(-\frac{1}{2}\left(\ln x_{1}-\ln x_{2}\right)^{2}\\
&  & +\frac{1}{3}\rho\left(1-2\delta\right)\left(\ln x_{1}-\ln x_{2}\right)^{3}+\frac{1}{4}\rho\nu\delta\left(1-\delta\right)\left(\ln x_{1}-\ln x_{2}\right)^{4}\Biggr)\nonumber \end{eqnarray}$$

157
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-20 14:23:05
%==============================Derivatives three inputs==========================
$$
y = \gamma \, e^{\lambda \, t} \, B^{-1/\rho}
$$
with
$$
\begin{align}
B = & \delta \; B_1^{\rho/\rho_1} + ( 1 - \delta ) x_3^{-\rho}\\
B_1 = & \delta_{1}x_{1}^{-\rho_{1}}+\left(1-\delta_{1}\right)x_{2}^{-\rho_{1}}
\end{align}
$$
For further simplification of the formulas in this section,
we make the following definition:
$$\begin{align}
L_1 &= \delta_1 \ln( x_1 ) + ( 1 - \delta_1 ) \ln( x_2 )
\end{align}$$

158
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-20 14:24:02
Limits for $\rho_1$ and/or $\rho$ approaching zero}

The limits of the three-input nested CES function
for $\rho_1$ and/or $\rho$ approaching zero are:
$$\begin{align}
\lim_{\rho_1 \rightarrow 0} y
   =& \gamma \, e^{\lambda \, t} \, \left( \delta
      \left\{ \exp( L_1 ) \right\}^{-\rho}
      + ( 1 - \delta ) x_3^{-\rho} \right)^{-\frac{\nu}{\rho}}\\
\lim_{\rho \rightarrow 0} y
   =& \gamma \, e^{\lambda \, t} \, \exp \left\{ \nu \left(
      \delta \left( - \frac{\ln( B_1 )}{\rho_1} \right)
      + ( 1 - \delta ) \ln x_3 \right) \right\}\\
\lim_{\rho_1 \rightarrow 0} \, \lim_{\rho \rightarrow 0} y
   =& \gamma \, e^{\lambda \, t} \, \exp \left\{ \nu \left( \delta \; L_1
      + ( 1 - \delta ) \ln x_3 \right) \right\}
\end{align}$$

159
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-20 14:26:04
he partial derivatives of the three-input nested CES function
with respect to the coefficients are:%
The partial derivatives with respect to $\lambda$ are always calculated
using Equation,
where $\partial y / \partial \gamma$ is calculated
according to Equation depending on the values of $\rho_1$ and $\rho$.
$$
\begin{align}
\frac{\partial y}{\partial \gamma} =& e^{\lambda \, t} \, B^{-\nu/\rho} \label{eq:derivY3Gamma}\\
\frac{\partial y}{\partial \delta_1} =& -\gamma \, e^{\lambda \, t} \, \frac{\nu}{\rho} B^{\frac{-\nu - \rho}{\rho}}
     \delta \frac{\rho}{\rho_1} B_1^{\frac{\rho-\rho_1}{\rho_1}}
     ( x_1^{-\rho_1} - x_2^{-\rho_1} )\\
\frac{\partial y}{\partial \delta} =& -\gamma \, e^{\lambda \, t} \, \frac{\nu}{\rho} B^{\frac{-\nu - \rho}{\rho}}
     \left[ B_1^{\frac{\rho}{\rho_1}} - x_3^{-\rho} \right]\\
\frac{\partial y}{\partial \nu} =& - \gamma \, e^{\lambda \, t} \, \frac{1}{\rho} \ln (B) B^{-\frac{\nu}{\rho}}\\  
\frac{\partial y}{\partial \rho_1} =& -\gamma \, e^{\lambda \, t} \, \frac{\nu}{\rho} B^{\frac{-\nu - \rho}{\rho}} \delta \\
     & \left[ \ln( B_1 )
     B_1^{\frac{\rho}{\rho_1}} \left( -\frac{\rho}{\rho_1^2} \right)
     + \frac{\rho}{\rho_1} B_1
     \left( -\delta_1 \ln x_1 x_1^{\rho_1} - (1- \delta_1) \ln x_2 x_2^{-\rho_1} \right) \right]^{\frac{\rho - \rho_1}{\rho_1}} \nonumber\\
\frac{\partial y}{\partial \rho} =& \gamma \, e^{\lambda \, t} \, \frac{\nu}{\rho^2} \ln( B ) B^{-\frac{\nu}{\rho}} -
     \gamma \, e^{\lambda \, t} \, \frac{\nu}{\rho} B^{\frac{-\nu - \rho}{\rho}}
     \left\{ \delta \ln( B_1 )
     B_1^{\frac{\rho}{\rho_1}} \frac{1}{\rho_1}
      -( 1 - \delta ) \ln x_3 x_3^{-\rho} \right\}
\end{align}
$$

160
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-20 14:28:10
%========================CES three inputs limits of rho=========================
Limits of the derivatives for rho approaching zero
   Limits of the derivatives for $\rho$ approaching zero
$$
\begin{align}
\lim_{\rho \rightarrow 0} \frac{\partial y}{\partial \gamma} =&
          e^{\lambda \, t} \, \exp \left\{ \nu \left( \delta \left(   
         - \frac{\ln( B_1 )}{\rho_1} \right)
         + ( 1 - \delta ) \ln x_3 \right) \right\}                                 
\label{eq:derivY3GammaRho0}
\end{align}$$
$$
\begin{align}
\lim_{\rho \rightarrow 0} \frac{\partial y}{\partial \delta_1} =&
         -\gamma \, e^{\lambda \, t} \, \nu \left[\left(\delta \frac{(x_1^{-\rho_1} - x_2^{-\rho_1})}{\rho_1
         B_1 }\right)
         \exp \left\{ -\nu \left(-\delta \left(
         - \frac{\ln( B_1 )}{\rho_1} \right)
         - ( 1 - \delta ) \ln x_3 \right) \right\} \right]
\end{align}
$$
$$
\begin{align}
\lim_{\rho \rightarrow 0} \frac{\partial y}{\partial \delta} =&
         -\gamma \, e^{\lambda \, t} \, \nu \left[ \left(  
         \frac{\ln( B_1 )}{\rho_1} + \ln x_3 \right)
         \exp \left\{ \nu \left(-\delta \left(  
         -\frac{\ln( B_1 )}{\rho_1} \right)
         - ( 1 - \delta ) \ln x_3 \right) \right\} \right]
\end{align}
$$
$$
\begin{align}
\lim_{\rho \rightarrow 0} \frac{\partial y}{\partial \nu} =& \gamma \, e^{\lambda \, t} \,
         \left( \delta \left( - \frac{\ln( B_1 )}{\rho_1} \right)
         + ( 1 - \delta ) \ln x_3 \right) \\
         & \cdot \exp \left\{-\nu \left(-\delta \left(  
         -\frac{\ln( B_1 )}{\rho_1} \right)
         - ( 1 - \delta ) \ln x_3 \right) \right\} \nonumber
\end{align}
$$
$$
\begin{align}
\lim_{\rho \rightarrow 0} \frac{\partial y}{\partial \rho_1} =&
        -\gamma \, e^{\lambda \, t} \, \nu \frac{\delta}{\rho_1} \left( -\frac{\ln( B_1 )}{\rho_1}
        + \frac{ - \delta_1 \ln x_1 x_1^{-\rho_1} - (1-\delta_1) \ln x_2 x_2^{-\rho_1} }
        { B_1 } \right) \\
        & \cdot \exp \left\{ -\nu \left(-\delta \left(  
        - \frac{\ln( B_1 )}{\rho_1} \right)
        - ( 1 - \delta ) \ln x_3 \right) \right\} \nonumber
\end{align}
$$
$$
\begin{align}
\lim_{\rho\rightarrow0}\frac{\partial y}{\partial\rho} =&
       \gamma \, e^{\lambda \, t} \, \nu\:\left[-\frac{1}{2}\left(\delta\left( \frac{\ln B_{1}}{\rho_{1}} \right)^{2}
       +\left(1-\delta\right)\left(\ln x_{3}\right)^{2}\right)
       +\frac{1}{2}\left(\delta \frac{\ln B_{1}}{\rho_{1}} -\left(1-\delta\right)\ln x_{3}\right)^{2}\right]\\
       & \cdot \exp\left(\nu\left(-\delta    \frac{\ln B_{1}}{\rho_{1}} +\left(1-\delta\right)\ln x_{3}\right)\right) \nonumber
\end{align}
$$

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