<p>I can provide answers to the three numbered questions.</p>
<p><span class="math-container">$\textbf{Point 1}$</span> 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 <span class="math-container">$t$</span> (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 <span class="math-container">$t$</span>. 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 <span class="math-container">$t$</span> may be a state variable, but this capital level was chosen as a control variable at time <span class="math-container">$t-1$</span> based on some investment decision.</p>
<p>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 <span class="math-container">$$\begin{align} \tilde{y}_t &amp;= \mathbb{E}_t[\tilde{y}_{t+1}]-\frac{1}{\sigma}(i_t-\mathbb{E}_t[\pi_{t+1}]-r_t^n) \\
\pi_t &amp;= \beta \mathbb{E}_t[\pi_{t+1}]+\kappa \tilde{y}_t \\
i_t &amp;= \gamma i_{t-1} + (1-\gamma)(\phi_{\pi}\pi_t +\phi_{y}\tilde{y}_t)+v_t\end{align}$$</span> where <span class="math-container">$r_t^n$</span> and <span class="math-container">$v_t$</span> are AR(1) processes. This is a linear system, so we can check uniqueness and stability with the Blanchard-Khan conditions.</p>
<p>To figure out which variables are predetermined and which are jump, we view things from a fixed time point <span class="math-container">$t$</span>. This is important as a variable may be a jump variable at time <span class="math-container">$t$</span> but a predetermined variable at time <span class="math-container">$t+1$</span>. At time <span class="math-container">$t$</span>, the interest rate at time <span class="math-container">$t-1$</span> is known and cannot be altered, so <span class="math-container">$i_{t-1}$</span> is predetermined at time <span class="math-container">$t$</span>. However, <span class="math-container">$\tilde{y}_t$</span>, <span class="math-container">$\pi_t$</span>, and <span class="math-container">$i_t$</span> are undetermined at time <span class="math-container">$t$</span>, 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 <span class="math-container">$i_{t-1}$</span> and <span class="math-container">$i_t$</span>, with one being a predetermined variable at time <span class="math-container">$t$</span> whilst one is a jump variable at time <span class="math-container">$t$</span>. This happens because <span class="math-container">$i_t$</span> is an endogenous state/predetermined variable: <span class="math-container">$i_t$</span> is chosen at time <span class="math-container">$t$</span> and is subsequently a state/predetermined variable at time <span class="math-container">$t+1$</span>.</p>
<p><span class="math-container">$\textbf{Point 2}$</span> The reason we don't have initial conditions on <span class="math-container">$P_t$</span> is because these are our control variables. Hence, <span class="math-container">$P_0$</span> 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 <span class="math-container">$$\begin{align*} \frac{c_t^{-\sigma}}{\beta c_{t+1}^{-\sigma}}&amp;=F'(k_{t+1})+1-\delta \\
k_{t+1}&amp;=AF(k_t)-\delta k_t -c_t \\
k_0&amp;=\hat{k}_0 \\
\lim_{\tau \to \infty}\beta^{\tau}c_{\tau}^{-\sigma}k_{\tau+1}&amp;=0 \end{align*}$$</span> Although we are given <span class="math-container">$k_0$</span> explicitly in this system, we are not given <span class="math-container">$c_0$</span>. However, we can theoretically calculate the optimal <span class="math-container">$c_0$</span> from this system. We do this by iterating from <span class="math-container">$k_0$</span> between budget constraint and the Euler equation before imposing the transversality boundary condition at infinity. This implicitly defines the optimal <span class="math-container">$c_0$</span> although it is generally impossible to get an exact analytical expression for <span class="math-container">$c_0$</span>.</p>
<p><span class="math-container">$\textbf{Point 3}$</span> Yes, if all variables are predetermined at time <span class="math-container">$t$</span>, 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 <span class="math-container">$$x_{t+1}=Ax_t +b_t \qquad x_0=\bar{x}_0$$</span> Therefore, for any <span class="math-container">$T\in \mathbb{N}\cup \{0\}$</span>, we can use backward induction to obtain <span class="math-container">$$x_T=A^{T}x_0+\sum_{i=0}^TA^ib_i$$</span> If any eigenvalue of <span class="math-container">$A$</span> lies outside the unit disk, then this blows up as <span class="math-container">$T\to \infty$</span>. If all eigenvalues of <span class="math-container">$A$</span> lie inside the unit disk, then <span class="math-container">$A^Tx_0$</span> converges and condition (1c) in the Blanchard-Khan paper ensures the second term converges also, meaning the system converges to a unique point as <span class="math-container">$T\to \infty$</span>.</p>

CC BY-SA 4.0
2024-05-13T11:02:04.797Z
Joseph, this is really helpful, thank you.
Maksim Tarasenko
CC BY-SA 4.0
2024-05-12T09:22:54.790Z
2024-05-12T09:22:54.790Z
Joseph Basford
<p>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 (<span class="math-container">$X_t$</span>) and jump variables (<span class="math-container">$P_t$</span>). Main feature of <span class="math-container">$X_t$</span> is that model knows its path in the future, i.e. <span class="math-container">$\mathbb{E}[X_{t+i}|\Omega_t] = X_{t+i}\quad \forall i = 0,...,\infty$</span> and <span class="math-container">$\Omega_t$</span> consists of paths of all variables for current and previous periods. There is also vector of exogenous variables <span class="math-container">$Z_t$</span>. We also have initial conditions for <span class="math-container">$X_{t=0}=X_0$</span> and do not have them for <span class="math-container">$P_t$</span>.</p>
<p><a href="https://i.sstatic.net/MKe8a6pB.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/MKe8a6pB.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$X_t \in \mathbb{R}^{n}, P_t \in \mathbb{R}^{m}$</span>.
<span class="math-container">$A$</span> can be decomposed into its eigenvalues and eigenvectors: <span class="math-container">$A = C^{-1}JC$</span>. A have <span class="math-container">$\bar{n}$</span> eigenvalues which are less than 1 in modulus and <span class="math-container">$\bar{m}$</span> eigenvalues which are greater than 1 in modulus.</p>
<p>Main result of paper is that:</p>
<ol>
<li>If <span class="math-container">$m = \bar{m}$</span> then model has a unique solution.</li>
<li>If <span class="math-container">$m &lt; \bar{m}$</span> then some tight linear constraints between variable occurs and model likely has no solution.</li>
<li>If <span class="math-container">$m &gt; \bar{m}$</span> model is underdetermined and there are lot of solutions.</li>
</ol>
<p>After reading the paper I have several questions:</p>
<ol>
<li><p>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?</p>
</li>
<li><p>Why we do not have initial conditions for <span class="math-container">$P_t$</span>? It does not seem plausible that we cannot get them.</p>
</li>
<li><p>If I have no jump variables in model does it follows that I should not have any eigenvalues of A greater than 1?</p>
</li>
</ol>
<p>We can linearly transform our variables with the decomposition of <span class="math-container">$A$</span> matrix:</p>
<p><a href="https://i.sstatic.net/KMvRMGyI.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/KMvRMGyI.png" alt="enter image description here" /></a></p>
<p>In BK (1980) proof relies on the fact that for <span class="math-container">$X_t$</span> there are boundary values and <span class="math-container">$Q_t$</span> is determined from equation</p>
<p><a href="https://i.sstatic.net/izUtPRj8.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/izUtPRj8.png" alt="enter image description here" /></a></p>
<p>Proof of the fact that &quot;good&quot; model should have <span class="math-container">$\bar{m} = m$</span> relies on equation:</p>
<p><a href="https://i.sstatic.net/oTSyXcvA.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/oTSyXcvA.png" alt="enter image description here" /></a></p>
<p>where <span class="math-container">$X_0$</span> is known, <span class="math-container">$Q_0$</span> is determined. Hence if we have <span class="math-container">$\bar{m} = m$</span>, then we can find unique <span class="math-container">$Y_0$</span> and then iterate them forward.</p>
<p>Then if we have <span class="math-container">$\bar{m} = m = 0$</span>, i.e. no jump variables, then
above equation becomes <span class="math-container">$X_0 = C^{-1} Y_0$</span>, term <span class="math-container">$B_2 Q_0$</span> vanishes and <span class="math-container">$Y_0$</span> is determined.</p>
<p>However, if <span class="math-container">$\bar{m} &gt; 0$</span>, then model seems to be underdetermined as in the paper, because again we have <span class="math-container">$X_0 = B_{1} Y_0 + B_{2} Q_0$</span>. There <span class="math-container">$B_{1}$</span> has more columns than rows which makes <span class="math-container">$Y_0$</span> overdetermined.</p>
<p>Am I right?</p>

CC BY-SA 4.0
2024-05-09T15:56:59.007Z
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.
mark leeds
CC BY-SA 4.0
2024-05-09T16:20:57.067Z
By jump variables I mean non-predetermined ones. The paper is "The solution of linear difference models under rational expectations". https://www.sfu.ca/~kkasa/blanchar.pdf
Maksim Tarasenko
CC BY-SA 4.0
2024-05-10T04:46:15.590Z
thanks for the link and explanation. maybe this will increase the chances of a response.
mark leeds
<macroeconomics>
Proof of Blanchar Kahn conditions questions about jump variables
Maksim Tarasenko
# Proof of Blanchar Kahn conditions questions about jump variables

(Asked by: username_0 on 2024-05-09)

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 $P\_t$.

[![enter image description here](https://i.sstatic.net/MKe8a6pB.png)](https://i.sstatic.net/MKe8a6pB.png)

$X\_t \in \mathbb{R}^{n}, P\_t \in \mathbb{R}^{m}$.
$A$ 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:

[![enter image description here](https://i.sstatic.net/KMvRMGyI.png)](https://i.sstatic.net/KMvRMGyI.png)

In BK (1980) proof relies on the fact that for $X\_t$ there are boundary values and $Q\_t$ is determined from equation

[![enter image description here](https://i.sstatic.net/izUtPRj8.png)](https://i.sstatic.net/izUtPRj8.png)

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

[![enter image description here](https://i.sstatic.net/oTSyXcvA.png)](https://i.sstatic.net/oTSyXcvA.png)

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?

**Tags:** macroeconomics
> **Comments:**
> **username_1**: 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.  
> **username_0**: By jump variables I mean non-predetermined ones. The paper is "The solution of linear difference models under rational expectations". https://www.sfu.ca/~kkasa/blanchar.pdf  
> **username_1**: thanks for the link and explanation. maybe this will increase the chances of a response.  

---

## 1 answers

### Answer by username_2 on 2024-05-12. Score: 2 | Accepted answer

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 $T\to \infty$.

> **Comments:**
> **username_0**: Joseph, this is really helpful, thank you.  

---

