<p>We have:</p>

<p>$$y_t = c_t − w^u(1 − N_t) + \kappa v_t$$</p>

<p>We take the log of both sides:</p>

<p>$$\ln y_t = \ln \left[ c_t − w^u(1 − N_t) + \kappa v_t \right]$$</p>

<p>And then linearize around the steady states:</p>

<p>$$\begin{align}
\ln \bar{y} + \frac{1}{\bar{y}}(y_t - \bar{y}) &amp; = \ln \left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right] + \frac{1}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}(c_t - \bar{c}) \\
&amp; + \frac{w_u}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}(N_t - \bar{N}) \\
&amp; + \frac{\kappa}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}(V_t - \bar{V})
\end{align}
$$</p>

<p>Cancel $\ln \bar{y}$ and $\ln \left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]$</p>

<p>$$\begin{align}
\frac{1}{\bar{y}}(y_t - \bar{y}) &amp; = \frac{1}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}(c_t - \bar{c}) \\
&amp; + \frac{w_u}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}(N_t - \bar{N}) \\
&amp; + \frac{\kappa}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}(V_t - \bar{V})
\end{align}
$$</p>

<p>Multiply the first term on the right hand side by $\frac{\bar{c}}{\bar{c}}$, and similarly for the other terms:</p>

<p>$$\begin{align}
\frac{1}{\bar{y}}(y_t - \bar{y}) &amp; = \frac{\bar{c}}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}\frac{(c_t - \bar{c})}{\bar{c}} \\
&amp; + \frac{w_u \bar{N}}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}\frac{(N_t - \bar{N})}{\bar{N}} \\
&amp; + \frac{\kappa \bar{V}}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}\frac{(V_t - \bar{V})}{\bar{V}}
\end{align}
$$</p>

<p>And we simplify, taking advantage of the fact that $\hat{v}_t = \hat{\theta}_t + \hat{u}_t$, to get what you have:</p>

<p>$$\hat{y}_t = \frac{\bar{C}}{\bar{Y}}\hat{c}_t + w^u \frac{\bar{N}}{\bar{Y}} \hat{n}_t + \left( \frac{K\bar{V}}{\bar{Y}} \right) (\hat{\theta}_t + \hat{u}_t) $$</p>

<p>I don't think this derivation is wrong. But you can see that $y_t$ and $N_t$ are directly proportional to each other. It's definitely plausible that $\frac{\bar{N}}{\bar{Y}} = 1$, but I've pored over the equations in the setup for 3 hours to no avail at showing that rigorously. It's probably something silly that I don't understand. I'll look back later and if nothing else, I'll turn this answer into a community wiki so anyone who has the right idea can edit it.</p>

CC BY-SA 3.0
2016-06-08T17:25:56.460Z
I forgot to mention that $\hat{z}_{t}$ is a stationary exogenous shock hitting all firms equally.
Ralle Kalle
CC BY-SA 3.0
2016-06-11T11:23:44.630Z
$\bar{z}_t $ is set 0 in the steady state.
Ralle Kalle
CC BY-SA 3.0
2016-05-22T12:15:12.090Z
Thank you for your effort.
I thought this through - again and again. Maybe I overread it or the authors do not point this one out, but could it be, that the technology has been left our of consideration? In this case $$y_{t}=n_{t} + a_{t} $$ would simplify to the identity  $$y_{t}=n_{t}$$.

This would lead to the log-linearized equation as mentioned in the paper, but I am not sure whether this simplification is feasible or not.
Ralle Kalle
CC BY-SA 3.0
2016-05-22T16:08:01.497Z
There's an $a_t$ in this model?? Or do you mean $z_t$. I thought about using $y_t = n_t + z_t$, but it's used in the next step rather than here, and I couldn't juut say $z_t$ was zero. (Not actually sure what $z_t$ is either.)
Kitsune Cavalry
CC BY-SA 3.0
2016-05-22T16:30:57.017Z
No, your right: Technology is not mentioned in the paper. I just thought that $y_{t} = n_{t} + a_{t}$ depicts one of the key elements of equilibirum and thus holds for every DSGE model...
$Z_{t}$ is defined as an exogenous stationary productivity shock common to all firms.
Ralle Kalle
CC BY-SA 3.0
2016-05-21T21:39:50.597Z
2016-05-21T21:39:50.597Z
Kitsune Cavalry
<p>I am dealing with a paper of Walsh &amp; Ravenna.</p>

<p>www.banque-france.fr/fondation/gb/telechar/bourses_recherche/Welfare-based_Ravenna.pdf</p>

<p>I am kind of confused by the euqation (19) on page 33.</p>

<p>The market clearing condition goes as follows:</p>

<p>$$Y_{t} = C_{t} - w^{u}(1 - N_{t}) + \kappa\upsilon_{t}$$</p>

<p>Log linearization around the steady state yields</p>

<p>$$\hat{y}_{t} = \frac{\bar{C}}{\bar{Y}}\hat{c}_{t} - w^{u}\hat{n}_{t} + \left( \frac{\kappa\bar{\upsilon}}{\bar{Y}}\right)(\hat{\Theta}_{t} + \hat{u}_{t} )    \;\;\; \mathbf{(19)} $$</p>

<p>I don't know how this formula (19) is derived?
Isn't there something missing? From my basic understanding of log linearization it ought to look like this:</p>

<p>$$\hat{y}_{t} = \frac{\bar{C}}{\bar{Y}}\hat{c}_{t} - \left( \mathbf{\frac{\bar{N}}{\bar{Y}}} \right) w^{u}\hat{n}_{t} + \left( \frac{\kappa\bar{\upsilon}}{\bar{Y}}\right)(\hat{\Theta}_{t} + \hat{u}_{t} )$$</p>

<p>with
$$ Y_{t}\;\;...\;\; output $$
$$ C_{t}\;\;...\;\;  consumption $$
$$ w^{u}\;\;...\;\;  wage\;of\;unmatched\;workers $$
$$ 1-N_{t}\;\;...\;\;  unmatched\;workers $$
$$ w^{u}(1-N_{t})\;\;...\;\;home\;production$$
$$ \kappa\;\;...\;\;  cost\;of\;posting\;vacancy $$
$$ v_{t}\;\;...\;\;  vacancies $$
$$ \hat{v}_{t} = (\hat{\Theta}_{t} + \hat{u}_{t} ) $$
$$ \omega = \frac{v_{t}}{u_{t}}\;\;...\;\;measure\;of\;labour\;market\;tightness$$
$$ \hat{\cdot}\;\;...\;\;log\;deviation\;from\;steady\;state\;value$$
$$ \bar{\cdot}\;\;...\;\;steady\;state\;value$$</p>

<p>Small letter with a hat: log deviation of a variable arount its steady state.
Big letter with a bar: steady state value.
K: cost of posting a job vancancy.
w^u: "wage" of unemployed workers.</p>

<p>I've read the paper and the appendix aswell, read both the papers in the bibliography of this one aswell as later ones basing on this publication, but I could not find a helpful hint.</p>

<p>Is there any special relationship between N and Y in the steady state that explains why this whole term vanishs? Or do I have a wrong understanding of log-linearization?</p>

<p>I have to apologyze for my rusty English. I am already taking care of this problem. But for the above mentioned I would like to have your help. Does anyone has a decisive hint?</p>

CC BY-SA 3.0
2016-05-21T17:23:55.923Z
Could you please 1. also state the equation just above in your question (not numbered in the paper) and 2. a precise definition of the variables involved?
user_newbie10
CC BY-SA 3.0
2016-05-22T11:59:54.640Z
@dugo: This way?
Ralle Kalle
<macroeconomics>
Log-linearization of the market clearing condition
Ralle Kalle
# Log-linearization of the market clearing condition

(Asked by: username_0 on 2016-05-21)

I am dealing with a paper of Walsh & Ravenna.

www.banque-france.fr/fondation/gb/telechar/bourses\_recherche/Welfare-based\_Ravenna.pdf

I am kind of confused by the euqation (19) on page 33.

The market clearing condition goes as follows:

$$Y\_{t} = C\_{t} - w^{u}(1 - N\_{t}) + \kappa\upsilon\_{t}$$

Log linearization around the steady state yields

$$\hat{y}\_{t} = \frac{\bar{C}}{\bar{Y}}\hat{c}\_{t} - w^{u}\hat{n}\_{t} + \left( \frac{\kappa\bar{\upsilon}}{\bar{Y}}\right)(\hat{\Theta}\_{t} + \hat{u}\_{t} ) \;\;\; \mathbf{(19)} $$

I don't know how this formula (19) is derived?
Isn't there something missing? From my basic understanding of log linearization it ought to look like this:

$$\hat{y}\_{t} = \frac{\bar{C}}{\bar{Y}}\hat{c}\_{t} - \left( \mathbf{\frac{\bar{N}}{\bar{Y}}} \right) w^{u}\hat{n}\_{t} + \left( \frac{\kappa\bar{\upsilon}}{\bar{Y}}\right)(\hat{\Theta}\_{t} + \hat{u}\_{t} )$$

with
$$ Y\_{t}\;\;...\;\; output $$
$$ C\_{t}\;\;...\;\; consumption $$
$$ w^{u}\;\;...\;\; wage\;of\;unmatched\;workers $$
$$ 1-N\_{t}\;\;...\;\; unmatched\;workers $$
$$ w^{u}(1-N\_{t})\;\;...\;\;home\;production$$
$$ \kappa\;\;...\;\; cost\;of\;posting\;vacancy $$
$$ v\_{t}\;\;...\;\; vacancies $$
$$ \hat{v}\_{t} = (\hat{\Theta}\_{t} + \hat{u}\_{t} ) $$
$$ \omega = \frac{v\_{t}}{u\_{t}}\;\;...\;\;measure\;of\;labour\;market\;tightness$$
$$ \hat{\cdot}\;\;...\;\;log\;deviation\;from\;steady\;state\;value$$
$$ \bar{\cdot}\;\;...\;\;steady\;state\;value$$

Small letter with a hat: log deviation of a variable arount its steady state.
Big letter with a bar: steady state value.
K: cost of posting a job vancancy.
w^u: "wage" of unemployed workers.

I've read the paper and the appendix aswell, read both the papers in the bibliography of this one aswell as later ones basing on this publication, but I could not find a helpful hint.

Is there any special relationship between N and Y in the steady state that explains why this whole term vanishs? Or do I have a wrong understanding of log-linearization?

I have to apologyze for my rusty English. I am already taking care of this problem. But for the above mentioned I would like to have your help. Does anyone has a decisive hint?

**Tags:** macroeconomics
> **Comments:**
> **username_1**: Could you please 1. also state the equation just above in your question (not numbered in the paper) and 2. a precise definition of the variables involved?  
> **username_0**: @dugo: This way?  

---

## 1 answers

### Answer by username_2 on 2016-05-21. Score: 3 | Accepted answer

We have:

$$y\_t = c\_t − w^u(1 − N\_t) + \kappa v\_t$$

We take the log of both sides:

$$\ln y\_t = \ln \left[ c\_t − w^u(1 − N\_t) + \kappa v\_t \right]$$

And then linearize around the steady states:

$$\begin{align}
\ln \bar{y} + \frac{1}{\bar{y}}(y\_t - \bar{y}) & = \ln \left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right] + \frac{1}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}(c\_t - \bar{c}) \\
& + \frac{w\_u}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}(N\_t - \bar{N}) \\
& + \frac{\kappa}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}(V\_t - \bar{V})
\end{align}
$$

Cancel $\ln \bar{y}$ and $\ln \left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]$

$$\begin{align}
\frac{1}{\bar{y}}(y\_t - \bar{y}) & = \frac{1}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}(c\_t - \bar{c}) \\
& + \frac{w\_u}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}(N\_t - \bar{N}) \\
& + \frac{\kappa}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}(V\_t - \bar{V})
\end{align}
$$

Multiply the first term on the right hand side by $\frac{\bar{c}}{\bar{c}}$, and similarly for the other terms:

$$\begin{align}
\frac{1}{\bar{y}}(y\_t - \bar{y}) & = \frac{\bar{c}}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}\frac{(c\_t - \bar{c})}{\bar{c}} \\
& + \frac{w\_u \bar{N}}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}\frac{(N\_t - \bar{N})}{\bar{N}} \\
& + \frac{\kappa \bar{V}}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}\frac{(V\_t - \bar{V})}{\bar{V}}
\end{align}
$$

And we simplify, taking advantage of the fact that $\hat{v}\_t = \hat{\theta}\_t + \hat{u}\_t$, to get what you have:

$$\hat{y}\_t = \frac{\bar{C}}{\bar{Y}}\hat{c}\_t + w^u \frac{\bar{N}}{\bar{Y}} \hat{n}\_t + \left( \frac{K\bar{V}}{\bar{Y}} \right) (\hat{\theta}\_t + \hat{u}\_t) $$

I don't think this derivation is wrong. But you can see that $y\_t$ and $N\_t$ are directly proportional to each other. It's definitely plausible that $\frac{\bar{N}}{\bar{Y}} = 1$, but I've pored over the equations in the setup for 3 hours to no avail at showing that rigorously. It's probably something silly that I don't understand. I'll look back later and if nothing else, I'll turn this answer into a community wiki so anyone who has the right idea can edit it.

> **Comments:**
> **username_0**: Thank you for your effort. I thought this through - again and again. Maybe I overread it or the authors do not point this one out, but could it be, that the technology has been left our of consideration? In this case $$y_{t}=n_{t} + a_{t} $$ would simplify to the identity  $$y_{t}=n_{t}$$.  This would lead to the log-linearized equation as mentioned in the paper, but I am not sure whether this simplification is feasible or not.  
> **username_2**: There's an $a_t$ in this model?? Or do you mean $z_t$. I thought about using $y_t = n_t + z_t$, but it's used in the next step rather than here, and I couldn't juut say $z_t$ was zero. (Not actually sure what $z_t$ is either.)  
> **username_0**: No, your right: Technology is not mentioned in the paper. I just thought that $y_{t} = n_{t} + a_{t}$ depicts one of the key elements of equilibirum and thus holds for every DSGE model... $Z_{t}$ is defined as an exogenous stationary productivity shock common to all firms.  
> **username_0**: I forgot to mention that $\hat{z}_{t}$ is a stationary exogenous shock hitting all firms equally.  
> **username_0**: $\bar{z}_t $ is set 0 in the steady state.  

---

