<p>Here's my guess.</p>
<p>Let use the notation
<span class="math-container">$$
\tilde x_t \approx \ln(x_t) - \ln(x) \approx \dfrac{x_t - x}{x}.
$$</span>
If we take logs on both sides we get:
<span class="math-container">$$
\ln(G_t) = \frac{1}{1 -\rho} \ln(p_t) + \ln(y_t) - \ln(1 + p_t^{\frac{\rho}{\rho-1}})
$$</span>
Subtracting the steady state gives:
<span class="math-container">$$
\tilde G_t = \frac{1}{1 - \rho} \tilde p_t + \tilde y_t - \left[\ln(1 + p^{\frac{\rho}{\rho - 1}}) - \ln(1 + p^{\frac{\rho}{\rho - 1}})\right]
$$</span></p>
<p>Taking a Taylor expansion of the last term gives:
<span class="math-container">$$
\begin{align*}
\ln(1 + p_t^{\frac{\rho}{\rho - 1}}) - \ln(1 + p^{\frac{\rho}{\rho - 1}}) &amp;\approx \frac{1}{1 + p^{\frac{\rho}{\rho - 1}}}\frac{\rho}{\rho - 1}p^{\frac{\rho}{\rho - 1}}\frac{(p_t- p)}{p},\\
&amp;\approx \frac{p^{\frac{\rho}{\rho- 1}}}{1 + p^{\frac{\rho}{\rho - 1}}} \frac{\rho}{\rho - 1} \tilde p_t
\end{align*}
$$</span>
So we get:
<span class="math-container">$$
\tilde G_t \approx \left(\frac{1}{1 - \rho}- \frac{p^{\frac{\rho}{\rho - 1}}}{1 + p^{\frac{\rho}{\rho - 1}}} \frac{\rho}{\rho - 1} \right) \tilde p_t + \tilde y_t,\\
= \left(\frac{1}{1 - \rho}- \frac{G}{Y} \frac{\rho}{\rho - 1} \right) \tilde p_t + \tilde y_t,\
$$</span></p>

CC BY-SA 4.0
2021-10-01T07:20:29.030Z
Thank you, that was really helpful!
An economict
CC BY-SA 4.0
2021-09-30T16:10:49.383Z
2021-09-30T16:10:49.383Z
tdm
<p>I have been trying to log-linearise the demand function that follows from a standard two-good CES-utility maximamalisation problem. That is:</p>
<p>Maximise
<span class="math-container">\begin{eqnarray}
U(h,c)= \left(G_1^{\rho}+ G_2^{\rho} \right)^{1/\rho}
   \end{eqnarray}</span>
Subjected to:
<span class="math-container">\begin{eqnarray}
Y = G_1 +  pG_2   
   \end{eqnarray}</span><br />
Where I normalised the price of the first good to 1. From this maximization it follows that the demand of good 2 can be written as:
<span class="math-container">\begin{eqnarray}\label{VAR}
 G_2  =  \frac{  p^{\frac{1}{\rho-1}}Y}{  1   +  p^ {\frac{\rho}{\rho-1}}   }
   \end{eqnarray}</span>
When I try to log-linearise this equation I am not sure how to work with the denominator. Thus far, I have this:</p>
<p><span class="math-container">\begin{eqnarray}
\ln G_2  = \frac{1}{\rho-1} \ln   p+ \ln y - \ln {  (1   +  p^ {\frac{\rho}{\rho-1}})   }
   \end{eqnarray}</span></p>
<p><span class="math-container">\begin{eqnarray}
\ln G_2^* +\frac{1}{G_2^*}(G_2-G_2^*)  = \frac{1}{\rho-1} \ln p^* +\frac{\frac{1}{\rho-1}}{p^*}(p-p^*)
+ \ln y^* +\frac{1}{y^*}(y-y^*) - ???
   \end{eqnarray}</span>
Yet, I fail to log-linearise the last term with the summation and exponent. I tried to find similar log-linearizations online, but I was only able to find papers that log-linearise the production function of a CES-function, which was not helpful for my problem.</p>

<macroeconomics><utility><log-linearization>
Log Linearising CES demand
An economict
# Log Linearising CES demand

(Asked by: username_0 on 2021-09-30)

I have been trying to log-linearise the demand function that follows from a standard two-good CES-utility maximamalisation problem. That is:

Maximise
\begin{eqnarray}
U(h,c)= \left(G\_1^{\rho}+ G\_2^{\rho} \right)^{1/\rho}
\end{eqnarray}
Subjected to:
\begin{eqnarray}
Y = G\_1 + pG\_2
\end{eqnarray}  
Where I normalised the price of the first good to 1. From this maximization it follows that the demand of good 2 can be written as:
\begin{eqnarray}\label{VAR}
G\_2 = \frac{ p^{\frac{1}{\rho-1}}Y}{ 1 + p^ {\frac{\rho}{\rho-1}} }
\end{eqnarray}
When I try to log-linearise this equation I am not sure how to work with the denominator. Thus far, I have this:

\begin{eqnarray}
\ln G\_2 = \frac{1}{\rho-1} \ln p+ \ln y - \ln { (1 + p^ {\frac{\rho}{\rho-1}}) }
\end{eqnarray}

\begin{eqnarray}
\ln G\_2^\* +\frac{1}{G\_2^\*}(G\_2-G\_2^\*) = \frac{1}{\rho-1} \ln p^\* +\frac{\frac{1}{\rho-1}}{p^\*}(p-p^\*)
+ \ln y^\* +\frac{1}{y^\*}(y-y^\*) - ???
\end{eqnarray}
Yet, I fail to log-linearise the last term with the summation and exponent. I tried to find similar log-linearizations online, but I was only able to find papers that log-linearise the production function of a CES-function, which was not helpful for my problem.

**Tags:** macroeconomics, utility, log-linearization


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## 1 answers

### Answer by username_1 on 2021-09-30. Score: 3 | Accepted answer

Here's my guess.

Let use the notation
$$
\tilde x\_t \approx \ln(x\_t) - \ln(x) \approx \dfrac{x\_t - x}{x}.
$$
If we take logs on both sides we get:
$$
\ln(G\_t) = \frac{1}{1 -\rho} \ln(p\_t) + \ln(y\_t) - \ln(1 + p\_t^{\frac{\rho}{\rho-1}})
$$
Subtracting the steady state gives:
$$
\tilde G\_t = \frac{1}{1 - \rho} \tilde p\_t + \tilde y\_t - \left[\ln(1 + p^{\frac{\rho}{\rho - 1}}) - \ln(1 + p^{\frac{\rho}{\rho - 1}})\right]
$$

Taking a Taylor expansion of the last term gives:
$$
\begin{align\*}
\ln(1 + p\_t^{\frac{\rho}{\rho - 1}}) - \ln(1 + p^{\frac{\rho}{\rho - 1}}) &\approx \frac{1}{1 + p^{\frac{\rho}{\rho - 1}}}\frac{\rho}{\rho - 1}p^{\frac{\rho}{\rho - 1}}\frac{(p\_t- p)}{p},\\
&\approx \frac{p^{\frac{\rho}{\rho- 1}}}{1 + p^{\frac{\rho}{\rho - 1}}} \frac{\rho}{\rho - 1} \tilde p\_t
\end{align\*}
$$
So we get:
$$
\tilde G\_t \approx \left(\frac{1}{1 - \rho}- \frac{p^{\frac{\rho}{\rho - 1}}}{1 + p^{\frac{\rho}{\rho - 1}}} \frac{\rho}{\rho - 1} \right) \tilde p\_t + \tilde y\_t,\\
= \left(\frac{1}{1 - \rho}- \frac{G}{Y} \frac{\rho}{\rho - 1} \right) \tilde p\_t + \tilde y\_t,\
$$

> **Comments:**
> **username_0**: Thank you, that was really helpful!  

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