<p>There is a typo in your (B.2.4). I believe the right-hand side should include a <span class="math-container">$P_{T,t-1}^H$</span> term. That is, it should be <span class="math-container">$P_{T,t}^H=\left[\theta_T (P_{T,t-1})^{1-\varepsilon}+(1-\theta_T)(\bar{P}_{T,t}^H)^{1-\varepsilon}\right]^{\frac{1}{1-\varepsilon}}$</span>. Divide this expression by <span class="math-container">$P_{T,t-1}^H$</span> to get <span class="math-container">$$\frac{P_{T,t}^H}{P_{T,t-1}^H}=\left[\theta_T+(1-\theta_T)\left(\frac{\bar{P}_{T,t}^H}{P_{T,t-1}^H}\right)^{1-\varepsilon}\right]^{\frac{1}{1-\varepsilon}}$$</span> Taking the exponent to the other side, <span class="math-container">$$\left(\frac{P_{T,t}^H}{P_{T,t-1}^H}\right)^{1-\varepsilon}=\theta_T+(1-\theta_T)\left(\frac{\bar{P}_{T,t}^H}{P_{T,t-1}^H}\right)^{1-\varepsilon}$$</span></p>
<p>Log-linearisation then proceeds like (with tildes indicating steady-state values), <span class="math-container">$$\begin{align*} 0&amp;=\theta_T+(1-\theta_T)\left(\frac{\bar{P}_{T,t}^H}{P_{T,t-1}^H}\right)^{1-\varepsilon}-\left(\frac{P_{T,t}^H}{P_{T,t-1}^H}\right)^{1-\varepsilon} \\
&amp;=\theta_T+(1-\theta_T) \left[\widetilde{\left(\frac{\bar{P}_{T,t}^H}{P_{T,t-1}^H}\right)}^{1-\varepsilon}\right]\exp\{(1-\varepsilon)(\bar{p}_{T,t}^H-p_{T,t-1}^H)\}- \left[\widetilde{\left(\frac{P_{T,t}^H}{P_{T,t-1}^H}\right)}^{1-\varepsilon}\right]\exp\{(1-\varepsilon)\pi_{T,t}^H\} \\
&amp;\approx \underbrace{\theta_T+(1-\theta_T) \left[\widetilde{\left(\frac{\bar{P}_{T,t}^H}{P_{T,t-1}^H}\right)}^{1-\varepsilon}\right] -\left[\widetilde{\left(\frac{P_{T,t}^H}{P_{T,t-1}^H}\right)}^{1-\varepsilon}\right]}_{=0}+\dots\\
&amp; \qquad \qquad \dots+(1-\varepsilon)\left[(1-\theta_T)(\bar{p}_{T,t}^H-p_{T,t-1}^H)\left[\widetilde{\left(\frac{\bar{P}_{T,t}^H}{P_{T,t-1}^H}\right)}^{1-\varepsilon}\right]-\pi_{T,t}^H\left[\widetilde{\left(\frac{P_{T,t}^H}{P_{T,t-1}^H}\right)}^{1-\varepsilon}\right]\right] \end{align*}$$</span></p>
<p>Therefore, <span class="math-container">$$\pi_{T,t}^H\left[\widetilde{\left(\frac{P_{T,t}^H}{P_{T,t-1}^H}\right)}^{1-\varepsilon}\right] =(1-\theta_T)(\bar{p}_{T,t}^H-p_{T,t-1}^H)\left[\widetilde{\left(\frac{\bar{P}_{T,t}^H}{P_{T,t-1}^H}\right)}^{1-\varepsilon}\right]$$</span></p>
<p>In steady state, <span class="math-container">$\left[\widetilde{\left(\frac{P_{T,t}^H}{P_{T,t-1}^H}\right)}^{1-\varepsilon}\right]=\left[\widetilde{\left(\frac{\bar{P}_{T,t}^H}{P_{T,t-1}^H}\right)}^{1-\varepsilon}\right]=1$</span>. So, our expression simplifies to the desired one: <span class="math-container">$$\pi_{T,t}^H =(1-\theta_T)(\bar{p}_{T,t}^H-p_{T,t-1}^H)$$</span></p>

CC BY-SA 4.0
2024-08-26T13:48:09.820Z
Thank you so much Joseph for the answer. Really appreciate the help.
Goldman Clarck
CC BY-SA 4.0
2024-08-26T14:23:53.803Z
@GoldmanClarck No problem! I realised after looking again that there is a typo in the (B.2.4) equation you stated which may have been what made the derivation difficult. I have edited my answer to account for this.
Joseph Basford
CC BY-SA 4.0
2024-08-29T10:23:32.857Z
You are absolutely right! Thank you.
Goldman Clarck
CC BY-SA 4.0
2024-08-25T09:33:25.100Z
2024-08-26T14:21:24.977Z
2024-08-26T14:21:24.977Z
Joseph Basford
<p>Can anyone please provide a step by step derivation of the log-linearized form of the following equation. I've tried to get from the first equation to the second but to no avail. I'm really struggling.<a href="https://i.sstatic.net/C22JFRrk.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/C22JFRrk.jpg" alt="enter image description here" /></a></p>

<macroeconomics><dsge><log-linearization>
Log-Linearization in Calvo Pricing
Goldman Clarck
# Log-Linearization in Calvo Pricing

(Asked by: username_0 on 2024-08-24)

Can anyone please provide a step by step derivation of the log-linearized form of the following equation. I've tried to get from the first equation to the second but to no avail. I'm really struggling.[![enter image description here](https://i.sstatic.net/C22JFRrk.jpg)](https://i.sstatic.net/C22JFRrk.jpg)

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

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## 1 answers

### Answer by username_1 on 2024-08-25. Score: 3 | Accepted answer

There is a typo in your (B.2.4). I believe the right-hand side should include a $P\_{T,t-1}^H$ term. That is, it should be $P\_{T,t}^H=\left[\theta\_T (P\_{T,t-1})^{1-\varepsilon}+(1-\theta\_T)(\bar{P}\_{T,t}^H)^{1-\varepsilon}\right]^{\frac{1}{1-\varepsilon}}$. Divide this expression by $P\_{T,t-1}^H$ to get $$\frac{P\_{T,t}^H}{P\_{T,t-1}^H}=\left[\theta\_T+(1-\theta\_T)\left(\frac{\bar{P}\_{T,t}^H}{P\_{T,t-1}^H}\right)^{1-\varepsilon}\right]^{\frac{1}{1-\varepsilon}}$$ Taking the exponent to the other side, $$\left(\frac{P\_{T,t}^H}{P\_{T,t-1}^H}\right)^{1-\varepsilon}=\theta\_T+(1-\theta\_T)\left(\frac{\bar{P}\_{T,t}^H}{P\_{T,t-1}^H}\right)^{1-\varepsilon}$$

Log-linearisation then proceeds like (with tildes indicating steady-state values), $$\begin{align\*} 0&=\theta\_T+(1-\theta\_T)\left(\frac{\bar{P}\_{T,t}^H}{P\_{T,t-1}^H}\right)^{1-\varepsilon}-\left(\frac{P\_{T,t}^H}{P\_{T,t-1}^H}\right)^{1-\varepsilon} \\
&=\theta\_T+(1-\theta\_T) \left[\widetilde{\left(\frac{\bar{P}\_{T,t}^H}{P\_{T,t-1}^H}\right)}^{1-\varepsilon}\right]\exp\{(1-\varepsilon)(\bar{p}\_{T,t}^H-p\_{T,t-1}^H)\}- \left[\widetilde{\left(\frac{P\_{T,t}^H}{P\_{T,t-1}^H}\right)}^{1-\varepsilon}\right]\exp\{(1-\varepsilon)\pi\_{T,t}^H\} \\
&\approx \underbrace{\theta\_T+(1-\theta\_T) \left[\widetilde{\left(\frac{\bar{P}\_{T,t}^H}{P\_{T,t-1}^H}\right)}^{1-\varepsilon}\right] -\left[\widetilde{\left(\frac{P\_{T,t}^H}{P\_{T,t-1}^H}\right)}^{1-\varepsilon}\right]}\_{=0}+\dots\\
& \qquad \qquad \dots+(1-\varepsilon)\left[(1-\theta\_T)(\bar{p}\_{T,t}^H-p\_{T,t-1}^H)\left[\widetilde{\left(\frac{\bar{P}\_{T,t}^H}{P\_{T,t-1}^H}\right)}^{1-\varepsilon}\right]-\pi\_{T,t}^H\left[\widetilde{\left(\frac{P\_{T,t}^H}{P\_{T,t-1}^H}\right)}^{1-\varepsilon}\right]\right] \end{align\*}$$

Therefore, $$\pi\_{T,t}^H\left[\widetilde{\left(\frac{P\_{T,t}^H}{P\_{T,t-1}^H}\right)}^{1-\varepsilon}\right] =(1-\theta\_T)(\bar{p}\_{T,t}^H-p\_{T,t-1}^H)\left[\widetilde{\left(\frac{\bar{P}\_{T,t}^H}{P\_{T,t-1}^H}\right)}^{1-\varepsilon}\right]$$

In steady state, $\left[\widetilde{\left(\frac{P\_{T,t}^H}{P\_{T,t-1}^H}\right)}^{1-\varepsilon}\right]=\left[\widetilde{\left(\frac{\bar{P}\_{T,t}^H}{P\_{T,t-1}^H}\right)}^{1-\varepsilon}\right]=1$. So, our expression simplifies to the desired one: $$\pi\_{T,t}^H =(1-\theta\_T)(\bar{p}\_{T,t}^H-p\_{T,t-1}^H)$$

> **Comments:**
> **username_0**: Thank you so much Joseph for the answer. Really appreciate the help.  
> **username_1**: @GoldmanClarck No problem! I realised after looking again that there is a typo in the (B.2.4) equation you stated which may have been what made the derivation difficult. I have edited my answer to account for this.  
> **username_0**: You are absolutely right! Thank you.  

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