Hi,

You cannot conclude that the model is correct, just that Dynare is able to solve the model. To understand why we need eigenvalues greater than one in models with leaded variables, you should go read some papers dealing with the solution algorithms for linear rational expectation models. A must read is the paper by Blanchard and Kahn (Econometrica, 1980), even if Dynare does not use the algorithm described in this paper (we use a more general approach).

I cannot explain this in detail here, so I only consider a simple example. Suppose we have a model of the form:

y_{t+1} = \rho y_{t}

and that we add the requirement to the model that the solution path must not explode (stability). This model admits only one variable, and this variable is non predetermined (*ie* its level is not inherited from the past, y_t is unknown). For any value of \rho, which in this univariate model plays the role of the eigenvalues, the steady state is y^{\star}=0 (note that if \rho happens to be equal to 1 we actually have an infinity of steady states, any real value of y is a steady state).

If \rho\leq 1, any value of y_t will generate a stable path converging to the steady state y^{\star} at a speed given by the autoregressive parameter (in the case where \rho is precisely equal to 1, y converges instantaneously to the steady state since the steady state is the initial condition). In this case the model admits an infinity of solution paths, this property is called indeterminacy.

If \rho>1, the model admits only one stable solution, y_t = y^{\star} for all t. Any other choice for the initial condition results in an explosive path for y.

Hence, the model admits a unique and stable solution if and only if the autoregressive parameter (the eigenvalue) \rho is strictly greater than one. The example is so simple here that the solution does not look quite interesting: the unique stable solution path is such that the endogenous variable jumps directly to the steady state. To obtain a transitional dynamic to the steady state we would have to add some persistence with lags on the endogenous variable in the model. Although minimalistic, this example illustrates why we need eigenvalues greater than 1 if the model has variables with leads (the general result is that we need as many eigenvalues greater than 1 as non predetermined variables in the model, see the Blanchard and Kahn paper).

Best,

Stéphane.