1
$\begingroup$

Consider the system of two equations: $$y_t=\beta\mathbb{E}_t[y_{t+1}+\gamma\cdot z_{t+1}]$$ $$x_t=\rho x_{t+1}+y_t$$ $$z_t=(z_0-Z_t)e^{-at}+z_T$$ Determine the steady state.

The solution manual says: $$y=\frac{\beta\gamma z}{1-\beta}$$ $$x=\frac{\beta\gamma z}{(1-\beta)(1-\rho)}$$ Now I have no idea how to derive these values. The lecture sheets don't mention anything, except that the steady state is solved when we have the value of the shock $z_t$ (intial and terminal). Could anyone explain me the mathematical derivation of the steady state?

| improve this question | | | | |
$\endgroup$
  • 1
    $\begingroup$ The system of two equations consists of three equations. It is also unclear what the variables are and whether time is discreet or continuous. Furthermore I do not know what you are asking. Do you know what a steady state is, just not in this system, or do you need the definition and its mathematical properties? $\endgroup$ – Giskard Jul 22 '16 at 15:55
  • $\begingroup$ I assume it is a discrete model, although not explicitly given in the excercise. $z_t$ is the shock variable, where $T$ is the 'terminal' time. I am mostely interested in the mathematical definition. Although I know it is the derivative w.r.t. $t$, I can't seem to understand how we obtain the results here - taking the derivative w.r.t. $t$ is not exactely straight forward in this case. $\endgroup$ – Holograph Jul 22 '16 at 21:42
  • $\begingroup$ Please edit this information into the body of the question. Also it may be wise to look up what a steady state is exactly and then rephrase the whole question because there seems to be some misunderstanding. $\endgroup$ – Giskard Jul 22 '16 at 23:54
  • $\begingroup$ @Holograph which manual are you using? $\endgroup$ – An old man in the sea. Oct 17 '16 at 14:39
  • $\begingroup$ This is a homework-type of question that shows no effort. $\endgroup$ – luchonacho Feb 24 '17 at 7:48
1
$\begingroup$

In general, a steady state in discrete time is that the first difference is zero at some $t=t_0$, and that it remains so: $$y_t-y_{t-1}=x_t-x_{t-1}=0$$ for all $t\geq t_0$.

Hence, in a steady state, $y_t$ and $x_t$ are constant at some values $y$ and $x$. Thus, $y=\beta(y+\gamma z)$, which implies $$y=\frac{\beta\gamma z}{1-\beta},$$ and $x=\rho x+y$, which implies $$x=\frac{y}{1-\rho}=\frac{\beta\gamma z}{(1-\rho)(1-\beta)}.$$

| improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.