Stability Conditions for VAR Processes

Why VAR Stability Matters

A VAR describes how a vector of variables depends on its own lags. But specifying those lag coefficients does not by itself tell us whether the system’s dynamics settle down, persist indefinitely, or grow. Stability answers this question: under suitable assumptions on the innovations, do the effects of past conditions decay so that the process can have a time-invariant distribution with finite second moments?

The key test uses the eigenvalues of a matrix that rewrites the VAR as a first-order system. Stability is a property of the full system, not of each coefficient considered separately. In particular, coefficients need not all be small in absolute value for a VAR to be stable; the relevant quantities are the moduli of the companion-matrix eigenvalues.

The Companion-Matrix Representation

Consider a zero-mean VAR(\(p\)) for a \(k\)-dimensional vector \(y_t\):

y_t=A_1y_{t-1}+A_2y_{t-2}+\cdots+A_py_{t-p}+u_t.

Here each \(A_j\) is a \(k\times k\) coefficient matrix and \(u_t\) is a \(k\times1\) innovation vector. For the stability results below, assume \(u_t\) has mean zero, finite covariance matrix \(\Sigma_u\), and is serially uncorrelated. Define a \(kp\times1\) state vector by stacking the current vector and its lags:

s_t= \begin{pmatrix} y_t\\ y_{t-1}\\ \vdots\\ y_{t-p+1} \end{pmatrix}, \qquad s_t=Fs_{t-1}+Gu_t,

where the companion matrix \(F\) and the impact matrix \(G\) are

F= \begin{pmatrix} A_1&A_2&\cdots&A_{p-1}&A_p\\ I_k&0&\cdots&0&0\\ 0&I_k&\cdots&0&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\cdots&I_k&0 \end{pmatrix}, \qquad G= \begin{pmatrix} I_k\\ 0\\ \vdots\\ 0 \end{pmatrix}.

The lower block rows of \(F\) simply shift the lagged values down one position. The top block row reproduces the VAR equation. Thus the original system is stable precisely when repeated application of \(F\) makes the effect of the initial state vanish.

Definition: A VAR(\(p\)) is stable if every eigenvalue \(\lambda\) of its companion matrix \(F\) lies strictly inside the unit circle: \(|\lambda|<1\). Equivalently, the spectral radius \(\rho(F)=\max_\lambda|\lambda|\) is less than one.

The condition is strict. An eigenvalue with modulus exactly one is not stable under this definition, and an eigenvalue with modulus greater than one is also unstable. For complex eigenvalues, compare the modulus, not the real part: for \(\lambda=a+bi\), its modulus is \(\sqrt{a^2+b^2}\).

How Eigenvalues Determine the Dynamics

For a homogeneous system with no new shocks, repeated substitution gives \(s_t=F^h s_{t-h}\) after \(h\) steps. The eigenvalues of \(F\) govern whether the powers \(F^h\) decay. If all eigenvalues have modulus less than one, then \(F^h\) tends to the zero matrix as \(h\) grows. If an eigenvalue has modulus greater than one, there are directions in the state space along which the homogeneous dynamics grow. An eigenvalue on the unit circle can produce non-decaying behavior.

The eigenvalue test can also be expressed using a matrix polynomial in a scalar \(z\). The characteristic equation for a VAR(\(p\)) is

\det\!\left(I_k-A_1z-A_2z^2-\cdots-A_pz^p\right)=0.

The VAR is stable if this equation has no roots with \(|z|\leq1\). To see the link, a nonzero companion eigenvalue \(\lambda\) satisfies

\det\!\left(\lambda^p I_k-A_1\lambda^{p-1}-\cdots-A_p\right)=0.

Putting \(z=1/\lambda\) turns this into the matrix-polynomial equation above. Therefore, eigenvalues inside the unit circle correspond to characteristic roots outside it. Remember which object is being discussed: the companion eigenvalues must be inside; the roots in \(z\) must be outside.

Worked Example: A Stable Scalar VAR(1)

Suppose a hypothetical one-variable process, measured in index points, is

y_t=0.7y_{t-1}+u_t,

where \(u_t\) is zero-mean white noise with variance \(4\) index-points squared. The companion matrix is the \(1\times1\) matrix \(F=(0.7)\), so its only eigenvalue is \(0.7\). Since \(|0.7|<1\), the process is stable.

Repeated substitution shows how the initial value fades:

y_t=0.7^h y_{t-h}+\sum_{j=0}^{h-1}0.7^j u_{t-j}.

The factor \(0.7^h\) tends to zero. Under the stated white-noise assumptions, the stable process has a stationary representation whose variance is

\operatorname{Var}(y_t) =4\sum_{j=0}^{\infty}0.7^{2j} =\frac{4}{1-0.7^2} =\frac{4}{0.51} \approx 7.843.

Thus shocks have effects that decay geometrically, and the stationary standard deviation is approximately \(\sqrt{7.843}=2.801\) index points. This variance calculation relies on stability as well as the assumed innovation variance.

Worked Stability Checks

Worked Example: A Bivariate VAR(1) With an Unstable Eigenvalue

Consider a hypothetical two-variable system, with both variables measured in their respective index units:

y_t= \begin{pmatrix} 0.55&0.20\\ 0&1.02 \end{pmatrix}y_{t-1}+u_t.

For a VAR(1), the coefficient matrix itself is the companion matrix. Since this matrix is triangular, its eigenvalues are its diagonal entries: \(0.55\) and \(1.02\). The second eigenvalue has modulus greater than one, so the system is unstable.

The second variable’s equation is \(y_{2,t}=1.02y_{2,t-1}+u_{2,t}\). Even without further shocks, a nonzero initial value in this direction is multiplied by \(1.02^h\) after \(h\) periods. Because \(1.02^h\) grows, this effect does not decay. The first eigenvalue being below one does not make the full VAR stable: every eigenvalue must satisfy the condition.

Worked Example: Checking a Scalar VAR(2)

Suppose a hypothetical process is

y_t=1.2y_{t-1}-0.32y_{t-2}+u_t.

Its companion state and matrix are

s_t= \begin{pmatrix} y_t\\ y_{t-1} \end{pmatrix}, \qquad F= \begin{pmatrix} 1.2&-0.32\\ 1&0 \end{pmatrix}.

The eigenvalues solve \(\det(\lambda I-F)=0\):

\lambda^2-1.2\lambda+0.32=0 =(\lambda-0.8)(\lambda-0.4).

The eigenvalues are \(0.8\) and \(0.4\), both strictly inside the unit circle. This VAR(2) is stable. Equivalently, its characteristic polynomial in \(z\) is \(1-1.2z+0.32z^2\), with roots \(1.25\) and \(2.5\), both outside the unit circle. This gives the same conclusion using the reciprocal-root convention.

Worked Example: A Unit Root Is Not Stability

Consider the hypothetical scalar process \(y_t=y_{t-1}+u_t\), with zero-mean white-noise innovations of variance \(1\) squared unit. Its companion matrix has the single eigenvalue \(1\), so it is not stable: \(|1|\) is not strictly less than one.

Starting from \(y_0=0\), substitution gives \(y_t=\sum_{j=1}^{t}u_j\). Serially uncorrelated innovations therefore imply \(\operatorname{Var}(y_t)=t\), which grows with time rather than remaining constant. This unit-root case contrasts with an explosive eigenvalue greater than one: both fail the stability condition, but their dynamics differ.

Why Stability Implies Stationary Dynamics

The link between stability and stationarity follows from iterating the companion system. After \(h\) steps,

s_t=F^h s_{t-h}+\sum_{j=0}^{h-1}F^jGu_{t-j}.

If \(\rho(F)<1\), then \(F^h\) tends to zero. More precisely, the Jordan form of \(F\) expresses its powers using terms that are a polynomial in \(h\) multiplied by powers of eigenvalues. When every eigenvalue has modulus below one, each such term tends to zero. The influence of the initial state therefore disappears, while the accumulated effects of past innovations converge under the finite-variance assumptions.

In the limit, the state has the causal moving-average representation

s_t=\sum_{j=0}^{\infty}F^jGu_{t-j}.

Because the innovations have a time-invariant covariance and are serially uncorrelated, this representation has a time-invariant covariance matrix. The state covariance \(P=\operatorname{Var}(s_t)\) satisfies

P=FPF^\mathsf{T}+G\Sigma_uG^\mathsf{T}.

This equation follows by applying \(s_t=Fs_{t-1}+Gu_t\), using the fact that \(u_t\) is uncorrelated with past states, and taking covariances. Stability ensures the convergent covariance sum \(P=\sum_{j=0}^{\infty}F^jG\Sigma_uG^\mathsf{T}(F^\mathsf{T})^j\). The top \(k\) components of this stationary state are \(y_t\), so the VAR has time-invariant first and second moments. Stability concerns the dynamics; it does not, by itself, guarantee that innovations are well behaved or that a particular dataset follows the model.

Common Mistakes and Practical Interpretation

Checking coefficients one at a time. A coefficient such as \(1.2\) does not, by itself, establish that a higher-order VAR is unstable. Construct the companion matrix and check all its eigenvalues.
Using the wrong root convention. Companion eigenvalues must be inside the unit circle. Roots of \(\det(I_k-A_1z-\cdots-A_pz^p)=0\) must be outside it.
Treating the boundary as stable. An eigenvalue with modulus exactly one fails the strict stability condition. A unit root can lead to persistent shocks and time-varying variance.
Checking only one eigenvalue. A multivariate system is stable only if every eigenvalue has modulus less than one, including complex eigenvalues.
Equating stability with causation. Stability describes the system’s dynamic behavior under its model assumptions. It does not establish that one variable causally affects another.

In practice, stability is a diagnostic for whether a fitted VAR describes mean-reverting dynamics and supports a stationary interpretation. An estimated eigenvalue close to the unit circle implies slow decay, even if it is technically below one. The distinction matters: a stable system can be highly persistent, while an eigenvalue at or beyond the boundary changes the long-run behavior.

Key takeaway: Rewrite the VAR as \(s_t=Fs_{t-1}+Gu_t\), calculate the eigenvalues of \(F\), and classify the system by their moduli. All eigenvalues strictly below one in modulus imply stability and decaying initial-state effects; unit or larger moduli do not.

Check Your Understanding

Answer each question by applying the eigenvalue condition and stating what it implies for the dynamics.

A VAR(1) has companion eigenvalues \(0.6\) and \(-0.9\). Is it stable? Explain why the negative eigenvalue does not change the conclusion.
A fitted VAR has one eigenvalue \(0.98\) and all others below \(0.7\) in modulus. Is it stable, and what does the eigenvalue near one suggest about persistence?
For a scalar VAR(2), the roots of its characteristic polynomial in \(z\) are \(1.4\) and \(-2\). Is the model stable under the characteristic-root convention?
Why does an eigenvalue equal to one fail the stability condition? Give one implication for the variance in the scalar random-walk example.
State the companion-matrix condition for stability and explain why checking only one eigenvalue is insufficient.