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I have the nonhomogenous systen of difference equations with $ 0<\beta<1$: $$\underbrace{\begin{bmatrix}d_{t+1}\\e_{t+1}\end{bmatrix}}_\text{$X_{t+1}$} =\underbrace{\begin{bmatrix}1&0\\-1&\frac{1}{\beta}\end{bmatrix}}_\text{$\Phi$}\underbrace{\begin{bmatrix}d_t\\e_t\end{bmatrix}}_\text{$X_t$}+\underbrace{\begin{bmatrix}0\\1\end{bmatrix}}_\text{=A}s_t$$ I want to decouple the system into: $$Y_{t+1}=\Lambda Y_t+B s_t $$ Where $Q$ are the eigenvectors of $\Phi$, $Y_t=Q^{-1}X_t$ and $B=Q^{-1}A$ and the eigenvalues:$$\Lambda=\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}$$ Then I want to solve the decoupled system with the larger eigenvalue.

EDIT:
I think I figured out a step: $$Y_{1,t+1}=\lambda_1 Y_{1,t} $$ $$Y_{2,t+1}=\lambda_2 Y_{2,t} + z_t$$ With $\lambda_2=\frac{1}{\beta}$ being larger I use: $$Y_{2,t+1}= \frac{1}{\beta}Y_{2,t} + z_t$$ But then how do I solve this, if it's still depending on two periods?


What I have thus far: $$\lambda_1=1,\lambda_2=\frac{1}{\beta}$$ $$Q=\begin{bmatrix}\frac{1-\beta}{\beta}&0\\1&1\end{bmatrix}$$ $$Q^{-1}=\begin{bmatrix}\frac{\beta}{1-\beta}&0 \\ -\frac{\beta}{1-\beta}&1\end{bmatrix}$$ And when I combine everything I get: $$Q^{-1}X_{t+1}=\begin{bmatrix}\frac{\beta}{1-\beta}&0 \\ -\frac{\beta}{1-\beta}&\frac{1}{\beta}\end{bmatrix}X_t+\begin{bmatrix}0 \\ 1\end{bmatrix}s_t$$

  • Is this system already decoupled?
  • How can I solve this with the larger eigenvalue (which is $\lambda_2=\frac{1}{\beta}$ since $\beta<1$)?
    Both eigenvalues are still contained in the equation, so I just don't get how you could solve it without both of them.

I'm very gratefull for any help or advice :)

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1 Answer 1

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With

$$ Q = \left( \begin{array}{cc} 0 & -\frac{\beta -1}{\beta } \\ 1 & 1 \\ \end{array} \right),\ \ \ \Lambda = \left( \begin{array}{cc} \frac{1}{\beta } & 0 \\ 0 & 1 \\ \end{array} \right),\ \ \ \delta_k = \left( \begin{array}{c} 0\\ 1 \end{array} \right)s_k $$

we have

$$ y_{k+1} = Q\cdot\Lambda\cdot Q^{-1}y_k + \delta_k $$

or

$$ Q^{-1}y_{k+1} = \Lambda\cdot Q^{-1}y_k + Q^{-1}\delta_k $$

or

$$ Y_{k+1} = \Lambda\cdot Y_k + d_k $$

with solution

$$ Y_k = \Lambda^{k-1}Y_0 + \Lambda^{k-1}\sum_{j=0}^{k-1}\Lambda^{-j}d_j $$

and finally

$$ y_k = Q\cdot\left( \Lambda^{k-1}Q^{-1}y_0 + \Lambda^{k-1}\sum_{j=0}^{k-1}\Lambda^{-j}Q^{-1}\delta_j\right) $$

or

$$ y_k = M^{k-1}y_0 + \sum_{j=0}^{k-1}M^{k-j-1}\delta_j $$

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