Q:

Solving for a difference equation for $s_{t}$

Given $f_{t}=u_{t} - \bar{P}$ and the law of motion for $u_{t} = \rho u_{t-1} + \epsilon_{t}$, where $0<\rho<1$, $\epsilon_{t}$ is mean-zero iid and can be interpreted as a domestic price level shock. 
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I want to calculate this difference equation: $s_{t} = (1-\beta)f_{t}+\beta E_{t}s_{t+1}$.
Using Law of Iterated Expectations, I iterate forward by one period and obtain $$s_{t} = (1-\beta) E_{t} \sum_{j=0}^{T} \beta^{j} f_{t+j} + \beta^{T+1} E_{t} s_{t+T+1}$$
$$ s_{t}=(1-\beta) E_{t} \sum_{j=0}^{\infty} \beta^{j} f_{t+j}$$ if $\lim_{T\rightarrow \infty } \beta^{T}E_{t} s_{t+T} = 0$.
Then how do I use the law of motion for $u_{t}$ to solve for $s_{t}$?

A:

By applying the above difference equation on $ s_{t}=(1-\beta) E_{t} \sum_{j=0}^{\infty} \beta^{j} f_{t+j}$, we obtain $$s_{t}=(1-\beta) E_{t} \sum_{j=0}^{\infty} \beta^{j} \left [ \bar{P} - u_{t+j}   \right ].$$
The given process for $u_{t}$ implies $E_{t} u_{t+j} = \rho ^{j} u_{t}$. Substituting this result into the previous equation and by using the formula of the geometric sum yields $$s_{t} = \bar{P} - \frac{1-\beta}{1 - \rho \beta} u_{t}.$$  

