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ARMA Estimation and Forecasting · Tutorial 373 of 952

Standard Errors and Wald Inference for ARMA Parameters

Use the curvature of an ARMA likelihood or a score-based sandwich estimate to quantify parameter uncertainty and conduct Wald inference.

Intermediate 12 min read

What You'll Learn

  • Obtain ARMA parameter standard errors from the observed likelihood Hessian.
  • Distinguish model-based covariance from a robust sandwich covariance.
  • Construct and interpret asymptotic confidence intervals for ARMA coefficients.
  • Test linear restrictions with a Wald statistic and its reference distribution.
  • Apply the delta method to a nonlinear function such as an ARMA long-run mean.
  • Recognize when boundary estimates and model selection limit ordinary Wald inference.

Why ARMA Estimates Need Standard Errors

Maximum likelihood produces estimates of an ARMA model’s parameters, but an estimate alone does not show how precisely the sample has determined them. Standard errors describe the estimated sampling variability; they support confidence intervals and tests of parameter restrictions. The calculations rely on a large-sample approximation, so the model and the conditions behind that approximation matter.

Write the parameter vector as \(\eta\), the log-likelihood based on \(n\) observations as \(\ell_n(\eta)\), and its maximizer as \(\hat{\eta}\). The vector may include autoregressive coefficients, moving-average coefficients, a mean or intercept, and the innovation variance. The likelihood may be exact Gaussian likelihood or a conditional likelihood, as appropriate to the fitted model. In either case, inference is based on the likelihood actually optimized.

Definition: The observed information matrix at the maximum-likelihood estimate is the negative Hessian of the log-likelihood: $$ \mathcal{J}_n(\hat{\eta})=-\left.\frac{\partial^2\ell_n(\eta)}{\partial\eta\,\partial\eta'}\right|_{\eta=\hat{\eta}}. $$ When this matrix is nonsingular and positive definite, the model-based estimated covariance matrix is \(\widehat{\operatorname{Var}}(\hat{\eta})=\mathcal{J}_n(\hat{\eta})^{-1}\). The standard error of a component is the square root of the corresponding diagonal entry.

The Hessian measures local curvature: a sharply peaked likelihood indicates relatively precise estimates, while a flatter likelihood indicates greater uncertainty. Because the likelihood sums information over observations, its curvature typically increases with sample size, and the resulting covariance typically decreases at rate \(1/n\). Numerical optimization software often returns the Hessian, its inverse, or a covariance estimate derived from it.

Conditions for Wald Inference

Ordinary likelihood-based Wald inference uses an asymptotic normal approximation. In an ARMA setting, its standard justification requires a correctly specified and identifiable model, suitable differentiability and moment conditions, a stable stationary and invertible parameter value in the interior of the admissible parameter region, and a nonsingular information matrix. The fitted solution must also be a genuine local maximum rather than a saddle point or a boundary solution. Under these conditions, the estimate is approximately multivariate normal around the true parameter, with covariance estimated by the inverse information matrix.

$$ \hat{\eta}\ \dot{\sim}\ N\!\left(\eta_0,\ \mathcal{J}_n(\hat{\eta})^{-1}\right) $$

The notation \(\dot{\sim}\) indicates an approximate large-sample distribution, not an exact finite-sample result. In particular, the standard errors do not account for uncertainty from choosing the ARMA orders or selecting a model after examining the data. They describe uncertainty conditional on the fitted model and its assumptions.

Worked Example: A Hessian-Based Standard Error and Interval

Suppose a hypothetical series is fitted with a stationary, invertible ARMA(1,1) model, and the numerical maximum is an interior solution. The autoregressive estimate is \(\hat{\phi}=0.62\). The optimizer reports an observed information matrix whose inverse has the following relevant entries: \(\widehat{\operatorname{Var}}(\hat{\phi})=0.0064\). Assume the likelihood is regular at the maximum and the sample is large enough for the normal approximation to be useful.

The estimated standard error is the square root of this variance:

$$ SE(\hat{\phi})=\sqrt{0.0064}=0.0800. $$

An approximate 95% confidence interval uses the standard normal critical value 1.96:

$$ \hat{\phi}\pm 1.96SE(\hat{\phi}) =0.62\pm 1.96(0.0800) =0.62\pm 0.1568 =[0.4632,\ 0.7768]. $$

Under the fitted model and the asymptotic approximation, this interval gives a range of plausible values for the autoregressive coefficient. Its endpoints are dimensionless, as is \(\phi\). The frequentist interpretation is that intervals constructed this way across repeated samples would cover the fixed true coefficient about 95% of the time when the assumptions hold; it is not a 95% probability statement about this particular interval containing the coefficient.

Sandwich Covariance for Score-Based Inference

The inverse Hessian is model-based: it relies on the likelihood curvature being an appropriate description of sampling variability. A sandwich covariance can instead combine curvature with the variability of the score. This can provide a more robust estimate when the likelihood’s variance assumptions are imperfect, provided the estimator still targets a well-defined parameter and the score covariance is estimated appropriately.

Let \(s_t(\eta)\) denote the contribution of observation \(t\) to the score, so the full score is \(S_n(\eta)=\sum_{t=1}^n s_t(\eta)\). Define the average negative Hessian \(A_n=-n^{-1}\partial^2\ell_n(\hat{\eta})/\partial\eta\partial\eta'\). Let \(B_n\) estimate the long-run covariance of the score contributions. The sandwich covariance estimate is

$$ \widehat{\operatorname{Var}}_{\mathrm{sand}}(\hat{\eta}) =\frac{1}{n}A_n^{-1}B_nA_n^{-1}. $$

The middle matrix \(B_n\) must reflect dependence across score contributions when such dependence is present. For example, a long-run covariance estimate accounts for autocovariances of the score sequence; simply summing individual score outer products is not generally adequate for dependent scores. Under a correctly specified likelihood with suitable conditions, score contributions may have a martingale-difference property, in which case their cross-time covariances vanish and a simpler score outer-product estimate can be appropriate. The method used should match the dependence and likelihood assumptions.

This distinction is important in time series. “Robust” does not mean assumption-free: the sandwich estimate still needs a consistent estimate of the score’s long-run covariance and regularity conditions for the estimator. If the model is correctly specified, the model-based and sandwich estimates should be close in large samples under the information identity. Large discrepancies can signal misspecification or an unstable estimate, but they do not identify the cause by themselves.

Worked Example: A Wald Test of a Linear Restriction

Continue with a hypothetical fitted ARMA(1,1) model. Suppose the moving-average estimate is \(\hat{\psi}=-0.20\), and the estimated covariance entries for \((\hat{\phi},\hat{\psi})\) are \(\widehat{\operatorname{Var}}(\hat{\phi})=0.0064\), \(\widehat{\operatorname{Var}}(\hat{\psi})=0.0100\), and \(\widehat{\operatorname{Cov}}(\hat{\phi},\hat{\psi})=0.0016\). Consider the restriction \(H_0:\phi+\psi=0\), tested against \(H_a:\phi+\psi\ne 0\). Assume the estimates are interior and the joint normal approximation is adequate.

The estimated restricted combination is \(0.62+(-0.20)=0.42\). Its estimated variance includes the covariance between the estimates:

$$ \widehat{\operatorname{Var}}(\hat{\phi}+\hat{\psi}) =0.0064+0.0100+2(0.0016)=0.0196, \qquad SE(\hat{\phi}+\hat{\psi})=\sqrt{0.0196}=0.1400. $$

The one-restriction Wald statistic is the squared estimate divided by its estimated variance:

$$ W=\frac{(0.42-0)^2}{0.0196}=9.00. $$

Under \(H_0\), \(W\) is approximately distributed as \(\chi^2_1\). The upper-tail p-value is about 0.0027, so at the 5% significance level the restriction is rejected. In context, the fitted coefficients provide evidence against their sum being zero under the stated model and approximation. This test does not establish why the restriction fails or that either coefficient has a causal interpretation.

Wald Tests and Confidence Intervals

For a scalar parameter \(\eta_j\), a two-sided Wald test of \(H_0:\eta_j=\eta_{j,0}\) uses

$$ Z=\frac{\hat{\eta}_j-\eta_{j,0}}{SE(\hat{\eta}_j)}\ \dot{\sim}\ N(0,1). $$

Equivalently, its square is approximately \(\chi^2_1\). For \(q\) linear restrictions \(R\eta=r\), where \(R\) is a known \(q\)-row matrix, use

$$ W=(R\hat{\eta}-r)' \left[R\widehat{\operatorname{Var}}(\hat{\eta})R'\right]^{-1} (R\hat{\eta}-r)\ \dot{\sim}\ \chi^2_q. $$

These formulae apply whether the covariance matrix is model-based or a valid sandwich estimate. The chosen covariance estimate must be used consistently in both the test and the corresponding interval. For a scalar coefficient, the usual approximate \(100(1-\alpha)\%\) interval is \(\hat{\eta}_j\pm z_{1-\alpha/2}SE(\hat{\eta}_j)\), where \(z_{1-\alpha/2}\) is the standard normal quantile.

Functions of ARMA Parameters: The Delta Method

Inference often concerns a function of several estimated coefficients rather than a single coefficient. If \(g(\eta)\) is differentiable near the true parameter and the estimate is asymptotically normal, its approximate variance is obtained by evaluating the gradient at \(\hat{\eta}\):

$$ \widehat{\operatorname{Var}}\{g(\hat{\eta})\} \approx \nabla g(\hat{\eta})' \widehat{\operatorname{Var}}(\hat{\eta}) \nabla g(\hat{\eta}). $$

The square root is the delta-method standard error. A normal confidence interval then uses \(g(\hat{\eta})\pm z_{1-\alpha/2}SE\{g(\hat{\eta})\}\). This approximation is most useful when the function is smooth near the estimate and the estimate is not close to a point where the function becomes unstable.

For example, in the ARMA(1,1) equation \(Y_t=c+\phi Y_{t-1}+\varepsilon_t+\psi\varepsilon_{t-1}\), with a stationary autoregressive coefficient, the long-run mean is \(\mu=c/(1-\phi)\). Its estimate depends on both \(c\) and \(\phi\), so its uncertainty must include their covariance.

Worked Example: A Long-Run Mean and Its Standard Error

Suppose a hypothetical fit gives \(\hat{c}=1.20\), \(\hat{\phi}=0.62\), with \(\widehat{\operatorname{Var}}(\hat{c})=0.0025\), \(\widehat{\operatorname{Var}}(\hat{\phi})=0.0064\), and \(\widehat{\operatorname{Cov}}(\hat{c},\hat{\phi})=0.0006\). Assume the autoregressive estimate is inside the stationary region and the likelihood-based covariance is appropriate. The estimated long-run mean is

$$ \hat{\mu}=\frac{\hat{c}}{1-\hat{\phi}} =\frac{1.20}{1-0.62} =\frac{1.20}{0.38} =3.158. $$

For \(g(c,\phi)=c/(1-\phi)\), the gradient is \(\nabla g=(1/(1-\phi),\ c/(1-\phi)^2)'\). At the estimates, the gradient is approximately \((2.6316,\ 8.3102)'\). The delta-method variance is therefore

$$ \begin{aligned} \widehat{\operatorname{Var}}(\hat{\mu}) &\approx (2.6316)^2(0.0025)+(8.3102)^2(0.0064)\\ &\quad+2(2.6316)(8.3102)(0.0006)\\ &\approx 0.4855, \qquad SE(\hat{\mu})\approx\sqrt{0.4855}=0.6968. \end{aligned} $$

The approximate 95% interval is \(3.158\pm1.96(0.6968)\), or \([1.792,\ 4.524]\). If \(Y_t\) is measured in a given unit, the long-run mean and both interval endpoints have that same unit. The covariance term matters: treating \(\hat{c}\) and \(\hat{\phi}\) as independent would omit part of the estimated uncertainty.

Derivation: Why Inverse Curvature Gives a Covariance

The key approximation follows by expanding the score around the true parameter \(\eta_0\). At an interior maximum the score is zero, so a first-order Taylor expansion gives \(0\approx S_n(\eta_0)+\partial S_n(\eta_0)/\partial\eta'(\hat{\eta}-\eta_0)\). Rearranging yields

$$ \sqrt{n}(\hat{\eta}-\eta_0) \approx A^{-1}\frac{S_n(\eta_0)}{\sqrt{n}}, $$

where \(A\) is the probability limit of the average negative Hessian. If the normalized score has limiting covariance \(B\), the covariance of this approximation is \(A^{-1}BA^{-1}/n\) on the scale of \(\hat{\eta}\). Under a correctly specified regular likelihood, the information identity gives \(A=B\), reducing the expression to \(A^{-1}/n\), the large-sample counterpart of the inverse observed information. This derivation explains both the Hessian formula and the sandwich formula, and why the sandwich middle matrix must represent score variability.

Common Pitfalls and Practical Limits

  • Ignoring covariance in a function or restriction. Variances of coefficient combinations include cross-covariance terms. Use the full estimated covariance matrix, not only its diagonal.
  • Calling a Wald interval exact. It is an asymptotic approximation. Small samples, weak identification, or highly skewed likelihoods can make its coverage inaccurate.
  • Using ordinary inference at a boundary. Near-unit roots or parameters close to the stationarity or invertibility boundary can produce nonstandard behavior and unreliable normal approximations. A positive-definite Hessian alone does not resolve this concern.
  • Treating a sandwich estimate as automatically robust. Its score covariance must account for relevant time dependence. A misspecified or inconsistent estimate of that covariance does not become valid merely by using the sandwich formula.
  • Interpreting a nonsignificant test as proof of the null. Failure to reject a restriction means the data did not provide sufficiently strong evidence against it under the test; it does not establish that the restriction is true.
  • Forgetting model-selection uncertainty. Standard errors computed after choosing the ARMA order generally condition on that chosen order and do not include the uncertainty from the selection process.
Key takeaway: ARMA standard errors quantify uncertainty through likelihood curvature or a suitable score-based sandwich covariance. Wald tests, coefficient intervals, and delta-method intervals are useful when the model is regular and the large-sample approximation is credible; near boundaries or after model selection, interpret them with particular care.

Check Your Understanding

Answer each question using the assumptions and calculations developed above.

  1. An estimated coefficient has variance 0.0144 from the inverse observed information. What is its standard error, and what additional approximation is needed to construct a normal-based 95% interval?
  2. Why can the variance of \(\hat{\phi}+\hat{\psi}\) not generally be calculated by adding only the two estimated variances?
  3. For a dependent score sequence, what feature should the middle matrix of a sandwich covariance estimate represent?
  4. In the ARMA(1,1) model with mean \(c/(1-\phi)\), which gradient components are needed for a delta-method standard error?
  5. Give one reason a Wald interval may be unreliable when an ARMA estimate is close to the boundary of the stationary or invertible region.
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