Why Seasonal Unit Roots Matter
A series can be nonstationary not only because its level wanders from one observation to the next, but also because each seasonal position follows its own accumulating path. For example, quarterly observations may have four linked paths: one for first quarters, one for second quarters, and so on. A seasonal unit root describes this kind of recurring nonstationarity. Seasonal differencing compares an observation with the one from the same season in the preceding cycle.
The distinction from ordinary differencing is important. Ordinary differencing compares adjacent observations and targets a unit root at frequency zero. Seasonal differencing compares observations one full seasonal period apart; its lag polynomial contains roots at all seasonal frequencies. It therefore removes more than a repeating seasonal level alone.
In an ARIMA or SARIMA specification, \(d\) denotes the ordinary difference order and \(D\) the seasonal difference order. The differencing operator is \((1-B)^d(1-B^s)^D\). This tutorial focuses on what one seasonal difference, \((1-B^s)\), does and how it differs from \((1-B)\).
The Seasonal Random Walk
A basic seasonal unit-root model is \(y_t=y_{t-s}+\varepsilon_t\), where \(\varepsilon_t\) is a mean-zero white-noise sequence with finite variance \(\sigma^2\). White noise has constant variance and zero covariance between distinct times. The model says that the current observation equals the corresponding observation from the previous seasonal cycle plus a new innovation.
For quarterly data, the model links each quarter to the same quarter a year earlier. The sequence of first-quarter values accumulates innovations over years; the same is true separately for second-, third-, and fourth-quarter values. Consequently, the series does not have a constant mean and covariance structure over time, as a weakly stationary series must.
The equation also shows the direct remedy: applying one seasonal difference returns the innovations. If those innovations are white noise, the differenced series is stationary. More generally, if \((1-B^s)y_t\) follows a stationary process, seasonal differencing removes the seasonal unit root and leaves that stationary process.
Why the Seasonal Difference Has Seasonal Roots
The lag polynomial \(1-B^s\) factors as \((1-B)(1+B+\cdots+B^{s-1})\). Its roots are the \(s\) complex numbers on the unit circle whose \(s\)th powers equal one. One of them is \(B=1\), the ordinary zero-frequency unit root; the others correspond to recurring seasonal frequencies. Thus seasonal differencing includes ordinary differencing algebraically, but it also removes the other seasonal roots.
This factorization explains why the two differences are not interchangeable. Applying \((1-B)\) removes the root at \(B=1\), but it does not remove the other roots of \(1-B^s\). Applying \((1-B^s)\) removes the full seasonal-root structure in one operation. For \(s=4\), for instance, the seasonal difference is \(1-B^4\), not merely \(1-B\).
Choose \(s\) from the data frequency: four for quarterly observations, twelve for monthly observations with annual seasonality.
A factor \(1-B^s\) indicates seasonal unit roots. A seasonal autoregressive coefficient near, but not equal to, one is not automatically a unit root.
Compute \(y_t-y_{t-s}\). Then assess whether the differenced series is stationary; differencing is not justified solely because a plot has a repeating pattern.
Worked Example: Removing a Quarterly Seasonal Unit Root
A hypothetical quarterly index follows \(y_t=y_{t-4}+\varepsilon_t\). The initial observations are \(y_1=100\), \(y_2=120\), \(y_3=90\), and \(y_4=110\). The innovations for quarters 5 through 8 are, in order, \(2,-1,3,0\). Calculate the next four observations and their seasonal differences. Assume the innovations are independent, mean zero, and have constant finite variance.
Method. The seasonal period is \(s=4\), so use \(\Delta_4y_t=y_t-y_{t-4}\). From the model, each new observation is its value four quarters earlier plus the stated innovation.
Quarter 5: \(y_5=y_1+\varepsilon_5=100+2=102\), so \(\Delta_4y_5=102-100=2\).
Quarter 6: \(y_6=y_2+\varepsilon_6=120+(-1)=119\), so \(\Delta_4y_6=119-120=-1\).
Quarter 7: \(y_7=y_3+\varepsilon_7=90+3=93\), so \(\Delta_4y_7=93-90=3\).
Quarter 8: \(y_8=y_4+\varepsilon_8=110+0=110\), so \(\Delta_4y_8=110-110=0\).
The differenced values are \(2,-1,3,0\), exactly the innovations. The result checks against \((1-B^4)y_t=\varepsilon_t\). The common mistake is to subtract the preceding quarter instead; that would compute \(\Delta_1 y_t\), not the seasonal difference.
Seasonal Versus Ordinary Differencing
Ordinary differencing is \(\Delta_1y_t=(1-B)y_t=y_t-y_{t-1}\). It compares consecutive observations, while seasonal differencing compares observations separated by \(s\) time steps. When a seasonal unit root is present, ordinary differencing alone leaves the nonzero seasonal-frequency roots. A resulting series may therefore remain nonstationary even if its level appears less persistent.
The reverse caution matters too: seasonal differencing is not a general-purpose substitute for ordinary differencing. If the only unit root is at frequency zero and there is no seasonal unit root, the seasonal difference applies additional factors that may be unnecessary. The model and evidence should support the difference used.
Worked Example: Comparing the Two Differences
A hypothetical quarterly series follows \(y_t=y_{t-4}+\varepsilon_t\). Its starting values are \(y_1=100\), \(y_2=120\), \(y_3=90\), and \(y_4=110\). For quarters 5 through 8, the innovations are \(2,-1,3,0\). Calculate the ordinary first differences and seasonal differences for quarters 5 through 8, and identify which sequence equals the innovations.
Method. First calculate the observations using the seasonal random-walk equation: \(y_5=102\), \(y_6=119\), \(y_7=93\), and \(y_8=110\).
For the ordinary differences, the preceding observations are \(y_4=110\), \(y_5=102\), \(y_6=119\), and \(y_7=93\). Therefore, \(\Delta_1y_5=102-110=-8\), \(\Delta_1y_6=119-102=17\), \(\Delta_1y_7=93-119=-26\), and \(\Delta_1y_8=110-93=17\).
For the seasonal differences, subtract the matching quarter from the previous year: \(\Delta_4y_5=102-100=2\), \(\Delta_4y_6=119-120=-1\), \(\Delta_4y_7=93-90=3\), and \(\Delta_4y_8=110-110=0\).
Only the seasonal differences equal the innovations \(2,-1,3,0\). The ordinary differences are \(-8,17,-26,17\); they combine changes in the recurring seasonal positions with the innovations and do not directly remove the seasonal unit-root factor. The common mistake is to assume any first difference removes every unit root. The factorization of \(1-B^4\) shows why that is false.
A Seasonal AR Term Is Not Necessarily a Unit Root
A seasonal autoregressive term can create strong dependence at lag \(s\) without making the series nonstationary. For example, consider \(x_t=\phi x_{t-s}+\varepsilon_t\), with mean-zero white-noise innovations of variance \(\sigma^2\). For the causal stationary model, the roots of \(1-\phi B^s=0\) must lie outside the unit circle. If \(|\phi|<1\), their modulus is \(|\phi|^{-1/s}>1\). At \(\phi=1\), the polynomial becomes \(1-B^s\), and the seasonal unit roots are present.
Worked Example: Distinguishing a Seasonal AR Term from a Unit Root
A hypothetical quarterly series is modeled as \(x_t=0.8x_{t-4}+\varepsilon_t\), where the innovations are independent with mean zero and variance \(9\). Determine whether the model has a seasonal unit root, calculate the root modulus, and find the stationary variance. Compare this with the same model when the coefficient is \(1\).
Method. For coefficient \(0.8\), the lag polynomial is \(1-0.8B^4\). Its roots satisfy \(B^4=1/0.8=1.25\), so each root has modulus \(1.25^{1/4}\), approximately \(1.0574\). Since this is greater than 1, the roots are outside the unit circle and the model is stationary; it has no seasonal unit root.
For each seasonal subsequence, the variance equation is \(\operatorname{Var}(x_t)=0.8^2\operatorname{Var}(x_{t-4})+9\). Under stationarity, the two variances are equal, so \(\operatorname{Var}(x_t)=0.64\operatorname{Var}(x_t)+9\). Thus \(0.36\operatorname{Var}(x_t)=9\), giving \(\operatorname{Var}(x_t)=25\).
If the coefficient is \(1\), the polynomial is \(1-B^4\), whose roots have modulus 1. It has seasonal unit roots, and the model is instead \(x_t=x_{t-4}+\varepsilon_t\). One seasonal difference yields \(\Delta_4x_t=\varepsilon_t\). The common mistake is to call a strong seasonal AR coefficient a unit root without checking whether it equals the unit-root value.
Derivation and Practical Interpretation
The seasonal-difference result follows directly from the model, not from an approximation. Starting from \(y_t=y_{t-s}+\varepsilon_t\), subtract \(y_{t-s}\) from both sides. This gives \(y_t-y_{t-s}=\varepsilon_t\). By definition, the left side is \((1-B^s)y_t\). Thus seasonal differencing removes the recursive link to the same season in the previous cycle and leaves the new innovation. If the innovations are white noise, the result has constant mean, constant variance, and zero covariance at every nonzero lag.
In applications, a seasonal pattern alone does not prove a seasonal unit root. A stable repeating mean pattern and a stochastic seasonal unit root are different features. The first can recur without accumulating random shocks; the second implies persistence within each seasonal position. Evidence for the latter must come from the model and time-series behavior, not simply from seeing similar peaks or troughs each year.
After differencing, check whether the transformed series is plausibly stationary and whether its residual structure is adequately represented. A seasonal difference can be unnecessary if the series is already stationary, and differencing does not guarantee that all sources of nonstationarity have been removed. In particular, if an ordinary unit root also remains, a seasonal difference’s factor \(1-B\) removes one ordinary root as part of \(1-B^s\), but the full model still needs to be assessed.
Common Exam Traps
- Using the wrong lag. For monthly data with annual seasonality, \(\Delta_{12}y_t=y_t-y_{t-12}\), not \(y_t-y_{t-1}\).
- Assuming ordinary differencing removes all seasonal roots. \((1-B)\) removes the root at \(B=1\), but seasonal roots at other frequencies can remain.
- Confusing a large coefficient with a unit root. A seasonal AR coefficient such as \(0.8\) is not equal to \(1\); check the roots of its lag polynomial.
- Calling a repeating pattern a stochastic seasonal unit root. Deterministic seasonal means and seasonal unit roots represent different mechanisms.
Check Your Understanding
Use the seasonal period and model specified in each problem. Answers are shown for checking.
- A hypothetical monthly series follows \(y_t=y_{t-12}+\varepsilon_t\). If \(y_{13}=54\) and \(y_1=50\), calculate \(\Delta_{12}y_{13}\) and state what it equals under the model. (Answer: \(4=\varepsilon_{13}\).)
- For quarterly data, write the seasonal-difference operator and state how many lags separate the two observations in the difference. (Answer: \(1-B^4\); four quarters, or one year.)
- A monthly model has seasonal factor \(1-B^{12}\). Does ordinary differencing \((1-B)\) remove every root of this factor? Explain briefly. (Answer: No; it removes the root \(B=1\), but not the other seasonal roots.)
- A quarterly seasonal AR model has polynomial \(1-0.5B^4\). Find the modulus of its roots and decide whether it has a seasonal unit root. (Answer: \(2^{1/4}\approx1.1892\); no, the roots are outside the unit circle.)
- A hypothetical quarterly series has \(y_1=40,y_2=55,y_3=45,y_4=60\), and innovations for quarters 5 and 6 equal \(-2\) and \(4\). Under \(y_t=y_{t-4}+\varepsilon_t\), calculate \(y_5,y_6,\Delta_4y_5,\Delta_4y_6\). (Answer: \(y_5=38,\ y_6=59,\ \Delta_4y_5=-2,\ \Delta_4y_6=4\).)