Calculate the elasticity of substitution of Epstein-Zin preferences

Question

$$ \newcommand{\E}{\mathbb{E}} $$ Let a consumption sequence be given $C=(C_0, C_1,...)$ and let $C_t^+ = (C_t, C_{t+1}, ...)$. Now, suppose I have Epstein-Zin preferences, \begin{align*} U_t(C_t^+) &= f(C_t, q(U_{t+1}(C_{t+1}^+))) \\ U_t &= \left \{(1-\beta) C_t^{1-\rho} + \beta \left(\E_t[U_{t+1}^{1-\gamma}]\right)^{\frac{1-\rho}{1-\gamma}} \right\}^{\frac{1}{1-\rho}}, \end{align*} where $f$ is the time aggregator and $q$ is the conditional certainty equivalent operator. That is, $$ f(c,q) = ((1-\beta) c^{1-\rho} + \beta q^{1-\rho})^{\frac{1}{1-\rho}} $$ and $$ q_t = q(U_{t+1}) = \left(\E_t[U_{t+1}^{1-\gamma}]\right)^{\frac{1}{1-\gamma}}. $$ How do I show that the intertemporal elasticity of substitution is $\rho^{-1}$?

Answer 1

$ \newcommand{\dd}{\, \mathrm{d}} \newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}} $ Here is my solution. Let me know if anybody has a cleaner/clearer approach.

Consider a fixed (non-random) consumption sequence $C = (C_0, C_1,...)$. Then, the elasticity of intertemporal substitution (EIS) is defined as $$ \text{EIS} = \left| \frac{\dd \ln(C_s/C_t)}{\dd \ln MRS_{s,t}}\right| = \left|\frac{\dd \ln \left(\frac{C_s}{C_t}\right)} {\dd \ln \left( \frac{\partial U/\partial C_s}{\partial U/\partial C_{t}}\right)} \right |. $$ To compute this figure, let's start by computing \begin{align*} \pd{U_t}{C_t} &= f_c(C_t, q_t(U_{t+1}(C_{t+1}^+))) \\ &= \frac{1}{1-\rho} \left( (1-\beta) C_t^{1-\rho} + \beta q^{1-\rho} \right)^{\frac{\rho}{1-\rho}} (1-\beta) (1-\rho) C_t^{-\rho} \\ &= (1-\beta) f_t^{\rho} C_t^{-\rho}. \end{align*} Also, \begin{align*} \pd{U_t}{C_{t+1}} &= f_q \cdot \frac{\dd q_t}{\dd U_{t+1}} \cdot \pd{U_{t+1}}{C_{t+1}}. \end{align*} It will be easier to compute these pieces in parts. First, $$ f_q = \beta f^{\rho} q_t^{-\rho}. $$ Next, consider $\frac{\dd q_t}{\dd U_{t+1}}$. This is simplified by the fact that $C$ is non-random, $$ \frac{\dd q_t}{\dd U_{t+1}} = q_t^\gamma U_{t+1}^{-\gamma} = 1. $$ Finally, $$ \pd{U_{t+1}}{C_{t+1}} = (1-\beta) f_{t+1}^{\rho} C_{t+1}^{-\rho}, $$ which follows from our earlier calculation. Thus, \begin{align*} \pd{U_t}{C_{t+1}} &= f_q \cdot \frac{\dd q_t}{\dd U_{t+1}} \cdot \pd{U_{t+1}}{C_{t+1}} \\ &= \beta f_t^{\rho} q_t^{-\rho} (1-\beta) f_{t+1}^{\rho} C_{t+1}^{-\rho}, \end{align*} where $f_t = f(C_t, q_t)$ and $q_t = q(U_{t+1}(C_{t+1}^+))$. Now, we can calculate \begin{align*} \frac{\partial U_t/\partial C_{t+1}}{\partial U_t/\partial C_t} &= \frac { \beta f_t^{\rho} q_t^{-\rho} (1-\beta) f_{t+1}^{\rho} C_{t+1}^{-\rho}} {(1-\beta) f_t^{\rho} C_t^{-\rho}} \\ &= \beta q_t^{-\rho} f_{t+1}^{\rho} \left(\frac{C_{t+1}}{C_t}\right)^{-\rho}\\ \end{align*} Now, let $$ \frac{p_{t+1}}{p_t} = \frac{\partial U_t/\partial C_{t+1}}{\partial U_t/\partial C_t}. $$ Then, taking the differentials, $$ \dd \left(\frac{p_t}{p_0}\right) = \frac{\partial U_t/\partial C_{t+1}}{\partial U_t/\partial C_t} \left(\frac{C_{t+1}}{C_t}\right)^{-1} (-\rho) \dd \frac{C_{t+1}}{C_t}. $$ So, \begin{align*} \frac{\dd\left(\frac{ C_{t+1}}{C_t}\right)}{\dd \left(\frac{p_{t+1}}{p_t}\right)} \cdot \frac{\frac{p_{t+1}}{p_t}}{\frac{C_{t+1}}{C_t}} &= \left( \frac{\partial U_t/\partial C_{t+1}}{\partial U_t/\partial C_t} \left(\frac{C_{t+1}}{C_t}\right)^{-1} (-\rho) \right)^{-1} \cdot \frac{\frac{p_{t+1}}{p_t}}{\frac{C_{t+1}}{C_t}} \\ &= -\frac{\frac{C_{t+1}}{C_t} \frac{p_{t+1}}{p_t}} {\frac{p_{t+1}}{p_t} \frac{C_{t+1}}{C_t}} \rho^{-1} = -\rho^{-1}. \end{align*} Now, plugging this into the definition of the EIS, \begin{align*} \text{EIS} &= \left|\frac{\dd \ln \left(\frac{C_s}{C_t}\right)} {\dd \ln \left( \frac{\partial U/\partial C_s}{\partial U/\partial C_{t}}\right)} \right | \\ &= \left| \frac{\dd\left(\frac{ C_{t+1}}{C_t}\right)}{\dd \left(\frac{p_{t+1}}{p_t}\right)} \cdot \frac{\frac{p_{t+1}}{p_t}}{\frac{C_{t+1}}{C_t}} \right| \\ &= \rho^{-1}. \end{align*}

Answer 2

By using compact notation, and a bold treatment of the differential symbol $\text{d}$, I think you can shortcut this significantly.

Mathematically, this is a bivariate CES function so we know that the elasticity of substitution will be constant between the two arguments, whatever they are.

$$f(c,q) = ((1-\beta) c^{1-\rho} + \beta q^{1-\rho})^{\frac{1}{1-\rho}} = \left[h(c,q\right]^{\frac{1}{1-\rho}}$$

Then

$$\frac {\partial f}{\partial c} = \frac{1}{1-\rho}\left[h(c,q\right]^{\frac{1}{1-\rho}-1}\cdot h_c$$

and

$$\frac {\partial f}{\partial q} = \frac{1}{1-\rho}\left[h(c,q\right]^{\frac{1}{1-\rho}-1}\cdot h_q$$

So

$$\frac{\partial f/\partial c}{\partial f/\partial q} = \frac{h_c}{h_q} = \frac {(1-\beta)}{\beta}\cdot (c/q)^{-\rho}$$

Further

$$\ln \left(\frac{\partial f/\partial c}{\partial f/\partial q}\right) = \ln\frac {(1-\beta)}{\beta} -\rho\ln (c/q)$$

$$\implies \text{d}\left[\ln \left(\frac{\partial f/\partial c}{\partial f/\partial q}\right)\right] = -\rho\cdot \text{d}\,\ln (c/q)$$

since the first term is a constant. So finally,

$$\text{EIS} = \left|\frac{\text{d}\, \ln \left(c/q\right)} {\text{d}\, \ln \left( \frac{\partial f/\partial c}{\partial f/\partial q}\right)} \right |$$

$$=\left|\frac{\text{d}\, \ln \left(c/q\right)} {-\rho\cdot \text{d}\,\ln (c/q)} \right | = \rho^{-1}$$

Don't tell your friend the mathematician, but tell Leibniz.