**Appendix A**

**Proof of Proposition 1.** Let *Xt* following (3) and (5). Assume that the innovations distribution has a non-zero absolutely continuous component, with a positive density on a Borel set with a non-empty interior. Examples of this are the normal and Student's *t* distribution.

Then *Xt* can be written as *Xt* = *<sup>H</sup>*(*Xt*−1,*εt*), where *H* is a non-linear continuous function for each *εt* fixed. Then, using the *Continuous Mapping Theorem* we obtain the weak convergence of *Xt*, namely, the conditional distribution *Xt* given *Xt*−<sup>1</sup> = *yk* converges to the conditional distribution of *Xt* given *Xt*−<sup>1</sup> = *y* if *yk* → *y*. So, the Markov Chain *Xt* that represents the PHARCH(*m,p*) process is also a *T*-chain.

**Proof of Proposition 2.** Define the function *V* as:

$$V\left(X\_t\right) := \sum\_{i=1}^{a\_m-1} \alpha\_i r\_{t-i}^2 + 2 \sum\_{i=1}^{a\_m-2} \sum\_{j=i+1}^{a\_m-1} \beta\_j r\_{t-i} r\_{t-j} + \sum\_{i=0}^{p-1} \gamma\_i r\_{t-i}^2.$$

A simple algebraic computation gives

$$\mathbb{E}\left[\sum\_{i=1}^{a\_m-1} \alpha\_i r\_{t+1-i}^2 \mid \mathbf{X}\_t\right] = \alpha\_1 \sigma\_t^2 + \sum\_{i=1}^{a\_m-2} \alpha\_{i+1} r\_{t-i}^2$$

$$\mathbb{E}\left[\mathbf{2} \sum\_{i=1}^{a\_m-2} \sum\_{j=i+1}^{a\_m-1} \beta\_j r\_{t+1-i} r\_{t+1-j} \mid \mathbf{X}\_t\right] = \mathbf{2} \sum\_{i=1}^{a\_m-3} \sum\_{j=i+1}^{a\_m-2} \beta\_{j+1} r\_{t-i} r\_{t-j} \mu\_j$$

and using (4) we have,

$$\begin{split} \mathbb{E}\left[\sum\_{l=0}^{p-1} \gamma\_{l} \sigma\_{t+1-l}^{2} | X\_{l} \right] &= \gamma\_{0} \left[ \mathbb{C}\_{0} + \sum\_{l=1}^{m} \mathbb{C}\_{l} \left[ \sigma\_{l}^{2} + \left( \sum\_{j=1}^{q\_{l}-1} r\_{t-j} \right)^{2} \right] + \sum\_{l=1}^{p} b\_{l} \sigma\_{t+1-l}^{2} \right] + \mathbb{E}\left[ \sum\_{l=1}^{p} \gamma\_{l} | \sigma\_{t+1-l}^{2}| \right] \\ &+ \sum\_{l=1}^{p-1} \gamma\_{l} \sigma\_{t+1-l}^{2} . \end{split}$$

Therefore, taking *<sup>α</sup>am* = *βam* = *γp* = 0 and *a*0 = 1 and grouping we have

$$\begin{split} V\left(\mathbf{X}\_{l}\right) - \mathbb{E}\left[V\left(\mathbf{X}\_{l+1}\right)|X\_{l}\right] &= \sum\_{k=1}^{m} \sum\_{j=a\_{l-1}}^{a\_{l}-1} \left(\alpha\_{j} - \alpha\_{j+1} - \gamma\_{0} \sum\_{l=k}^{m} \mathbb{C}\_{l}\right) r\_{l-j}^{2} + \\ &+ \sum\_{l=1}^{m} \sum\_{j=1}^{a\_{l}-2} \sum\_{k=a\_{l-1}+j}^{a\_{l}-1} \left(\beta\_{k} - \beta\_{k+1} - \gamma\_{0} \sum\_{l=l}^{m} \mathbb{C}\_{l}\right) r\_{l-j} r\_{l-k} + \\ &+ \sum\_{l=1}^{p-1} \left(\gamma\_{l} - \gamma\_{l+1} - \gamma\_{0} b\_{l+1}\right) \sigma\_{l-l}^{2} + \\ &+ \left(\gamma\_{0} - \gamma\_{1} - \gamma\_{0} b\_{1} - \alpha\_{1} - \gamma\_{0} \sum\_{l=1}^{m} \mathbb{C}\_{l}\right) \sigma\_{l}^{2} - \gamma\_{0} \mathbb{C}\_{0}. \end{split}$$

We choose *k* ∈ Z+, *al*−1 < *k* < *al*, *l* ∈ {1, . . . , *<sup>m</sup>*}, and *βk* = *β<sup>k</sup>*+<sup>1</sup> + ∑*mi*=*<sup>l</sup> Ci*. If we take *αj* > *βj*, for all *j*, then we have *V* (*Xt*) ≥ 0, and we can take *γ*0 = 1, so

$$\begin{split} \left[ V(\mathbf{X}\_{t}) - \mathbb{E} \left[ V \left( \mathbf{X}\_{t+1} \right) \mid \mathbf{X}\_{t} \right] \right] &= \sum\_{k=1}^{m} \sum\_{j=a\_{k-1}}^{a\_{k}-1} \left( a\_{j} - a\_{j+1} - \sum\_{i=k}^{m} \mathbb{C}\_{i} \right) r\_{t-j}^{2} \\ &\quad + \sum\_{i=1}^{p-1} \left( \gamma\_{i} - \gamma\_{i+1} - b\_{i+1} \right) \sigma\_{t-i}^{2} \\ &\quad + \left( 1 - \gamma\_{1} - b\_{1} - a\_{1} - \sum\_{i=1}^{m} \mathbb{C}\_{i} \right) \sigma\_{t}^{2} - \mathbb{C}\_{0} . \end{split} \tag{A1}$$

Similarly, we can choose *k* ∈ Z+, *al*−1 < *k* < *al*, *l* ∈ {1, . . . , *<sup>m</sup>*}, *αk* > *<sup>α</sup>k*+<sup>1</sup> + ∑*mi*=*<sup>l</sup> Ci* and *γi* > *γi*+1 + *bi*+1.

If ∑*mi*=<sup>1</sup> *aiCi* + ∑*pi*=<sup>1</sup> *bi* < 1, then we have for the expressions in Equation (A1): *ξi* = *αi* − *<sup>α</sup>i*+1 − <sup>∑</sup>*mj*=*<sup>k</sup> Cj* > 0, *i* = 1, . . . , *am* − 1 and *k* is chosen such that *i* ∈ (*ak*−1, *ak*); *ξam*−1+*<sup>i</sup>* = (*<sup>γ</sup>i* − *γi*+1 − *bi*+<sup>1</sup>) > 0, *i* = 1, . . . , *p* − 1; and

$$
\zeta\_{a\_m + p - 1} = \left(1 - \gamma\_1 - b\_1 - a\_1 - \sum\_{i=1}^m \mathbb{C}\_i\right) > 0.
$$

So, *V*(*Xt*) − E [*V* (*Xt*+<sup>1</sup>)|*Xt*] can be as large as we want if *Xt* ∈ *Cc*.

Then, using Lemma 1 we have that there exists an invariant measure, finite on compact sets of Ω. Choosing min = min(*ξi*), for *i* = 1, . . . , *am* + *p* − 1, we have,

$$\begin{aligned} \mathbb{E}\left[\left(\mathbf{X}\_{t}\right) - \mathbb{E}\left[V\left(\mathbf{X}\_{t+1}\right)\mid \mathbf{X}\_{t}\right] &\geq \mathcal{O}\_{\text{min}} \sum\_{i=1}^{a\_{m}-1} r\_{t-j}^{2} + \mathcal{O}\_{\text{min}} \sum\_{i=1}^{p} \sigma\_{t-i}^{2} - \mathcal{C}\_{0} \\ &\geq 1 - \left(\mathcal{C}\_{0} + 1\right) I\_{B} \left(\mathbf{0}, \sqrt{\frac{c\_{0}+1}{\sigma\_{\text{min}}^{2}}}\right)' \end{aligned}$$

where *B* (*<sup>c</sup>*,*<sup>r</sup>*) is the ball with center *c* and radius *r*.

Therefore, the Markov chain *Xt* = *rt*−1,...,*rt*−*am*+1, *σt*,..., *<sup>σ</sup>t*−*p*+<sup>1</sup> that represents the PHARCH(*m,p*)processisrecurrent,withaninvariantprobabilitymeasure(stationarydistribution).

 Now, if we consider *f <sup>x</sup>*1,..., *xam*<sup>+</sup>*p*−<sup>1</sup> = *x*21 + ... + *<sup>x</sup>*2*am*+*p*−1, then,

$$\begin{split} V\left(X\_{t}\right) - \mathbb{E}\left[V\left(X\_{t+1}\right)|X\_{t}\right] &\geq \frac{\mathcal{O}\_{\text{min}}}{2} f\left(r\_{t-1}, \dots, r\_{t-\mathfrak{a}\_{m}+1}, \sigma\_{t}, \dots, \sigma\_{t-p+1}\right) + \mathbb{E}\left[V\left(X\_{t}\right)|X\_{t}\right] \\ &+ \mathbb{C}\_{0} \mathbf{1}\_{B\left(0, \sqrt{\frac{2\zeta\_{0}}{\mathfrak{a}\_{\text{min}}}}\right)}. \end{split}$$

We conclude, using Lemmas 2 and 3, that the process *Xt* is a *T* chain having a stationary distribution with finite second order moments.
