Deterministic Components In Cointegrated VARs

Why Deterministic Terms Matter

A cointegrated system can contain stochastic trends and deterministic components at the same time. An intercept or trend may shift the equilibrium relation among variables, create drift in their levels, or produce a trend in the cointegrating relation itself. These are distinct possibilities: placing a deterministic term in the wrong part of the model changes the economic interpretation as well as the reference distribution for a rank test.

Johansen analysis organizes these possibilities into five commonly used specifications. The distinction is not simply whether a regression includes a constant or a time trend. It is whether that term is restricted to the cointegrating space or is unrestricted in the equations for changes. The restrictions determine which deterministic movements can occur in the levels and which are ruled out.

Definition: Let $y_t$ be a $p$-dimensional vector of variables, with $I(1)$ components and cointegration rank $r$, where $0<r<p$. A reduced-rank error-correction representation is $$ \Delta y_t=\alpha\beta' y_{t-1}+\sum_{i=1}^{k-1}\Gamma_i\Delta y_{t-i}+d_t+\varepsilon_t, $$ where $\alpha$ and $\beta$ are $p\times r$ matrices of rank $r$, $d_t$ collects deterministic terms, and $\varepsilon_t$ is a zero-mean innovation vector. The columns of $\beta$ define the cointegrating relations; $\alpha$ determines how changes respond to departures from those relations.

The usual asymptotic rank-test analysis assumes the lag structure adequately captures short-run dependence, the innovations have a nonsingular covariance matrix and satisfy suitable regularity conditions, and the system has no $I(2)$ components. The deterministic specification must also be appropriate. The five cases below describe alternative restrictions on $d_t$; they do not change the definition of rank itself.

The Five Johansen Deterministic Cases

Write $\rho$ and $\delta$ as $r$-vectors, and $\mu$ and $\nu$ as $p$-vectors. The following forms use the common five-case convention. In the table, “restricted” means the term enters through $\alpha$, so it affects the cointegrating space; “unrestricted” means it can also act outside that space.

Case	Deterministic term in the VECM	Main implication
I	None	No intercept or trend in the specified system.
II	$\alpha\rho$	Constant restricted to the cointegrating relations.
III	$\mu$	Unrestricted intercept; levels can have drift.
IV	$\mu+\alpha\delta t$	Unrestricted intercept and trend restricted to the cointegrating relations.
V	$\mu+\nu t$	Unrestricted intercept and unrestricted linear trend.

For example, Case II can be written as $\alpha(\beta'y_{t-1}+\rho)$, making the restriction visible: the constant changes the location of the equilibrium relation, rather than adding an unrestricted constant to every change equation. Case IV similarly includes a trend inside the relation, $\beta'y_{t-1}+\delta t$, as well as an unrestricted intercept. In Case V, the time trend can act outside the cointegrating space as well.

These specifications imply different possible deterministic paths, not guaranteed paths. The realized levels also depend on the initial values, innovations, and estimated parameters. In particular, the presence of an unrestricted trend permits certain movements; it does not establish that those movements are large or substantively important.

Reading the Restrictions Through the Cointegrating Space

The distinction between restricted and unrestricted terms becomes clearer by decomposing unrestricted vectors. For $0<r<p$, let $\alpha_\perp$ be a $p\times(p-r)$ matrix whose columns span the directions orthogonal to the columns of $\alpha$. An unrestricted intercept can be decomposed as

\mu=\alpha\rho+\alpha_\perp\kappa.

The first component acts through the adjustment directions and can be represented as a constant in the cointegrating relation. The second component acts outside those directions and can generate drift in the common stochastic trends. Likewise, an unrestricted trend vector can be decomposed as

\nu=\alpha\delta+\alpha_\perp\xi.

The $\alpha\delta t$ component can be represented as a trend in the cointegrating relation. The $\alpha_\perp\xi t$ component acts outside that relation; because it enters the equation for $\Delta y_t$ and grows with time, it can generate quadratic deterministic movement in levels. This decomposition explains why “a trend is included” is not enough information to interpret a specification.

Worked Example: A Constant Restricted to the Long-Run Relation

Suppose a two-variable system has $\beta=(1,-1)'$, $\alpha=(-0.30,0.20)'$, and Case II constant $\rho=2$. Ignore short-run lag terms for this illustration. What equilibrium relation does the model specify, and how does an observed deviation affect the changes?

The error-correction term is

\beta'y_{t-1}+\rho =y_{1,t-1}-y_{2,t-1}+2.

Setting this term to zero gives $y_1-y_2=-2$: the constant shifts the equilibrium difference from zero to $-2$. If the term equals $0.50$, its contribution to the changes is

\alpha(0.50) = \begin{pmatrix}-0.30\\0.20\end{pmatrix}(0.50) = \begin{pmatrix}-0.15\\0.10\end{pmatrix}.

Thus, the error-correction component lowers $y_1$ by $0.15$ units and raises $y_2$ by $0.10$ units for that period. These are model-implied contributions, not necessarily the total changes, which also include short-run terms and innovations. No unrestricted intercept or trend has been specified in this example.

Worked Example: Separating an Unrestricted Intercept

Consider a two-variable Case III system with $\alpha=(-0.30,0.20)'$, $\alpha_\perp=(0.20,0.30)'$, and unrestricted intercept $\mu=(0.01,0.08)'$. Decompose the intercept into a component in the adjustment space and one outside it.

With $\rho=0.10$ and $\kappa=0.20$, direct substitution gives

\alpha\rho+\alpha_\perp\kappa = \begin{pmatrix}-0.30\\0.20\end{pmatrix}(0.10) + \begin{pmatrix}0.20\\0.30\end{pmatrix}(0.20) = \begin{pmatrix}-0.03\\0.02\end{pmatrix} + \begin{pmatrix}0.04\\0.06\end{pmatrix} = \begin{pmatrix}0.01\\0.08\end{pmatrix} =\mu.

The adjustment-space component corresponds to a constant in the cointegrating relation, whose zero-error condition is $\beta'y+0.10=0$. The orthogonal component is not restricted to the cointegrating relation and permits drift in the common-trend direction. Case III therefore allows a different level behavior from Case II, even though both can accommodate a nonzero mean in a cointegrating relation.

Worked Example: Restricted Versus Unrestricted Trend

Use the same $\alpha$ and $\alpha_\perp$ as in the preceding example. In Case IV, let $\delta=0.02$ and retain the unrestricted intercept $\mu=(0.01,0.08)'$. What trend enters the long-run relation, and does the model also permit drift in levels?

The deterministic error-correction term is $\beta'y_{t-1}+0.02t$, so its zero-error path is

\beta'y_{t-1}=-0.02t.

The relation therefore has a linear deterministic trend of $0.02$ units per time period, in the units of $\beta'y$. The intercept still decomposes as $\alpha(0.10)+\alpha_\perp(0.20)$, as calculated above. Its component outside the adjustment space allows drift in the common-trend direction. Case IV thus combines a trend in the cointegrating relation with an unrestricted intercept; it does not impose the same level behavior as Case II.

In Case V, a term $\nu t$ is unrestricted. For instance, if $\nu=\alpha(0.02)+\alpha_\perp(0.03)$, then

\nu = \begin{pmatrix}-0.30\\0.20\end{pmatrix}(0.02) + \begin{pmatrix}0.20\\0.30\end{pmatrix}(0.03) = \begin{pmatrix}0\\0.013\end{pmatrix}.

The $\alpha(0.02)t$ part can enter the cointegrating relation as a trend. The $\alpha_\perp(0.03)t$ part acts outside it and grows in the change equation. Summing a linearly growing contribution to changes over time can produce quadratic deterministic movement in levels. Case V permits this behavior; whether it is supported by an application is a modeling and empirical question.

Rank Tests Depend on the Deterministic Case

Johansen trace and maximum-eigenvalue tests assess restrictions on the cointegration rank. Their statistics are constructed from estimated eigenvalues, but under unit roots their null distributions are generally nonstandard. The relevant limiting distributions depend on the rank hypothesis and on how deterministic terms are restricted. Consequently, a critical value for one case cannot be transferred to another simply because the sample and eigenvalues are the same.

For ordered estimated eigenvalues $\hat\lambda_1\geq\cdots\geq\hat\lambda_p$, the trace statistic for the null that the rank is at most $r$ is

\operatorname{Trace}(r)=-T\sum_{i=r+1}^{p}\ln(1-\hat\lambda_i),

where $T$ is the effective sample size. The maximum-eigenvalue statistic for testing rank $r$ against rank $r+1$ is

\operatorname{MaxEig}(r)=-T\ln(1-\hat\lambda_{r+1}).

These formulas describe the computed statistics, not their critical values. Under the usual cointegration null, the statistics do not generally have ordinary chi-square reference distributions. A short way to see the role of the eigenvalues is that the likelihood-ratio comparison combines the unexplained fractions $1-\hat\lambda_i$ for eigenvalues excluded by the rank restriction. Taking minus twice the log of that likelihood ratio gives the sum of negative log terms above, scaled by $T$. The asymptotic distribution of that sum still depends on the deterministic restrictions.

Worked Example: Calculating a Trace Statistic and Using the Correct Reference

Suppose a hypothetical two-variable Johansen analysis uses $T=120$ observations and reports $\hat\lambda_1=0.16$ and $\hat\lambda_2=0.04$. Calculate the trace statistics for $r=0$ and $r=1$. For illustration only, suppose a reference table matched to the fitted deterministic case gives 5% critical values of 18.4 for $r=0$ and 5.6 for $r=1$.

For the null $r=0$, both eigenvalues enter:

\operatorname{Trace}(0) =-120\{\ln(1-0.16)+\ln(1-0.04)\} =-120\{\ln(0.84)+\ln(0.96)\} \approx 25.82.

Since $25.82>18.4$, reject $r=0$ at the 5% level in this illustrative calculation. For $r=1$, only the second eigenvalue enters:

\operatorname{Trace}(1) =-120\ln(1-0.04) =-120\ln(0.96) \approx 4.899.

Since $4.899<5.6$, do not reject the null that the rank is at most one. The sequential trace-test conclusion is rank one, given these illustrative critical values. The values 18.4 and 5.6 are hypothetical and are not universal Johansen cutoffs: actual inference requires critical values appropriate to the deterministic case, rank null, and chosen reference procedure.

Specification Choice and Common Pitfalls

Deterministic terms should be chosen by considering plausible behavior in both the levels and the long-run relations. A case that is too restrictive can leave systematic patterns unexplained; a more permissive case changes the limiting reference distribution and may reduce the power of a rank test. The five cases are not a mechanical ladder in which the most elaborate specification is automatically safest.

Case I is appropriate only when the model excludes both an intercept and a trend.
Case II allows a constant equilibrium offset, but no unrestricted intercept outside the cointegrating space.
Case III allows an unrestricted intercept, including a component that can induce drift in levels.
Case IV permits an unrestricted intercept and a trend within the cointegrating relations.
Case V permits an unrestricted trend, including components that can induce quadratic trends in levels.

A common mistake is to include an intercept or trend in a software command and then treat the resulting critical values as interchangeable with those from another deterministic case. The incorrect reasoning is that the rank statistic formula is the same, so its null distribution must also be the same. The correction is that the formula alone does not determine the reference distribution: the deterministic restrictions change the asymptotic process used to obtain that distribution. Report the deterministic case, and compare the statistic only with critical values or p-values calibrated for that case and rank hypothesis.

Key takeaway: In Johansen analysis, ask where each deterministic term enters. A restricted term changes the cointegrating relation; an unrestricted term can also affect common-trend behavior. Since these choices alter the null distribution of rank tests, the model specification and test reference must match.

Check Your Understanding

Answer each question by identifying both the restriction and its implication.

In Case II, where does the constant enter, and what does it change about the cointegrating relation?
Why can an unrestricted intercept in Case III permit drift in levels even if the cointegrating relation has a constant offset?
What distinguishes the trend in Case IV from the trend in Case V?
For a two-variable system with $T=100$ and $\hat\lambda_2=0.05$, calculate $\operatorname{Trace}(1)$.
Why is it incorrect to use a critical value from Case II for a statistic estimated under Case V?

Case	Deterministic term in the VECM	Main implication
I	None	No intercept or trend in the specified system.
II	\(\alpha\rho\)	Constant restricted to the cointegrating relations.
III	\(\mu\)	Unrestricted intercept; levels can have drift.
IV	\(\mu+\alpha\delta t\)	Unrestricted intercept and trend restricted to the cointegrating relations.
V	\(\mu+\nu t\)	Unrestricted intercept and unrestricted linear trend.