Why AR(2) Stationarity Has a Coefficient Test
For an AR(2) model, stationarity depends on both autoregressive coefficients, not on either coefficient considered alone. The coefficient conditions give a direct test: they classify a parameter pair without requiring numerical root calculations. The strictness of the inequalities matters, especially when a parameter lies on a boundary.
Use the centered AR(2) model
Assume that \(\{\varepsilon_t\}\) is white noise with mean zero and finite, positive variance \(\sigma^2\): innovations at distinct times are uncorrelated and each has variance \(\sigma^2\). The stationarity test below concerns the causal solution, in which \(X_t-\mu\) is built from present and past innovations. The earlier material on characteristic roots gives the root criterion: the roots of the AR polynomial must lie outside the unit circle.
The reciprocal-root form is convenient because a quadratic test gives inequalities directly in \(\phi_1\) and \(\phi_2\). A covariance-stationary process has a constant mean and variance, and its covariance between two observations depends only on their time separation. The coefficient test determines whether the causal AR(2) solution has this property; it does not assert that an arbitrarily initialized sequence is stationary from its first observation.
The Three Coefficient Inequalities
The first two inequalities also imply \(\phi_2<1\). Adding them gives \(2\phi_2<2\), hence \(\phi_2<1\). Thus the three-condition form is equivalent to requiring \(|\phi_2|<1\) together with the first two inequalities. In the usual three-condition test, the upper bound on \(\phi_2\) is already implied; it need not be added as a fourth condition.
Here is a short justification of the test. For a real quadratic \(q(w)=w^2+aw+b\), the transformation \(w=(1+s)/(1-s)\) maps roots with \(|w|<1\) to roots with negative real part in \(s\). Substitution and multiplication by \((1-s)^2\) give
A real quadratic \(As^2+Bs+C\) has both roots with negative real part exactly when \(A>0\), \(B>0\), and \(C>0\). To see why, if the roots are real, positive coefficients imply their sum is negative and their product positive, so both are negative; if they are nonreal conjugates, their real part is \(-B/(2A)<0\). The converse follows from the same sum and product conditions. Applying this criterion to the transformed quadratic gives \(1-a+b>0\), \(1-b>0\), and \(1+a+b>0\). With \(a=-\phi_1\) and \(b=-\phi_2\), these become
which are exactly the three stated inequalities. This argument explains why the conditions are strict: a zero coefficient in the relevant test is not an interior, stable case.
Worked Classifications
Worked Example: An Interior Stationary Pair
A hypothetical monthly index is modeled by \(X_t=0.5X_{t-1}+0.2X_{t-2}+\varepsilon_t\), with mean \(\mu=0\) and independent innovations \(\varepsilon_t\sim N(0,4)\). Classify the causal model. As a check, calculate its reciprocal roots.
Method. First apply the three coefficient inequalities with \(\phi_1=0.5\) and \(\phi_2=0.2\).
The first quantity is \(\phi_1+\phi_2=0.5+0.2=0.7<1\). The second is \(\phi_2-\phi_1=0.2-0.5=-0.3<1\). Finally, \(\phi_2=0.2>-1\). All three conditions hold, so the model has a causal covariance-stationary solution.
For the root check, form \(q(w)=w^2-0.5w-0.2\). The quadratic formula gives \(w=(0.5\pm\sqrt{0.25+0.8})/2=(0.5\pm\sqrt{1.05})/2\). Since \(\sqrt{1.05}\approx1.0247\), the roots are approximately \(0.7623\) and \(-0.2623\). Both have absolute value less than one, agreeing with the inequality test.
Check. The two reciprocal roots are inside the unit circle, so the corresponding roots of \(\Phi(z)\) are their reciprocals and lie outside it. A common mistake avoided here is testing only whether \(|\phi_1|<1\); stationarity is determined by the joint coefficient conditions.
Worked Example: One Failed Inequality Rules Out Stationarity
A hypothetical quarterly claim-count index follows \(X_t=1.5X_{t-1}-0.2X_{t-2}+\varepsilon_t\), with \(\mu=0\) and independent \(\varepsilon_t\sim N(0,1)\). Decide whether it has a causal stationary solution and verify the result using reciprocal roots.
Method. Substitute \(\phi_1=1.5\) and \(\phi_2=-0.2\). The first condition gives \(\phi_1+\phi_2=1.5-0.2=1.3\), which is not less than 1. The second gives \(\phi_2-\phi_1=-0.2-1.5=-1.7<1\). The third gives \(-0.2>-1\). Because the first condition fails, the pair is not stationary in the causal AR(2) sense.
For verification, the reciprocal-root polynomial is \(q(w)=w^2-1.5w+0.2\). Its discriminant is \(1.5^2-4(0.2)=2.25-0.8=1.45\), and \(\sqrt{1.45}\approx1.2042\). Thus \(w=(1.5\pm1.2042)/2\), giving roots approximately \(1.3521\) and \(0.1479\). One reciprocal root is outside the unit circle, so the root condition fails.
Check. The coefficient test and the root calculation agree. A common mistake would be to notice that \(\phi_2\) is between \(-1\) and \(1\) and conclude stationarity; that condition alone is insufficient.
Worked Example: Interpreting the \(\phi_2=-1\) Equality Carefully
Consider two hypothetical AR(2) models, each with \(\mu=0\) and independent innovations \(\varepsilon_t\sim N(0,1)\). The first has \((\phi_1,\phi_2)=(3,-1)\); the second has \((\phi_1,\phi_2)=(0,-1)\). Classify them and determine whether their reciprocal roots lie on the unit circle.
First pair: \((3,-1)\). The first inequality fails because \(\phi_1+\phi_2=3-1=2\), which is not less than 1. The second quantity is \(\phi_2-\phi_1=-1-3=-4<1\), while the third strict condition fails because \(\phi_2=-1\) is not greater than \(-1\). This pair is not stationary. Its reciprocal polynomial is \(q(w)=w^2-3w+1\), with roots \(w=(3\pm\sqrt{9-4})/2=(3\pm\sqrt5)/2\), approximately \(2.6180\) and \(0.3820\). Neither root has absolute value 1. Thus equality \(\phi_2=-1\), by itself, does not imply a unit-circle root.
Second pair: \((0,-1)\). Here \(\phi_1+\phi_2=-1<1\) and \(\phi_2-\phi_1=-1<1\), but the strict condition \(\phi_2>-1\) fails. The reciprocal polynomial is \(q(w)=w^2+1\), whose roots are \(w=i\) and \(w=-i\), both of absolute value 1. This pair is on the unit-root boundary, not in the stationary region.
Check. On the line \(\phi_2=-1\), the other two strict inequalities hold exactly when \(-2<\phi_1<2\); on this segment the reciprocal roots lie on the unit circle. At \(\phi_1=\pm2\), the roots are repeated at \(1\) or \(-1\). Outside that segment, equality \(\phi_2=-1\) alone does not settle whether a unit-circle root exists. The common mistake is to infer a unit root from \(\phi_2=-1\) without checking \(\phi_1\).
What the Equalities Mean
The strict inequalities separate the stationary region from parameter values that fail the causal root test. Two equalities have a particularly direct interpretation. If \(\phi_1+\phi_2=1\), then \(q(1)=1-\phi_1-\phi_2=0\), so \(w=1\) is a reciprocal root. If \(\phi_2-\phi_1=1\), then \(q(-1)=1+\phi_1-\phi_2=0\), so \(w=-1\) is a reciprocal root. These equalities identify unit roots; when the remaining conditions place the pair on the boundary of the stationary region, they are its corresponding unit-root boundaries.
The third equality needs more care. When \(\phi_2=-1\), \(q(w)=w^2-\phi_1w+1\), whose roots have product 1. For \(-2<\phi_1<2\), its discriminant \(\phi_1^2-4\) is negative, so the roots are a complex-conjugate pair. Their product is their squared modulus, hence each has modulus 1. But when \(|\phi_1|>2\), the roots are real and reciprocal; one can be greater than 1 in absolute value and the other less than 1, as the first pair in the example demonstrates. Therefore, do not treat the entire line \(\phi_2=-1\) as a unit-root boundary.
These distinctions matter when classifying a proposed AR(2) model. A point satisfying all three strict inequalities is in the causal stationary region. A point failing at least one is outside that region; it may lie on a unit-root boundary or farther away from the region. The inequalities classify stationarity, while the root calculation can clarify the specific reason a boundary claim is or is not appropriate.
Common Exam Traps
- Using non-strict inequalities. Replacing \(<\) or \(>\) with \(\leq\) or \(\geq\) incorrectly includes a boundary point as stationary.
- Checking only one coefficient. The size of \(\phi_1\) or \(\phi_2\) alone does not determine AR(2) stationarity.
- Confusing the two root conventions. Roots of \(\Phi(z)\) must be outside the unit circle; roots of \(q(w)\) must be inside it. They are reciprocals.
- Overgeneralizing a boundary equality. The equality \(\phi_2=-1\) gives unit-circle roots only on the segment where \(-2\leq\phi_1\leq2\), with the endpoints also having a root at \(1\) or \(-1\). Outside that segment, inspect the other conditions or calculate the roots.
A reliable exam procedure is to write the three inequalities, evaluate each separately, and then state the conclusion. If the question asks about the roots or a boundary case, use \(q(w)=w^2-\phi_1w-\phi_2\) to verify which root behavior applies. This tutorial is aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.
Check Your Understanding
For each parameter pair, apply all three inequalities. Where requested, use the reciprocal-root polynomial to interpret a boundary case.
- A hypothetical monthly process has \(\mu=0\), \(\phi_1=0.3\), \(\phi_2=0.4\), and independent \(N(0,2)\) innovations. Is its causal solution stationary? (Answer: Yes; \(0.7<1\), \(0.1<1\), and \(0.4>-1\).)
- A hypothetical quarterly process has \(\mu=0\), \(\phi_1=1.1\), \(\phi_2=-0.2\), and independent \(N(0,1)\) innovations. Check all three conditions, including the one that may appear closest to failing. (Answer: Yes; \(0.9<1\), \(-1.3<1\), and \(-0.2>-1\).)
- For \((\phi_1,\phi_2)=(-0.8,-0.3)\), classify the causal AR(2) model by evaluating each inequality. (Answer: Stationary; \(-1.1<1\), \(0.5<1\), and \(-0.3>-1\).)
- For \((\phi_1,\phi_2)=(0,1)\), determine whether the pair is stationary and identify the reciprocal roots. (Answer: Not stationary; \(\phi_1+\phi_2=1\) and \(\phi_2-\phi_1=1\), and \(q(w)=w^2-1\) has roots \(1\) and \(-1\).)
- For \((\phi_1,\phi_2)=(2.4,-1)\), determine stationarity and whether equality \(\phi_2=-1\) implies a unit-circle root here. (Answer: Not stationary; \(\phi_1+\phi_2=1.4\) fails, and \(q(w)=w^2-2.4w+1\) has reciprocal roots approximately \(1.8633\) and \(0.5367\), neither on the unit circle.)