<p>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.</p>

CC BY-SA 3.0
2016-01-22T07:30:53.880Z
2016-01-22T07:30:53.880Z
Giskard
<blockquote>
  <p>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 $&gt;0$. </p>
</blockquote>

<p>I need to prove that this equation has a unique steady state.</p>

<p>This is what I have done so far; </p>

<p>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? </p>

<macroeconomics><dynamic-programming><steady-state><difference-equations>
Prove the uniqueness of steady state
Sher Afghan
# Prove the uniqueness of steady state

(Asked by: username_0 on 2016-01-22)

> 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?

**Tags:** macroeconomics, dynamic-programming, steady-state, difference-equations

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## 1 answers

### Answer by username_1 on 2016-01-22. Score: 5 | Accepted answer

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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