<p>I found the error in my derivation: I mistakenly supposet that the steady state of $p_t+1$ equals $\bar{\rho}$. 
I did not recognize that p was different from $\rho$ because of the poor quality of my printout. 
Reading the paper again on a screen I immediately spotted the mistake. </p>

CC BY-SA 3.0
2016-07-07T14:20:14.527Z
Arithmetic strikes again!
Kitsune Cavalry
CC BY-SA 3.0
2016-07-06T17:42:25.290Z
2016-07-06T17:42:25.290Z
Ralle Kalle
<p>The following equations are taken from Ravenna, walsh: "Optimal Monetary Policy with Unemployment and Sticky Prices" (2011).</p>

<p>(i)$$\frac{Z_{t}}{\mu_{t}} = w_t + \frac{\kappa}{q_{t}} - (1-\rho)E_{t}(\frac{1}{R_{t}})(\frac{\kappa}{q_{t+1}}) $$</p>

<p>(ii) $$ w_{t} = w^u + (\frac{b_t}{1-b_t})(\frac{\kappa}{q_t}) - (1-\rho)E_t(\frac{1}{R_t})(1-p_{t+1})(\frac{b_{t+1}}{1-b_{t+1}})(\frac{\kappa}{q_{t+1}}) $$</p>

<p>$ \phi $...  wage replacement rate</p>

<p>$b_t$... surplus bargaining share of a worker</p>

<p>$Z_t$... exogenous productivity shock ($\bar{Z}=1$)</p>

<p>$\mu_t$... retail price mark up</p>

<p>$\kappa$... post of posting a vacancy</p>

<p>$\rho$... Share of matches $N_{t-1}$ that loses its jobs in t</p>

<p>$q_t$... probability that a firm will fill its vacancy</p>

<p>$\lambda_t$... marginal utility of consumption</p>

<p>$\frac{1}{R_t} \equiv \beta(\frac{\lambda_{t+1}}{\lambda_t})$</p>

<p>$p_{t+1} \equiv \frac{m_t}{u_t}$... job finding probability with $m_t$ matches and $u_t$ unemployed ones</p>

<p>$\beta$... discount factor</p>

<p>$w^u = \phi w $</p>

<p>Given the last identity, (i) and (ii) could be used to jointly solve  for $\kappa$ and $w$. At least this is suggested in the paper by the atuhors.</p>

<p>(ii) $w_t = \phi w_t + (\frac{b_t}{1-b_t})(\frac{\kappa}{q_t}) - (1-\rho)E_t(\frac{1}{R_t})(1-p_{t+1})(\frac{b_{t+1}}{1-b_{t+1}})(\frac{\kappa}{q_{t+1}})$</p>

<p>&lt;=> $w_t = \frac{(\frac{b_t}{1-b_t})(\frac{\kappa}{q_t}) - (1-\rho)E-t(\frac{1}{R_t})(1-p_{t+1})(\frac{b_{t+1}}{1-b_{t+1}})(\frac{\kappa}{q_{t+1}})}{1-\phi}$</p>

<p>Put into (i) to eliminate $w_t$</p>

<p>$\frac{Z_{t}}{\mu_{t}} = \frac{(\frac{b_t}{1-b_t})(\frac{\kappa}{q_t}) - (1-\rho)E_t(\frac{1}{R_t})(1-p_{t+1})(\frac{b_{t+1}}{1-b_{t+1}})(\frac{\kappa}{q_{t+1}})}{1-\phi} +\frac{\kappa}{q_{t}} - (1-\rho)E_{t}(\frac{1}{R_{t}})(\frac{\kappa}{q_{t+1}}) $</p>

<p>&lt;=> $\frac{Z_{t}}{\mu_{t}}(1-\phi) = (\frac{b_t}{1-b_t})(\frac{\kappa}{q_t}) - (1-\rho)E_t(\frac{1}{R_t})(1-p_{t+1})(\frac{b_{t+1}}{1-b_{t+1}})(\frac{\kappa}{q_{t+1}}) +\frac{\kappa}{q_{t}}(1-\phi) - (1-\rho)E_{t}(\frac{1}{R_{t}})(\frac{\kappa}{q_{t+1}})(1-\phi) $</p>

<p>Rearrange and solve for $\kappa$:</p>

<p>$\frac{Z_{t} (1-\phi)}{\mu_{t} ((\frac{b_t}{1-b_t})(\frac{1}{q_t}) - (1-\rho)E_t(\frac{1}{R_t})(1-p_{t+1})(\frac{b_{t+1}}{1-b_{t+1}})(\frac{1}{q_{t+1}}) +\frac{1}{q_{t}}(1-\phi) - (1-\rho)E_{t}(\frac{1}{R_{t}})(\frac{1}{q_{t+1}})(1-\phi)} = \kappa $</p>

<p>&lt;=> $\kappa = \frac{Z_{t} (1-\phi)}{\mu_{t} \big(\big(\frac{1}{q_t}\big)\big[(\frac{b_t}{1-b_t})+(1-\phi)\big] - (1-\rho)\big(\frac{1}{q_{t+1}}\big)E_t\big(\frac{1}{R_t}\big)\Big[(1-p_{t+1})\big(\frac{b_{t+1}}{1-b_{t+1}}\big)+(1-\phi)\big]\big)}$</p>

<p>Evaluating this formula arount the steady state yields</p>

<p>$\bar{\kappa} = \frac{(1-\phi)}{\bar{\mu} \big(\big(\frac{1}{\bar{Q}}\big)\big[(\frac{b}{1-b})+(1-\phi)\big] - (1-\rho)\big(\frac{1}{\bar  {Q}}\big)\beta\Big[(1-\frac{\bar{M}}{\bar{U}})\big(\frac{b}{1-b}\big)+(1-\phi)\big]\big)}$</p>

<p>Since all this parameters and steady state values are well defined I am able to calculate the steady state value of $\kappa$ and hence the one of $w_t$ as well. </p>

<p>I wanted to simulate the related model in dynare. Having a look at the dynare code provided by the authors, I noticed that they made use of a totally diferent approach to calculate the steady state value of $\kappa$:</p>

<p>$AAk = \left[{\begin{array}{cc} 1-\phi(1-b) &amp; -\phi b(1-\rho)\beta \bar{\theta}\\ 1-b &amp; (1-\beta(1-\rho))(\frac{1}{\bar{Q}})+ b\beta(1-\rho)\bar{\theta}  \end{array} } \right]$</p>

<p>$BBk = \left[ {\begin{array}{cc} \frac{\phi b}{\bar{\mu}} \\ \frac{(1-b)}{\bar{\mu}} \end{array} } \right] $</p>

<p>$CCk = AAk^{-1}BBk$</p>

<p>$\bar{\kappa}=CCk(2)$</p>

<p>Where</p>

<p>$\theta_t=\frac{v_t}{u_t}$ measures the labor market tightness (relation of vacancies and job seekers)</p>

<p>I would like to know whether my derivation of the steady state value for $\kappa$ does make sense and how the authors of the paper derive it?
I have read the appendix to the paper aswell as the dynare code, but to be hones I do not have an idea of how they calculate $\kappa$.</p>

<p>The results of my calculation and the one of Ravenna and Walsh differ significantly.</p>

CC BY-SA 3.0
2016-07-06T17:43:00.527Z
I found the error in my derivation: I mistakenly supposet that the steady state of $p_t+1$ equals $\bar{\rho}$. 
I did not recognize that p was different from $\rho$ because of the poor quality of my printout. 
Reading the paper again on a screen I immediately spotted the mistake.
Kind of embarrassing :'D
Ralle Kalle
<steady-state><dsge><dynare>
Evaluation around steady state for a specific DSGE Model
Ralle Kalle
# Evaluation around steady state for a specific DSGE Model

(Asked by: username_0 on 2016-07-05)

The following equations are taken from Ravenna, walsh: "Optimal Monetary Policy with Unemployment and Sticky Prices" (2011).

(i)$$\frac{Z\_{t}}{\mu\_{t}} = w\_t + \frac{\kappa}{q\_{t}} - (1-\rho)E\_{t}(\frac{1}{R\_{t}})(\frac{\kappa}{q\_{t+1}}) $$

(ii) $$ w\_{t} = w^u + (\frac{b\_t}{1-b\_t})(\frac{\kappa}{q\_t}) - (1-\rho)E\_t(\frac{1}{R\_t})(1-p\_{t+1})(\frac{b\_{t+1}}{1-b\_{t+1}})(\frac{\kappa}{q\_{t+1}}) $$

$ \phi $... wage replacement rate

$b\_t$... surplus bargaining share of a worker

$Z\_t$... exogenous productivity shock ($\bar{Z}=1$)

$\mu\_t$... retail price mark up

$\kappa$... post of posting a vacancy

$\rho$... Share of matches $N\_{t-1}$ that loses its jobs in t

$q\_t$... probability that a firm will fill its vacancy

$\lambda\_t$... marginal utility of consumption

$\frac{1}{R\_t} \equiv \beta(\frac{\lambda\_{t+1}}{\lambda\_t})$

$p\_{t+1} \equiv \frac{m\_t}{u\_t}$... job finding probability with $m\_t$ matches and $u\_t$ unemployed ones

$\beta$... discount factor

$w^u = \phi w $

Given the last identity, (i) and (ii) could be used to jointly solve for $\kappa$ and $w$. At least this is suggested in the paper by the atuhors.

(ii) $w\_t = \phi w\_t + (\frac{b\_t}{1-b\_t})(\frac{\kappa}{q\_t}) - (1-\rho)E\_t(\frac{1}{R\_t})(1-p\_{t+1})(\frac{b\_{t+1}}{1-b\_{t+1}})(\frac{\kappa}{q\_{t+1}})$

<=> $w\_t = \frac{(\frac{b\_t}{1-b\_t})(\frac{\kappa}{q\_t}) - (1-\rho)E-t(\frac{1}{R\_t})(1-p\_{t+1})(\frac{b\_{t+1}}{1-b\_{t+1}})(\frac{\kappa}{q\_{t+1}})}{1-\phi}$

Put into (i) to eliminate $w\_t$

$\frac{Z\_{t}}{\mu\_{t}} = \frac{(\frac{b\_t}{1-b\_t})(\frac{\kappa}{q\_t}) - (1-\rho)E\_t(\frac{1}{R\_t})(1-p\_{t+1})(\frac{b\_{t+1}}{1-b\_{t+1}})(\frac{\kappa}{q\_{t+1}})}{1-\phi} +\frac{\kappa}{q\_{t}} - (1-\rho)E\_{t}(\frac{1}{R\_{t}})(\frac{\kappa}{q\_{t+1}}) $

<=> $\frac{Z\_{t}}{\mu\_{t}}(1-\phi) = (\frac{b\_t}{1-b\_t})(\frac{\kappa}{q\_t}) - (1-\rho)E\_t(\frac{1}{R\_t})(1-p\_{t+1})(\frac{b\_{t+1}}{1-b\_{t+1}})(\frac{\kappa}{q\_{t+1}}) +\frac{\kappa}{q\_{t}}(1-\phi) - (1-\rho)E\_{t}(\frac{1}{R\_{t}})(\frac{\kappa}{q\_{t+1}})(1-\phi) $

Rearrange and solve for $\kappa$:

$\frac{Z\_{t} (1-\phi)}{\mu\_{t} ((\frac{b\_t}{1-b\_t})(\frac{1}{q\_t}) - (1-\rho)E\_t(\frac{1}{R\_t})(1-p\_{t+1})(\frac{b\_{t+1}}{1-b\_{t+1}})(\frac{1}{q\_{t+1}}) +\frac{1}{q\_{t}}(1-\phi) - (1-\rho)E\_{t}(\frac{1}{R\_{t}})(\frac{1}{q\_{t+1}})(1-\phi)} = \kappa $

<=> $\kappa = \frac{Z\_{t} (1-\phi)}{\mu\_{t} \big(\big(\frac{1}{q\_t}\big)\big[(\frac{b\_t}{1-b\_t})+(1-\phi)\big] - (1-\rho)\big(\frac{1}{q\_{t+1}}\big)E\_t\big(\frac{1}{R\_t}\big)\Big[(1-p\_{t+1})\big(\frac{b\_{t+1}}{1-b\_{t+1}}\big)+(1-\phi)\big]\big)}$

Evaluating this formula arount the steady state yields

$\bar{\kappa} = \frac{(1-\phi)}{\bar{\mu} \big(\big(\frac{1}{\bar{Q}}\big)\big[(\frac{b}{1-b})+(1-\phi)\big] - (1-\rho)\big(\frac{1}{\bar {Q}}\big)\beta\Big[(1-\frac{\bar{M}}{\bar{U}})\big(\frac{b}{1-b}\big)+(1-\phi)\big]\big)}$

Since all this parameters and steady state values are well defined I am able to calculate the steady state value of $\kappa$ and hence the one of $w\_t$ as well.

I wanted to simulate the related model in dynare. Having a look at the dynare code provided by the authors, I noticed that they made use of a totally diferent approach to calculate the steady state value of $\kappa$:

$AAk = \left[{\begin{array}{cc} 1-\phi(1-b) & -\phi b(1-\rho)\beta \bar{\theta}\\ 1-b & (1-\beta(1-\rho))(\frac{1}{\bar{Q}})+ b\beta(1-\rho)\bar{\theta} \end{array} } \right]$

$BBk = \left[ {\begin{array}{cc} \frac{\phi b}{\bar{\mu}} \\ \frac{(1-b)}{\bar{\mu}} \end{array} } \right] $

$CCk = AAk^{-1}BBk$

$\bar{\kappa}=CCk(2)$

Where

$\theta\_t=\frac{v\_t}{u\_t}$ measures the labor market tightness (relation of vacancies and job seekers)

I would like to know whether my derivation of the steady state value for $\kappa$ does make sense and how the authors of the paper derive it?
I have read the appendix to the paper aswell as the dynare code, but to be hones I do not have an idea of how they calculate $\kappa$.

The results of my calculation and the one of Ravenna and Walsh differ significantly.

**Tags:** steady-state, dsge, dynare
> **Comments:**
> **username_0**: I found the error in my derivation: I mistakenly supposet that the steady state of $p_t+1$ equals $\bar{\rho}$.  I did not recognize that p was different from $\rho$ because of the poor quality of my printout.  Reading the paper again on a screen I immediately spotted the mistake. Kind of embarrassing :'D  

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## 1 answers

### Answer by username_0 on 2016-07-06. Score: 2 | Accepted answer

I found the error in my derivation: I mistakenly supposet that the steady state of $p\_t+1$ equals $\bar{\rho}$.
I did not recognize that p was different from $\rho$ because of the poor quality of my printout.
Reading the paper again on a screen I immediately spotted the mistake.

> **Comments:**
> **username_1**: Arithmetic strikes again!  

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