I am trying to understand BK model stability conditions (1980). Want to remind setup for these conditions. We have linear model with vector of predetermined ($X_t$) and jump variables ($P_t$). Main feature of $X_t$ is that model knows its path in the future, i.e. $\mathbb{E}[X_{t+i}|\Omega_t] = X_{t+i}\quad \forall i = 0,...,\infty$ and $\Omega_t$ consists of paths of all variables for current and previous periods. There is also vector of exogenous variables $Z_t$. We also have initial conditions for $X_{t=0}=X_0$ and do not have them for $P_t$.

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$X_t \in \mathbb{R}^{n}, P_t \in \mathbb{R}^{m}$. $A$ can be decomposed into its eigenvalues and eigenvectors: $A = C^{-1}JC$. A have $\bar{n}$ eigenvalues which are less than 1 in modulus and $\bar{m}$ eigenvalues which are greater than 1 in modulus.

Main result of paper is that:

  1. If $m = \bar{m}$ then model has a unique solution.
  2. If $m < \bar{m}$ then some tight linear constraints between variable occurs and model likely has no solution.
  3. If $m > \bar{m}$ model is underdetermined and there are lot of solutions.

After reading the paper I have several questions:

  1. What is economical difference between predetermined and jump variables? For example, if I want to build simple multivariate model of the economy how do I decide which variable will enter equations with expectation operator and which won't?

  2. Why we do not have initial conditions for $P_t$? It does not seem plausible that we cannot get them.

  3. If I have no jump variables in model does it follows that I should not have any eigenvalues of A greater than 1?

We can linearly transform our variables with the decomposition of $A$ matrix:

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In BK (1980) proof relies on the fact that for $X_t$ there are boundary values and $Q_t$ is determined from equation

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Proof of the fact that "good" model should have $\bar{m} = m$ relies on equation:

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where $X_0$ is known, $Q_0$ is determined. Hence if we have $\bar{m} = m$, then we can find unique $Y_0$ and then iterate them forward.

Then if we have $\bar{m} = m = 0$, i.e. no jump variables, then above equation becomes $X_0 = C^{-1} Y_0$, term $B_2 Q_0$ vanishes and $Y_0$ is determined.

However, if $\bar{m} > 0$, then model seems to be underdetermined as in the paper, because again we have $X_0 = B_{1} Y_0 + B_{2} Q_0$. There $B_{1}$ has more columns than rows which makes $Y_0$ overdetermined.

Am I right?

  • $\begingroup$ I can't help ( never heard of jump variables ) but it might be useful for others if you stated the exact title of the paper or provided a link. Also, is this a rational expectations framework, etc. $\endgroup$
    – mark leeds
    Commented May 9 at 15:56
  • $\begingroup$ By jump variables I mean non-predetermined ones. The paper is "The solution of linear difference models under rational expectations". sfu.ca/~kkasa/blanchar.pdf $\endgroup$ Commented May 9 at 16:20
  • $\begingroup$ thanks for the link and explanation. maybe this will increase the chances of a response. $\endgroup$
    – mark leeds
    Commented May 10 at 4:46

1 Answer 1


I can provide answers to the three numbered questions.

$\textbf{Point 1}$ The difference between predetermined and jump variables is not whether they have an expectation operator around them. Rather, jump variables - also known as control variables - are ones which we can alter at time $t$ (causing them to 'jump' in value, hence the name). On the other hand, predetermined variables - also known as state variables - cannot be changed at time $t$. It is important to emphasise that whether a variable is a state or a control or state depends on the time period from which we view it. Capital at time $t$ may be a state variable, but this capital level was chosen as a control variable at time $t-1$ based on some investment decision.

The distinction between state and control is usually clear from the context of the model. I think an example would help. Take the following three equation NK model $$\begin{align} \tilde{y}_t &= \mathbb{E}_t[\tilde{y}_{t+1}]-\frac{1}{\sigma}(i_t-\mathbb{E}_t[\pi_{t+1}]-r_t^n) \\ \pi_t &= \beta \mathbb{E}_t[\pi_{t+1}]+\kappa \tilde{y}_t \\ i_t &= \gamma i_{t-1} + (1-\gamma)(\phi_{\pi}\pi_t +\phi_{y}\tilde{y}_t)+v_t\end{align}$$ where $r_t^n$ and $v_t$ are AR(1) processes. This is a linear system, so we can check uniqueness and stability with the Blanchard-Khan conditions.

To figure out which variables are predetermined and which are jump, we view things from a fixed time point $t$. This is important as a variable may be a jump variable at time $t$ but a predetermined variable at time $t+1$. At time $t$, the interest rate at time $t-1$ is known and cannot be altered, so $i_{t-1}$ is predetermined at time $t$. However, $\tilde{y}_t$, $\pi_t$, and $i_t$ are undetermined at time $t$, with each being chosen to satisfy the household and firm optimality conditions, along with the central bank's policy response. Notice there is a distinction between $i_{t-1}$ and $i_t$, with one being a predetermined variable at time $t$ whilst one is a jump variable at time $t$. This happens because $i_t$ is an endogenous state/predetermined variable: $i_t$ is chosen at time $t$ and is subsequently a state/predetermined variable at time $t+1$.

$\textbf{Point 2}$ The reason we don't have initial conditions on $P_t$ is because these are our control variables. Hence, $P_0$ should follow from optimality conditions. For example, in a simple (non-linearised) RBC model, optimality conditions are the Euler equation for consumption, the budget constraint, and two boundary conditions $$\begin{align*} \frac{c_t^{-\sigma}}{\beta c_{t+1}^{-\sigma}}&=F'(k_{t+1})+1-\delta \\ k_{t+1}&=AF(k_t)-\delta k_t -c_t \\ k_0&=\hat{k}_0 \\ \lim_{\tau \to \infty}\beta^{\tau}c_{\tau}^{-\sigma}k_{\tau+1}&=0 \end{align*}$$ Although we are given $k_0$ explicitly in this system, we are not given $c_0$. However, we can theoretically calculate the optimal $c_0$ from this system. We do this by iterating from $k_0$ between budget constraint and the Euler equation before imposing the transversality boundary condition at infinity. This implicitly defines the optimal $c_0$ although it is generally impossible to get an exact analytical expression for $c_0$.

$\textbf{Point 3}$ Yes, if all variables are predetermined at time $t$, then all eigenvalues must lie in the unit disk for uniqueness and stability. If all variables are predetermined then our system is a standard recurrence relation of the form $$x_{t+1}=Ax_t +b_t \qquad x_0=\bar{x}_0$$ Therefore, for any $T\in \mathbb{N}\cup \{0\}$, we can use backward induction to obtain $$x_T=A^{T}x_0+\sum_{i=0}^TA^ib_i$$ If any eigenvalue of $A$ lies outside the unit disk, then this blows up as $T\to \infty$. If all eigenvalues of $A$ lie inside the unit disk, then $A^Tx_0$ converges and condition (1c) in the Blanchard-Khan paper ensures the second term converges also, meaning the system converges to a unique point as $T\to \infty$.

  • $\begingroup$ Joseph, this is really helpful, thank you. $\endgroup$ Commented May 13 at 11:02

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