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PARAMETER ESTIMATION OF BILINEAR PROCESSES USING ABC 139
the errors are Pareto(↵), ↵ > 0, with ↵ = 2.5 and ↵ = 1.5, the process can generate very large values (center and right panels). It is known that, for ↵ = 2.5, the Pareto distribution, with distribution function given by F (x) = 1 x ↵, x > 1 and ↵ > 0, has both finite mean and variance, but when ↵ = 1.5, although the mean is finite, the variance is not. The situation can be more extreme if ↵  1 because neither the mean nor the variance are finite.
As mentioned before, the conditions that guarantee invertibility and sta- tionarity of bilinear process can only be derived for some particular models (see Subba Rao and Gabr [36]), such as the simple bilinear process which has only one bilinear term. Invertibility is a fundamental requirement if the model is to be used for prediction. Let Yt be a simple first order bilinear process, BL(1,0,1,1), given by
Yt = aYt 1 + bYt 1✏t 1 + ✏t, (2.2)
where {✏t} are iid rvs. A su cient condition for the process to be stationarity is given by Pham and Tran [30]: the parameters a, b and  2 must satisfy the condition a2 + b2 2 < 1, where |a| < 1. A su cient condition for invertibility is that the parameters a and b are such that b2E(Yt2)  1, as proved by Pham and Tran [30]. They also show that, provided the {et} are Gaussian rvs, the condition
2(1+a)b4 4 +2(1 a)b2 2  (1 a)2(1+a)0 (2.3)
is su cient for the simple bilinear process to be invertible. Additional condi- tions on a, b and  2 are required if we want the bilinear process to have up to the fourth moment finite. In this case, besides | a |< 1 and a2 + b2 2 < 1, the following (necessary) conditions | a3 +3ab2 2 |< 1 and a4 +6a2b2 2 +3b4 4 < 1 have to be satisfied (Sesay and Subba Rao [34], Kim et al. [24], as referred by Leon-Gonzalez and Yang [25]).
All these conditions severely restrict the parameter space while searching for admissible estimators for the parameters.
Although bilinear processes do not satisfy the Markov property, it is possible to write them as state space models (SSM) satisfying a Markovian property, which may be very helpful in several aspects, namely for using SMC algo- rithms. If the observable process Yt is a BL(1,0,1,1) given in (2.2) then, solving iteratively for Yt, we get, after n iterations,
n n 1 " j #
Yt =Y(a+b✏t i)Yt n +X Y(a+b✏t i) ✏t j +✏t.
i=1 j=1 i=1 If Xt = (a + b✏t)Yt, then
Xt = (a + b✏t)Xt 1 + (a + b✏t)✏t


































































































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