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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.

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  • $\begingroup$ 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 $\endgroup$ – Ralle Kalle Jul 6 '16 at 17:43
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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.

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  • $\begingroup$ Arithmetic strikes again! $\endgroup$ – Kitsune Cavalry Jul 7 '16 at 14:20

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