What is the explicit solution for the DSGE model with one unique solution?

[Charlotte87]

Homework Statement

I have understood the point with the Blanchard Kahn condition, my problem is to find the explicit solution when I know there exists one unique solution to the problem. The problem comes from a DSGE model.

Homework Equations

$\begin{pmatrix} p_{t} \\ m_{t} \\ \end{pmatrix} = \begin{pmatrix} \beta & \gamma/\lambda \\ 0 & 1/\gamma \\ \end{pmatrix} \begin{pmatrix} p_{t+1} \\ m_{t+1} \\ \end{pmatrix} + \begin{pmatrix} \gamma/\lambda & \beta \\ 1/\lambda & 0 \\ \end{pmatrix} \begin{pmatrix} u_{t+1} \\ w_{t+1} \\ \end{pmatrix}$

I have found the diagonal matrix of the matrix in front of p(t+1) and m(t+1). Let Q be the matrix of eigenvectors from A and $\Lambda$ be the diagonal matrix of A, we then have $A=Q\Lambda Q^{-1}$:

$\begin{pmatrix} \beta & \gamma/\lambda \\ 0 & 1/\gamma \\ \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & (1-\lambda\beta)/\gamma \\ \end{pmatrix} \begin{pmatrix} \beta & 0 \\ 0 & 1/\lambda \\ \end{pmatrix} \begin{pmatrix} 1 & -\gamma/(1-\lambda\beta) \\ 0 & \gamma/(1-\lambda\beta) \\ \end{pmatrix}$.

The Attempt at a Solution

I have followed the lecture notes where the lecturer starts by taking expectations such that:

$\begin{pmatrix} p_{t} \\ m_{t} \\ \end{pmatrix} = A E_{t} \begin{pmatrix} p_{t+1} \\ m_{t+1} \\ \end{pmatrix}$

Then we premultiply by $Q^{-1}$ and define $Q^{-1} \begin{pmatrix} p_{t} \\ m_{t} \\ \end{pmatrix} = \begin{pmatrix} z_{t}^{1} \\ z_{t}^{2} \\ \end{pmatrix}$. We then get:

$\begin{pmatrix} z_{t}^{1} \\ z_{t}^{2} \\ \end{pmatrix} = \begin{pmatrix} \beta & 0 \\ 0 & 1/\lambda \\ \end{pmatrix} E_{t} \begin{pmatrix} z_{t+1}^{1} \\ z_{t+1}^{2} \\ \end{pmatrix}$

The stable eigenvalue in the diagonal matrix of A is $\beta$ and the unstable one $1/\lambda$. I have tried to follow theoretical papers I have found online but I just do not manage to do it.

The solution is supposed to be:
$p_{t} = \frac{-\gamma}{1-\lambda\beta}m_{t}$

 

[Asim Riaz]

+ \beta^t \left( \frac{\gamma}{1-\lambda\beta} u_{t+1} + p_{0} \right) and m_{t} = \frac{\lambda}{1-\lambda\beta}p_{t+1} + \left(\frac{1}{1-\lambda\beta} \right)^t \left( \frac{1}{\lambda} w_{t+1} + m_{0} \right) .