Friday, June 5, 2015

linearization of Solution of Producer's Problem

From the previous post (here), we know that the solution of producer’s problem is
\[{{X}_{ij}}={{\alpha }_{ij}}{{X}_{j}}Q\prod\limits_{i=1}^{4}{P_{i}^{{{\alpha }_{ij}}}}/{{P}_{i}}\] Where, \[Q=\left( \frac{1}{{{A}_{j}}}\prod\limits_{i=1}^{4}{\alpha _{ij}^{-{{\alpha }_{ij}}}} \right)\]


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Now, Let’s linearize this solution, for that lets take log on both sides
\[\begin{align}
  & Ln({{X}_{ij}})=Ln\left( {{\alpha }_{ij}}{{X}_{j}}Q\prod\limits_{i=1}^{4}{P_{i}^{{{\alpha }_{ij}}}}/{{P}_{i}} \right) \\
 & Ln({{X}_{ij}})=Ln\left( {{\alpha }_{ij}}{{X}_{j}}QP_{1}^{{{\alpha }_{1j}}}P_{2}^{{{\alpha }_{ij}}}P_{3}^{{{\alpha }_{3j}}}P_{4}^{{{\alpha }_{4j}}}/{{P}_{i}} \right) \\
 & or,Ln({{X}_{ij}})=Ln({{\alpha }_{ij}})+Ln({{X}_{j}})+Ln(Q)+Ln(P_{1}^{{{\alpha }_{1j}}})+Ln(P_{2}^{{{\alpha }_{ij}}})+Ln(P_{3}^{{{\alpha }_{3j}}})+Ln(P_{4}^{{{\alpha }_{4j}}})-Ln({{P}_{i}}) \\
 & or,Ln({{X}_{ij}})=Ln({{\alpha }_{ij}})+Ln({{X}_{j}})+Ln(Q)+{{\alpha }_{1j}}Ln({{P}_{1}})+{{\alpha }_{2j}}Ln({{P}_{2}})+{{\alpha }_{3j}}Ln({{P}_{3}})+{{\alpha }_{4j}}Ln({{P}_{4}})-Ln({{P}_{i}}) \\
 & or,Ln({{X}_{ij}})=Ln({{\alpha }_{ij}})+Ln({{X}_{j}})+Ln(Q)+\sum\limits_{i=1}^{4}{{{\alpha }_{ij}}Ln({{P}_{i}})}-Ln({{P}_{i}}) \\
\end{align}\]
Now taking differentiation w.r.t${{X}_{ij}}=\frac{{{\alpha }_{ij}}\lambda {{X}_{j}}}{{{P}_{i}}}$
\[\begin{align}
  & \frac{1}{{{X}_{ij}}}\frac{d{{X}_{ij}}}{d{{X}_{ij}}}=\frac{1}{{{\alpha }_{ij}}}\frac{d{{\alpha }_{ij}}}{d{{X}_{ij}}}+\frac{1}{{{X}_{j}}}\frac{d{{X}_{j}}}{d{{X}_{ij}}}+\frac{1}{Q}\frac{dQ}{d{{X}_{ij}}}+\left( \sum\limits_{i=1}^{4}{{{\alpha }_{ij}}\frac{1}{{{P}_{i}}}\frac{d{{P}_{i}}}{d{{X}_{ij}}}} \right)-\frac{1}{{{P}_{i}}}\frac{d{{P}_{i}}}{d{{X}_{ij}}} \\
 & here, \\
 & \frac{d{{X}_{ij}}}{d{{X}_{ij}}}=1, \\
 & \frac{d{{\alpha }_{ij}}}{d{{X}_{ij}}}=0, \\
 & \frac{dQ}{d{{X}_{ij}}}=0 \\
 & so, \\
 & \frac{1}{{{X}_{ij}}}=0+\frac{1}{{{X}_{j}}}\frac{d{{X}_{j}}}{d{{X}_{ij}}}+0+\left( \sum\limits_{i=1}^{4}{{{\alpha }_{ij}}\frac{1}{{{P}_{i}}}\frac{d{{P}_{i}}}{d{{X}_{ij}}}} \right)-\frac{1}{{{P}_{i}}}\frac{d{{P}_{i}}}{d{{X}_{ij}}} \\
 & or,\frac{1}{{{X}_{ij}}}=\frac{1}{{{X}_{j}}}\frac{d{{X}_{j}}}{d{{X}_{ij}}}+\left( \sum\limits_{i=1}^{4}{{{\alpha }_{ij}}\frac{1}{{{P}_{i}}}\frac{d{{P}_{i}}}{d{{X}_{ij}}}} \right)-\frac{1}{{{P}_{i}}}\frac{d{{P}_{i}}}{d{{X}_{ij}}} \\
\end{align}\]
Taking $\frac{1}{d{{X}_{ij}}}$ as common, then
\[\begin{align}
  & or,\frac{1}{{{X}_{ij}}}=\frac{1}{d{{X}_{ij}}}\left( \frac{d{{X}_{j}}}{{{X}_{j}}}+\left( \sum\limits_{i=1}^{4}{{{\alpha }_{ij}}\frac{d{{P}_{i}}}{{{P}_{i}}}} \right)-\frac{d{{P}_{i}}}{d{{P}_{i}}} \right) \\
 & or,\frac{d{{X}_{ij}}}{{{X}_{ij}}}=\frac{d{{X}_{j}}}{{{X}_{j}}}+\left( \sum\limits_{i=1}^{4}{{{\alpha }_{ij}}\frac{d{{P}_{i}}}{{{P}_{i}}}} \right)-\frac{d{{P}_{i}}}{{{P}_{i}}} \\
\end{align}\]
Now, $\frac{d{{X}_{ij}}}{{{X}_{ij}}}$ is the change in ${{X}_{ij}}$, $\frac{d{{X}_{j}}}{{{X}_{j}}}$ is change in ${{X}_{j}}$ and $\frac{d{{P}_{i}}}{{{P}_{i}}}$ is change in ${{P}_{i}}$. Lets represent them by ${{x}_{ij}}$, ${{x}_{j}}$ and ${{p}_{i}}$ respectively. Then our equation will be:
\[\therefore {{x}_{ij}}={{x}_{j}}+\sum\limits_{i=1}^{4}{{{\alpha }_{ij}}{{p}_{i}}}-{{p}_{i}}\]
This is the final linear solution, which you will see in many research paper while the CD production function is assumed.

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