Exercise 7.3

Special Cases of the CES Armington Aggregator

Problem

Consider the CES Armington aggregator

\[ A(a^m,a^x) = \left[ \chi \left(a^m\right)^{1-\frac1{\mu}} + (1-\chi) \left(a^x\right)^{1-\frac1{\mu}} \right]^{\frac{1}{1-\frac{1}{\mu}}}. \]

Show that

\[ \lim_{\mu\rightarrow1}A(a^m,a^x)= (a^m)^{\chi}(a^x)^{1-\chi}. \]

\[ \lim_{\mu\rightarrow0}A(a^m,a^x)= \min\{ a^m,a^x\}. \]

and

\[ \lim_{\mu\rightarrow\infty}A(a^m,a^x)= \chi a^m + (1-\chi)a^x. \]

Note that the share parameter $\chi$ drops from the Armington aggregator as $\mu\rightarrow 0$. This may be an undesirable property for certain applications. Provide an alternative specification of the CES aggregator such that

\[ \lim_{\mu\rightarrow0}A(a^m,a^x)= \min\{ \chi a^m,(1-\chi)a^x\}. \]

Answer

We first show that

Use

\[ \ln A(a^m,a^x) = \frac{\ln \left[ \chi \left(a^m\right)^{1-\frac1{\mu}} + (1-\chi) \left(a^x\right)^{1-\frac1{\mu}} \right]}{1-\frac{1}{\mu}} \]

For $\mu=1$ this is not defined because the denominator is $0$. Thus let’s apply the l’Hospital rule.

\[ \begin{eqnarray*} &\lim_{\mu\rightarrow 1} \frac{\ln \left[ \chi \left(a^m\right)^{1-\frac1{\mu}} + (1-\chi) \left(a^x\right)^{1-\frac1{\mu}} \right]}{1-\frac{1}{\mu}} \\ =&\lim_{\mu\rightarrow 1} \frac{\frac1{ \left[ \chi \left(a^m\right)^{1-\frac1{\mu}} + (1-\chi) \left(a^x\right)^{1-\frac1{\mu}} \right]} \left[\chi \ln a^m \frac1{\mu^2} \left(a^m\right)^{1-\frac1{\mu}} + (1-\chi) \ln a^x \frac1{\mu^2} \left(a^x\right)^{1-\frac1{\mu}}\right]}{\frac{1}{\mu^2}}\\ =&\chi \ln a^m + (1-\chi) \ln a^x \\ =&\ln\left[ \left(a^m\right)^{\chi}\left(a^x\right)^{(1-\chi)} \right] \end{eqnarray*} \]

Next, we show that

Note that

\[ \lim_{\mu\rightarrow 0} \frac{1}{1-\frac{1}{\mu}} = \lim_{\mu\rightarrow 0} \frac{\mu}{\mu -1} = \frac{0}{-1} =0 \]

and that

\[ \lim_{\mu\rightarrow 0} \left(1-\frac{1}{\mu}\right) = -\infty \]

Suppose that $a^m/a^x>1$, so that $(a^m/a^x)^{-\infty} =0$ Take out $a^x$

\[ A(a^m,a^x) = a^x \left[ \chi \left(a^m/a^x \right)^{1-\frac1{\mu}} + (1-\chi) \right]^\frac1{{1-\frac{1}{\mu}}} \]

Now the part in the square brackets is well defined as $\mu\rightarrow 0$ so that

\[ \lim_{\mu\rightarrow 0} \left[ \chi \left(a^m/a^x \right)^{1-\frac1{\mu}} + (1-\chi) \right]^\frac1{{1-\frac{1}{\mu}}} = \left[ (1-\chi) \right]^0 = 1 \]

and hence if $a^m/a^x>1$, then

\[ \lim_{\mu \rightarrow 0} A(a^m,a^x) = a^x \]

By the same argument we have that if if $a^m/a^x<1$, then

\[ \lim_{\mu \rightarrow 0} A(a^m,a^x) = a^m \]

It follows that

\[ \lim_{\mu \rightarrow 0} A(a^m,a^x) = \min\{a^m, a^x\}, \]

which is what we were asked to show.

Then, we show that

The claim follows from the facts that

\[ \lim_{\mu\rightarrow \infty} \frac{1}{1-\frac{1}{\mu}} = 1 \]

and

\[ \lim_{\mu\rightarrow \infty} \left(1-\frac{1}{\mu}\right) = 1. \]

Finally, we propose an alternative specification such that

Let

\[ A(a^m,a^x) = \left[ \left( \chi a^m \right)^{1-\frac1{\mu}} + \left( (1-\chi) a^x \right)^{1-\frac1{\mu}} \right] ^\frac1{{1-\frac{1}{\mu}}} \]

Suppose $\chi a^m<(1-\chi) a^x$, then we have:

\[ A(a^m,a^x) = \chi a^m \left[ 1 + \left(\frac{ (1-\chi) a^x } {\chi a^m} \right)^{1-\frac1{\mu}} \right] ^\frac1{{1-\frac{1}{\mu}}} \]

For the rest proceed as above.

--- title: "Exercise 7.3" subtitle: "Special Cases of the CES Armington Aggregator" --- <a class="return-button" href="index.qmd">⬅ Return</a> ## Problem Consider the CES Armington aggregator $$ A(a^m,a^x) = \left[ \chi \left(a^m\right)^{1-\frac1{\mu}} + (1-\chi) \left(a^x\right)^{1-\frac1{\mu}} \right]^{\frac{1}{1-\frac{1}{\mu}}}. $$ Show that $$ \lim_{\mu\rightarrow1}A(a^m,a^x)= (a^m)^{\chi}(a^x)^{1-\chi}. $$ $$ \lim_{\mu\rightarrow0}A(a^m,a^x)= \min\{ a^m,a^x\}. $$ and $$ \lim_{\mu\rightarrow\infty}A(a^m,a^x)= \chi a^m + (1-\chi)a^x. $$ Note that the share parameter $\chi$ drops from the Armington aggregator as $\mu\rightarrow 0$. This may be an undesirable property for certain applications. Provide an alternative specification of the CES aggregator such that $$ \lim_{\mu\rightarrow0}A(a^m,a^x)= \min\{ \chi a^m,(1-\chi)a^x\}. $$ ## Answer We first show that Use $$ \ln A(a^m,a^x) = \frac{\ln \left[ \chi \left(a^m\right)^{1-\frac1{\mu}} + (1-\chi) \left(a^x\right)^{1-\frac1{\mu}} \right]}{1-\frac{1}{\mu}} $$ For $\mu=1$ this is not defined because the denominator is $0$. Thus let's apply the l'Hospital rule. $$ \begin{eqnarray*} &\lim_{\mu\rightarrow 1} \frac{\ln \left[ \chi \left(a^m\right)^{1-\frac1{\mu}} + (1-\chi) \left(a^x\right)^{1-\frac1{\mu}} \right]}{1-\frac{1}{\mu}} \\ =&\lim_{\mu\rightarrow 1} \frac{\frac1{ \left[ \chi \left(a^m\right)^{1-\frac1{\mu}} + (1-\chi) \left(a^x\right)^{1-\frac1{\mu}} \right]} \left[\chi \ln a^m \frac1{\mu^2} \left(a^m\right)^{1-\frac1{\mu}} + (1-\chi) \ln a^x \frac1{\mu^2} \left(a^x\right)^{1-\frac1{\mu}}\right]}{\frac{1}{\mu^2}}\\ =&\chi \ln a^m + (1-\chi) \ln a^x \\ =&\ln\left[ \left(a^m\right)^{\chi}\left(a^x\right)^{(1-\chi)} \right] \end{eqnarray*} $$ Next, we show that Note that $$ \lim_{\mu\rightarrow 0} \frac{1}{1-\frac{1}{\mu}} = \lim_{\mu\rightarrow 0} \frac{\mu}{\mu -1} = \frac{0}{-1} =0 $$ and that $$ \lim_{\mu\rightarrow 0} \left(1-\frac{1}{\mu}\right) = -\infty $$ Suppose that $a^m/a^x>1$, so that $(a^m/a^x)^{-\infty} =0$ Take out $a^x$ $$ A(a^m,a^x) = a^x \left[ \chi \left(a^m/a^x \right)^{1-\frac1{\mu}} + (1-\chi) \right]^\frac1{{1-\frac{1}{\mu}}} $$ Now the part in the square brackets is well defined as $\mu\rightarrow 0$ so that $$ \lim_{\mu\rightarrow 0} \left[ \chi \left(a^m/a^x \right)^{1-\frac1{\mu}} + (1-\chi) \right]^\frac1{{1-\frac{1}{\mu}}} = \left[ (1-\chi) \right]^0 = 1 $$ and hence if $a^m/a^x>1$, then $$ \lim_{\mu \rightarrow 0} A(a^m,a^x) = a^x $$ By the same argument we have that if if $a^m/a^x<1$, then $$ \lim_{\mu \rightarrow 0} A(a^m,a^x) = a^m $$ It follows that $$ \lim_{\mu \rightarrow 0} A(a^m,a^x) = \min\{a^m, a^x\}, $$ which is what we were asked to show. Then, we show that The claim follows from the facts that $$ \lim_{\mu\rightarrow \infty} \frac{1}{1-\frac{1}{\mu}} = 1 $$ and $$ \lim_{\mu\rightarrow \infty} \left(1-\frac{1}{\mu}\right) = 1. $$ Finally, we propose an alternative specification such that Let $$ A(a^m,a^x) = \left[ \left( \chi a^m \right)^{1-\frac1{\mu}} + \left( (1-\chi) a^x \right)^{1-\frac1{\mu}} \right] ^\frac1{{1-\frac{1}{\mu}}} $$ Suppose $\chi a^m<(1-\chi) a^x$, then we have: $$ A(a^m,a^x) = \chi a^m \left[ 1 + \left(\frac{ (1-\chi) a^x } {\chi a^m} \right)^{1-\frac1{\mu}} \right] ^\frac1{{1-\frac{1}{\mu}}} $$ For the rest proceed as above.