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I was working on a model and arrived to the following difference equation system: \begin{align} x_{t+1} &= x_{t} \alpha \beta (y_{t} - 2 x_{t})^{\alpha-1} \\ y_{t+1} &= (y_{t} - 2 x_{t})^{\alpha} \end{align} where $\alpha$ and $\beta$ are parameters and $\alpha \in (0,1)$ and $\beta \in (0,1)$. These equations are obtained from a dynamic optimization problem and hence they are Euler equations. When I solve the program by using Bellman equation, I obtain the following solution: $$x_{t} = \frac{1}{2}(1 - \alpha \beta) y_{t} $$ So we know that above system of difference equations have this solution. How to obtain the same result from the system above? I tried iteration technique but seems that it doesn't work here.

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Let $g_t = x_t/y_t$. Dividing two recurrence equations one readily gets: $$ \frac{x_{t+1}}{y_{t+1}} = \alpha \beta \frac{x_t}{y_t - 2 x_t} = \alpha \beta \frac{x_t/y_t}{1 - 2 x_t/y_t} $$ Seeking a solution such that $x_t = c y_t$, the above recurrence gets $1 = \frac{\alpha \beta}{1 - 2 c}$, implying $c = \frac{1}{2} \left(1- \alpha \beta\right)$. –  Sasha Jan 28 '13 at 22:54
    
Thanks a lot Sasha! –  BarisV Jan 28 '13 at 23:26
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Getting the above comment into an answer.

Let $g_t = x_t/y_t$. Dividing two recurrence equations one readily gets: $$ \frac{x_{t+1}}{y_{t+1}} = \alpha \beta \frac{x_t}{y_t - 2 x_t} = \alpha \beta \frac{x_t/y_t}{1 - 2 x_t/y_t} $$ Seeking a solution such that $x_t = c y_t$, the above recurrence gets $1 = \frac{\alpha \beta}{1 - 2 c}$, implying $c = \frac{1}{2} \left(1- \alpha \beta\right)$.

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