I have a difference equation $$ p_t^{1-\alpha}=\alpha\sigma \left(y-p_t-\frac{(\sigma p_{t-1}^\alpha+b)p_t^{1-\alpha}}{\alpha\sigma} \right) $$ where $\alpha \in [0,1]$ and everything else is $>0$.

I need to prove that this equation has a unique steady state.

This is what I have done so far;

Simplified the expression to write it in the closed form as follows; $$ p_{t-1}=\left[\frac{\alpha y}{p_{t}^{1-\alpha}}-\alpha p_{t}^{\alpha}-\frac{a+1}{\sigma}\right]^{1/\alpha} $$ Substituted $p_{t-1}=p_t=\overline{p}$ in the closed form, this gave. $$ \overline{p}^{\alpha}=\alpha y\overline{p}^{\alpha-1}- \alpha\overline{p}^{\alpha}-\frac{a+1}{\sigma} $$ I'm stuck here. How can I prove that $\overline{p}$ has a unique solution?


1 Answer 1


Rearranging the steady state equation $$ \overline{p}^{\alpha}=\alpha y\overline{p}^{\alpha-1}- \alpha\overline{p}^{\alpha}-\frac{a+1}{\sigma} $$ we get $$ (1 + \alpha)\overline{p}^{\alpha}=\alpha y\overline{p}^{\alpha-1}- \frac{a+1}{\sigma}. $$ As $\alpha \in [0,1]$, the left hand side of the equation is increasing in $\overline{p}$ and the right hand side is decreasing. At least one of these is strictly monotonic because $\alpha$ cannot be $0$ and $1$ at the same time. Hence at most one solution is possible.


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