Hello everyone,

I have found a replication of a labour market model that uses a notation of this kind:

model;

(1/c) = beta*(1/c(+1))*(1+r(+1)-delta);*

However, I want to integrate it with another NK model that uses a notation of this kind:

model;

(1/exp©) = beta(1/exp(c(+1)))*(1+exp(r(+1))-delta);

I thought I could solve by re-writing the first piece of code in in exp-logs (as the second one) but when I do, dynare says it cannot find the steady state anymore.

Does anyone understand why that might happen?

Cheers

Then you must have introduced a mistake when doing this. You must consistently replace all variables that have been redefined.

Thank you for the reply.

I have tried very hard to debug it, but there must be something that I systematically rewrite wrong thinking it’s right. I wasn’t sure if, for instance c^ALPHA became exp©^ALPHA or (exp©)^ALPHA, so I tried both ways and neither worked. I still think the second one is correct. Except for this, it’s really just 9 short equations.

Do you or anyone else spot any consistent mistake?

From:

```
var U V M S u e v t z;
varexo zi;
parameters A a B x l K b c rho sigma;
l = 0.039; % obsolescence separation rate
a = 0.5; % matching function exponent
A = 0.636; % matching function scale parameter
B = 0.9967; % discount factor
x = 0.5; % worker bargaining weight
rho = 0.975; % persistence of productivity process
sigma = 0.0076; % variance of productivity process
b = 0.9; % value of leisure
c = 0.17; % vacancy posting cost
K = 26.94; % creation cost
model;
% Ausiliary definitions for matching function and flow probabilities
e = 1-u;
t = v/u;
% Unemployment value function (eq 16)
U = b + B*( A * t^(1-a)*x*S(+1) + U(+1) );
% Vacancy value function (eq 17)
V = -c + B*( A * t^(-a)*(1-x)*S(+1) + V(+1) );
% Output value function (eq 18)
M = z + B*( (1-l)*S(+1) + U(+1) + V(+1) );
% Matching surplus definition (S=M-U-V) (eq 19)
S = z - b + c + B*((1-l) - A * t^(1-a)*x - A * t^(-a)*(1-x))*S(+1);
% FONC: V=K*n (eq 21)
c = B*( A * t^(-a)*(1-x)*S(+1));
% Law of motion for (eq 22)
u = u(-1) + l*(1-u(-1)) - A*(t(-1)^(1-a)) * u(-1);
% Log productivity follows an exogenous AR(1) process
log(z) = rho*log(z(-1)) + zi;
end;
initval;
U = 294.865;
V = 0.543609;
M = 295.738;
S = 0.32959;
u = 0.0798989;
e = 0.920101;
v = 0.04076282;
t = 0.510245;
z = 1;
end;
shocks;
var zi = sigma^2;
end;
steady;
stoch_simul(
nograph,
order = 1,
periods = 30300,
drop = 300,
IRF = 90
);
```

To:

```
var U V M S u e v t z;
varexo zi;
parameters A a B x l K b c rho sigma;
l = 0.039; % obsolescence separation rate
a = 0.5; % matching function exponent
A = 0.636; % matching function scale parameter
B = 0.9967; % discount factor
x = 0.5; % worker bargaining weight
rho = 0.975; % persistence of productivity process
sigma = 0.0076; % variance of productivity process
b = 0.9; % value of leisure
c = 0.17; % vacancy posting cost
K = 26.94; % creation cost
model;
% Auxiliary definitions for matching function and flow probabilities
exp(e) = 1-exp(u);
exp(t) = exp(v-u);
% Unemployment value function (eq 16)
exp(U) = b + B*( A*((exp(t))^(1-a))*x*exp(S(+1)) + exp(U(+1)) );
% Vacancy value function (eq 17)
exp(V) = -c + B*( A*((exp(t))^(-a))*(1-x)*exp(S(+1)) + exp(V(+1)) );
% Output value function (eq 18)
exp(M) = exp(z) + B*( (1-l)*exp(S(+1)) + exp(U(+1)) + exp(V(+1)) );
% Matching surplus definition (S=M-U-V) (eq 19)
exp(S) = exp(z) - b + c + B*((1-l) - A*((exp(t))^(1-a))*x - A*((exp(t))^(-a))*(1-x))*exp(S(+1));
% FONC: V=K*n (eq 21)
c = B*( A*((exp(t))^(-a))*(1-x)*exp(S(+1)));
% Law of motion for (eq 22)
exp(u) = exp(u(-1)) + l*(1-exp(u(-1))) - A*((exp(t(-1)))^(1-a)) * exp(u(-1));
% Log productivity follows an exogenous AR(1) process
z = rho*z(-1) + zi;
end;
initval;
U = 294.865;
V = 0.543609;
M = 295.738;
S = 0.32959;
u = 0.0798989;
e = 0.920101;
v = 0.04076282;
t = 0.510245;
z = 1;
end;
shocks;
var zi = sigma^2;
end;
steady(solve_algo=0);
stoch_simul(
order = 1,
periods = 30300,
drop = 300,
IRF = 90
);
```

The original code’s source is michael-droste.com/replications/

Cheers

When doing an exp()-transformation, you need to adjust the initial values as well. When using

[code] var U V M S u e v t z;

varexo zi;

```
parameters A a B x l K b c rho sigma;
l = 0.039; % obsolescence separation rate
a = 0.5; % matching function exponent
A = 0.636; % matching function scale parameter
B = 0.9967; % discount factor
x = 0.5; % worker bargaining weight
rho = 0.975; % persistence of productivity process
sigma = 0.0076; % variance of productivity process
b = 0.9; % value of leisure
c = 0.17; % vacancy posting cost
K = 26.94; % creation cost
model;
% Auxiliary definitions for matching function and flow probabilities
exp(e) = 1-exp(u);
exp(t) = exp(v-u);
% Unemployment value function (eq 16)
exp(U) = b + B*( A*((exp(t))^(1-a))*x*exp(S(+1)) + exp(U(+1)) );
% Vacancy value function (eq 17)
exp(V) = -c + B*( A*((exp(t))^(-a))*(1-x)*exp(S(+1)) + exp(V(+1)) );
% Output value function (eq 18)
exp(M) = exp(z) + B*( (1-l)*exp(S(+1)) + exp(U(+1)) + exp(V(+1)) );
% Matching surplus definition (S=M-U-V) (eq 19)
exp(S) = exp(z) - b + c + B*((1-l) - A*((exp(t))^(1-a))*x - A*((exp(t))^(-a))*(1-x))*exp(S(+1));
% FONC: V=K*n (eq 21)
c = B*( A*((exp(t))^(-a))*(1-x)*exp(S(+1)));
% Law of motion for (eq 22)
exp(u) = exp(u(-1)) + l*(1-exp(u(-1))) - A*((exp(t(-1)))^(1-a)) * exp(u(-1));
% Log productivity follows an exogenous AR(1) process
z = rho*z(-1) + zi;
end;
initval;
U = log(294.865);
V = log(0.543609);
M = log(295.738);
S = log(0.32959);
u = log(0.0798989);
e = log(0.920101);
v = log(0.04076282);
t = log(0.510245);
z = log(1);
end;
shocks;
var zi = sigma^2;
end;
resid;
steady(solve_algo=4,maxit=1000);
stoch_simul(
order = 1,
periods = 30300,
drop = 300,
IRF = 90
);
```

[/code]

it works.

Dear Professor,

Thank you so much, I was stuck I couldn’t get why and thus could not proceed with my work.

Your work here in this blog is precious.

Best,

Fabio

Dear Prof.,

One thing remains obscure about the explog version of the model.

The steady-state value of unemployment becomes log(0.0798989) = -2.5270,

how should this be interpreted?

In the normal model 0.08 should be the equilibrium fraction of the household that remains unemployed, but I don’t really know how to interpret the transformed value.

Best,

Fabio

I am not following. After substitution, the new variables are actually the logarithms of the old variables. Steady state unemployment is therefore still 0.0798989.