# How to determine the Steady State of this model?

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?

• 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? Jul 22, 2016 at 15:55
• 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. Jul 22, 2016 at 21:42
• 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. Jul 22, 2016 at 23:54
• @Holograph which manual are you using? Oct 17, 2016 at 14:39
• This is a homework-type of question that shows no effort. Feb 24, 2017 at 7:48

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 $$x=\frac{y}{1-\rho}=\frac{\beta\gamma z}{(1-\rho)(1-\beta)}.$$