# Proof of Blanchar Kahn conditions questions about jump variables

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 $$P_t$$.

$$X_t \in \mathbb{R}^{n}, P_t \in \mathbb{R}^{m}$$. $$A$$ 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:

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

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

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?

• 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. Commented May 9 at 15:56
• 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 Commented May 9 at 16:20
• thanks for the link and explanation. maybe this will increase the chances of a response. Commented May 10 at 4:46

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 $$T\to \infty$$.

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