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	\section*{Supplement to Lecture 3: Derivation of CES Special Cases}
	\begin{center} Macroeconomics, EC2B1 \\ Benjamin Moll \end{center}
	In the lecture notes we defined the CES production function
	$$Y = \left(\alpha^{\frac{1}{\sigma}} (A_K K)^\frac{\sigma-1}{\sigma} +  (1-\alpha)^{\frac{1}{\sigma}} (A_N N)^\frac{\sigma-1}{\sigma} \right)^\frac{\sigma}{\sigma-1}, \quad \sigma = \mbox{elasticity of substitution}$$
	and asserted that there are three special cases
	\begin{enumerate}
		\item Case $\sigma = 1$: Cobb-Douglas
		$$Y = \left(\frac{A_K K}{\alpha}\right)^\alpha \left(\frac{A_N N}{1-\alpha}\right)^{1-\alpha} $$
		\item Case $\sigma \rightarrow \infty$: perfect substitutes
		$$Y = A_K K + A_N N$$
		\item Case $\sigma = 0$: perfect complements, fixed proportions, or ``Leontief"
		$$Y = \min\left\{\frac{A_K K}{\alpha},\frac{A_N N}{1-\alpha} \right\} $$
	\end{enumerate}
	This supplement to the lecture notes provides the corresponding derivations.
	
	
	\section{Case $\sigma = 1$: Cobb-Douglas}
	
	In order to make computations easier, we can define $\rho = \frac{1}{\sigma}-1 \Leftrightarrow \sigma = \frac{1}{1+\rho}$:
	$$Y = \left(\alpha^{1+\rho}(A_KK)^{-\rho} + (1-\alpha)^{1+\rho}(A_NN)^{-\rho}\right)^{-\frac{1}{\rho}}$$
	Therefore studying the case of $\sigma = 1$ is the same as $\rho = 0$. Taking the limit as $\rho \rightarrow 0$:
	$$  Y = \exp\left\{ \underset{\rho \rightarrow 0}{\lim} - \dfrac{1}{\rho}\ln\left(\alpha^{1+\rho}(A_KK)^{-\rho} + (1-\alpha)^{1+\rho}(A_NN)^{-\rho}\right)\right\} $$
Taking the limit yields $\frac{0}{0}$. Therefore we use L'Hôpital's rule:
	\scriptsize
	$$  Y = \exp\left\{ \underset{\rho \rightarrow 0}{\lim} - \dfrac{\alpha^{1+\rho}\ln(\alpha)(A_KK)^{-\rho} - \alpha^{1+\rho}(A_KK)^{-\rho}\ln(A_KK) + (1-\alpha)^{1+\rho}\ln(1-\alpha)(A_NN)^{-\rho}-(1-\alpha)^{1+\rho}(A_NN)^{-\rho}\ln(A_NN)}{\alpha^{1+\rho}(A_KK)^{-\rho}+(1-\alpha)^{1+\rho}(A_NN)^{-\rho}}\right\} $$
	\normalsize
Taking the limit and using a bit of algebra:
	$$ Y = \exp\left\{- \dfrac{\alpha\ln(\alpha) - \alpha\ln(A_KK) + (1-\alpha)\ln(1-\alpha)-(1-\alpha)\ln(A_NN)}{\alpha+(1-\alpha)}\right\} = \left(\frac{A_K K}{\alpha}\right)^\alpha \left(\frac{A_N N}{1-\alpha}\right)^{1-\alpha}.$$
	
	\section{Case $\sigma \rightarrow \infty$: perfect substitutes}
	
	Again, taking the limit:
	$$ Y = \underset{\sigma \rightarrow \infty}{\lim} \left(\alpha^{\frac{1}{\sigma}} (A_K K)^\frac{\sigma-1}{\sigma} +  (1-\alpha)^{\frac{1}{\sigma}} (A_N N)^\frac{\sigma-1}{\sigma} \right)^\frac{\sigma}{\sigma-1} $$
	$$ Y = \left(\underset{\sigma \rightarrow \infty}{\lim} \left[\alpha^{\frac{1}{\sigma}} (A_K K)^\frac{\sigma-1}{\sigma} +  (1-\alpha)^{\frac{1}{\sigma}} (A_N N)^\frac{\sigma-1}{\sigma}\right] \right)^{\underset{\sigma \rightarrow \infty}{\lim} \frac{\sigma}{\sigma-1}} $$
	$$ Y = A_KK+A_NN $$
	
	\section{Case $\sigma = 0$: perfect complements, fixed proportions, or ``Leontief"}
	
	Going back to the alternative representation of the CES as a function of $\rho$ and taking the limit as $\rho \rightarrow \infty$:
	$$ Y = \exp\left\{ \underset{\rho \rightarrow \infty}{\lim} - \dfrac{1}{\rho}\ln\left(\alpha^{1+\rho}(A_KK)^{-\rho} + (1-\alpha)^{1+\rho}(A_NN)^{-\rho}\right)\right\} $$
This time taking the limit $\rho \rightarrow \infty$ yields $\frac{\infty}{\infty}$. Therefore again using L'Hôpital's rule:
	\scriptsize
	$$ Y = \exp\left\{ \underset{\rho \rightarrow \infty}{\lim} - \dfrac{\alpha^{1+\rho}\ln(\alpha)(A_KK)^{-\rho} - \alpha^{1+\rho}(A_KK)^{-\rho}\ln(A_KK) + (1-\alpha)^{1+\rho}\ln(1-\alpha)(A_NN)^{-\rho}-(1-\alpha)^{1+\rho}(A_NN)^{-\rho}\ln(A_NN)}{\alpha^{1+\rho}(A_KK)^{-\rho}+(1-\alpha)^{1+\rho}(A_NN)^{-\rho}}\right\} $$
	\normalsize
	Let $x=\min\left\{\frac{A_KK}{\alpha},\frac{A_N N}{1-\alpha}\right\}$, $\theta_K = \frac{A_KK}{\alpha x}$ and  $\theta_N = \frac{A_NN}{(1-\alpha) x}$. Take the previous expression and divide the numerator and denominator by $x^{-\rho}$. Simplifying, we get:
	$$ 	Y = \exp\left\{ \underset{\rho \rightarrow \infty}{\lim} - \dfrac{\alpha\ln(\alpha)\theta_K^{-\rho} - \alpha\ln(A_KK)\theta_K^{-\rho} + (1-\alpha)\ln(1-\alpha)\theta_N^{-\rho} - (1-\alpha)\ln(A_NN)\theta_N^{-\rho}}{\alpha\theta_N^{-\rho}+(1-\alpha)\theta_N^{-\rho}}\right\} $$
Using the definitions of $x,\theta_K$ and $\theta_N$ we have $1=\min\left\{\theta_K,\theta_N\right\}$. Therefore we must have either $\theta_K = 1 \Leftrightarrow x = \frac{A_KK}{\alpha}$ or  $\theta_N = 1 \Leftrightarrow x = \frac{A_NN}{1-\alpha}$, since one of $\theta_K$ or $\theta_N$ has to be the smaller of the two. It also follows that $\theta_i \ge 1, \forall i = \{K,N\}$, given that one of $\theta_K$ or $\theta_N$ has to be the smaller of the two. We have that:
	\begin{equation*}
		\underset{\rho \rightarrow \infty}{\lim}\theta_i ^{-\rho }= \begin{cases}
			1 & \text{ if } \theta_i = 1 \\
			0 & \text{ if } \theta_i > 1 \\
		\end{cases}
	\end{equation*}
	Applying this result to the limit of interest: 
	\begin{equation*}
		\begin{split}
			Y & = \begin{cases}
				\exp\left\{ - \tfrac{\alpha\ln(\alpha) - \alpha\ln(A_KK)}{\alpha}\right\} & \text{ if } \theta_K = 1 \\ 
				\exp\left\{ - \tfrac{(1-\alpha)\ln(1-\alpha) - (1-\alpha)\ln(A_NN)}{1-\alpha}\right\} & \text{ if } \theta_N = 1 \end{cases} \\
			& = \begin{cases}
				\tfrac{A_KK}{\alpha} & \text{ if } x = \tfrac{A_KK}{\alpha} \\ 
				\tfrac{A_NN}{\alpha} & \text{ if } x = \tfrac{A_NN}{\alpha} \\
			\end{cases} \\
			& = x = \min\left\{\frac{A_K K}{\alpha},\frac{A_N N}{1-\alpha} \right\}
		\end{split}
	\end{equation*}
	
	
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