*Mathematical Proofs*

**Proof of Proposition 2.** Since *μ* = <sup>E</sup>[*<sup>ε</sup>i*] and the process *ηi* = (ln *Xi* − *μ*) − ln *σi* is a martingale difference sequence, <sup>E</sup>[*ηi*|F*<sup>i</sup>*−<sup>1</sup>] = 0 almost surely (a.s.). Replacing (16) in the equation ln *Xi* = ln *σi* + *ηi* + *μ*, note that the GBS-ACD(*<sup>r</sup>*,*<sup>s</sup>*) model can be written as

$$\ln X\_i = a + \mu \left( 1 + \sum\_{j=1}^r \beta\_j \right) - \sum\_{j=1}^r \beta\_j \eta\_{i-j} + \sum\_{j=1}^s \gamma\_j \left[ \frac{X\_{i-j}}{\sigma\_{i-j}} \right] + \sum\_{j=1}^r \beta\_j \ln X\_{i-j} + \eta\_i. \tag{A1}$$

Note that

$$\mathbb{E}\left[\frac{\chi\_{i-j}}{\sigma\_{i-j}}\right] = \mathbb{E}[\varrho\_{i-j}] = \frac{1}{2}(2 + \mu\_1 \kappa^2),\tag{A2}$$

where, in the first equality, we use the relation *Xi* = *<sup>σ</sup>iϕi*; in the second equality, the identity <sup>E</sup>[*ϕi*]=(<sup>2</sup><sup>+</sup> *<sup>u</sup>*1*κ*<sup>2</sup>)/2 is used, where *ur* = *ur*(*g*) = E[*Ur*] with *U* ∼ <sup>G</sup>*χ*<sup>2</sup>(1, *g*), because *ϕi* = exp(*<sup>ε</sup>i*) ∼ GBS(*<sup>κ</sup>*, 1, *g*).

Provided that {*Xi*} is a strictly stationary process, the transformed process {ln *Xi*} is always strictly stationary, too. Using this fact, taking expectation on both sides in (A1), and using the identity (A2), we obtain, after some algebra, that

$$\operatorname{E}[\ln X\_i] = \frac{2[\alpha + \mu(1 + \sum\_{j=1}^r \beta\_j)] + (2 + \mu\_1 \kappa^2) \sum\_{j=1}^o \gamma\_j}{2(1 - \sum\_{j=1}^r \beta\_j)}$$

whenever ∑*rj*=<sup>1</sup> *βj* = 1. The proof is complete.

**Proof of Proposition 3.** From Proposition 2 follows the expression for E[ln *Xi*]. In what follows, we find the expression for E[(ln *Xi*)<sup>2</sup>]. Indeed, since *Xi* = *<sup>σ</sup>iϕi*, *σi* is F*i*−1-measurable and *ϕi* ∼ GBS(*<sup>κ</sup>*, 1, *g*), it follows that

$$\begin{aligned} \operatorname{E}[\ln X\_{i}] &= \mu + \operatorname{E}[\ln \sigma\_{i}], \\ \operatorname{E}[(\ln X\_{i})^{2}] &= \mu(2 + \mu) + \operatorname{E}[(\ln \sigma\_{i})^{2}] + 2\mu \operatorname{E}[\ln \sigma\_{i}], \\ \operatorname{E}\left[\ln \sigma\_{i}\left(\frac{\mathbf{X}\_{i}}{\sigma\_{i}}\right)\right] &= \operatorname{E}\left[\ln \sigma\_{i}\operatorname{E}[\varphi\_{i}|\mathcal{F}\_{i-1}]\right] = \frac{1}{2}(2 + u\_{1}\kappa^{2})\operatorname{E}[\ln \sigma\_{i}], \\ \operatorname{E}\left[\frac{\mathbf{X}\_{i-1}}{\sigma\_{i-1}}\right] &= \frac{1}{2}(2 + u\_{1}\kappa^{2}) \\ \operatorname{E}[(\frac{\mathbf{X}\_{i-1}}{\sigma\_{i-1}})^{2}] &= \operatorname{Var}[\varphi\_{i-1}] + \operatorname{E}^{2}[\varphi\_{i-1}] \stackrel{(\mathcal{D})}{=} \frac{1}{2}(u\_{2}k^{4} + 4u\_{1}k^{2} + 2). \end{aligned} \tag{A3}$$

Taking the square of ln *σi* in (23) and after the expectation, by a strictly stationary process, we have

$$\begin{split} \mathbb{E}[(\ln \sigma\_{i})^{2}] &= \alpha^{2} + \beta^{2} \mathbb{E}[(\ln \sigma\_{i})^{2}] + 2\alpha\beta \mathbb{E}[\ln \sigma\_{i}] \\ &\quad + \gamma^{2} \mathbb{E}\left[(\frac{\underline{X}\_{i-1}}{\sigma\_{i-1}})^{2}\right] + 2\gamma\alpha \mathbb{E}\left[\frac{\underline{X}\_{i-1}}{\sigma\_{i-1}}\right] + 2\gamma\beta \mathbb{E}\left[\ln \sigma\_{i-1}(\frac{\underline{X}\_{i-1}}{\sigma\_{i-1}})\right]. \end{split} \tag{A4}$$

Combining the Equation (A3) with (A4),

$$(1 - \beta^2) \mathcal{E}[(\ln \sigma\_i)^2] = a^2 - 2a\beta + \frac{\gamma^2}{2} (u\_2 \kappa^4 + 4u\_1 \kappa^2 + 2) + \gamma a (2 + u\_1 \kappa^2) - \gamma \beta (2 + u\_1 \kappa^2) \{ \mathcal{E}[\ln X\_i] - \mu \}.$$

Using this identity and Proposition 2 in the second identity for E[(ln *Xi*)<sup>2</sup>] in (A3), the proof follows.
