Obtaining impulse response for a variable outside the model

[jrubin]

I'm working with a simple RBC model specified specified in levels so that Dynare linearizes it.
I already know the solutions for capital (K) and labour (L).
The output (Y) equation is not specified in the set of model equations as Y was substituted out.

How can I obtain the impulse response of Y without adding it to the model equations since I don't need the first order approximation to Y as I already know K and L?

Any suggestion will be much appreciated!

[Daniel Bendel]

I would do the following:

Y is computed by using the production function Y= K^(1-alpha)*L^(alpha).

From your model you get the irf for K and L and the corresponding steady state values save in oo_.steady_state().
Then just use these values to calculate the irf for Y:

Y_irf = (K_eps_YOURSHOCK + oo_.steady_state(HERE THE NUMBER WHERE DYNARE HAS SAVED THE SS FOR K).*ones(size(K_eps_YOURSHOCK)))^(1-alpha)* (L_eps_YOURSHOCK + oo_.steady_state(HERE THE NUMBER WHERE DYNARE HAS SAVED THE SS FOR L).*ones(size(L_eps_YOURSHOCK)))^(alpha)

[jrubin]

Thanks for this helpful suggestion!

[jpfeifer]

Be careful about your definition of variables. If they are in logs, you cannot use the output formula for levels.

[jrubin]

Thanks. Yes, the model I have is linearized (not log-linearized).

[jrubin]

Hello Daniel and Johannes:

1. Seems there is an error in the expression: should coefficients be .^(1-alpha) and .^(alpha)?
2. Is the product of the two terms involving K and L conformable to give Y_irf of dimension size(K_eps_YOURSHOCK) x 1?

Y_irf = (K_eps_YOURSHOCK + oo_.steady_state(HERE THE NUMBER WHERE DYNARE HAS SAVED THE SS FOR K).*ones(size(K_eps_YOURSHOCK)))^(1-alpha)* (L_eps_YOURSHOCK + oo_.steady_state(HERE THE NUMBER WHERE DYNARE HAS SAVED THE SS FOR L).*ones(size(L_eps_YOURSHOCK)))^(alpha)

3. If Y=AK^(1-alpha)*L^(alpha), where A is the AR(1) shock, then one need to multiply the expression above with (A_eps_YOURSHOCK +oo_.steady_state(HERE THE NUMBER WHERE DYNARE HAS SAVED THE SS FOR A))?


Thanks, again!!

[jpfeifer]

Please provide the file to replicate the issue.

[jrubin]

Please see the attached file.
Lines 166 is the output expression.
My attempt based on the above suggestion is in lines 170-176.

Thanks, once again!

[jpfeifer]

Your code should most probably be

Code: Select all

//Constructing the irf for Output and M for the A1 (technology) shock:
Y1_irf_A1 = (K_t0_eps_A1 + oo_.steady_state(strmatch('K_t0',M_.endo_names,'exact'))).^(1-alpha);
Y2_irf_A1 = (U_t0_eps_A1 + oo_.steady_state(strmatch('U_t0',M_.endo_names,'exact'))).^(1-alpha);
Y3_irf_A1 = A1_t0_eps_A1 + oo_.steady_state(strmatch('A1_t0',M_.endo_names,'exact'));
Y4_irf_A1 = (1-d0)-(d2*(X_t0_eps_A1 +  oo_.steady_state(strmatch('X_t0',M_.endo_names,'exact')))+d1*(X_t0_eps_A1 +  oo_.steady_state(strmatch('A1_t0',M_.endo_names,'exact'))));
Y_irf_A1 = Y1_irf_A1.*Y2_irf_A1.*Y3_irf_A1.*Y4_irf_A1;
M_irf_A1  = 1 -(mu_t0_eps_A1 + oo_.steady_state(strmatch('mu_t0',M_.endo_names,'exact'))).*Y_irf_A1.^(1-gamma);


[jrubin]

Thanks Johannes, much appreciated!