Deterministic Terms in VAR Models

Why Deterministic Terms Matter

A VAR can describe feedback among several time series, but its forecasts and mean behavior also depend on which predictable components appear in each equation. An intercept, a linear trend, a seasonal pattern, and a one-time intervention are not interchangeable: each implies a different path for the conditional expectation. In particular, an intercept is generally not the long-run mean of a VAR.

The choice of deterministic terms is part of the model specification. Omitting a systematic trend or seasonal pattern can leave predictable structure in the residuals; adding unnecessary terms can make the model harder to interpret and estimate. The key is to distinguish a constant average level from a changing or recurring mean path.

Definition: A VAR($p$) with deterministic regressors $x_t$ can be written as $$ y_t=c+\sum_{i=1}^{p} A_i y_{t-i}+B x_t+u_t. $$ Here $y_t$ is a $k$-vector of variables, $A_i$ are $k\times k$ coefficient matrices, $c$ is a $k$-vector intercept, and $B x_t$ represents specified deterministic terms. The innovation $u_t$ has conditional mean zero, $E(u_t\mid\mathcal F_{t-1})=0$, and finite covariance. When discussing a stable VAR, assume its autoregressive dynamics satisfy the stability condition and that the innovations have constant mean zero. The regressors $x_t$ are specified as known calendar or intervention values.

For convenience, write $A_\Sigma=\sum_{i=1}^{p}A_i$. Stability makes $I-A_\Sigma$ nonsingular, so the mean equations below have unique solutions. A deterministic component is a fixed, specified input—not an innovation. Conditional expectations therefore depend on its future values when those values are known.

Intercepts and the Stationary Mean

For a stable VAR with an intercept and no other deterministic regressors, the stationary mean $\mu=E(y_t)$ is constant over time. Taking expectations of the model and using $E(u_t)=0$ gives a linear system for $\mu$. The intercept $c$ is the model’s input at each time period; lagged values feed back into that input, so $c$ need not equal $\mu$.

\mu=c+\sum_{i=1}^{p}A_i\mu, \qquad (I-A_\Sigma)\mu=c, \qquad \mu=(I-A_\Sigma)^{-1}c.

Worked Example: Finding the Mean of a Bivariate VAR

Suppose a hypothetical stable VAR(1) is $y_t=c+A_1y_{t-1}+u_t$, with $$ c=\begin{pmatrix}1\\2\end{pmatrix}, \qquad A_1=\begin{pmatrix}0.5&0.1\\0.2&0.4\end{pmatrix}. $$ Assume the innovations have mean zero. Find the stationary mean.

The mean solves $(I-A_1)\mu=c$. In this case, $$ \begin{pmatrix}0.5&-0.1\\-0.2&0.6\end{pmatrix} \begin{pmatrix}\mu_1\\\mu_2\end{pmatrix} = \begin{pmatrix}1\\2\end{pmatrix}. $$ The determinant is $0.5(0.6)-(-0.1)(-0.2)=0.28$, so $$ \mu_1=\frac{0.6(1)+0.1(2)}{0.28} =\frac{0.8}{0.28}=2.857, \qquad \mu_2=\frac{0.2(1)+0.5(2)}{0.28} =\frac{1.2}{0.28}=4.286. $$

Thus the two series have stationary means of approximately 2.857 and 4.286, in their respective measurement units. Substitution checks the result: $1+0.5(2.857)+0.1(4.286)\approx2.857$, and $2+0.2(2.857)+0.4(4.286)\approx4.286$. Neither mean is simply the corresponding intercept.

The stability condition matters: the displayed mean is a stationary, time-invariant mean for a stable process, not a claim that every realized series stays near that value in every period. Innovations continue to move the series around its mean.

Linear Trends: A Time-Varying Mean Path

A linear trend enters as a deterministic regressor, often $t$, with coefficient vector $g$. Consider $y_t=c+\sum_i A_i y_{t-i}+g t+u_t$. Even if the autoregressive dynamics are stable, a nonzero $g$ generally means the unconditional expectation changes over time. A useful particular solution is a linear mean path $m_t=a+b t$, where $m_t=E(y_t)$.

b=(I-A_\Sigma)^{-1}g, \qquad a=(I-A_\Sigma)^{-1}\left(c-\sum_{i=1}^{p}iA_i b\right).

These expressions follow by substituting $m_t=a+bt$ into the mean recurrence $m_t=c+\sum_i A_i m_{t-i}+g t$. Since $m_{t-i}=a+b(t-i)$, the right side becomes $c+A_\Sigma a+A_\Sigma b\,t-\sum_i iA_i b+g t$. Matching the coefficient of $t$ gives $b=A_\Sigma b+g$; matching constants gives $a=c+A_\Sigma a-\sum_i iA_i b$. Rearranging yields the formulas above. Stability ensures the required inverse exists.

Worked Example: A Trend in a Scalar AR(1)

Suppose a hypothetical scalar process satisfies $y_t=2+0.5y_{t-1}+0.4t+u_t$, with zero-mean innovations. The coefficient $0.5$ has absolute value below one, so the autoregressive part is stable. Find its deterministic mean path.

Here $A_1=0.5$, $c=2$, and $g=0.4$. The slope is $$ b=\frac{g}{1-A_1}=\frac{0.4}{0.5}=0.8. $$ For the intercept of the mean path, $$ a=\frac{c-A_1b}{1-A_1} =\frac{2-0.5(0.8)}{0.5} =\frac{1.6}{0.5}=3.2. $$ Therefore $m_t=3.2+0.8t$, in the units of $y_t$. For a direct check at $t=1$, $m_0=3.2$, and the mean recurrence gives $2+0.5(3.2)+0.4=4.0$, matching $m_1=3.2+0.8=4.0$.

The slope of the process’s mean is $0.8$, not the trend coefficient $0.4$: persistence carries part of the deterministic rise into later periods. A stable autoregressive component does not make a process with a continuing linear trend stationary around a constant mean.

Seasonal Indicators and Periodic Means

A seasonal pattern can be represented by calendar indicators. For $q$ seasons, include $q-1$ indicators and an intercept, taking one season as the reference; each indicator is one in its season and zero otherwise. The coefficient on an indicator is its direct effect in that equation, not necessarily the complete difference in the process’s mean for that season. Lagged values transmit seasonal effects into later periods.

For a stable model with a repeating deterministic seasonal pattern, the mean is generally periodic rather than constant. If $s$ labels the season in a cycle, the phase-specific mean $m_s$ satisfies the mean recurrence using the seasonally appropriate deterministic input and the preceding phase means. Those $q$ equations can be solved together. Including all $q$ seasonal indicators alongside an intercept causes exact collinearity because the indicators sum to one in every period; omit a reference indicator or omit the intercept.

Worked Example: Solving for Quarterly Means

Consider a hypothetical stable scalar quarterly model $y_t=1+0.5y_{t-1}+\delta_s+u_t$. Let the seasonal effects be $\delta_1=0$, $\delta_2=0.4$, $\delta_3=-0.2$, and $\delta_4=0.6$, with quarter 1 as the reference. Assume zero-mean innovations and a repeating quarterly calendar. Find the periodic means.

Writing $m_s$ for the expected value in quarter $s$, with quarter 4 preceding quarter 1, gives $$ m_1=1+0.5m_4,\quad m_2=1.4+0.5m_1,\quad m_3=0.8+0.5m_2,\quad m_4=1.6+0.5m_3. $$ Substitute successively: $$ m_2=1.9+0.25m_4,\qquad m_3=1.75+0.125m_4,\qquad m_4=2.475+0.0625m_4. $$ Thus $0.9375m_4=2.475$, so $m_4=2.64$. Back-substitution gives $m_3=1.75+0.125(2.64)=2.08$, $m_2=1.9+0.25(2.64)=2.56$, and $m_1=1+0.5(2.64)=2.32$.

The means, in the units of $y_t$, are 2.32, 2.56, 2.08, and 2.64 for quarters 1 through 4. For example, the quarter 2 check is $1.4+0.5(2.32)=2.56$. These are periodic means, not four independent intercepts: each depends on the previous quarter’s mean through the autoregressive term.

Intervention Dummies: Pulses and Steps

An intervention dummy represents a specified event or regime change. A pulse is one in a single period $T$ and zero otherwise. A step is zero before $T$ and one from $T$ onward. If the dummy enters a VAR equation through $B x_t$, its coefficient is the direct effect in that period, while the autoregressive dynamics determine how the effect carries forward.

A pulse is not the same as a step. A pulse supplies a temporary input, but a stable VAR can respond for several periods after that input disappears. A step supplies an ongoing input, so its effect can approach a new persistent level. In a multivariate system, these responses can propagate across variables as well as through time.

Worked Example: Comparing a Pulse With a Step

Suppose the hypothetical scalar model is $y_t=4+0.6y_{t-1}+\delta D_t+u_t$. Its baseline mean is $4/(1-0.6)=10$. Assume the process is at that baseline mean before an intervention at time $T$, and set $\delta=3$ for a pulse or $\delta=2$ for a step. The autoregressive coefficient is stable, and future innovations have conditional mean zero.

For a pulse, the impact at $T$ is 3 units. At horizon $h\geq0$, its contribution to the forecast relative to baseline is $3(0.6)^h$: 3 at $h=0$, 1.8 at $h=1$, and 1.08 at $h=2$. The pulse itself is over after $T$, but its effect decays through the lagged value.

For a step, the added input is 2 units in every period from $T$ onward. At horizon $h$, its contribution relative to baseline is $$ 2\sum_{j=0}^{h}(0.6)^j =2\frac{1-0.6^{h+1}}{1-0.6}. $$ The contributions are 2 at $h=0$, 3.2 at $h=1$, and 3.92 at $h=2$, approaching $2/(1-0.6)=5$ units. Thus the new limiting mean is 15, compared with the baseline mean of 10. These are forecast differences in the units of $y_t$, conditional on the specified intervention.

Placement, Forecasts, and Interpretation

The location of a deterministic term determines what it means. An intercept enters every equation in every period. A trend coefficient multiplies time, so it adds a gradually changing input. A seasonal coefficient applies when its indicator is active. An intervention coefficient applies according to the specified pulse or step. In a VAR, each input first affects the equation where it is placed and then can influence other variables through the lag matrices.

For forecasting, future deterministic values must be specified. A calendar indicator or known trend value can be carried into the forecast period. An intervention forecast must state whether a pulse has ended or a step continues. The expected future innovations are zero under the stated conditional-mean assumption; the known deterministic inputs remain in the forecast recursion. Consequently, changing the assumed future path of a dummy changes the forecast even when the estimated lag matrices and current data stay the same.

Key takeaway: A VAR intercept is not generally its stationary mean; solve $(I-A_\Sigma)\mu=c$. Trends produce a changing mean path, seasonal indicators produce phase-specific means, and interventions produce effects that propagate through the VAR dynamics. Interpret each deterministic coefficient together with its placement and the associated lag structure.

Common Mistakes

Calling the intercept the long-run mean. That is only generally correct in the special case with no lagged feedback. With feedback, solve the mean equation using $I-A_\Sigma$.
Reading a seasonal indicator coefficient as the full seasonal mean difference. Lagged values carry previous seasonal effects forward. Solve the cycle of mean equations when the phase-specific means are needed.
Using an intercept and all seasonal indicators together. Their columns are exactly linearly dependent. Omit a reference-season indicator or omit the intercept.
Treating a pulse as a one-period effect on the series. It is a one-period input; lagged dynamics can make its effect persist after the pulse ends.
Forecasting a step as though it ended after one period. A step remains active from its start date onward unless the scenario explicitly says otherwise.

Check Your Understanding

Use the definitions and calculations above to test your understanding.

For a stable VAR with intercept $c$ and lag matrices $A_i$, write the equation that determines its stationary mean. Why is the intercept usually not that mean?
In a scalar AR(1) with coefficient $0.4$ and trend coefficient $0.3$, what equation determines the slope of the mean path?
Why can a seasonal indicator coefficient differ from the difference between the corresponding phase-specific means?
What is the collinearity problem with an intercept and all $q$ seasonal indicators?
In a stable scalar AR(1), explain why the effect of a pulse can persist after its dummy returns to zero, and contrast this with the eventual effect of a continuing step.