IRF and derivative

Dear all,

I need to get the derivative of some variable at the steady-state for my model and I was wondering the following :

If my stoch_simul command is used at first order, do I have the equality for a variable X and a shock E:

((dX/dE)|E=0) = X_E(1)/Sigma

Where X_E(1) is the first value of my IRF vector after a shock of standard deviation Sigma.

In other words, can I use the first value of the IRF vector as the derivative evaluated at the steady-state*standard deviation.

Thank you very much for your help.

Best regards,


Which derivative exactly are you interested in?

The derivative of a variable X around the steady-state. For example the output of a sector j conditionnaly on the shock in sector i :

Capture d’écran 2022-01-14 à 11.39.13

Here, epsilon is the shock and I computed the variable log(Yj) in dynare. I’m wondering if this expression is equal to logYj_esp_i(1) (first value of my IRF after simulation)

That above looks to me like the very definition of a generalized IRF.

Excuse-me professor, what do you mean by “generalized” ? Do you mean it’s indeed the good formula ?

See Impulse response analysis in nonlinear multivariate models - ScienceDirect
In a linear model, the initial point does not matter. In a nonlinear one, it does. The \epsilon=0 seems to be about selecting a point in the state space.

Thanks professor, but I’m not sure I get what you want to explain. In a system linearized at first order with a Taylor expansion (such as in Dynare if the option order=1 is put in the stoch_simul command), isn’t the first value of the IRF vector of a variable the same thing as the derivative evaluated at the steady state ?

Yes, it is the derivative w.r.t the current shock. But due to linearity it does not matter under which initial condition you evaluate the partial derivative. In other words, the partial effect of a shock is always the same in a linear model.

Dear Max1,

Thank you very much for your answer ! I get it now.

Best regards,