2. STATIONARITY AND ERGODICITY OF A FAMILY OF GARCH PROCESSES

In this paper, we always write z = z₀.

THEOREM 2.1. Assume that E ln⁺(g(z)) < ∞ and E ln⁺(c(z)) < ∞. Then the family of GARCH processes defined by (1.1) and (1.2) has a unique strictly stationary and ergodic solution if and only if E ln(c(z)) < 0.

Proof. We assume that E ln(c(z)) < 0. Because

is an i.i.d. random variable sequence with zero mean and unit variance, applying the strong law of large numbers, we can obtain that

as n → ∞, where →_a.s. denotes the convergence with probability 1. Hence,

as n → ∞. Moreover, note that {g(z_t−1−n),n ≥ 1} is i.i.d. and E ln⁺(g(z)) < ∞; hence

It follows that

for some b ∈ (0,|E ln(c(z))|) and sufficiently large n, with probability 1. Therefore,

Thus,

is well defined. Moreover,

i.e., h_t^δ defined as in (2.5) is a solution of (1.2). Let

Then ε_t = z_t h_t is a strictly stationary and ergodic solution of the family of GARCH processes defined by (1.1) and (1.2).

For uniqueness, assume that e_t is an arbitrary stationary solution to the family of GARCH processes defined by (1.1) and (1.2) and that h_t(e) satisfying z_t h_t(e) = e_t is a solution of equation (1.2). Iterating equation (1.2) yields

Note that the distribution of (h_−m^δ(e) − h_−m^δ) is independent of m; combining this with (2.2) and (2.7) implies

as m → ∞. Because of the stationarity of h_t^δ and h_t^δ(e), we conclude that h_t^δ = h_t^δ(e), a.s. This shows uniqueness. The “if” part of the theorem is completed.

Conversely, suppose that there exists a strictly stationary solution

of the family of GARCH processes defined as in (1.1) and (1.2). Iterating equation (1.2), we have, for any m > 0, that

Therefore, for any m > 0,

This shows that the series

converges a.s. Thus,

as n → ∞. Because {g(z_−n−1),n ≥ 1} is i.i.d., then (2.8) implies

as n → ∞. Moreover, noting that E ln⁺(c(z)) < ∞ and applying Lemma 2.1 in Bougerol and Picard (1992), we know that E ln(c(z)) < 0. This completes the proof of the theorem.

COROLLARY 2.1 (Ling and McAleer, 2002, Thm. 2.1). If E|z|^αδ < ∞, E [g(z)]^α < ∞, and E [c(z)]^α < 1 for some α ∈ (0,1], then there exists a unique αδ-order stationary solution to (1.1) and (1.2).

Proof. Note that E [c(z)]^α < 1 yields E ln(c(z)) < 0. Again according to E [g(z)]^α < ∞ and E [c(z)]^α < 1, we have E ln⁺(g(z)) < ∞ and E ln⁺(c(z)) < ∞. Thus, by Theorem 2.1, there exists a unique stationary solution to (1.1) and (1.2); i.e., (2.6) holds. Moreover, using (2.6),

where

. This concludes the proof.

In what follows we are concerned with the conditions for the existence of the moments of the family of GARCH processes defined by (1.1) and (1.2).

LEMMA 2.1. Assume that E ln(c(z)) < 0, P(c(z) > 1) > 0, and there exists u₀ ≤ ∞ such that E [c(z)]^u < ∞ for all u < u₀ and E [c(z)]^u₀ = ∞. Define

Then the function f is strictly convex in u, and there is a unique solution κ > 0 to the equation f (u) = 1.

Proof. Notice that the function f (u) = E [c(z)]^u is continuous and convex in u ≥ 0 and f (0) = 1. Because E ln(c(z)) < 0 it assumes values smaller than 1 in some neighborhood of the origin. Moreover, for sufficiently large u, because P(c(z) > 1) > 0 and f (u₀) = ∞, f (u) ≥ 1. Hence f (u) = 1 has a unique positive solution.

THEOREM 2.2. Let

be the GARCH process defined as in (1.1) and (1.2). Assume that the conditions in Lemma 2.1 are satisfied. Moreover, let κ be the unique positive solution of the equation f (u) = 1, where f is defined as in (2.9). Then, for 0 ≤ u < κ, if E|z|^δu < ∞ and E [g(z)]^u < ∞, we have that

Denote by p the largest integer strictly less than κ; then, for m = 1,2,…, p, if E|z|^δm < ∞ and E [g(z)]^m < ∞, we can obtain that

Proof. Write ∥ζ∥_q = (E|ζ|^q)^1/q, where ζ is any random variable and q ≥ 1. By Lemma 2.1, the function f (u) is strictly convex and satisfies f (κ) = f (0) = 1. Hence f (u) < 1 for 0 < u < κ. Moreover, note that the conditions in Theorem 2.1 are satisfied; thus (2.6) holds. For 0 < u < κ, if u ≥ 1, by (2.6),

if u < 1,

implies (2.10).

Finally, note that h_t^δ =_d g(z_t) + c(z_t)h_t^δ, where =_d denotes equality in distribution. By the binomial expression, it is easy to obtain

Thus, (2.12) and E|ε_t|^δl = E|z_t|^δlEh_t^δl, l = 1,…, m, yield (2.11).

3. TAIL BEHAVIORS OF A FAMILY OF GARCH PROCESSES

First recall that a random variable X (resp., its law) is said to be arithmetic (lattice) if there exist a,d ≥ 0 such that

.

In the following result we describe the tail of the marginal distribution of a family of processes. It gives a precise meaning to the statement “light-tailed input causes heavy-tailed output.”

THEOREM 3.1. Let

be the GARCH process defined by (1.1) and (1.2). Assume that the law of ln(c(z)) is nonarithmetic and the conditions in Lemma 2.1 are satisfied and let κ be the unique positive solution of the equation f (u) = 1, where f is defined by (2.9). Moreover, suppose that E [g(z)]^κ+σ < ∞ and E|z|^δκ+σ < ∞ for some σ > 0; then

where

Before we can prove this result we have to ensure that the constant c₀ in (3.2) is well defined. Indeed, its numerator is the expected difference of two quantities each of which has infinite expectation.

LEMMA 3.1. Under the assumptions of Theorem 3.1,

Proof. Because f (x) = (g(z) + x)^κ − x^κ is a decreasing function for κ ≤ 1 and x ≥ 0, then

Thus (3.3) holds for κ ≤ 1. For κ > 1, the function f (x) = x^κ is strictly convex on

, and hence

Taking p = κ + σ > 1 and q = (κ − 1 + σ)⁻¹(κ + σ) and applying Hölder's inequality, we know that

Note that (κ − 1 + σ)⁻¹(κ − 1)(κ + σ) < κ; by the preceding inequality and Theorem 2.2, (3.3) is true.

We also need the following two lemmas.

LEMMA 3.2. Let X ≥ Y a.s. be nonnegative random variables. Then

for the right-hand side finite or infinite.

Proof.

LEMMA 3.3. Assume that X and Y are independent nonnegative random variables such that P(X > x) = L(x)x^−κ for some slowly varying function L and EY^κ+σ < ∞ for some σ > 0. Then

Lemma 3.3 is a consequence of Proposition 3 in Breiman (1965) and can be also found in Mikosch and Starica (2000, p. 1433).

Proof of Theorem 3.1. Define the random walk

generated by the i.i.d. sequence

:

Now consider

Define the measures ν_n by

and write

Because h_t^δ ≥ c(z_t−1)h_t−1^δ, hence m is nonnegative. Moreover, we have

where * denotes the convolution. Also, define the measure ξ by

Furthermore, by the definition of κ, we have

Therefore, ξ defines a probability measure on

(not lattice) with mean

Let ν be the renewal measure corresponding to ξ, i.e.,

Note that

for

, where ν_n is defined in (3.4).

By Lemmas 3.1 and 3.2, it is easy to prove that m is integrable and bounded and lim_x→∞ m(x) = 0.

Because of μ(ξ) > 0, the renewal measure ν has the property

But,

Moreover, note that E ln(c(z)) < 0. Therefore S_n has a negative drift and

Combined with (3.5) and (3.6), this implies that

Note that m is integrable and bounded, lim_x→∞ m(x) = 0, and ξ is absolutely continuous; these conditions allow us to apply a version of the key renewal theorem (Alsmeyer, 1991, Satz 3.1.7), which in turn, together with Lemma 3.2, leads to the relation

with c₀ as in (3.2). Therefore, we get that

i.e.,

Moreover, note that ε_t = z_t h_t. Applying Lemma 3.3, we can obtain that

This completes the proof.

COROLLARY 3.1. Under the assumptions of Theorem 3.1,

Proof. Notice that P(|ε_t| > x) ∼ c₀ E|z|^δκx^−δκ as x → ∞; the conclusion is obvious.