Thursday, June 4, 2015

Consumer's Problem (Solution and linearization)

In our previous post on IO table, we assumed that consumer can consume two goods ${{S}_{1}}$ and ${{S}_{2}}$ which he can purchase at ${{P}_{1}}$ and ${{P}_{2}}$ prices respectively. Please read the previous post (here) before this. And, if you find any mistakes please kindly make aware of them.

I have also video post the solutions, so get ready with your paper, eraser and pencil. Follow my steps.


A consumer is a rational agent who tries to maximize his/her utility under his/her given budget. Let’s suppose the utility function of the consumer is Cobb Douglas in nature. Then consumer’s problem can be written as: \[{{U}_{\max }}=X_{10}^{{{\alpha }_{10}}}X_{20}^{{{\alpha }_{20}}}\] subjected to budget $Y={{P}_{1}}{{X}_{10}}+{{P}_{2}}{{X}_{20}}$ where ${{\alpha }_{10}}+{{\alpha }_{20}}=1$ and ${{\alpha }_{10}}$ and ${{\alpha }_{10}}$ are positive.Let’s write this in a very fancy mathematical way, that you probably would see in the research papers.

$\begin{align}
  & {{U}_{\max }}=\prod\limits_{i=1}^{2}{X_{i0}^{{{\alpha }_{i0}}}} \\
 & s.t. \\
 & Y=\sum\limits_{i=1}^{2}{{{P}_{i}}{{X}_{i0}}} \\
\end{align}$

Where, $\sum\limits_{i=1}^{2}{{{\alpha }_{i0}}}=1$ and ${{\alpha }_{i0}}>0$.

Can you find which cell above equation represents? 


Now, let’s solve this problem using Lagrange function,
\[L=X_{10}^{{{\alpha }_{10}}}X_{20}^{{{\alpha }_{20}}}+\lambda [Y-({{P}_{1}}{{X}_{10}}+{{P}_{2}}{{X}_{20}})]\]
Taking first order derivative w.r.t ${{X}_{10}}$
\[\frac{dL}{d{{X}_{10}}}={{\alpha }_{10}}X_{10}^{{{\alpha }_{10}}-1}X_{20}^{{{\alpha }_{20}}}-\lambda {{P}_{1}}\]
Setting this equation to zero, for first order condition
\[0={{\alpha }_{10}}X_{10}^{{{\alpha }_{10}}-1}X_{20}^{{{\alpha }_{20}}}-\lambda {{P}_{1}}\]
\[\lambda {{P}_{1}}={{\alpha }_{10}}X_{10}^{{{\alpha }_{10}}-1}X_{20}^{{{\alpha }_{20}}}\](i)
With similar logic, taking first order derivative w.r.t. ${{X}_{20}}$
\[\frac{dL}{d{{X}_{20}}}={{\alpha }_{20}}X_{20}^{{{\alpha }_{20}}-1}X_{10}^{{{\alpha }_{10}}}-\lambda {{P}_{2}}\]
Setting this equation to zero
\[0={{\alpha }_{20}}X_{20}^{{{\alpha }_{20}}-1}X_{10}^{{{\alpha }_{10}}}-\lambda {{P}_{2}}\]
\[\lambda {{P}_{2}}={{\alpha }_{20}}X_{20}^{{{\alpha }_{20}}-1}X_{10}^{{{\alpha }_{10}}}\] (ii)
Now, taking first order derivative w.r.t $\lambda $
\[\frac{dL}{d\lambda }=Y-({{P}_{1}}{{X}_{10}}+{{P}_{2}}{{X}_{20}})\]
Setting it to zero
\[0=Y-({{P}_{1}}{{X}_{10}}+{{P}_{2}}{{X}_{20}})\]
And we get,
\[{{P}_{2}}{{X}_{20}}=Y-{{P}_{1}}{{X}_{10}}\]   (iii)

Now, lets divide equation (i) by (ii)
\[\begin{align}
  & \frac{\lambda {{P}_{1}}}{\lambda {{P}_{2}}}=\frac{{{\alpha }_{10}}X_{10}^{{{\alpha }_{10}}-1}X_{20}^{{{\alpha }_{20}}}}{{{\alpha }_{20}}X_{20}^{{{\alpha }_{20}}-1}X_{10}^{{{\alpha }_{10}}}} \\
 & or,\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{{{\alpha }_{10}}X_{20}^{{{\alpha }_{20}}-{{\alpha }_{20}}+1}}{{{\alpha }_{20}}X_{10}^{{{\alpha }_{10}}-{{\alpha }_{10}}+1}} \\
 & or,\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{{{\alpha }_{10}}{{X}_{20}}}{{{\alpha }_{20}}{{X}_{10}}} \\
 & or,{{\alpha }_{20}}{{P}_{1}}{{X}_{10}}={{\alpha }_{10}}{{P}_{2}}{{X}_{20}} \\
\end{align}\]
From the equation (ii) we know ${{P}_{2}}{{X}_{20}}=Y-{{P}_{1}}{{X}_{10}}$, so
\[\begin{align}
  & or, {{\alpha }_{20}}{{P}_{1}}{{X}_{10}}={{\alpha }_{10}}(Y-{{P}_{1}}{{X}_{10}}) \\
 & or, {{\alpha }_{20}}{{P}_{1}}{{X}_{10}}={{\alpha }_{10}}Y-{{\alpha }_{10}}{{P}_{1}}{{X}_{10}} \\
 & or, {{\alpha }_{20}}{{P}_{1}}{{X}_{10}}+{{\alpha }_{10}}{{P}_{1}}{{X}_{10}}={{\alpha }_{10}}Y \\
 & or, {{P}_{1}}{{X}_{10}}({{\alpha }_{10}}+{{\alpha }_{20}})={{\alpha }_{10}}Y \\
\end{align}\]
Further, $\sum\limits_{i=1}^{2}{{{\alpha }_{i0}}}=1$, so
\[\begin{align}
  & {{P}_{1}}{{X}_{10}}={{\alpha }_{10}}Y \\
 & or, {{X}_{10}}=\frac{{{\alpha }_{10}}Y}{{{P}_{1}}} \\
\end{align}\]
With similar logic,
\[{{X}_{20}}=\frac{{{\alpha }_{20}}Y}{{{P}_{2}}}\]
In our example $i=1, 2$ but while we do research we may need to extent the concept for consumption of $n$ different commodities $i=1,2,3,...,n$. In the Generalized form we can write no matter how many products a consumer consumes.
\[{{X}_{i0}}=\frac{{{\alpha }_{i0}}Y}{{{P}_{i}}}\]
In, the research paper we directly see ${{X}_{i0}}={{\alpha }_{i0}}Y/{{P}_{i}}$ as the solution or optimum quantity that the consumer takes. Since, in CGE we linearize the solution so, taking log on both sides in equation (iv), we get
\[\begin{align}
  & \ln ({{X}_{i0}})=\ln \left( \frac{{{\alpha }_{i0}}Y}{{{P}_{i}}} \right) \\
 & or,\ln ({{X}_{i0}})=\ln ({{\alpha }_{i0}}Y)-\ln ({{P}_{i}}) \\
 & or,\ln ({{X}_{i0}})=\ln ({{\alpha }_{i0}})+\ln (Y)-\ln ({{P}_{i}}) \\
\end{align}\]
Let’s differentiate w.r.t  ${{X}_{i0}}$
\[\begin{align}
  & \frac{1}{{{X}_{i0}}}\frac{d{{X}_{i0}}}{d{{X}_{i0}}}=\frac{1}{{{\alpha }_{i0}}}\frac{d{{\alpha }_{i0}}}{d{{X}_{i0}}}+\frac{1}{Y}\frac{dY}{d{{X}_{i0}}}-\frac{1}{{{P}_{i}}}\frac{d{{P}_{i}}}{d{{X}_{i0}}} \\
 & and \\
 & \frac{d{{\alpha }_{i0}}}{d{{X}_{i0}}}=0 \\
 & \frac{d{{X}_{i0}}}{d{{X}_{i0}}}=1 \\
 & so, \\
 & \frac{1}{{{X}_{i0}}}=\frac{1}{Y}\frac{dY}{d{{X}_{i0}}}-\frac{1}{{{P}_{i}}}\frac{d{{P}_{i}}}{d{{X}_{i0}}} \\
 & \frac{1}{{{X}_{i0}}}=\frac{1}{d{{X}_{i0}}}\frac{dY}{Y}-\frac{1}{d{{X}_{i0}}}\frac{d{{P}_{i}}}{{{P}_{i}}} \\
 & \frac{d{{X}_{i0}}}{{{X}_{i0}}}=\frac{dY}{Y}-\frac{d{{P}_{i}}}{{{P}_{i}}} \\
\end{align}\]
Here, $\frac{d{{X}_{i0}}}{{{X}_{i0}}}$ is change in ${{X}_{i0}}$,$\frac{dY}{Y}$ is change in income $Y$ and $\frac{d{{P}_{i}}}{{{P}_{i}}}$ is change in price level $P$. Let’s represent them as ${{x}_{i0}}$, $y$ and ${{p}_{i}}$. Now our linearize equation is given as:
\[{{x}_{i0}}=y-{{p}_{i}}\]
You must and always remember these two equation. You may notice in research paper (which assumes the CD utility) that these equations will pop out from nowhere. But after these solution and derivation you must be comfortable.
\[{{X}_{i0}}=\frac{{{\alpha }_{i0}}Y}{{{P}_{i}}}\]
\[{{x}_{i0}}=y-{{p}_{i}}\]
In next blog, I will solve the producer's problem. See you.

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