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6 votes

First order condition of log functions in general and interpretation

The $\gamma$ on the RHS comes from applying chain rule when differentiating the second term with respect to $a_i$. Regarding elasticity, note that with a differentiable function $f$, the ratio $f'(x)/...
Herr K.'s user avatar
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6 votes
Accepted

First order condition of log functions in general and interpretation

It just comes from the derivative of profit function. I assume that $a_i$ is the choice variable here so the derivative of $\pi$ wrt $a_i$ is (step by step): $$\frac{\partial \pi}{ \partial a_i} = \...
1muflon1's user avatar
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3 votes
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Log-linearizing a second order term around the steady-state

If $\Pi_t$ is gross inflation then indeed $(\Pi_t-1)^2$ is a second-order term and is approximately zero. For example, for a reasonable quarterly steady state gross inflation rate of 1.005, the term ...
BrsG's user avatar
  • 1,652
3 votes

Log linearization with sums

First you need the value $w_0$ of $w$ for $s=0$: $$ 0=\alpha(1-w_0)^{\zeta-1}-(1-\alpha)w_0^{\zeta-1}\;,\\ (1-\alpha)w_0^{\zeta-1}=\alpha(1-w_0)^{\zeta-1}\;,\\ (1-\alpha)^\frac1{\zeta-1}w_0=\alpha^\...
joriki's user avatar
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3 votes
Accepted

Log Linearising CES demand

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) - ...
tdm's user avatar
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3 votes
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Log linearising EUler equation

As you say the first step is to take log of both sides after that you are just applying the rules for logarithms and rearrange. For example: $$\ln (XZ)=\ln X + \ln Z$$ $$\ln X/Z= \ln X - \ln Z$$ $$\...
1muflon1's user avatar
  • 57k
2 votes

Constant elasticity proof for log-linear demand curve

$\frac{dlnQ}{dp}=\frac{dlnQ}{dQ} \frac{dQ}{dp}$ thus: $\frac{dQ}{dp}=\frac{dlnQ}{dp} \frac{dQ}{dlnQ}$ Since we know that if $f(x)=lnx \Rightarrow f'(x)=\frac{1}{x}\Rightarrow \frac{1}{f'(x)}=x$ We ...
user18214's user avatar
  • 805
2 votes

How do you use a Log-linear model when you have negative Xs?

The elasticity of Y with respect to X is often estimated by running a regression like this: $$ \ln Y = \alpha + \beta \ln X + error$$ However, this isn't always applicable because it is perfectly ...
BKay's user avatar
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2 votes

a question about second-order log-linear approximation

If you set the partial derivative of (OA5) wrt $y_i$ to zero, the division by 2 can be simplified, and disappears: $$ \frac{ \partial v}{\partial y_i}(y_i,y, a_i,\omega)=0 \iff v_1 + v_{11}y_i+v_{12}y+...
Bertrand's user avatar
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1 vote
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Log-linearization of the additive habit formation model

Your second step, $X^b e^{b \hat{X}_t} \big( \approx X^b (1 + b \hat{X}_t) \big)$, is wrong here. Specifically, you linearized the first term in RHS, $(C_t - b C_{t-1})^{-\sigma}$, as \begin{equation} ...
JiYong Jung's user avatar
1 vote

Problem with Optimizing Profit in Log-Linear Demand Model

Your objective function is additive separable between $myPrice$ and $competitorPrice$. This means that you can write: $$ Objective = f(myPrice) + g(competitorPrice). $$ For some functions $f$ and $g$. ...
tdm's user avatar
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1 vote

log linearize exponential expressions

The aim of log-linearization is to get an expression that is linear in the deviation from steady state $x_t$, where $x_t:=\log(X_t/X)$, $X$ is the steady state of $X_t$ and we have $X_t = Xe^{x_t} \...
BrsG's user avatar
  • 1,652
1 vote
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New Keynesian Phillips Curve: log-linearization

[Not a full solution] Unlike what the name suggests, log-linearization does not involve taking logs of individual terms in an equation. Otherwise, the equation t wouldn't hold. The general approach is ...
BrsG's user avatar
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