3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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: $$ \tilde G_t \approx \left(\frac{1}{1 - \rho}- \frac{p^{\frac{\rho}{\rho - 1}}}{1 + p^{\frac{\rho}{\rho - 1}}} \frac{\rho}{\rho - 1} \right) \tilde p_t + \tilde y_t,\\ = \left(\frac{1}{1 - \rho}- \frac{G}{Y} \frac{\rho}{\rho - 1} \right) \tilde p_t + \tilde y_t,\ $$

$\endgroup$
1
  • $\begingroup$ Thank you, that was really helpful! $\endgroup$
    – Rosa
    Oct 1 at 7:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.