<p>$
\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.</p>

<p>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*}</p>

CC BY-SA 3.0
2016-06-21T17:04:20.697Z
Following your derivation except for one piece: $\frac{\dd q_t}{\dd U_{t+1}}$. Can you explain why the expectation operator disappears? Specifically why is it not this: $\frac{\dd q_t}{\dd U_{t+1}} = q_t^\gamma E_{t}U_{t+1}^{-\gamma}$
Alex
CC BY-SA 3.0
2016-08-27T01:24:02.553Z
@Alex. It's because I've assumed that a fixed (non-random) consumption sequence. When I get some time, I'd like to try this again without making that assumption.
jmbejara
CC BY-SA 3.0
2015-11-20T17:22:49.877Z
2015-11-20T22:47:12.197Z
2015-11-20T22:47:12.197Z
jmbejara
<p>By using compact notation, and a bold treatment of the differential symbol $\text{d}$, I think you can shortcut this significantly.</p>

<p>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.</p>

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

<p>Then</p>

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

<p>and </p>

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

<p>So</p>

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

<p>Further</p>

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

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

<p>since the first term is a constant. So finally,</p>

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

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

<p>Don't tell your friend the mathematician, but tell Leibniz.</p>

CC BY-SA 3.0
2016-02-25T18:45:59.690Z
This is definitely cleaner. My only concern was that, technically, this appears to give the elasticity of substitution between consumption today $c$ and the continuation value $q$. While this turns out to be the same thing as the EIS between consumption today $C_t$ and consumption tomorrow $C_{t+1}$, I don't feel like I know enough about Epstein-Zin preferences (and recursive preferences, in general) to comment as to when this will or will not be the case generally.
jmbejara
CC BY-SA 3.0
2016-02-25T19:08:59.930Z
@jmbejara Indeed, given the CES specification (which is a mathematical structure, irrespective of the interpretation we attach to the symbols), the "elasticity of substitution" for it is _necessarily_ calculated between its two arguments, which are indeed $C_t$ and $q_t$.
Alecos Papadopoulos
CC BY-SA 3.0
2015-11-20T21:59:31.857Z
2015-11-20T22:04:41.660Z
2015-11-20T22:04:41.660Z
Alecos Papadopoulos
<p>$$
\newcommand{\E}{\mathbb{E}}
$$
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
$\rho^{-1}$?</p>

CC BY-SA 3.0
2016-02-25T14:42:27.920Z
Hi there. In your self-answer you wrote "Let me know if anybody has a cleaner/clearer approach", but you haven't responded at all at the approach I proposed. Any comments?
Alecos Papadopoulos
CC BY-SA 3.0
2016-02-25T18:46:08.923Z
I have commented below.
jmbejara
<utility><elasticity>
Calculate the elasticity of substitution of Epstein-Zin preferences
jmbejara
# Calculate the elasticity of substitution of Epstein-Zin preferences

(Asked by: username_0 on 2015-11-20)

$$
\newcommand{\E}{\mathbb{E}}
$$
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
$\rho^{-1}$?

**Tags:** utility, elasticity

> **Comments:**
> **username_1**: Hi there. In your self-answer you wrote "Let me know if anybody has a cleaner/clearer approach", but you haven't responded at all at the approach I proposed. Any comments?  
> **username_0**: I have commented below.  

---

## 2 answers

### Answer by username_0 on 2015-11-20. Score: 4

$
\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\*}

> **Comments:**
> **username_2**: Following your derivation except for one piece: $\frac{\dd q_t}{\dd U_{t+1}}$. Can you explain why the expectation operator disappears? Specifically why is it not this: $\frac{\dd q_t}{\dd U_{t+1}} = q_t^\gamma E_{t}U_{t+1}^{-\gamma}$  
> **username_0**: @username_2. It's because I've assumed that a fixed (non-random) consumption sequence. When I get some time, I'd like to try this again without making that assumption.  

---

### Answer by username_1 on 2015-11-20. Score: 1

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.

> **Comments:**
> **username_0**: This is definitely cleaner. My only concern was that, technically, this appears to give the elasticity of substitution between consumption today $c$ and the continuation value $q$. While this turns out to be the same thing as the EIS between consumption today $C_t$ and consumption tomorrow $C_{t+1}$, I don't feel like I know enough about Epstein-Zin preferences (and recursive preferences, in general) to comment as to when this will or will not be the case generally.  
> **username_1**: @username_0 Indeed, given the CES specification (which is a mathematical structure, irrespective of the interpretation we attach to the symbols), the "elasticity of substitution" for it is _necessarily_ calculated between its two arguments, which are indeed $C_t$ and $q_t$.  

---

