A weakly stationary process has finite second moments, a constant expected value , and covariance depending only on the time difference:A strictly stationary process, also called strongly stationary, has every finite-dimensional probability distribution invariant under a common shift: for every finite choice of times and every shift , and have the same law. strict stationarity with finite second moments implies weak stationarity. weak stationarity alone does not determine the full joint law, and strict stationarity alone does not guarantee finite moments.
Two useful features of autoregressive conditional heteroscedasticity are persistent changes in conditional scale and excess unconditional kurtosis. Its time-varying conditional variance can explain volatility clustering, where large absolute returns occur in groups even when signed returns have little autocorrelation. A homoscedastic autoregressive moving-average model has a fixed innovation variance.
Also, a conditional normal distribution with a random scale is a Gaussian scale mixture. Its unconditional kurtosis can exceed , or its fourth moment can be infinite. A Gaussian autoregressive moving-average model remains jointly Gaussian and cannot reproduce this effect. In the particular lag-two ARCH process, persistence of the squared scale occurs within each parity subsequence; the two parity subsequences are independent.
First justify the dependence structure rather than assume that adjacent squares behave like an ordinary lag-one ARCH model. Write . Iterating its nonnegative recurrence givesThe product coefficient has expected value , so it tends to zero in probability. The stationary are a tight family; consequently the remainder tends to zero in probability, without needing independence between that remainder's factors. The increasing partial sums therefore give the representationThis proves the parity decomposition of a lag-two ARCH process: even and odd observations are functions of disjoint noise families. It also shows that the positive scale is a function of past noise, independent of . The conditional expectation of given past noise is zero, and the second-moment recursion is . independence of the parity families gives the adjacent-square covariance. ThusTo establish finiteness of the fourth moment before using its recursion, note that the norm of the th term in the positive series for is . The triangle inequality in L2 space makes the series square-integrable when . independence of the current noise and past scale then yieldsand henceIf , a finite would make , which is impossible. Its fourth moment is then infinite. The adjacent-square product remains integrable by parity independence, even in that case.
The useful autoregressive model is for the squares, not for signed observations. Put andThenThe errors form a martingale difference sequence relative to the noise history, because the current standardized noise is independent of the past. When the fourth moment is finite they have finite variance and are uncorrelated across distinct times, although their conditional variance depends on the regressor. In centered form, .
For ordinary least squares, use the response-regressor pairs , , . With their separate means and , minimize . If the regressor sum of squares is positive, the estimators areThe two means use the matched pairs; replacing them indiscriminately by a single full-sample mean is not the exact least-squares formula. The noise-series representation supplies an ergodic stationary process. If , finite regressor second moments and the error's zero conditional expectation justify the usual population regression and statistical consistency argument. The observations still define a finite-sample least-squares fit outside that moment range, but the ordinary finite-variance justification must not be claimed there. If parameter constraints are required, minimize the same criterion subject to and , rather than assert that unconstrained estimates automatically satisfy them.
The covariance is zero. Let . The stationary noise-series representation makes measurable with respect to , whereas the martingale difference sequence property gives . The law of total expectation therefore givesSince , this is the required covariance. All products are integrable by the Cauchy-Schwarz inequality, using finite second moments of and the stipulated square integrability of .
This covariance is also zero, but symmetry is essential to the proof. Write , where is an independent fair sign. The normal distribution makes that sign independent of its magnitude and of every other driving variable. Conditional on the magnitudes and all noise except this sign, changing flips and leaves unchanged. Every future conditional scale uses only squared past observations, so is unchanged for .
Thus is independent of the remaining fair sign in . Averaging that sign gives , and henceThe Cauchy-Schwarz inequality again guarantees integrability. For odd , the parity decomposition of a lag-two ARCH process also gives independence directly. For even , the sign symmetry of an ARCH process supplies the argument; the martingale difference sequence property alone would not justify this reversed covariance.
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