Deriving The AR(2) Stationarity Region

Why Derive the AR(2) Stationarity Region?

For an AR(2) model, stationarity is often checked by testing whether the roots of its characteristic equation lie outside the unit circle. The coefficient inequalities are quicker to apply, but they are easier to remember correctly when their connection to those roots is clear. The derivation also helps distinguish a genuine boundary case from an algebraic feature that only looks like one.

Use the centered model

where \(\mu\) is constant and \(\{\epsilon_t\}\) is zero-mean white noise with finite variance \(\sigma_\epsilon^2\). The conventional AR(2) stationarity condition is the condition for a causal, weakly stationary solution: the roots of the AR polynomial must lie strictly outside the unit circle. White noise here means the errors are uncorrelated over time, have mean zero, and share the same finite variance; independence is not required for this second-moment condition.

From Characteristic Roots to Reciprocal Roots

The roots \(z\) of the AR polynomial can be awkward to test directly. Instead, take their reciprocals \(w=1/z\). Multiplying \(A(1/w)=0\) by \(w^2\) gives a quadratic in \(w\):

A nonzero root \(z\) is outside the unit circle exactly when its reciprocal \(w\) is inside it: \(|z|>1\) if and only if \(|w|<1\). Thus the stationarity problem is equivalent to asking when both roots of this reciprocal-root quadratic lie strictly inside the unit circle.

Deriving the Inequalities

For a real quadratic \(q(w)=w^2+aw+b\), its two roots lie strictly inside the unit circle if and only if

Here is why these three tests work. If the roots are a complex-conjugate pair, their product is their squared modulus. The condition \(|b|<1\) then says their common modulus is less than one; also, \(q(1)\) and \(q(-1)\) are positive because they are squared moduli. If the roots are real, \(q(1)>0\) means both roots are below \(1\) or both are above \(1\); \(q(-1)>0\) means both are above \(-1\) or both are below \(-1\). Together these signs either place both roots in \((-1,1)\) or place both beyond the same endpoint. The latter possibilities have a product with absolute value greater than one, so \(|b|<1\) rules them out. This proves the criterion, including the strictness of its inequalities.

For the reciprocal-root polynomial, \(a=-\phi_1\) and \(b=-\phi_2\). Substitution gives the three stationarity inequalities:

The first inequality is equivalent to \(-1<\phi_2<1\). The second gives \(\phi_1<1-\phi_2\), while the third gives \(\phi_1>\phi_2-1\). Therefore an equivalent coefficient-region description is

The strict region is triangular in the \((\phi_1,\phi_2)\) plane. Its closure has vertices \((-2,-1)\), \((2,-1)\), and \((0,1)\), but those boundary points are not part of the stationary region. The inequalities must be checked strictly; equality is not enough.

What the Boundaries Mean

The two inequalities involving \(\phi_1\) have a direct root interpretation. If \(1-\phi_1-\phi_2=0\), the reciprocal polynomial has \(q(1)=0\), so \(w=1\) is a root and the characteristic polynomial has a root at \(z=1\). If \(1+\phi_1-\phi_2=0\), then \(q(-1)=0\), giving a reciprocal root \(w=-1\) and a characteristic root \(z=-1\). In either case, a root lies on the unit circle and strict stationarity fails.

The product condition also matters: \(|\phi_2|<1\) is necessary for both reciprocal roots to lie inside the unit circle. At an admissible boundary where \(|\phi_2|=1\), the strict inside-unit-circle condition is lost. Do not infer that every coefficient equal to zero has this meaning. A zero coefficient in an intermediate transformed polynomial is not, by itself, proof of a unit-circle root.

Degree Reduction and a Transformation Caution

The reciprocal-root argument also covers \(\phi_2=0\), but this case deserves care. The AR polynomial is then \(A(z)=1-\phi_1z\), which has degree one when \(\phi_1\ne0\). Its characteristic root is \(z=1/\phi_1\), so it is outside the unit circle exactly when \(|\phi_1|<1\). The reciprocal polynomial is \(w^2-\phi_1w=w(w-\phi_1)\); its roots are \(0\) and \(\phi_1\), and both are inside the unit circle exactly when \(|\phi_1|<1\). The same three inequalities give this result: \(|0|<1\), \(1-\phi_1>0\), and \(1+\phi_1>0\). If both coefficients are zero, \(A(z)=1\) is a degree-zero constant polynomial, and the model is white noise.

A related check uses the change of variable \(w=(1+s)/(1-s)\), which maps the left half-plane in \(s\) to the unit disk in \(w\). Substituting into the reciprocal polynomial and multiplying by \((1-s)^2\) gives

For a real quadratic to have both roots strictly in the left half-plane, its leading, linear, and constant coefficients must all be positive. These coefficient conditions give the same region: their middle condition is \(\phi_2>-1\), and adding the leading and constant coefficients gives \(2(1-\phi_2)>0\), or \(\phi_2<1\). The leading and constant conditions are the other two inequalities.

However, a zero linear coefficient alone does not imply a unit-circle root. For example, with \(\phi_1=3\) and \(\phi_2=-1\), the transformed polynomial is \(5s^2-1\). Its roots are \(s=\pm1/\sqrt{5}\), not roots on the imaginary axis. In the reciprocal variable, the roots are \((3\pm\sqrt{5})/2\), neither of modulus one; the pair is simply outside the strict stationarity region. The coefficient conditions, not one isolated coefficient, determine whether the roots satisfy the required location.

A Practical Check for Exam Problems

When a problem gives AR(2) coefficients, the inequalities are usually the fastest reliable test. Keep the signs attached to their expressions rather than trying to recall a shape from memory.

This derivation is aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus. The essential distinction is between roots of the AR polynomial, which must be outside the unit circle, and roots of the reciprocal polynomial, which must be inside it.