Deterministic Terms In Unit-Root Specifications

Why Deterministic Terms Matter

A series can move over time because shocks accumulate, because its expected change is nonzero, or because that expected change itself changes over time. These possibilities are represented by deterministic terms such as an intercept and a time trend. In a unit-root specification, those terms affect not only the alternative model but also the path of the series under the unit-root null.

The central distinction is between a stochastic trend, generated by accumulated shocks, and a deterministic component, specified as a function of time. A unit-root process may contain both. Thus, choosing whether a test regression includes no deterministic term, a constant, or a constant and trend is a modeling decision about the kinds of paths allowed under the null and alternative.

Definition: A unit root in the basic autoregressive model \(y_t=\alpha+\beta t+\rho y_{t-1}+u_t\) means \(\rho=1\). Writing \(\gamma=\rho-1\), the equivalent first-difference regression is \(\Delta y_t=\alpha+\beta t+\gamma y_{t-1}+u_t\), where \(\Delta y_t=y_t-y_{t-1}\). The unit-root null is \(H_0:\gamma=0\), or equivalently \(H_0:\rho=1\). The deterministic terms included in this equation specify which forms of the process are being considered.

For the basic models below, take \(u_t\) to be an innovation with mean zero, constant finite variance, and no serial correlation. These assumptions let us focus on how the deterministic terms and the autoregressive coefficient determine the conditional mean dynamics. They do not by themselves determine the appropriate test critical values.

The Three Basic Specifications

The usual specifications differ in whether the change in the series has no deterministic component, a constant component, or a constant plus a linear time trend. Their null models are not interchangeable.

Regression specification	Unit-root null, \(\gamma=0\)	Stationary alternative for the basic AR(1)
\(\Delta y_t=\gamma y_{t-1}+u_t\)	Random walk without drift	Stationary around zero when \(-2<\gamma<0\)
\(\Delta y_t=\alpha+\gamma y_{t-1}+u_t\)	Random walk with drift \(\alpha\)	Stationary around a constant mean when \(-2<\gamma<0\)
\(\Delta y_t=\alpha+\beta t+\gamma y_{t-1}+u_t\)	Unit root with a time-varying drift; levels have a quadratic deterministic component if \(\beta\ne0\)	Stationary around a linear deterministic trend when \(-2<\gamma<0\)

The restriction on the stationary alternatives matters. Since \(\rho=1+\gamma\), covariance stationarity for the basic AR(1) requires \(|\rho|<1\), which is equivalent to \(-2<\gamma<0\). Saying only \(\gamma<0\) is not enough: sufficiently negative values imply an autoregressive coefficient at or below \(-1\), not a stationary process. In the conventional unit-root testing setup, the left-sided alternative \(\gamma<0\) is understood within a parameter space that excludes such nonstationary negative values.

What the Null Implies for the Level Path

Under the null, \(\rho=1\), so the first-difference equation directly describes the increments of \(y_t\). Summing those increments reveals how the deterministic terms accumulate in levels.

Result: With \(\gamma=0\), the three null specifications imply, respectively, \(\Delta y_t=u_t\), \(\Delta y_t=\alpha+u_t\), and \(\Delta y_t=\alpha+\beta t+u_t\). The first has no deterministic drift in levels; the second has a linear deterministic component; the third has a quadratic deterministic component when \(\beta\ne0\). In all three, shocks also accumulate in the level.

For the trend specification, the accumulation can be shown explicitly. Starting with \(y_0\) and summing from \(s=1\) to \(t\), under the unit-root null:

\begin{aligned} y_t &= y_0+\sum_{s=1}^{t}\Delta y_s \\ &= y_0+\sum_{s=1}^{t}(\alpha+\beta s+u_s) \\ &= y_0+\alpha t+\beta\frac{t(t+1)}{2}+\sum_{s=1}^{t}u_s. \end{aligned}

The \(\alpha t\) term is linear in time, and the term involving \(\beta\) is quadratic in time. The final sum is the stochastic component: each innovation remains in every subsequent level. This calculation explains why a linear trend in the difference equation does not mean merely a linear trend in levels when the process has a unit root.

Worked Example: No Deterministic Term

Suppose the hypothetical process satisfies \(\Delta y_t=u_t\), starts at \(y_0=2\), and has innovations \(u_1=1\), \(u_2=-0.5\), and \(u_3=0.5\). Find its first three levels under the unit-root null.

Here each change equals the current innovation. Thus:

\begin{aligned} y_1 &= y_0+u_1=2+1=3,\\ y_2 &= y_1+u_2=3-0.5=2.5,\\ y_3 &= y_2+u_3=2.5+0.5=3. \end{aligned}

The series has no deterministic drift in this specification, but it does not stay fixed: its level changes as innovations accumulate. The return to 3 at \(t=3\) reflects the particular hypothetical innovations, not a mean-reverting mechanism.

Worked Example: A Constant and Drift Under the Null

Suppose \(\Delta y_t=0.4+u_t\), \(y_0=1\), and the hypothetical innovations are \(0.1\), \(-0.2\), and \(0.1\). Calculate the levels and separate their deterministic contribution from the innovations.

The increments are \(0.4+0.1=0.5\), \(0.4-0.2=0.2\), and \(0.4+0.1=0.5\). Therefore:

y_1=1+0.5=1.5,\qquad y_2=1.5+0.2=1.7,\qquad y_3=1.7+0.5=2.2.

After \(t\) periods, the deterministic contribution to the level is \(\alpha t=0.4t\), while the stochastic contribution is \(\sum_{s=1}^{t}u_s\). At \(t=3\), these are \(1.2\) and \(0.1+(-0.2)+0.1=0\), respectively. Including the initial value gives \(y_3=1+1.2+0=2.2\). The constant in the difference equation is drift under the unit-root null.

Worked Example: A Trend in the Difference Equation

Consider the hypothetical unit-root specification \(\Delta y_t=0.2+0.1t+u_t\), with \(y_0=0\) and innovations \(u_1=0\), \(u_2=0.05\), and \(u_3=-0.05\). Find the first three levels and verify the deterministic component at \(t=3\).

The increments are \(0.2+0.1(1)+0=0.3\), \(0.2+0.1(2)+0.05=0.45\), and \(0.2+0.1(3)-0.05=0.45\). Hence \(y_1=0.3\), \(y_2=0.75\), and \(y_3=1.2\).

The deterministic part of the level at \(t=3\) is:

\alpha t+\beta\frac{t(t+1)}{2} =0.2(3)+0.1\frac{3(4)}{2} =0.6+0.6=1.2.

The innovations sum to \(0+0.05-0.05=0\), so the calculated level equals the deterministic component plus \(y_0=0\). With a nonzero sum of innovations, the level would also include that accumulated stochastic contribution.

How the Stationary Alternatives Differ

When \(-2<\gamma<0\), the coefficient \(\rho=1+\gamma\) satisfies \(|\rho|<1\). The effect of a shock then decays geometrically rather than persisting permanently. What the process fluctuates around depends on the deterministic terms retained in its specification.

With no deterministic term, the model is \(y_t=\rho y_{t-1}+u_t\); its stationary mean is zero. With a constant, \(y_t=\alpha+\rho y_{t-1}+u_t\); its stationary mean \(\mu\) solves \(\mu=\alpha+\rho\mu\), giving \(\mu=\alpha/(1-\rho)=-\alpha/\gamma\). With a constant and a linear trend, the stationary alternative has a deterministic mean that is linear in time, with deviations around that trend governed by a stable autoregressive process.

For the last case, write the mean path as \(m_t=c+dt\). Substituting it into \(m_t=\alpha+\beta t+\rho m_{t-1}\) and equating the coefficients of \(t\) and the constant gives \(d=\beta/(1-\rho)\) and \(c=(\alpha-\rho d)/(1-\rho)\). Thus the trend coefficient in the difference equation is not generally the slope of the stationary mean in levels; the level slope is \(\beta/(1-\rho)\).

Worked Example: Interpreting a Stationary Trend Alternative

Suppose a hypothetical trend specification has \(\alpha=0.2\), \(\beta=0.1\), and \(\gamma=-0.5\). Determine whether the autoregressive part is stationary and find the associated linear mean path.

First, \(\rho=1+\gamma=0.5\), so \(|\rho|=0.5<1\). Equivalently, \(-2<-0.5<0\); the stationary-alternative condition holds. The slope of the linear mean path is:

d=\frac{\beta}{1-\rho}=\frac{0.1}{1-0.5}=0.2.

Its intercept is \(c=(\alpha-\rho d)/(1-\rho)=(0.2-0.5(0.2))/0.5=0.1/0.5=0.2\). Thus the mean path is \(m_t=0.2+0.2t\). This is a stationary process around a linear deterministic trend in the sense that deviations from this path follow a stable AR(1) process; it is not stationary around a fixed level.

Reading the Null and Alternative Together

In the three regression specifications, the shared unit-root restriction is \(\gamma=0\), but the deterministic terms remain part of each model. For example, in the constant specification, setting \(\gamma=0\) leaves \(\Delta y_t=\alpha+u_t\), not \(\Delta y_t=u_t\). In the trend specification, setting \(\gamma=0\) leaves \(\Delta y_t=\alpha+\beta t+u_t\). The null therefore permits the deterministic behavior implied by the included terms.

The corresponding alternatives also differ: zero-mean stationarity, stationarity around a constant mean, and stationarity around a linear trend. Including an unnecessary deterministic term can reduce the precision of inference, while omitting a relevant one can make the modeled null or alternative inappropriate for the series. The precise testing distributions and critical values depend on the specification; they are not interchangeable with ordinary stationary-regression critical values.

Common Mistakes

Calling every negative \(\gamma\) stationary. The condition is not merely \(\gamma<0\). In the basic AR(1), stationarity requires \(-2<\gamma<0\), because \(\rho=1+\gamma\) must satisfy \(|\rho|<1\).
Confusing drift with a trend in the difference equation. A constant \(\alpha\) gives a linear deterministic component in levels under the unit-root null. A linear trend \(\beta t\) in the differences gives a quadratic component in levels.
Treating the three nulls as the same model. All impose a unit root, but the constant and trend terms determine what deterministic changes are allowed under that null.
Interpreting a stationary trend alternative as stationary around a constant. When the trend term is included and the autoregressive coefficient is stable, the deviations are stationary around a linear mean path, not around one fixed mean.

Key takeaway: Always read the deterministic terms together with the unit-root restriction. No term, a constant, and a constant plus trend imply different null paths and different stationary alternatives; in the basic AR(1), the stable alternative requires \(-2<\gamma<0\).

Check Your Understanding

Answer each question using the distinction between deterministic behavior in differences and in levels.

Under \(\Delta y_t=\alpha+u_t\) with \(\gamma=0\), what deterministic component accumulates in the level by time \(t\)?
For the basic AR(1), explain why \(\gamma=-2.5\) does not describe a stationary alternative.
Under \(\Delta y_t=\alpha+\beta t+u_t\) and the unit-root null, what is the deterministic contribution to \(y_t\), apart from \(y_0\) and accumulated innovations?
In the constant-only specification, derive the stationary mean in terms of \(\alpha\) and \(\gamma\).
When the trend specification has a stable autoregressive coefficient, is its alternative stationary around a constant or a linear trend? State the condition on \(\gamma\).