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I'm doing economic modelling where $x_t$ is the intertemporal cash-flow variable. I need to solve the following recurrence relation

$$x_{t+1}=\frac{x_{t+8}}{x_{t+1}}$$

My problem is that I don't exactly know how to linearize this function. Maybe I should introduce a new variable $y_{t} = x_{t+1}$ and get

$$y_{t}^2={x_{t+8}}$$

Could you help me out with the correct linearization technique, please? I'm looking forward to your answer.

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    $\begingroup$ Linear in what exactly? You have $x_{t+1}^2=x_{t+8}$. Take log on both sides: $2\ln(x_{t+1})=\ln(x_{t+8})$. Is this linear enough? Or do you want the function to be linear in $x_t$? $\endgroup$ – Herr K. Dec 12 '15 at 23:24
  • $\begingroup$ You can take the log of both sides. $\endgroup$ – user6695 Dec 13 '15 at 0:11
  • $\begingroup$ Thanks, I've already thought about loglinearization. But what to do after that? I'm looking for the $x_{t}$=... form solution. $\endgroup$ – Übel Yildmar Dec 13 '15 at 11:01
  • $\begingroup$ Do you have some boundary condition(s) as part of the definition of the difference equation, e.g. $x_0$ or $x_T$ given? From the log-linearized form, you'll have $x_t=\sqrt{x_{t+7}}$. So if you know $x_0$, then you'll know $x_7$. $\endgroup$ – Herr K. Dec 13 '15 at 11:23
  • $\begingroup$ I have a basic condition like $x_{0} > 0$. Maybe this stands for to make available the loglinearization. Thank you @Herr K. $\endgroup$ – Übel Yildmar Dec 13 '15 at 11:30
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We have the recurrence relation

$$x_{k+1} = \frac{x_{k+8}}{x_{k+1}}$$

If the denominator is nonzero, this recurrence relation can be rewritten as follows

$$x_{k+7} = x_k^2$$

Assuming positivity and taking the logarithm of both sides, we obtain a linear recurrence relation

$$\ln (x_{k+7}) = 2 \ln (x_k)$$

Shifting,

$$\ln (x_{k+1}) = 2 \ln (x_{k-6})$$

Let

$$\eta_k := \left( \ln (x_k), \ln (x_{k-1}), \dots, \ln (x_{k-6}) \right)$$

be a $7$-dimensional state vector. In matrix form,

$$\eta_{k+1} = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 2\\ 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0\end{bmatrix} \eta_k$$

Hence, we need $7$ initial conditions. We recover $x_k$ via $x_k = \exp( \mathrm e_1^{\top}\eta_k )$.

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