Why a Unit Root Determines Differencing
An autoregressive equation can appear to have a familiar form and still describe a nonstationary series. The key is not simply whether an autoregressive coefficient is large; it is the location of the roots of the autoregressive polynomial. A root at one means that the series retains a stochastic trend, so the level series is not stationary under the usual nondegenerate innovation assumptions. Ordinary first differencing removes one such root.
The root’s multiplicity matters too. A polynomial with a repeated root at one requires repeated differencing before the remaining autoregressive part can be stationary. This gives a direct algebraic connection between a polynomial’s factorization and the \(d\) in an ARIMA\((p,d,q)\) representation.
This convention is worth checking carefully: the roots are values of the lag-polynomial argument \(B\), not the autoregressive coefficients themselves. For example, the AR(1) polynomial \(1-\phi B\) has root \(B=1/\phi\). It has a root at one when \(\phi=1\), not whenever \(\phi\) is merely close to one.
Factorization and the Difference Order
The ordinary difference operator is \(\Delta=1-B\), so \(\Delta X_t=X_t-X_{t-1}\). A factor \(1-B\) in \(\phi(B)\) can therefore be removed by applying one ordinary difference to the series. If the factor occurs \(m\) times, then \(\phi(B)\) contains \((1-B)^m\), and \(m\) ordinary differences remove that factor.
In the standard ARIMA representation, deterministic terms are omitted here to focus on the root structure:
Here \(\varepsilon_t\) is white noise with mean zero and positive variance, \(\phi(B)\) is the stationary autoregressive polynomial that remains after differencing, and \(\theta(B)\) is a moving-average polynomial. The integer \(d\) is the ordinary differencing order. The order \(d\) is the multiplicity of the factor \(1-B\) in the reduced autoregressive polynomial, provided the remaining autoregressive part satisfies the stationarity root condition. If an autoregressive and moving-average polynomial share a factor, cancel common factors first; otherwise the apparent unit-root multiplicity can overstate the order in the reduced representation.
Use the stated signs and coefficients to form \(\phi(B)\) exactly.
Factor out \(1-B\) as many times as possible, or verify the root and its multiplicity by polynomial division or differentiation.
Confirm that the remaining autoregressive roots are outside the unit circle. If not, removing only the root at one has not produced a stationary autoregressive remainder.
The number of ordinary factors \(1-B\) removed is \(d\), subject to reducing any common AR and MA factors.
Why the Unit Root Causes Nonstationarity
A short variance calculation shows the central issue in the simplest case. Suppose \(X_t=X_{t-1}+\varepsilon_t\), where the innovations are independent with mean zero and variance \(\sigma^2>0\), and \(X_0\) is fixed. Repeated substitution gives \(X_t=X_0+\sum_{j=1}^{t}\varepsilon_j\). Independence makes the variance of the sum equal to the sum of its variances, so \(\operatorname{Var}(X_t)=t\sigma^2\). It increases with \(t\), violating the constant-variance requirement for weak stationarity.
The same issue has a general frequency-domain explanation. For a stationary ARMA process with white-noise innovation variance \(\sigma^2>0\), the spectral density is proportional to \(\sigma^2|\theta(e^{-i\lambda})|^2/|\phi(e^{-i\lambda})|^2\). If \(\phi(B)\) has a root at \(B=1\) of multiplicity \(m\), its value at \(e^{-i\lambda}\) approaches zero at order \(|\lambda|^m\) as \(\lambda\) approaches zero. If the moving-average polynomial does not cancel that factor, the spectral density behaves like a positive constant times \(|\lambda|^{-2m}\) near zero. This is not integrable for any positive integer \(m\), whereas a stationary process with finite variance must have an integrable spectral density. Thus the undifferenced series cannot be a stationary ARMA process under these assumptions.
Differencing moves the factor into the left-hand side of the ARIMA equation: applying \((1-B)^d\) to the levels cancels the \(d\) factors at one. It does not, by itself, guarantee that every other root problem has disappeared. The remaining autoregressive polynomial still needs to satisfy the stationarity condition.
Worked Example: One Unit Root in a Quadratic Polynomial
A hypothetical monthly claim-count index is modeled with autoregressive polynomial \(\phi(B)=1-1.3B+0.3B^2\). Assume there is no AR–MA common factor and the innovation variance is positive. Find the ordinary differencing order and the remaining autoregressive polynomial.
Method. Factor the polynomial and count its factors \(1-B\). This avoids deciding the order from the size of an individual coefficient.
First, factor by multiplying candidate linear terms: \( (1-B)(1-0.3B)=1-0.3B-B+0.3B^2=1-1.3B+0.3B^2\). Thus \(\phi(B)=(1-B)(1-0.3B)\), so there is exactly one factor \(1-B\), and \(d=1\).
The remaining polynomial is \(1-0.3B\). Its root solves \(1-0.3B=0\), so \(B=1/0.3=10/3\), which is greater than one in modulus. The differenced series therefore has a stationary AR(1) remainder under the stated model. The representation is \((1-0.3B)\Delta X_t=\varepsilon_t\), or \(\Delta X_t=0.3\Delta X_{t-1}+\varepsilon_t\).
Check. Expanding \((1-B)(1-0.3B)\) recovers the given polynomial, and its two roots are \(1\) and \(10/3\). The first is the sole unit root. A common mistake would be to call this an AR(2) requiring two differences; polynomial degree is not the differencing order.
Worked Example: A Repeated Root at One
A hypothetical quarterly exposure index has autoregressive polynomial \(\phi(B)=1-2.2B+1.4B^2-0.2B^3\). Determine the multiplicity of the root at one, the difference order, and whether the remaining autoregressive factor is stationary.
Method. Factor out repeated \(1-B\) terms, then test the root of the remaining factor.
Multiply the factors \((1-B)^2(1-0.2B)\). First, \((1-B)^2=1-2B+B^2\). Then \((1-2B+B^2)(1-0.2B)=1-0.2B-2B+0.4B^2+B^2-0.2B^3\). Collecting terms gives \(1-2.2B+1.4B^2-0.2B^3\), the stated polynomial. Thus the root at one has multiplicity two and the ordinary difference order is \(d=2\).
The remaining autoregressive polynomial is \(1-0.2B\). Solving \(1-0.2B=0\) gives \(B=5\), outside the unit circle. Hence the twice-differenced series has a stationary AR(1) remainder: \((1-0.2B)\Delta^2 X_t=\varepsilon_t\).
Check. Expanding the factors reproduces every coefficient, and the remaining root has modulus five. The trap is to identify only the presence of a unit root and stop at \(d=1\); the repeated factor shows that one difference is not enough.
Worked Example: A Unit-Circle Root That Is Not at One
A hypothetical annual index has autoregressive polynomial \(\phi(B)=(1+B)(1-0.25B)\). Is the ordinary differencing order \(d=1\), and does ordinary differencing alone produce a stationary autoregressive remainder?
Method. Locate each root and distinguish a root at one from a different unit-circle root.
The first factor gives \(1+B=0\), so \(B=-1\). The second gives \(1-0.25B=0\), so \(B=4\). The root at \(-1\) has modulus one, but it is not the root \(B=1\) removed by ordinary differencing. Therefore this polynomial has no factor \(1-B\), and its ordinary difference order based on roots at one is \(d=0\).
Applying one ordinary difference would multiply the polynomial by \(1-B\); it would not cancel the factor \(1+B\). The unit-circle root at \(-1\) would remain, so ordinary differencing alone does not produce a stationary autoregressive remainder.
Check. Substitution confirms \((1+B)|_{B=1}=2\), so \(B=1\) is not a root, while \((1+B)|_{B=-1}=0\). The exam trap is to treat every root with modulus one as a factor removable by \(1-B\). Ordinary differencing removes only the root at one.
What the Order Does—and Does Not—Tell You
For the ordinary, nonseasonal difference operator, the practical rule is exact: count the multiplicity of \(B=1\) in the reduced autoregressive polynomial. A single factor calls for first differencing; a repeated factor calls for the corresponding repeated ordinary differences. Then test the remaining AR roots rather than assuming that the series is stationary simply because the root at one has been removed.
A root at one also differs from a near-unit root. For example, \(1-0.98B\) has root \(B=1/0.98\), not exactly one. It may lead to highly persistent behavior, but it does not contain the exact factor \(1-B\), so it does not establish a positive integer ordinary differencing order. Root location is an algebraic condition, not a judgment based on whether a coefficient looks close to one.
Keep the following distinctions in view when interpreting an ARIMA representation:
- The degree of \(\phi(B)\) is the AR order \(p\); it does not equal the differencing order \(d\).
- The multiplicity of the factor \(1-B\), not the total number of roots or factors, determines ordinary \(d\).
- A root at \(B=-1\) is on the unit circle but is not removed by ordinary differencing.
- After differencing, the remaining autoregressive roots must still satisfy the stationarity condition.
- Common AR–MA factors should be canceled before interpreting the reduced model’s orders.
Check Your Understanding
For each polynomial, factor it or solve for its roots before deciding the ordinary difference order. Check whether the autoregressive remainder is stationary.
- Let \(\phi(B)=1-1.6B+0.6B^2\). Find its factors, \(d\), and the root of the remaining AR factor. (Answer: \((1-B)(1-0.6B)\); \(d=1\); remaining root \(B=5/3\), so the remainder is stationary.)
- Let \(\phi(B)=1-2.5B+2B^2-0.5B^3\). Factor it and find the ordinary difference order. (Answer: \((1-B)^2(1-0.5B)\); \(d=2\); remaining root \(B=2\), so the remainder is stationary.)
- For \(\phi(B)=1-0.7B\), determine whether ordinary differencing is indicated by a root at one. (Answer: the root is \(B=10/7\), so there is no root at one and \(d=0\); the AR(1) polynomial is stationary.)
- Let \(\phi(B)=(1-B)(1+0.25B)\). Find \(d\) and test the remaining root. (Answer: \(d=1\); the remaining root is \(B=-4\), so the differenced series has a stationary AR remainder.)
- Let \(\phi(B)=(1-B)(1+B)(1-0.2B)\). Does one ordinary difference produce a stationary AR remainder? (Answer: \(d=1\) removes the root at \(B=1\), but the root at \(B=-1\) remains; ordinary differencing alone does not yield a stationary AR remainder.)