# Impulse response functions?

I have a very basic model about a small open economy. This is a summary:

$$E_0\sum_{t=0}^{\infty}\theta_t U(c_t, h_t)$$ utility function: $$d_t = (1 + r_{t-1})d_{t-1} - y_t + c_t + i_t + \Phi(k_{t-1} - k_t)$$ evolution of foreign debt: $$y_t = A_t F(k_t, h_t)$$ output: $$k_{t+1} = i_t + (1 - \delta)k_t$$ stock of capital evolves according to:

The first order conditions of the household's maximization problem are: \begin{align} \lambda_t = \beta(c_t, h_t)(1 + r_t)E_t\lambda_{t+1}\\ \lambda_t = U_c(c_t,h_t) - \eta_t\beta_c(c_t,h_t)\\ \eta_t = -E_t U(c_{t+1}, h_{t+1}) + E_t \eta_{t+1} \beta(c_{t+1}, h_{t+1}) \\ -U_h(c_t,h_t) + \eta_t\beta_h(c_t,h_t) = \lambda_tA_tF_h(k_t,h_t)\\ \lambda_t[1 + \Phi'(k_{t+1} - k_t)] = \beta(c_t,h_t)E_t\lambda_{t+1}[A_{t+1}F_k(k_{t+1},h_{t+1}) + 1 - \delta + \Phi'(k_{t+2} - k_{t+1})] \end{align} Law of motions of the productivity shock: $$\ln A_{t+1} = \rho \ln A_t + \epsilon_{t+1}; ~~~~ \epsilon_{t+1} \sim { NIID}(0,\sigma_\epsilon^2); ~~~~ t\ge 0$$ we parametrize this as \begin{align} U(c,h) = \frac{[c - \omega^{-1}h^\omega]^{1 - \gamma} - 1}{1 - \gamma}\\ \beta(c,h) = [1 + c - \omega^{-1}h^\omega]^{-\psi_1} \\ F(k,h) = k^{\alpha}h^{1-\alpha} \\ \Phi(x) = \frac{\phi}{2}x^2; ~~~~ \phi>0 \end{align} $$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \gamma & \omega & \psi_1 & \alpha & \phi & r & \delta & \rho & \sigma_{\epsilon} \\\hline 2 & 1.455 & 0.11 & 0.32 & 0.028 & 0.04 & 0.1 & 0.42 & 0.0129 \\ \hline \end{array}$$

I have computed the steady state, but how can I compute the impulse response function?

• congratulations & thanks to @caverac on an epic edit latex-ifying a giant image Jan 5, 2018 at 9:37

1. Start by establishing $\epsilon _1 = \epsilon , \epsilon_{t>1}=0$ . Now, epsilon will be your impulse.
2. You can calculate an expression for $A_t,\bar{A}_t\ \forall t$ . From here, you can either:
Notice that, if you log-linearized your equations, $\epsilon=1$ would mean a 1% deviation from the steady state.