Calculate the elasticity of substitution of Epstein-Zin preferences $$
\newcommand{\E}{\mathbb{E}}
$$
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^+) &amp;= f(C_t, q(U_{t+1}(C_{t+1}^+))) \\
U_t  &amp;= \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
$\rho^{-1}$?
$
\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} &amp;= f_c(C_t, q_t(U_{t+1}(C_{t+1}^+)))  \\
  &amp;= \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} \\
  &amp;= (1-\beta) f_t^{\rho} C_t^{-\rho}.
\end{align*}
Also,
\begin{align*}
\pd{U_t}{C_{t+1}} &amp;=  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}} &amp;=  f_q \cdot \frac{\dd q_t}{\dd U_{t+1}}
  \cdot \pd{U_{t+1}}{C_{t+1}} \\
  &amp;= \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}
  &amp;= \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}} \\
  &amp;= \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}}
  &amp;= \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}} \\
  &amp;= -\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} &amp;= \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 | \\
  &amp;= \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| \\
  &amp;= \rho^{-1}.
\end{align*}