Proof of Theorem 5 (Low interaction). Considering the process \(B_{t,j}=K_{t,j}/(\theta+t)\) and denoting \(B^{*}_{t,j}=\zeta_{t}(1-(1-\iota)w)B_{t,j}\), from the dynamics 13 we obtain \[\label{eq-dyn-B-star} B^{*}_{t+1,j}=\Big(1+O\big(\tfrac{1}{t^2}\big)\Big)B^{*}_{t,j} +\tfrac{\zeta_{t+1}(1-(1-\iota)w)}{\theta+t+1} (\Delta M_{t+1,j} + \iota \overline{P}_t)\tag{15}\] and so, by the asymptotics for \(\overline{P}_t\) (see Table 1) and the classical Theorem 22, the process \((B^{*}_{t,j})_t\) is a non-negative almost super-martingale that converges a.s.to a finite r.v. \(B^{*}_\infty\). However, this is not enough for our scopes. Indeed, in that way, we do not know that such random limit \(B^{*}_\infty\) is a.s.strictly positive, and this is crucial for proving the right rate of convergence. For this reason, we are going to apply the more sophisticated Theorem 13 (case i) together with Remark 6 with \(\tau=\tau_j\)) to dynamics 12 with \(X_t=B_{t,j}\), \(Y_{t+1}=\Delta K_{t+1,j}\in \{0,1\}\), \(\delta=(1-\iota ) w\in (0, 1]\). To this purpose, we note that \(\rho_t=0\) when \(\iota=0\); while \(\rho_t=\iota \overline{P}_t > 0\) when \(\iota>0\). To handle the case \(\iota>0\), from Table 1 we get \(\overline{P}_t=O(1/t^{1-w+\beta})\) when \(0<\beta<w\leq 1\), \(\overline{P}_t=O(\ln(t)/t)\) when \(0<\beta=w\leq 1\) and \(\overline{P}_t=O(1/t)\) when \(0<w<\beta\leq 1\) (the case \(\beta=0\) is not possible with \(\iota>0\) because \(\beta>\iota w\) in the case of low interaction). Therefore, setting \(\delta=(1-\iota)w\), we have \[\sum_t \frac{\rho_t}{(\theta+t)^{\delta-\epsilon}} <+\infty\;a.s.\;for some 0<\epsilon<\delta\,,\] provided that \(\iota=0\) or \(0<\iota <\beta/w\). Moreover, we trivially (since \(Y_{t+1}\in \{0,1\}\)) have \[E[Y_{t+1}^2\mid {\mathcal{F}}_t]\leq E[Y_{t+1}\mid {\mathcal{F}}_t]\,.\] Hence, under our conditions, we find \[t^{1-(1-\iota)w}\, B_{t,j}\stackrel{a.s.}\longrightarrow B^*_{\infty,j}\in (0,+\infty)\] and so the first statement of the theorem follows with \(K^*_{\infty,j}=B^*_{\infty,j}\). Finally, it follows, by 6 , that \[t^{1-w(1-\iota)}\,P_{t,j}\stackrel{a.s.}\longrightarrow (1-\iota)w\,B^*_{\infty,j}=(1-\iota)w\,K^*_{\infty,j}\in (0,+\infty)\,,\] which is the second statement of the theorem. ◻

Proof of Theorem 6 (High interaction). We will prove the two cases separately.
From 6 we immediately get \(P_{t,j}=\overline{P}_t\) for each \(j\in{\mathcal{O}}_t\) and so the asymptotic behavior of \(P_{t,j}\) is described in Table 1, while, by Lemma 20, for \(K_{t,j}\) we have \(K_{t,j}/\sum_{n=1}^t\overline{P}_{n-1}\stackrel{a.s.}\longrightarrow 1\) because \(K_{t,j}=\sum_{n=1}^t X_{n,j}\) with \(E[X_{n+1,j}\mid{\mathcal{F}}_{n}]=P_{n,j}=\overline{P}_n\) and \(\sum_n \overline{P}_n=+\infty\) a.s..
Recall that we have \(B_{t,j}=K_{t,j}/(\theta+t)\) with dynamics 13 : \[B_{t+1,j}=B_{t,j}-\frac{(1-(1-\iota)w)B_{t,j}-\iota\overline{P}_t}{\theta+t+1} +\frac{\Delta M_{t+1,j}}{\theta+t+1}\,,\] where \[\Delta M_{t+1,j}=\Delta K_{t+1,j}-P_{t,j}=\Delta K_{t+1,j}-(1-\iota)wB_{t,j}-\iota\overline{P}_t\,.\] When \(\iota w>\beta>0\), the term \(\iota \overline{P}_t\) is not negligible as in Theorem 5 and so we have to study the dynamics of \(B_{t,j}\) paired with the one of \(\overline{P}_t\). From Table 1 we know that, since \(w>\beta\), we have \(t^{1-w+\beta}\overline{P}_t\stackrel{a.s.}\rightarrow {P}^*_{\infty}>0\). Now, let \(B_{t,j}^*=\zeta_t(1-w+\beta)B_{t,j}\), \(\overline{P}_t^*=\zeta_t(1-w+\beta)\overline{P}_t\) and \(\Delta M_{t+1,j}^*=\zeta_{t+1}(1-w+\beta)\Delta M_{t+1,j}\). Then, using 20 , we obtain \[\begin{align} B_{t+1,j}^*&=\Big(B_{t,j}^{*}-\tfrac{(1-(1-\iota)w)B_{t,j}^{*}-\iota\overline{P}_t^{*}}{\theta+t+1}\Big) \tfrac{\zeta_{t+1}(1-w+\beta)}{\zeta_t(1-w+\beta)} + \tfrac{\Delta M_{t+1,j}^*}{\theta+t+1}\\ &= \Big(B_{t,j}^{*}-\tfrac{(1-(1-\iota)w)B_{t,j}^{*}-\iota\overline{P}_t^{*}}{\theta+t+1}\Big) \Big(1+\tfrac{1-w+\beta}{\theta+t+1}\Big) + \tfrac{\Delta M_{t+1,j}^*}{\theta+t+1}\\ &=B_{t,j}^{*} -\tfrac{(\iota w-\beta)B_{t,j}^{*}-\iota\overline{P}_t^{*}}{\theta+t+1}+ \tfrac{\Delta M_{t+1,j}^*}{\theta+t+1}+\rho_t\,, \end{align}\] with \(\rho_t=O\big(\frac{B_{t,j}+\overline{P}_t}{t^{1+w-\beta}}\big)\) being \({\mathcal{F}}_t\)-measurable. This suggests that \((B^{*}_t)_t\) is not a non-negative almost super-martingale, as in Theorem 5, but instead it evolves according to a stochastic approximation dynamics, which should convergence to \(\iota {P}_\infty^{*}/(\iota w-\beta)\). For proving this convergence, we are going to apply Theorem 11 with \(X_t=B^*_t\), together with Lemma 12. To this end, we observe that, by 9 and since \(0\leq P_{t,j}\leq 1\), \[\begin{align} E[(\Delta M^*_{t+1,j})^2\mid{\mathcal{F}}_t]&= \zeta(1-w+\beta)^2P_{t,j}(1-P_{t,j}) \leq \zeta(1-w+\beta)^2P_{t,j}\\ &=\zeta(1-w+\beta)[(1-\iota) w B^*_{t,j} +\iota \overline{P}^*_t] =O(t^{1-w+\beta+\epsilon}) \quad\forall \epsilon>0\,, \end{align}\] where we have used \(B^*_t=o(t^\epsilon)\) that follows by Lemma 12 applied to \(X_t=B^*_t\) with \(A_t=\overline{P}^*_t\) and \(b=(\iota w-\beta)>0\). (Indeed, \(A_t=\overline{P}^*_t\) is a.s.convergent by Table 1 and, since both \(B_{t,j}\) and \(\overline{P}_t\) are bounded by one and \(w>\beta\), we a.s. have \(\sum_t X_t/t^2=\sum_t O(B_{t,j}/t^{1-w+\beta})<+\infty\) and \(\sum_t |\rho_{t}|=\sum_t O\left(\frac{B_{t,j}+\overline{P}_t}{t^{1+w-\beta}}\right) < +\infty\).) Hence, choosing a suitable \(\epsilon\), we get \[\sum_t \frac{1}{(\theta+t+1)^2}E[(\Delta M^*_{t+1,j})^2\mid{\mathcal{F}}_t]<+\infty \quada.s.\] and so, by Theorem 11, we have \[B_{t,j}^{*} = \zeta_t(1-w+\beta)B_{t,j}\stackrel{a.s.}\longrightarrow \frac{\iota}{(\iota w - \beta)}\, P_{\infty}^*\,.\] From Table 1 (case \(w>\beta>0\)) we get \(P_{\infty}^*=\frac{\beta}{\alpha}Z_\infty^*>0\) a.s.and so the first statement of case (2) of the theorem follows. Finally, by 6 , we get \[\begin{align} t^{1-w+\beta}\,P_{t,j}\stackrel{a.s.}\longrightarrow \tfrac{(1-\iota)w\,\iota}{(w\iota - \beta)}\, P_{\infty}^*+\iota P_{\infty}^* =\tfrac{\iota(w-\beta)}{(\iota w-\beta)}\, P_{\infty}^* = \tfrac{\iota(w-\beta)}{(\iota w-\beta)}\frac{\beta}{\alpha}Z_\infty^*\,, \end{align}\] which is the second statement of the case (2) of the theorem.
When \(\iota w>\beta=0\), analogously to case (2), the term \(\iota \overline{P}_t\) is not negligible in the dynamics 13 of \(B_{t,j}\) and so we have to study the dynamics of \(B_{t,j}\) paired with the one of \(\overline{P}_t\). From Table 1 we know that, since \(w>\beta=0\), we have \(t^{1-w}\ln(t)\overline{P}_t\stackrel{a.s.}\rightarrow P_{\infty}>0\), so the dynamics we need to consider is \[B_{t+1,j}=B_{t,j}-\frac{(1-(1-\iota)w)B_{t,j}-\iota\overline{P}_t}{\theta+t+1} +\frac{\Delta M_{t+1,j}}{\theta+t+1}\,,\] where \(\Delta M_{t+1,j}=\Delta K_{t+1,j}-P_{t,j}=\Delta K_{t+1,j}-(1-\iota)wB_{t,j}-\iota\overline{P}_t\). Now, let \(B_{t,j}^*=\zeta_t(1-w)\ln(\theta+t)B_{t,j}\), \(\overline{P}_t^*=\zeta_t(1-w)\ln(\theta+t)\overline{P}_t\) and \(\Delta M_{t+1,j}^*=\zeta_{t+1}(1-w)\ln(\theta+t+1)\Delta M_{t+1,j}\). Then, using 22 from Appendix 7, we obtain \[\begin{align} B_{t+1,j}^*&=\Big(B_{t,j}^{*}-\tfrac{(1-(1-\iota)w)B_{t,j}^{*}-\iota\overline{P}_t^{*}}{\theta+t+1}\Big) \tfrac{\zeta_{t+1}(1-w)}{\zeta_t(1-w)} \tfrac{\ln(\theta+t+1)}{\ln(\theta+t)} + \tfrac{\Delta M_{t+1,j}^*}{\theta+t+1}\\ &= \Big(B_{t,j}^{*}-\tfrac{(1-(1-\iota)w)B_{t,j}^{*}-\iota\overline{P}_t^{*}}{\theta+t+1}\Big) \Big(1+\tfrac{1-w}{\theta+t+1} -\tfrac{1}{(\theta+t+1)\ln(\theta+t+1)}+ O\big(\tfrac{1}{t^2\ln(t)}\big)\Big) +\tfrac{\Delta M_{t+1,j}^*}{\theta+t+1}\\ &=B_{t,j}^{*} -\iota \tfrac{wB_{t,j}^{*}-\overline{P}_t^{*}}{\theta+t+1}- \tfrac{B_{t,j}^{*}}{(\theta+t+1)\ln(\theta+t+1)} + \tfrac{\Delta M_{t+1,j}^*}{\theta+t+1} + \rho_t\\ &=B_{t,j}^{*} + \frac{1}{\theta+t+1}\Big(\iota\overline{P}_t^{*}-\Big(\iota w+\tfrac{1}{\ln(\theta+t+1)}\Big)B_{t,j}^{*}\Big)+ \tfrac{\Delta M_{t+1,j}^*}{\theta+t+1} + \rho_t\,, \end{align}\] with \(\rho_t=O\big(\frac{B_{t,j}+\overline{P}_t}{t^{1+w}}\big)\) being \({\mathcal{F}}_t\)-measurable. We want to apply Theorem 11 with \(X_t=B^{*}_t\), together with Lemma 12. To this end, analogously to the proof of case (2), we have that \[\begin{align} E[(\Delta M^*_{t+1,j})^2\mid{\mathcal{F}}_t]&= \zeta(1-w)^2\ln^2(\theta+t+1)P_{t,j}(1-P_{t,j})\\ &\leq \zeta(1-w)^2\ln^2(\theta+t+1)P_{t,j}\\ &=\zeta(1-w)\ln(\theta+t+1)[(1-\iota) w B^*_{t,j} +\iota \overline{P}^*_t]\\ &=O(t^{1-w+\epsilon}\ln(t)) \quad\forall \epsilon>0\,, \end{align}\] where we have used \(B^*_t=o(t^\epsilon)\) that follows by Lemma 12 applied to \(X_t=B^*_t\), with \(A_t=\iota\overline{P}^*_t\) and \(b_t=\iota w + 1/\ln(\theta+t+1)\) (note that we a.s. have \(\sum_t X_t/t^2=\sum_t O(B_{t,j}\ln(t)/t^{1+w})<+\infty\) and \(\sum_t|\rho_{t}|=\sum_t O\big(\frac{B_{t,j}+\overline{P}_t}{t^{1+w}}\big) < +\infty\) as \(w>0\)). Hence, choosing a suitable \(\epsilon\), we get \[\sum_t \frac{1}{(\theta+t+1)^2}E[(\Delta M^*_{t+1,j})^2\mid{\mathcal{F}}_t]<+\infty \quada.s.\,.\] Then, by Theorem 11, we have \[B_{t,j}^{*}= \zeta_t(1-w)\ln(\theta +t)B_{t,j}\stackrel{a.s.}\longrightarrow \frac{1}{w}\, P_{\infty}^*.\] From Table 1 (case \(w>\beta=0\)) we get \(P_{\infty}^*=\frac{1}{\alpha}Z_\infty^*>0\) a.s.and so the first statement of the case (3) of the theorem follows. Finally, by 6 , we get \[t^{1-w}\ln(t)\,P_{t,j}\stackrel{a.s.}\longrightarrow (1-\iota)w \frac{1}{w}\, P_{\infty}^*+\iota P_{\infty}^* =P_{\infty}^*=\frac{1}{\alpha}Z_\infty^*\,,\] which is the second statement of the case (3) of the theorem. ◻

Proof of Theorem 7 (Critical case). The structure of this proof is analogous to the proof of Theorem 6 as, also in the case \(\iota w=\beta>0\), we have to study the dynamics 13 of \(B_{t,j}=K_{t,j}/(\theta+t)\) paired with the one of \(\overline{P}_t\). From Table 1 we know that, since \(w>\beta\) (because \(0<\iota<1\) and \(\iota w=\beta\)), we have \(t^{1-w+\beta}\overline{P}_t\stackrel{a.s.}\rightarrow P^*_{\infty}>0\). Now, let \(B_{t,j}^{*}=\frac{\zeta_t(1-w+\beta)}{\ln(\theta+t)}B_{t,j}\), \(\overline{P}_t^*=\zeta_t(1-w+\beta)\overline{P}_t\) and \(\Delta M_{t+1,j}^{*}=\zeta_{t+1}(1-w+\beta)\Delta M_{t+1,j}\). Then, using 20 and 22 from Appendix 7, we obtain \[\begin{align} B_{t+1,j}^{*} &=\left(1-\tfrac{(1-(1-\iota)w)}{\theta+t+1}\right)B_{t,j}^{*} \tfrac{\zeta_{t+1}(1-w+\beta)}{\ln(\theta+t+1)} \tfrac{\ln(\theta+t)}{\zeta_t(1-w+\beta)} \\ & \quad + \tfrac{\iota\overline{P}_t^{*}}{(\theta+t+1)\ln(\theta+t+1)} \tfrac{\zeta_{t+1}(1-w+\beta)}{\zeta_t(1-w+\beta)} + \tfrac{\Delta M_{t+1,j}^{*}}{(\theta+t+1)\ln(\theta+t+1)}\\ &=\left(1-\tfrac{(1-(1-\iota)w)}{\theta+t+1}\right)B_{t,j}^{*} \Big(1+\tfrac{1-w+\beta}{\theta+t+1}- \tfrac{1}{(\theta+t+1)\ln(\theta+t+1)} + O\big(\tfrac{1}{t^2\ln(t)}\big)\Big)\\ &+ \tfrac{\iota\overline{P}_t^{*}}{(\theta+t+1)\ln(\theta+t+1)} \left(1+\tfrac{1-w+\beta}{\theta+t+1}\right) + \tfrac{\Delta M_{t+1,j}^{*}}{(\theta+t+1)\ln(\theta+t+1)}\\ &=B_{t,j}^{*} -\tfrac{B_{t,j}^{*}-\iota\overline{P}_t^{*}}{(\theta+t+1)\ln(\theta+t+1)}+ \tfrac{\Delta M_{t+1,j}^{*}}{(\theta+t+1)\ln(\theta+t+1)} + \rho_t\,, \end{align}\] with \(\rho_t=O\big(\frac{B_{t,j}+\overline{P}_t}{t^{1+w-\beta}\ln(t)}\big)=O\big(\frac{B_{t,j}+\overline{P}_t}{t^{1+w-\beta}}\big)\) being \({\mathcal{F}}_t\)-measurable. We want to apply Theorem 11 with \(X_t=B^{*}_t\), together with Lemma 12. To this end, analogously to the proof of Theorem 6, we have that \[E[(\Delta M^*_{t+1,j})^2\mid{\mathcal{F}}_t] \leq \zeta(1-w+\beta)[(1-\iota) w B^*_{t,j} +\iota \overline{P}^*_t] =O(t^{1-w+\beta+\epsilon}),\] for any \(\epsilon>0\), where we have used \(B^*_t=o(t^\epsilon)\) that follows by Lemma 12 applied to \(X_t=B^*_{t,j}\), with \(A_t=\overline{P}^*_t\) and \(b=0\) (note that we a.s. have \(\sum_t X_t/t^2=\sum_tO(B_{t,j}/t^{1+w-\beta})<+\infty\) and \(\sum_t|\rho_{t}|=\sum_t O\left(\frac{B_{t,j}+\overline{P}_t}{t^{1+w-\beta}}\right) < +\infty\) as \(w>\beta\)). Hence, choosing a suitable \(\epsilon\), we get \[\sum_t \frac{1}{(\theta+t+1)^2\ln^2(\theta+t+1)}E[(\Delta M^*_{t+1,j})^2\mid{\mathcal{F}}_t]<+\infty \quada.s.\,.\] Then, by Theorem 11, we have \[B_{t,j}^{*} = \frac{\zeta_t(1-w+\beta)}{\ln(\theta+t)}B_{t,j}\stackrel{a.s.}\longrightarrow \iota P_{\infty}^*\,.\] From Table 1 (case \(w>\beta>0\)) we get \(P_{\infty}^*=\frac{\beta}{\alpha}Z_\infty^*>0\) a.s.and so the first statement of the theorem follows. Finally, by 6 , we get \[\frac{t^{1-w+\beta}}{\ln(\theta+t)}\,P_{t,j}\stackrel{a.s.}\longrightarrow (1-\iota)w \iota P_{\infty}^* = (1-\iota)w \iota\frac{\beta}{\alpha}Z_\infty^*\,,\] which is the second statement of the theorem. ◻

Proof of Theorem 8. We will first prove the CLT for the process \(B_{t,j}=K_{t,j}/(\theta+t)\) and then we will derive from it the one for \(K_{t,j}\). Indeed, since the dynamics of \(B_{t,j}\) in 13 , i.e. \[B_{t+1,j}= \left(1-\frac{(1-(1-\iota)w)}{\theta+t+1}\right)B_{t,j} +\frac{(\Delta K_{t+1,j}-P_{t,j})}{\theta+t+1} +\frac{\iota\overline{P}_t}{\theta+t+1}\] is equal to the general dynamics 28 in Section 8.3 of the Appendix, with \(\delta=w(1-\iota)\in (0,1]\), \(\Delta M_{t+1}=(\Delta K_{t+1,j}-P_{t,j})\), \(\rho_{t+1}=\iota\overline{P}_t\), \[\rho=\begin{cases} w-\beta, & \text{if w>\beta AND 0<\iota<1}\,;\\ w(1-\iota)\varepsilon_0 & \text{if w\leq \beta OR \iota=0}\,; \end{cases}\] for an arbitrary \(\varepsilon_0\in (0,1)\), and \[\rho_\infty=\begin{cases} \iota(\beta/\alpha)Z^*_\infty, & \text{if w>\beta AND 0<\iota<1}\,;\\ 0 & \text{if w\leq \beta OR \iota=0}\,; \end{cases}\] (because, by Theorem 1 and Theorem 2, \(t^{1-\rho}\,\iota\overline{P}_t\) converges a.s.to \(\rho_\infty\)), and since we are assuming \(w\iota<\beta\), in all cases we have \(0<\rho<\delta\) and so the CLT for \(B_{t,j}\) can be obtained by applying Theorem 14 with \(X_t=B_{t,j}\) and \(X^*_\infty=K^*_{\infty,j}\). Indeed, by Theorem 5, we have \(t^{1-(1-\iota)w}B_{t,j}\stackrel{a.s.}\sim t^{-(1-\iota)w} K_{t,j}\stackrel{a.s.}\to K^*_{\infty,j}\) and, in addition, condition ?? of Theorem 14 follows by ?? in Lemma 18, because for any \(p\geq1\) and \(\epsilon>0\), we have \[E\Big[ \sum_{t=1}^{\infty} \frac{|\Delta {M}_{t+1}|^p}{t^{(1-\iota)w+\epsilon}} \Big] = \sum_{t=1}^{\infty} \frac{1}{t^{1+\epsilon}} E[ t^{1-(1-\iota)w}|\Delta {K}_{t+1}|^p] <+\infty\,.\] Moreover, condition ?? of Theorem 14 follows by Theorem 5, because \[t^{1-(1-\iota)w} E[(\Delta M_{t+1})^2|\mathcal{F}_t]= t^{1-(1-\iota)w}P_{t,j}(1-P_{t,j})\stackrel{a.s.}\longrightarrow V_\infty^{*} = w(1-\iota)K^*_{\infty,j}\,.\] Then, from Theorem 14, choosing \(\varepsilon_0<1/2\) (so that we always have \(\rho<\delta/2\) when \(w\leq \beta\) OR \(\iota=0\)), we obtain that:

• For \(\rho\leq \delta/2=(1-\iota)w/2\), i.e. for \(w\leq \beta\) OR \(\iota=0\) OR \(w>\beta\) and \(0<\iota<2\beta/w-1\) (that implies \(\iota<\beta/w\)) OR \(0<\iota=2\beta/w-1<1\) (that implies \(\iota<\beta/w<1\)), we have \[\begin{align} &t^{(1-\iota)w/2}\left(t^{1-(1-\iota)w}B_{t,j} + \tfrac{\rho_\infty}{(1-\iota)w-\rho}\,t^{-[(1-\iota)w-\rho]}-K^*_{\infty,j}\right) \stackrel{stably}\longrightarrow {\mathcal{N}}(0,\, K^*_{\infty,j})\,, \end{align}\] where \(\rho\) and \(\rho_\infty\) take different values according to the different cases and so we obtain ?? . Moreover, the convergence also holds true in the sense of the a.s. conditional convergence with respect to \(\mathcal{F}\). (Note that the above four cases can be grouped as in point \((i)\) of the statement: indeed, the case \(w\leq \beta\) is covered by the union of case \(\iota=0\) and case \(0<\iota<\min\{2\beta/w-1, 1\}\), that implies \(\iota<\min\{\beta/w,1\}\).)

• For \((1-\iota)w/2=\delta/2<\rho<\delta\), i.e. for \(\max\{0,(2\beta/w-1)\}<\iota<\beta/w<1\) (that obviously implies \(\beta<w\) and it is possible only if \(\beta>0\)), we have \[{t}^{\beta-w\iota}(t^{1-w(1-\iota)}B_{t,j}-K^*_{\infty,j})\stackrel{P}\longrightarrow -\tfrac{\iota\beta/\alpha}{\beta-w\iota}Z^*_\infty\,.\] Moreover, since \(\rho_{t+1}=\iota\overline{P}_t\), by the limit relation ?? in Corollary 4, we have \[\begin{align} &t^{\rho-\delta/2}(t^{1-\rho}\rho_{t+1}-\rho_\infty)=\\ &\iota \tfrac{t^{(w-\beta)-(1-\iota) w/2}}{t^{\beta/2}} t^{\beta/2}(t^{1-(w-\beta)}\overline{P}_t+\tfrac{\beta w}{(w-\beta)}t^{-(w-\beta)}-\tfrac{\beta}{\alpha}Z^*_\infty) \\ &-\tfrac{\beta\iota w}{(w-\beta)}t^{-(1-\iota)w/2} \stackrel{P}\longrightarrow 0\,, \end{align}\] provided \(\rho-\delta/2=(w-\beta)-(1-\iota)w/2< \beta/2\), and so, by Theorem 14, in this case we have \[t^{(1-\iota)w/2}\left(t^{1-(1-\iota)w}B_{t,j} + \tfrac{\iota(\beta/\alpha)}{(\beta-w\iota)}\,Z^*_\infty \,t^{-(\beta-w\iota)}-K^*_{\infty,j}\right)\stackrel{stably}\longrightarrow {\mathcal{N}}(0,\, K^*_{\infty,j})\,.\]

Finally, the thesis follows because \(B_{t,j}=\frac{K_{t,j}}{(\theta+t)}=\frac{K_{t,j}}{t}(1+O(\frac{1}{t}))\). ◻

Proof of Lemma 9. The proof is based on the continuous integral approximation of the series with proper integrals, namely \[\begin{gather} \label{eq:approx95series95variance950} \Big| b_n \sum_{k>n} \tfrac{1}{a_k\,b_k^2} - \tfrac{1}{\delta_0} \Big| \leq \Big| b_n \Big( \sum_{k>n} \tfrac{1}{a_k\,b_k^2} - \textstyle{\int}_n^{\infty} \overline{c}(t) dt \Big) \Big| + b_n \textstyle{\int}_n^{\infty} |\overline{c}(t) - c(t) | dt + \Big| b_n \textstyle{\int}_n^{\infty} c(t) dt - \tfrac{1}{\delta_0} \Big|\,, \end{gather}\tag{23}\] where \(\overline{c}(t) = r_t/(\theta+t+1)^2\) is the convex function that interpolates \(c_k = (b_k^2a_k)^{-1} = r_k/(\theta+k+1)^2\) and \(c(t) = (t^{1+\delta_0})^{-1}\) if \(\delta\neq\rho\); \(c(t) = (t^{1+\delta_0}\ln t)^{-1}\) if \(\delta=\rho=\delta_0\).

For the first term in 23 , the trapezoidal rule overestimates \(\int_{n}^{n+M} \overline{c}(t) dt < \tfrac{c_n}{2} + \sum_{n+1}^{N-1} c_k + \tfrac{c_M}{2}\) and hence \[0 < b_n \Big( \textstyle{\int}_n^{\infty} \overline{c}(t) dt - \sum_{k>n} \tfrac{1}{b_k^2a_k} \Big) \leq b_n\tfrac{c_n}{2} = \tfrac{\theta+n}{2(\theta+n+1)^2} = O(n^{-1})\,.\] For the second term in 23 , by 18 , \(|\overline{c}(t) -c(t)|\leq C t^{-2-w_0}\), and hence \[b_n\textstyle{\int}_n^{\infty} |\overline{c}(t) -c(t)| dt = O(n^{-1})\,.\]

The second term in 23 is immediate when \(\delta\neq \rho\), since \(\int_n^{\infty} c(t) dt = n^{-\delta_0}/\delta_0\). When \(\delta=\rho=\delta_0\), \[b_n \textstyle{\int}_n^{\infty} c(t)dt = n^{\delta_0}\ln(n) \textstyle{\int}_{\delta_0\ln n}^\infty \tfrac{e^{-y}}{y} dy \to \tfrac{1}{\delta_0}\,,\] since, for any \(x>0\) [36] \[\tfrac{1}{2} \ln\Big( 1+\tfrac{2}{x} \Big) < e^{x}\textstyle{\int}_x^\infty \tfrac{e^{-y}}{y}dy < \ln\Big( 1+\tfrac{1}{x} \Big)\,.\]

The proof of \(\tfrac{1}{b_n} \sum_{k\leq n} \tfrac{1}{r_k} \to \tfrac{1}{\delta_0}\) is trivial for \(\delta_0=1\), since \(r_k \equiv 1\) when \(\delta\neq\rho\) and \(\sum_{k\leq n}r_k^{-1}\sim n\ln(n)\) for \(\delta=\rho=1\). For \(\delta_0<1\), the proof is again made on continuous integral approximations, here \[\begin{gather} \label{eq:approx95series95variance951} \Big| \tfrac{1}{b_n} \sum_{k\leq n} \tfrac{1}{r_k} - \tfrac{1}{\delta_0} \Big| \leq \Big| \tfrac{1}{b_n} \Big( \sum_{k\leq n} \tfrac{1}{r_k} - \textstyle{\int}_0^n \tfrac{1}{r_t} dt \Big) \Big| + \tfrac{1}{b_n} \textstyle{\int}_0^n |\tfrac{1}{r_t} - c(t) | dt + \Big| \tfrac{1}{b_n} \textstyle{\int}_0^n c(t) dt - \tfrac{1}{\delta_0} \Big|\,, \end{gather}\tag{24}\] where \(c(t) = (t^{-1+w_0})\) if \(w\neq\beta\); \(c(t) = (t^{-1+\delta_0}\ln t)\) if \(\delta=\rho=\delta_0\). Since \(\tfrac{1}{r_t}\) is eventually decreasing by 20 and 22 , the first term in 24 is vanishing, since \[0 < \tfrac{1}{b_n} \Big(\sum_{k\leq n} \tfrac{1}{r_k} -\textstyle{ \int}_{0}^{n} \tfrac{1}{r_t} dt \Big) \leq \tfrac{1}{b_n}\tfrac{2}{r_{\min}} \to 0\,.\] Moreover, by 19 , \(|\tfrac{1}{r_t} -c(t)|\leq C t^{-2+\delta_0}\), and hence also the second term in 24 is vanishing: \[\tfrac{1}{b_n} \textstyle{\int}_{0}^{n} |\tfrac{1}{r_t} -c(t)| dt \to 0\,.\] When \(\delta\neq \rho\), \(\int_{0}^{n} c(t) dt = n^{\delta_0}/\delta_0\) and the proof is done. When \(\delta=\rho=\delta_0\), with \(e^y = t^{\delta_0}\), we obtain \[\textstyle{\int}_{0}^{n} c(t) dt = \textstyle{\int}_{0}^{n} \tfrac{t^{\delta_0}\ln t}{t} dt = \textstyle{\int}_{-\infty}^{\delta_0\ln n} \tfrac{e^yy}{\delta_0^2} dy = \tfrac{n^{\delta_0}(\delta_0\ln(n)-1)}{\delta_0^2}\,. \qedhere\] ◻

Proof of Theorem 11. Without loss of generality, since \(b_t\to b>0\), we can assume \(b_t>0\) for all \(t\). Moreover, we can focus on proving the thesis with \(b_t\) replaced by its limit \(b>0\). Indeed, since \((b_t/b)\to 1\), the thesis with a time-dependent sequence \((b_t)_t\) follows directly with the one with constant \(b\) by setting \(\widetilde{\eta}_t=(b_t/b)\eta_t\geq 0\) for \(t\) large enough, \(\widetilde{A}_t= (b/b_t)A_t\) and \(\Delta \widetilde{M}_{t+1}=(b/b_t)\Delta M_{t+1}\).
Assume \(b_t=b>0\) for all \(t\). Without loss of generality, since \(\eta_t\geq 0\) for \(t\) large enough and \(\eta_t \to 0\), we can assume \(\eta_t\) such that \(\eta_t\geq 0\) and \((1-b\eta_t)>0\) for all \(t\). The rest of the proof is divided in two steps.
For any \(a\in\mathbb{R}\) and \(t_0\in\mathbb{N}\) we can define the process \((X^{a,t_0}_{t})_{t\geq 0}\) as follows: \(X^{a,t_0}_{0}=X_{0}\) and \(X^{a,t_0}_{t}=\sum_{n=0}^{t-1} \Delta X^{a,t_0}_{n+1}\) for \(t\geq1\), with \[\Delta X^{a,t_0}_{n+1} = \begin{cases} \eta_n (A_n - b X_{n}^{a,t_0}) + \eta_n \Delta M_{n+1} + \rho_{n+1} & \text{if n\leq t_0}\,; \\ \eta_n (a - b X_{n}^{a,t_0}) + \eta_n \Delta M_{n+1} + \rho_{n+1} & \text{if n> t_0}\,; \end{cases}\] so that \(X^{a,t_0}_{t}\equiv X_{t}\) for any \(t\leq t_0\). Moreover notice that, for any \(n> t_0\), \[\Delta X^{a,t_0}_{n+1}-\Delta X_{n+1} = \eta_n ((a-A_n) - b (X^{a,t_0}_{n}-X_n))\] and then \[(X^{a,n_0}_{n+1} - X_{n+1}) = \eta_n (a-A_n) + ( 1- b\eta_n ) (X^{a,t_0}_{n}-X_n)\,.\] This means that

• \(\cap_{t\geq t_0} \{X^{a,t_0}_{t} \leq X_{t}\}\) with probability one on \(\cap_{t\geq t_0} \{a < A_t\}\),

• \(\cap_{t\geq t_0} \{X^{a,t_0}_{t} \geq X_{t}\}\) with probability one on \(\cap_{t\geq t_0} \{a > A_t\}\).

Therefore, on the set \(\{a > \limsup_{t\rightarrow\infty} A_t\}\), with probability one we have that \[\limsup_{t\rightarrow\infty} X_t = \inf_{t_0} \sup_{t\geq t_0} X_t \leq \inf_{t_0} \sup_{t\geq n_0} X^{a,t_0}_{t} = \limsup_{t\rightarrow\infty} X^{a,t_0}_{t}\,.\] Analogously, on the set \(\{a < \liminf_{t\rightarrow\infty} A_t\}\), with probability one we have that \[\liminf_{t\rightarrow\infty} X_t = \sup_{t_0} \inf_{t\geq t_0} X_t \geq \sup_{t_0} \inf_{t\geq t_0} X^{a,t_0}_{t} = \liminf_{t\rightarrow\infty} X^{a,t_0}_{t}\,.\] We will now show that, for any fixed \(a\in\mathbb{R}\) and \(t_0\in\mathbb{N}\), \[\limsup_{t\rightarrow\infty} X^{a,t_0}_{t}=\liminf_{t\rightarrow\infty} X^{a,t_0}_{t}=\tfrac{a}{b} \quad a.s.\,.\] We define \(\widetilde{\rho}_{t_0}=0\) and \(\widetilde{\rho}_{t+1}=(1-b\eta_t)\widetilde{\rho}_t+\rho_{t+1}\) for each \(t\geq t_0\), so that we have \(|\widetilde{\rho}_{t+1}|\leq (1-b\eta_t)|\widetilde{\rho}_t|+|\rho_{t+1}|\) and hence, by Lemma 10, we get \(\widetilde{\rho}_t\stackrel{a.s.}\to 0\). Set \(Y_t=X^{a,t_0}_t-a/b -\widetilde{\rho}_t\) for any \(t\geq t_0\). After some computations, we arrive to \[Y_{t+1}=(1-b\eta_t)Y_t+\eta_t\Delta M_{t+1}\] and so \[\begin{align} E[Y_{t+1}^2\mid{\mathcal{F}}_t]&= (1 +b^2\,\eta_t^2) Y_t^2 + \eta_t^2 E[(\Delta M_{t+1})^2\mid{\mathcal{F}}_t] -2b\, \eta_t\,Y_t^2 \,. \end{align}\] Hence, \((Y_t^2)_t\) is a non-negative almost super-martingale, which converges a.s. since \(\sum_t \eta^2_t<+\infty\) and ?? (see Theorem 22). We also get \(\sum_t \eta_t Y_t^2<+\infty\) a.s.and so, since \(\sum_t\eta_t=+\infty\), we necessarily have \(Y_t^2\stackrel{a.s.}\to 0\), that is \(X^{a,t_0}_t\stackrel{a.s.}\to a/b\).
We have shown that, for any pair \(a_1<a_2\), on the set \[\{a_1 < \liminf_{t\rightarrow\infty} A_t<\limsup_{t\rightarrow\infty} A_t < a_2\}\,,\] we have that \[\tfrac{a_1}{b}\leq \liminf_{t\rightarrow\infty} X_t < \limsup_{n\rightarrow\infty} X_t \leq \tfrac{a_2}{b} \quad a.s.\,.\] Finally, since \(a_1<a_2\) are arbitrary, we get the thesis. ◻

Proof. Set \(X_t^*=X_t/\zeta_t(\epsilon)\). \[X^*_{t+1} =\big(1-\tfrac{b_t}{\theta+t+1}\big)\tfrac{\zeta_{t}(\epsilon)}{\zeta_{t+1}(\epsilon)}X^*_t + \tfrac{\Delta M_{t+1}}{(\theta+t+1)\zeta_{t+1}(\epsilon)} +\tfrac{A_t/(\theta+t+1)+\rho_t}{\zeta_{t+1}(\epsilon)}\,.\] Since 21 , we get \[\begin{align} X^*_{t+1} &=\big(1-\tfrac{b_t}{\theta+t+1}\big) \big(1-\tfrac{\epsilon}{\theta+t+1}+O(\tfrac{1}{t^2})\big) X^*_t + \tfrac{\Delta M_{t+1}}{(\theta+t+1)\zeta_{t+1}(\epsilon)} +\tfrac{\tfrac{A_t}{(\theta+t+1)}+\rho_t}{\zeta_{t+1}(\epsilon)}\\ &=X^*_t - \tfrac{b_t+\epsilon}{\theta+t+1}X_t^* + O(1/t^2)X^*_t +\tfrac{\Delta M_{t+1}}{(\theta+t+1)\zeta_{t+1}(\epsilon)} +\tfrac{\tfrac{A_t}{(\theta+t+1)}+\rho_t}{\zeta_{t+1}(\epsilon)}\\ &\leq X^*_t +\tfrac{\Delta M_{t+1}}{(\theta+t+1)\zeta_{t+1}(\epsilon)} +\tfrac{\tfrac{A_t}{(\theta+t+1)}+\rho_t}{\zeta_{t+1}(\epsilon)} + O(1/t^2)X^*_t\,. \end{align}\] Hence, we have \[E[X^*_{t+1}\mid {\mathcal{F}}_t]\leq X^*_t + \tfrac{|A_t|/(\theta+t+1)+|\rho_t|}{\zeta_{t+1}(\epsilon)}+ O(1/t^2)X^*_t\] and so \((X^*_t)\) is a non-negative almost super-martingale (see Theorem 22) and, since we have \[\sum_t \Big[ \tfrac{|A_t|/(\theta+t+1)+|\rho_t|}{\zeta_{t+1}(\epsilon)}+ O(1/t^2)X^*_t \Big]<+\infty \quada.s.\,,\] we can conclude that \(X^*_t\), and so \(X_t/t^\epsilon\), converges a.s. to a finite random variable. Since \(\epsilon>0\) is arbitrary, the limit random variable has to be equal to zero, and so we can conclude that \(X_t=o(t^{\epsilon})\). ◻

Proof of Theorem 13. Let us set \(H_t=(\theta + t) X_t\) as we have done above so that, from 25 , we have \[H_{t+1}=H_t+Y_{t+1}\] and so \(H_{t}=H_0+\sum_{n=1}^t (H_n-H_{n-1})=H_0+\sum_{n=1}^t Y_{n}\). Hence, \((H_t)_t\) is a non-decreasing sequence of random variables with \(\sup_t H_t=H_0+\sum_t Y_t\). Therefore, if \(\sum_t Y_t=+\infty\) a.s., then \((H_t)\), and so \((t\, X_t)\), diverges a.s.to \(+\infty\) (and so \(X_t>0\) for \(t\) large enough).
Now, we set \(X^*_t=\zeta_t(1-\delta)X_t\) and \(\Delta M_{t+1}=Y_{t+1}-E[Y_{t+1}\mid {\mathcal{F}}_t]\) so that we have \[X^*_{t+1} =\big(1-\tfrac{(1-\delta)}{\theta+t+1}\big)\tfrac{\zeta_{t+1}(1-\delta)}{\zeta_t(1-\delta)}X^*_t+ \tfrac{\zeta_{t+1}(1-\delta)\Delta M_{t+1}}{\theta+t+1} +\tfrac{\zeta_{t+1}(1-\delta)}{\theta+t+1}\rho_t\,.\] Since 20 , we get \[\begin{align} \label{eq:X95star} X^*_{t+1} =\big(1-\tfrac{(1-\delta)}{\theta+t+1}\big)\big(1+\tfrac{(1-\delta)}{\theta+t+1}\big) X^*_t+\tfrac{\zeta_{t+1}(1-\delta)\Delta M_{t+1}}{\theta+t+1} +\tfrac{\zeta_{t+1}(1-\delta)}{\theta+t+1}\rho_t\\&=X^*_t - \tfrac{(1-\delta)^2}{(\theta+t+1)^2}X^*_t +\tfrac{\zeta_{t+1}(1-\delta)\Delta M_{t+1}}{\theta+t+1} +\tfrac{\zeta_{t+1}(1-\delta)}{\theta+t+1}\rho_t\\ &\leq X^*_t + \tfrac{\zeta_{t+1}(1-\delta)\Delta M_{t+1}}{\theta+t+1}+ \tfrac{\zeta_{t+1}(1-\delta)}{\theta+t+1}\rho_t\,. \\ \end{align}\tag{26}\] Hence, we have \[E[X^*_{t+1}\mid {\mathcal{F}}_t]\leq X^*_t + \tfrac{\zeta_{t+1}(1-\delta)}{\theta+t+1}\rho^+_t -\tfrac{\zeta_{t+1}(1-\delta)}{\theta+t+1}\rho^-_t\] and so \((X^*_t)\) is a non-negative almost super-martingale (see Theorem 22) and, since we have \[\sum_t \tfrac{\zeta_{t+1}(1-\delta)}{(\theta+t+1)}\rho^+_t \leq \sum_t \tfrac{\rho^+_t}{(\theta+t)^\delta}+ \sum_t \tfrac{O((\theta+t)^{-\delta})}{(\theta+t)}\rho_t^+<+\infty \quada.s.\,,\] we can conclude that \(X^*_t\), and so \(t^{1-\delta}X_t\), converges a.s. to a finite random variable \(X^*_\infty\) (and so \(X_t\stackrel{a.s.}\longrightarrow 0\)) and also \(\sum_t \tfrac{\zeta_{t+1}(1-\delta)}{(\theta+t)}\rho^-_t<+\infty\) a.s. so that we get \(\sum_t |\rho_t|/(\theta+t)^\delta<+\infty\) a.s..
Finally, we assume conditions ?? , ?? and one of the conditions i)-iii) and we are going to prove that \(X_\infty^*>0\) a.s. To this purpose we proceed in some steps.

We set \(Q_t=(\theta+t)/H_t^\gamma\) with \(\gamma>0\) and we observe that we have \[\begin{align} \tfrac{Q_{t+1}}{Q_t}- 1=\tfrac{(\theta+t+1)}{(\theta+t)}\tfrac{H_t^\gamma}{H_{t+1}^\gamma}-1 =\left(1+\tfrac{1}{\theta+t}\right)\tfrac{H_t^\gamma}{H_{t+1}^\gamma}-1\\&=\big(\tfrac{H_t^\gamma}{H_{t+1}^\gamma}\big)^\gamma-1+ \tfrac{1}{(\theta+t)}\big(\tfrac{H_t}{H_{t+1}}\big)^\gamma =\big(1-\tfrac{Y_{t+1}}{H_{t+1}}\big)^\gamma-1+ \tfrac{1}{(\theta+t)}\big(\tfrac{H_t}{H_{t+1}}\big)^\gamma\\ &\leq -\gamma \tfrac{Y_{t+1}}{H_{t+1}}+\tfrac{1}{(\theta+t)} = -\gamma \tfrac{Y_{t+1}}{H_t}+\gamma Y_{t+1}\big(\tfrac{1}{H_t}-\tfrac{1}{H_{t+1}}\big) +\tfrac{1}{(\theta+t)}\\ &=-\gamma \tfrac{Y_{t+1}}{H_t}+\gamma \tfrac{Y_{t+1}^2}{H_tH_{t+1}} +\tfrac{1}{(\theta+t)} \leq -\gamma \tfrac{Y_{t+1}}{H_t}+\gamma \tfrac{Y_{t+1}^2}{H_t^2} +\tfrac{1}{(\theta+t)}\,, \\ \end{align}\] where the first inequality follows since \((1-x)^\gamma\leq 1-\gamma x\) for \(0\leq x\leq 1\) and \(H_t\leq H_{t+1}\) and this last relation implies also the second inequality. If we take the conditional expectation given the past \({\mathcal{F}}_t\) and use the equality \(H_t=(\theta+t)X_t\) and \(E[Y_{t+1}\mid {\mathcal{F}}_t]=\delta X_t+\rho_t\), we obtain \[\begin{align} \label{eq-key} E\bigl[\tfrac{Q_{t+1}}{Q_t}- 1\mid {\mathcal{F}}_t\bigr]\leq -\gamma \tfrac{(\delta X_t +\rho_t)}{H_t}+\gamma \tfrac{E[Y_{t+1}^2\mid{\mathcal{F}}_t]}{H_t^2} +\tfrac{1}{(\theta+t)}\\&\leq -\gamma \tfrac{(\delta X_t +\rho_t)}{H_t}+\gamma \tfrac{O\left(E[Y_{t+1}\mid{\mathcal{F}}_t]\right)}{H_t^2}+\tfrac{1}{(\theta+t)}\\ &=\tfrac{(1-\gamma \delta)}{(\theta+t)} -\tfrac{\gamma}{(\theta+t)} \tfrac{(\theta+t)\rho_t}{H_t}+ \tfrac{1}{(\theta+t)}O\big(\tfrac{1}{H_t}\big) + \tfrac{1}{(\theta+t)}O\big(\tfrac{(\theta+t)|\rho_t|}{H_t^2}\big)\\ &=\tfrac{1}{(\theta+t)}\Big[(1-\gamma \delta) +O\big(\tfrac{1}{H_t}\big) +\tfrac{(\theta+t)|\rho_t|}{H_t}\big( -\gamma sign(\rho_t) + O\big(\tfrac{1}{H_t}\big)\big) \Big]\,. \\ \end{align}\tag{27}\] If we prove that the above last quantity into the square brackets is a.s.eventually non-positive for \(\gamma \delta>1\), then we get that for \(\gamma\delta>1\) we a.s. have \[E \big[Q_{t+1}- Q_t\mid {\mathcal{F}}_t \big]= Q_t E\big[\tfrac{Q_{t+1}}{Q_t}- 1\mid {\mathcal{F}}_t\big]\leq 0 \quadeventually.\] This proves that, for each \(\gamma>1/\delta\), the stochastic process \((Q_t)_t\), i.e. \(((\theta+t)/H_t^{\gamma})_t\) is eventually a (non-negative) super-martingales and so, for each \(\gamma>1/\delta\), it converges a.s.to a real random variable. Since \(\gamma>1/\delta\) is arbitrary, we necessarily have that \((\theta+t)/H_t^{\gamma}\) converges a.s. to zero. This fact concludes the proof of Step 1 because it implies that \((\theta+t)^{1/\gamma}/H_t\) converges a.s. to zero and we can set \(\eta=1/\gamma<\delta\leq 1\). Therefore the point is to have additional assumptions that entail the last term in the square brackets of 27 to be non-positive for \(\gamma\delta>1\). When \(\gamma\delta>1\), we have \((1-\gamma\delta)<0\) and so, since \(H_t\to +\infty\), the first quantity \(O(1/H_t)\) can be neglected. Hence the crucial term is \[\tfrac{(\theta+t)|\rho_t|}{H_t}\Big( -\gamma sign(\rho_t) + O\big(\tfrac{1}{H_t}\big)\Big)\,.\] In the case i), that is when \(\rho_t\geq 0\) eventually with probability one, we are done because the above quantity is equal to zero when \(\rho_t=0\) and, when \(\rho_t>0\), the quantity \(-\gamma sign(\rho_t)+O(1/H_t)=-\gamma+O(1/H_t)\) is eventually non-positive with probability one, because \(-\gamma<0\) and \(O(1/H_t)\to 0\) a.s.. Another simple cases are case ii), that is when \(\limsup_t (t |\rho_t|)<+\infty\) a.s., and case iii), that is when \(|\rho_t|=o(X_t)\). Indeed, in these cases we a.s. have \[\lim_t \tfrac{(\theta+t) |\rho_t|}{H_t}=\lim_t \tfrac{|\rho_t|}{X_t}=0\,.\]
From Step 1 and assumption ?? , we a.s. get \[\sum_t \tfrac{|\rho_t|}{H_t}= \sum_t \tfrac{(\theta+t)^{\eta}}{H_t}\,\tfrac{|\rho_t|}{(\theta+t)^{\eta}}\\ =\sum_t o(1)\tfrac{|\rho_t|}{(\theta+t)^{\eta}}<+\infty\,,\] where \(0<\eta=\delta-\epsilon<\delta\). Moreover, we a.s. have \[\sum_t \tfrac{1}{t H_t}=\sum_t\tfrac{t^\eta}{H_t} \tfrac{1}{t^{1+\eta}}= \sum_t o(1) \tfrac{1}{t^{1+\eta}}<+\infty\,,\] where \(0<\eta<\delta\).
Set \(L_t=\ln(H_t/(\theta+t)^{\delta})\) and \(U_t = E[L_{t+1} - L_t \mid \mathcal{F}_t ]\) and \(V_t = E[(L_{t+1} - L_t )^2 \mid {\mathcal{F}}_t]\). If we prove that \(\sum_t U_t\) and \(\sum_t V_t\) are a.s. convergent, then \(L_t\) converges a.s.to a real random variable (see Lemma 21). This fact implies that \((\theta+t)^{1-\delta}X_t=(\theta+t)X_t/(\theta+t)^{\delta}=H_t/(\theta+t)^{\delta}\) converges to a random variable with values in \((0, +\infty)\) and so \(P(X^*_\infty>0)=1\). The rest of the proof is devoted to verify that \(\sum_t |U_t | < +\infty\) and \(\sum_t V_t < +\infty\) a.s.. To this regard, we note that \[\begin{align} U_t &=E[\ln(H_{t+1})-\ln(H_t)\mid \mathcal{F}_t]-\delta\left(\ln(\theta+t+1)-\ln(\theta+t)\right) \\ &=E[\ln(H_{t}+Y_{t+1})-\ln(H_t)\mid \mathcal{F}_t]-\delta\ln(1+1/(\theta+t)) \\ &=E\Big[\textstyle{\int}_0^{Y_{t+1}}\tfrac{1}{H_t+x}\,dx\mid {\mathcal{F}}_t\Big]-\delta\ln(1+1/(\theta+t))\,. \end{align}\] Since \(1/(H_t+x)\leq 1/H_t\) and \(\ln(1+1/(\theta+t))\geq 1/(\theta+t)-1/(2(\theta+t)^2)\) for each \(x\geq 0\) and each \(t\), the last term of the above equalities is smaller than or equal to \[\begin{gather} \tfrac{E[Y_{t+1}|\mathcal{F}_t]}{H_t}-\tfrac{\delta}{\theta+t}+\tfrac{\delta}{2 (\theta+t)^2} =\tfrac{\delta X_t+\rho_t}{H_t}-\tfrac{\delta}{\theta+t}+\tfrac{\delta}{2(\theta+t)^2}= \tfrac{\delta}{\theta+t}+\tfrac{\rho_t}{H_t}-\tfrac{\delta}{\theta+t}+\tfrac{\delta}{2 (\theta+t)^2} =\tfrac{\rho_t}{H_t}+O(1/t^2)\,. \end{gather}\] Now, we note that \(-U_t=\delta\ln(1+1/(\theta+t))-\ln(H_{t+1})+\ln(H_t)\). Using \(\ln(1+1/(\theta+t))\leq 1/(\theta+t)\) and \(1/(H_t + x) \geq 1/H_t - x/H_t^2\) for each \(x\geq 0\) and each \(t\), we find that \(-U_t\) is smaller than or equal to \[\begin{align}\tfrac{\delta}{(\theta+t)}-\tfrac{1}{H_t}E[Y_{t+1}\mid \mathcal{F}_t]+ \tfrac{1}{2H_t^2}E[Y_{t+1}^2\mid \mathcal{F}_t] &\leq \tfrac{\delta}{(\theta+t)}-\tfrac{\delta X_t+\rho_t}{H_t}+ \tfrac{O(\delta X_t +\rho_t)}{2H_t^2} =\tfrac{\delta}{\theta+t}-\tfrac{\delta}{\theta+t}-\tfrac{\rho_t}{H_t}+ \tfrac{O(\delta X_t +\rho_t)}{2H_t^2}\\ &=-\tfrac{\rho_t}{H_t}+O(1/(tH_t))+o(\rho_t/H_t)\,.\end{align}\] Thus, by Step 2, we can derive that \(\sum_t|U_t|<+\infty\) a.s.. Finally, we observe we have \[\begin{align} E[( \ln(H_{t+1})-\ln(H_t)-\delta\ln(\theta+t+1)+\delta\ln(\theta+t) )^2 \mid \mathcal{F}_t]\\&\leq 2\left\{ E[(\ln(H_{t+1})-\ln(H_t))^2\mid \mathcal{F}_t]+ \delta^2(\ln(1+1/(\theta+t))^2 \right\} \\ &\leq 2 E\bigg[\Big(\textstyle{\int}_0^{Y_{t+1}}\tfrac{1}{H_t+x}\,dx\Big)^2\,\Big\vert \mathcal{F}_t\bigg] +2\delta^2/(\theta+t)^2 \\ & \leq 2 E[(Y_{t+1}/H_t)^2\mid \mathcal{F}_t]+ O(1/t^2) \leq \tfrac{2}{H_t^2} E[Y_{t+1}^2\mid\mathcal{F}_t]+O(1/t^2)\\ &= \tfrac{1}{H_t^2} O(\delta X_t+\rho_t) +O(1/t^2)\\ &= O(1/(tH_t)) + o(\rho_t/H_t) + O(1/t^2)\,. \\ \end{align}\] Therefore we have also \(\sum_t V_t<+\infty\) a.s.and we can conclude. ◻

Proof of Theorem 14. Let us consider the process \(X^*=(X^*_t)_t\) defined as \(X^*_t = \zeta_t(1-\delta)X_t\), where \(\zeta_t(x) \sim t^x\) is defined in 17 , so that \(X^*_t\stackrel{a.s.}\sim t^{1-\delta}X_t\stackrel{a.s.}\to X^*_\infty\) by assumption. Starting from dynamics 28 and recalling that \(r_k=\zeta_k(1-\delta)\) (by assumption \(\delta>\rho\)), we can follow the analogous calculations as in 26 to get \[\label{eq-sum} X^*_{k+1}-X^*_k = A_k+B_{k+1}+C_{k+1}+D_{k+1}\,,\tag{32}\] where \[\left\{\begin{align} A_k&=-\Big(\tfrac{1-\delta}{\theta+k+1}\Big)^2 X^*_k\quad(which is =0 if \delta=1)\,;\\ B_{k+1}&=\tfrac{\zeta_{k+1}(1-\delta)\rho_{k+1}}{\theta+k+1}\,;\\ C_{k+1}&=\Big(\tfrac{\zeta_{k+1}(1-\delta)}{\zeta_k(1-\delta)}-1\Big)\Delta \widehat{M}_{k+1}\quad(which is =0 if \delta=1)\,;\\ D_{k+1}&=\Delta \widehat{M}_{k+1}\,. \end{align}\right.\] Note that \(\sum_{k\geq t}(X_{k+1}^*-X^*_k)=(X^*_\infty-X^*_t)\). Therefore, it is enough to study the asymptotic behavior of the four terms \(\sum_{k\geq t}A_k,\,\sum_{k\geq t}B_{k+1}\), \(\sum_{k\geq t}C_{k+1}\) and \(\sum_{k\geq t} D_{k+1}\).

Let us start with the last term \(\sum_{k\geq t}D_{k+1}\). First of all we observe that, since \(\delta>\rho\) by assumption, we have \(r_t\sim t^{1-\delta}\) and so \[E[\,|\Delta \widehat{M}_{t+1}|^p\,] = E[\,|\Delta M_{t+1}|^p\,] \tfrac{r_t^p}{(\theta+t+1)^p} = E[\,|\Delta M_{t+1}|^p\,]O(t^{-p\delta})\,.\] Hence, relation ?? with \(\delta_0=\delta\) follows from ?? . Moreover, condition ?? equals condition ?? . It follows by Theorem 15 and the fact that \(b_t\sim t^\delta\), that we have \(\widehat{M}_t\stackrel{a.s.}\longrightarrow \widehat{M}_\infty\) for a suitable real random variable \(\widehat{M}_\infty\) and \[t^{\delta/2} \sum_{k\geq t} D_{k+1}=t^{\delta/2}\sum_{k\geq t} \Delta \widehat{M}_{k+1} = t^{\delta/2}(\widehat{M}_\infty-\widehat{M}_t) \stackrel{stably}\longrightarrow {\mathcal{N}}\Big(0,V^*_\infty/\delta\Big)\,,\] where the convergence is also in the sense of the a.s. conditional convergence with respect to the filtration \(\mathcal{F}\).
Let us now focus on the term \(\sum_{k\geq t}C_{k+1}\) when \(\delta<1\). Let us define \(Y_{k}=\zeta_{k}(1-\delta)\Delta M_{k+1}\) and \(\mathcal{G}_k=\mathcal{F}_{k+1}\), then \(E[Y_{k}\mid\mathcal{G}_{k-1}] = 0\). Now, let \(\tilde{a}_k=k^{-\delta}(\theta+k+1)^2\) and \(\tilde{b}_k=k^{\delta/2}\). Then \(\tilde{b}_k\uparrow +\infty\) and \[\begin{align} \sum_{k=1}^{\infty}\tfrac{E[Y_{k}^2]}{\tilde{a}_k^2\tilde{b}_k^2} &= \sum_{k=1}^{\infty}\tfrac{\zeta_k(1-\delta)^2}{\tilde{a}_k^2\,\tilde{b}_k^2} E[(\Delta M_{k+1})^2] = \sum_{k=1}^{\infty}\tfrac{\zeta_k(1-\delta)^2(\theta+k+1)^2}{\tilde{a}_k^2\,\tilde{b}_k^2\,r_k^2}E[(\Delta \widehat{M}_{k+1})^2] \\ &=\sum_{k=1}^{\infty}\tfrac{\zeta_k(1-\delta)^2(\theta+k+1)^2}{\tilde{a}_k^2\,\tilde{b}_k^2\,\zeta_k(1-\delta)^2}E[(\Delta \widehat{M}_{k+1})^2] =\sum_{k=1}^{\infty} \tfrac{k^{\delta}}{(\theta+k+1)^2}E[(\Delta \widehat{M}_{k+1})^2]\\ &\leq \sum_{k=1}^{\infty} k^{2\delta}\tfrac{E[(\Delta \widehat{M}_{k+1})^2]}{k^{\delta+\epsilon}} \,, \end{align}\] which is finite, by ?? with \(p=2\), \(\delta_0=\delta\) and some \(0<\epsilon< \delta\). Since \(\tilde{b}_t\sum_{k\geq t} \tfrac{1}{\tilde{a}_k\tilde{b}_k^2} \to 0\), then by Lemma 23, we obtain \[\begin{gather} t^{\delta/2}\sum_{k\geq t} C_{k+1}= t^{\delta/2} \sum_{k\geq t} \Big(\tfrac{\zeta_{k+1}(1-\delta)}{\zeta_{k}(1-\delta)}-1\Big) \Delta \widehat{M}_{k+1} = t^{\delta/2} \sum_{k\geq t} \tfrac{1-\delta}{\theta+k+1} \Delta \widehat{M}_{k+1} \\ = (1-\delta) t^{\delta/2} \sum_{k\geq t} \tfrac{\zeta_{k}(1-\delta)\Delta M_{k+1}}{(\theta+k+1)^2} = (1-\delta) \tilde{b}_t\sum_{k\geq t} \tfrac{Y_{k}}{\tilde{a}_k\,\tilde{b}_k^2}\stackrel{a.s.} \longrightarrow 0\,. \end{gather}\] Let us now focus on the term \(\sum_{k\geq t} A_k\) when \(\delta<1\). Since \(X^*_t\) converges a.s. to a real random variable \(X^*_\infty\), we have \[t^{\delta/2} \sum_{k\geq t} A_k= t^{\delta/2} \sum_{k\geq t} \Big(\tfrac{1-\delta}{\theta+k+1}\Big)^2 X^*_k = t^{\delta/2} \sum_{k\geq t} \tfrac{O(1)}{k^{1+\delta}}\stackrel{a.s.}\longrightarrow 0\,.\] Finally, regarding the term \(\sum_{k\geq t}B_{k+1}\), since 18 , we have \[\label{eq:rho95t} {t}^{\delta-\rho}\zeta_{k+1}(1-\delta)\rho_{k+1} = \rho_\infty + o(1)\,,\tag{33}\] and hence \[\sum_{k\geq t} B_{k+1}= \sum_{k\geq t} \tfrac{\zeta_{k+1}(1-\delta)\rho_{k+1}}{\theta+k+1} = \sum_{k\geq t} \tfrac{\rho_\infty}{k^{1+\delta-\rho}} + o\Big(\tfrac{1}{t^{\delta-\rho}}\Big) \sim \tfrac{\rho_{\infty}}{(\delta-\rho)} \tfrac{1}{t^{\delta-\rho}}\,,\] so that \[\begin{align} t^{\delta/2} \sum_{k\geq t}B_{k+1} = \tfrac{t^{\delta/2}}{t^{\delta-\rho}}\, t^{\delta-\rho} \sum_{k\geq t} B_{k+1} & \stackrel{a.s.}\longrightarrow \begin{cases} 0 & \text{if \rho<\delta/2}\,; \\ \rho_\infty/\rho=2\rho_\infty/\delta & \text{if \rho = \delta/2}\,; \end{cases} \\ t^{\delta-\rho} \sum_{k\geq t} B_{k+1} & \stackrel{a.s.}\longrightarrow \tfrac{\rho_\infty}{\delta-\rho}\quad \text{if \rho > \delta/2}\,. \end{align}\] Recalling that the stable convergence combines well with convergence in probability (e.g. Theorem 2.4.2 in [38]), and so with the a.s.convergence, and the a.s. conditional convergence combines well with the a.s. convergence (see Lemma 25), we are now ready to conclude:

• if \(\rho\leq \delta/2\), then, we have to combine the stable convergence and the a.s.conditional convergence of the suitable rescaled sequence \(\sum_{k\geq t} D_{k+1}=(\widehat{M}_\infty-\widehat{M}_t)\) with the a.s.convergences of the suitably rescaled sequences \(\sum_{k\geq t} A_k\) (only when \(\delta<1\)), \(\sum_{k\geq t} B_{k+1}\) and \(\sum_{k\geq t} C_{k+1}\) (only when \(\delta<1\)) so that we obtain 29 (with \(X^*_t\) instead of \(t^{1-\delta}X_t\)), in the sense of the stable convergence and in the sense of the a.s. conditional convergence with respect to \(\mathcal{F}\).

• if \(\delta/2 < \rho < \delta\), then the leading term is \(\sum_{k\geq t} B_{k+1}\) and the a.s.convergence of \(t^{\delta-\rho} \sum_{k\geq t} B_{k+1}\) has to be combined with the a.s.limit results for the other terms involving \(A_k\) and \(C_{k+1}\) (only when \(\delta<1\)) and, above all, with the convergence in probability to zero of \(\tfrac{t^{\delta-\rho}}{t^{\delta/2}}t^{\delta/2}(\widehat{M}_\infty-\widehat{M}_t)\), where \(\tfrac{t^{\delta-\rho}}{t^{\delta/2}}=t^{-(\rho-\delta/2)}\to 0\).

Furthermore, the 30 (with \(X^*_t\) instead of \(t^{1-\delta}X_t\)) is obvious when \(\rho\leq \delta/2\), because the stable convergence and the a.s.conditional convergence combine well with the a.s. convergence of \(t^{-(\delta/2-\rho)}\rho_\infty/(\delta-\rho)\) to zero when \(\rho<\delta/2\) and to \(2\rho_\infty/\delta\) when \(\rho=\delta/2\). In the case \(\delta/2<\rho<\delta\), if we assume \((*)\; t^{\rho-\delta/2}(t^{1-\rho}\rho_{t+1}-\rho_\infty)\stackrel{P}\longrightarrow 0\), then, equation 33 becomes \[{k}^{\delta-\rho}\zeta_{k+1}(1-\delta)\rho_{k+1} = \rho_\infty + o_P(k^{-(\rho-\delta/2)})\,,\] so that \[t^{\delta/2} \Big( \sum_{k\geq t} B_{k+1} - \tfrac{\rho_\infty}{(\delta-\rho)}\tfrac{1}{t^{\delta-\rho}} \Big)=o_P(1) \stackrel{P}\longrightarrow 0\] and this convergence in probability combines well with the stable convergence of \(t^{\delta/2}\sum_{k\geq t} D_{k+1}=t^{\delta/2}(\widehat{M}_\infty-\widehat{M}_t)\). If the additional assumed convergence \((*)\) is a.s., then we can replace \(o_P()\) with \(o()\) and the above limit relation is in the sense of a.s. convergence and it combines well with the a.s. conditional converge of \(t^{\delta/2}\sum_{k\geq t} D_{k+1}\).
Finally, we can obtain the results for \(t^{1-\delta} X_t\) from those proven for \(X^*_t\), because of 18 with \(x=1-\delta\). ◻

Proof of Theorem 15. First of all we observe that relation ?? with \(p=4\) implies that \(( \widehat{M}_t)_t\) is a martingale bounded in \(L^4\) and so it converges a.s.and in \(L^4\) (in particular, in mean) to a real (\({\mathcal{F}}_\infty\)-measurable) random variable \(\widehat{M}_\infty\) with \(E[\widehat{M}_\infty^4]<+\infty\). The sequel of the proof is based on Theorem 24. To this aim, for any \(n\geq 1\), define \(M_{n,0}=0\) and \(M_{n,t} = \sqrt{b_n} (\widehat{M}_{n}-\widehat{M}_{n+t-1})\) for \(t\geq 1\) (which is \(\mathcal{F}_{n+t-1}\)-measurable). We have \(M_{n,t}\stackrel{L^1}\longrightarrow \sqrt{b_n}(\widehat{M}_n-\widehat{M}_\infty)\) as \(t\to +\infty\). Moreover, we have \(X_{n,1}=M_{n,1}-M_{n,0}=0\) and, for each \(t\geq 2\) \[X_{n,t} = M_{n,t}-M_{n,t-1}= \sqrt{b_n} (\widehat{M}_{n+t-2}-\widehat{M}_{n+t-1})\,,\] so that \[U_{n} = b_n \sum_{t\geq n} (\widehat{M}_{t+1}-\widehat{M}_t)^2 , \quad X_{n}^* = \sqrt{b_n} \sup_{t\geq n} |\widehat{M}_{t+1}-\widehat{M}_t|\,.\] First step: \(X_n^* \stackrel{a.s.}\longrightarrow 0\): From ?? , with \(p=4\) and some \(0<\epsilon < \delta_0\), we have \[\label{AA:eq:inProof95mbound} \begin{align} E\Big[\sum_{t} (\sqrt{b_t})^4 |\Delta \widehat{M}_{t+1}|^{4} \Big] &\leq E\Big[\sum_{t} t^{2\delta_0+(\delta_0-\epsilon)} |\Delta \widehat{M}_{t+1}|^{4} \Big]\\ &= E\Big[\sum_{t} t^{p\delta_0} \tfrac{|\Delta \widehat{M}_{t+1}|^{p}}{t^{\delta_0+\epsilon}} \Big] <+\infty\,, \end{align}\tag{34}\] which implies \(\sum_{t} (\sqrt{b_t})^4 (\Delta \widehat{M}_{t+1})^{4}<+\infty\) a.s.. Then we have \[(X_n^*)^{4}\leq \sup_{t\geq n} (\sqrt{b_t})^4 |\Delta \widehat{M}_{t+1}|^{4}\leq \sum_{t\geq n} (\sqrt{b_t})^4 |\Delta \widehat{M}_{t+1}|^{4}\stackrel{a.s.}\to 0\quadasn\to +\infty\,.\] Second step: \((X_n^*)_n\) dominated in \(L^1\): We are going to show that \((X_n^*)_n\) is dominated in \(L^4\). In fact, with \(p=4\) and some \(0<\epsilon<\delta_0\), we have \[E\Big[(\sup_n X_n^*)^{4}\Big]\leq E\Big[\sup_n \sup_{t\geq n} (\sqrt{b_t})^4 |\Delta \widehat{M}_{t+1}|^{4}\Big] \leq E\Big[ \sum_t (\sqrt{b_t})^4 |\Delta \widehat{M}_{t+1}|^{4}\Big] < +\infty\] as shown in 34 in the first step.
Third step: \(U_n \stackrel{a.s.}\longrightarrow V^*_\infty/\delta_0\): Let \(a_k\) as defined in ?? (so that \(a_kb_k^2=(\theta+k+1)^2/r_k\)) and set \(Y_{k} = a_kb_k^2 (\Delta \widehat{M}_{k+1})^2=r_k(\Delta M_{k+1})^2\). Then, we have \(E[Y_{k}\mid\mathcal{F}_{k-1}]\to V^*_\infty\) by assumption ?? . Moreover, we have \(b_k\uparrow +\infty\) and, by ?? with \(p=4\) and some \(0<\epsilon<\delta_0\), we get \[\sum_{k=1}^{\infty}\tfrac{E[Y_k^2]}{a_k^2b_k^2}= \sum_{k=1}^{\infty}b_k^2E[(\Delta\widehat{M}_{k+1})^4] \leq C \sum_{k=1}^{\infty} k^{4\delta_0} \tfrac{E[(\Delta \widehat{M}_{k+1})^4]}{k^{\delta_0+\epsilon}}<+\infty\,.\] Since \(b_t\sum_{k\geq t} \tfrac{1}{a_kb_k^2}\to \tfrac{1}{\delta_0}\) by the technical Lemma 9, then \(U_t = b_t\sum_{k\geq t} \tfrac{Y_k}{a_kb_k^2}\stackrel{a.s.} \longrightarrow V^*_\infty/\delta_0\) by Lemma 23. ◻

Proof of Theorem 1. In the case \(w > \beta\) the term involving \(\lambda_t\) in the dynamics 8 of \((Z_t)_t\) behaves essentially as a remainder and hence we can focus on the process \(S_t=\sum_{j\in {\mathcal{O}}_t} P_{t,j}=Z_t-\lambda_t\), so that from 8 we get \[\begin{align} S_{t+1}-S_t&=Z_{t+1}-Z_t-(\lambda_{t+1}-\lambda_t)\\ &=w\tfrac{T_{t+1}-Z_t}{\theta+t+1}-(1-w)\tfrac{Z_t}{\theta+t+1} + \tfrac{\lambda_t}{\theta+t+1}\\ &=-\tfrac{Z_t}{\theta+t+1}+w\tfrac{T_{t+1}}{\theta+t+1} +\tfrac{\lambda_t}{\theta+t+1} =-\tfrac{S_t}{\theta+t+1}+\tfrac{wT_{t+1}}{\theta+t+1}\,. \end{align}\] The above dynamics for \(S_t\) can be written as \[\label{eq-dyn-S} S_{t+1} =\big(1-\tfrac{1}{\theta+t+1}\big)S_t+w\tfrac{T_{t+1}}{\theta+t+1} =\big(1-\tfrac{1-w}{\theta+t+1}\big)S_t+\tfrac{\Delta M_{t+1}}{\theta+t+1}+ \tfrac{w\lambda_t}{\theta+t+1}\,,\tag{35}\] with \(\Delta M_{t+1}=w(T_{t+1}-Z_t)\). Denoting \(S^{*}_{t}=\zeta_{t}(1-w)S_t\), where \(\zeta_t(x)\sim t^x\) is defined in 17 , we can obtain \[S^{*}_{t+1}=\big(1-\tfrac{(1-w)^2}{(\theta+t+1)^2}\big)S^{*}_t +\tfrac{\zeta_{t+1}(1-w)\Delta M_{t+1}}{\theta+t+1} +\tfrac{w\zeta_{t+1}(1-w)\lambda_t}{\theta+t+1}\,,\] and so, by the classical Theorem 22, the process \((S^{*}_t)_t\) is a non-negative almost super-martingale that converges a.s.to a finite r.v. \(S^{*}_\infty\). However, this is not enough for our scopes. Indeed, in that way, we do not know that such random limit \(S^{*}_\infty\) is a.s.strictly positive, and this is crucial for proving the right rate of convergence. For this reason we are going to apply the more sophisticated Theorem 13 with \(X_t=S_t\), \(Y_{t+1}=wT_{t+1}\) and \(\delta=w\in (0,1]\). To this end, we observe that we have \[\sum_t Y_{t}=w\sum_t T_t\geq w \sum_t N_t=w\sup_t \sum_{n=1}^t N_n= w\sup_t D_t=+\infty\quada.s.\,\] (by 10 ) and \[E[Y_{t+1}\mid {\mathcal{F}}_t] = E[w T_{t+1}\mid {\mathcal{F}}_t]=w Z_t=w S_t + w \lambda_t=\delta X_t +\rho_t\,,\] with \(\rho_t=w\lambda_t>0\) and \[\sum_t \tfrac{\rho_t}{(\theta+t)^{\delta-\epsilon}}= O\big( \textstyle{\sum_t} \tfrac{1}{(\theta+t)^{1+w-(\beta+\epsilon)}}\big)<+\infty\] for some \(\epsilon>0\) such that \(w-(\beta+\epsilon)>0\) (note that this \(\epsilon\) exists because \(w>\beta\) by assumption). Therefore, by Theorem 13, we firstly obtain that \(S_t=O(1/t^{1-w})\) and so \(Z_t=S_t+\lambda_t=O(1/t^{1-w})+O(1/t^{1-\beta})=O(1/t^{1-w})\) because \(\beta<w\). Hence, by 9 , we get \[E[Y_{t+1}^2\mid{\mathcal{F}}_t]=w^2(Z_t+Z_t^2 -R_t) =O(Z_t) =O(E[Y_{t+1}\mid {\mathcal{F}}_t])\,.\] Therefore, we can apply Theorem 13 (case \(i)\)) and we obtain that \[t^{1-w}S_t\stackrel{a.s.}\longrightarrow S_\infty^*\in (0,+\infty)\,.\] As before, since \(Z_t=S_t+\lambda_t\) and \(\beta<w\), we get \[t^{1-w} Z_t= t^{1-w} S_t +t^{1-w}\lambda_t \stackrel{a.s.}\longrightarrow S_\infty^*\] and so the first convergence holds true with \(Z_\infty^*=S^*_\infty\). ◻

Proof of Theorem 2. In the case \(w \leq \beta\) the term involving \(\lambda_t\) in the dynamics 8 of \((Z_t)_t\) cannot be considered as a remainder term as in the proof of Theorem 1. Indeed, letting \[Z^*_t = r_t Z_t \qquadwith\qquad r_t=\begin{cases} \zeta_t(1-\beta)\quad &whenw<\beta\,,\\ \tfrac{\zeta_t(1-\beta)}{\ln(\theta+t)} \quad &whenw=\beta\,, \end{cases}\] where \(\zeta_t(x) \sim t^x\) is defined in 17 . From 8 , taking into account 20 and 22 , we get \[\label{eq:dynamics95Z95star} Z^*_{t+1} - Z^*_{t} = \eta_{t+1}(\alpha\beta-bZ^*_{t}) + \eta_{t+1}\Delta M^*_{t+1}+ \rho_{t+1}\,,\tag{36}\] where \(\rho_{t+1}=Z^*_t O(\tfrac{1}{t^2}) + O(\tfrac{1}{t^{1+\beta}})\) (here \(O(\tfrac{1}{t^2})\) and \(O(\tfrac{1}{t^{1+\beta}})\) are deterministic terms and \(\beta>0\)), \(\Delta M^*_{t+1}=w\,\zeta_{t+1}(1-\beta)(T_{t+1}-Z_t)\) and \[\begin{cases} b=\beta-w\quadand\quad \eta_{t+1}=\tfrac{1}{(\theta+t+1)} \quad&ifw< \beta\,; \\ b=1\quadand\quad \eta_{t+1}=\tfrac{1}{(\theta+t+1)\ln(\theta+t+1)} \quad & ifw= \beta\,. \end{cases}\] This suggests that \((Z^{*}_t)_t\) is not a non-negative almost super-martingale, as in the case \(w>\beta\). Instead, it evolves according to the stochastic approximation dynamics in 36 , which suggests convergence toward a deterministic limit given by \(\alpha\beta/b\). For this reason, we apply Theorem 11 to the process \((Z^{*}_t)_t\) governed by 36 . To this end, note that from Lemma 16 we have that \(\sup_t E[r_t Z_t]<+\infty\) and hence, since by 9 , \[\begin{align} E[(\Delta M^*_{t+1})^2\mid{\mathcal{F}}_t]&= w^2\zeta_{t+1}(1-\beta)^2\left(E[T_{t+1}^2\mid\mathcal{F}_t]-Z_t^2\right)\\ &=w^2\zeta_{t+1}(1-\beta)^2(Z_t-R_t)\\ &\leq w^2\zeta_{t+1}(1-\beta)^2 Z_t\,, \end{align}\] we get, for any \(w\leq \beta\), that \[\begin{align} \sum_t \eta_{t+1}^2 E[(\Delta M_{t+1}^*)^2] \leq \sum_t \tfrac{w^2 r_t}{(\theta+t+1)^2} E[r_t Z_t] <+\infty\,, \end{align}\] which implies \[\sum_t \eta_{t+1}^2 E[(\Delta M^*_{t+1})^2\mid{\mathcal{F}}_t]<+\infty \quada.s.\,.\] In addition, since \(\sup_t E[Z_t^*]=\sup_t E[r_t Z_t]<+\infty\) (by Lemma 16), the expression for \(\rho_{t+1}\) implies that \(\sum_t E[\,|\rho_{t+1}|\,]= \sum_t O(E[Z^*_t]/t^2) + O(1/t^{1+\beta})\) is convergent, and so \(\sum_t |\rho_{t+1}|<+\infty\) with probability one. Then, we can apply Theorem 11 to the dynamics 36 so obtaining \[Z^*_{t} \stackrel{a.s.}\to \tfrac{\alpha\beta}{b} = \begin{cases} \tfrac{\alpha\beta}{\beta-w} \quad & ifw< \beta\,; \\ \alpha\beta \quad & ifw= \beta\,. \end{cases}\] ◻

Proof of Theorem 3. We will first prove the CLT for the process \(S_t=\sum_{j\in\mathcal{O}_t}P_{t,j}\) and then, since \(S_t=Z_t-\lambda_t\), we will derive from it the one for \(Z_t\). Indeed, since the dynamics of \(S_t\) in 35 , i.e. \[S_{t+1}=\left(1-\tfrac{1-w}{\theta+t+1}\right)S_t+\tfrac{w(T_{t+1}-Z_t)}{\theta+t+1}+ \tfrac{w\lambda_t}{\theta+t+1},\] is equal to the general dynamics in 28 with \(\delta=w\in (0,1]\), \(\Delta M_{t+1}=w(T_{t+1}-Z_t)\), \(\rho_{t+1}=w\lambda_t\), \(\rho=\beta\in [0,1]\) and \(\rho_\infty=w\alpha\) (because \(t^{1-\beta}\lambda_t\) trivially converges to \(\alpha\)), and since we are assuming \(\beta<w\), that is \(\rho<\delta\), the CLT for \(S_t\) can be obtained by applying Theorem 14 with \(X_t=S_t\) and \(X^*_\infty=Z^*_\infty\) (recall that, by Theorem 1, \(t^{1-w}S_t\stackrel{a.s.}\to Z_\infty^*\)). Indeed, condition ?? follows by ?? in Lemma 16 as for any \(p\geq1\) and \(\epsilon>0\) we have \[E\Big[ \sum_{t=1}^{\infty} \tfrac{|\Delta {M}_{t+1}|^p}{t^{w+\epsilon}} \Big] = \sum_{t=1}^{\infty} \tfrac{1}{t^{1+\epsilon}} E[ t^{1-w}|\Delta {M}_{t+1}|^p] <+\infty\,.\] Moreover, condition ?? follows by 9 , Theorem 1 and Lemma 17 (in the case \(\beta<w\)): \[t^{1-w} E[(\Delta M_{t+1})^2|\mathcal{F}_t]= w^2 t^{1-w}(Z_t-R_t)\stackrel{a.s.}\longrightarrow V_\infty^{*} = w^2 \Sigma_\infty\,.\] Then, since \((t^{1-\beta}\lambda_t-\alpha)=\alpha[1-1/(t+1)]^{1-\beta}-\alpha=O(t^{-1})=o(t^{-(\beta-w/2)})\), also condition \((*)\) of Theorem 14 is verified with a.s.convergence and so, from that theorem, we get \[t^{w/2}(t^{1-w}S_t+\tfrac{ w\alpha }{(w-\beta)}t^{-(w-\beta)} -Z^*_\infty)\stackrel{stably}\longrightarrow {\mathcal{N}}(0, w\Sigma_\infty)\,.\] For what concerns the process \((Z_t)_t\) recall that \(Z_t=S_t+\lambda_t\) and so \(t^{1-w}Z_t=t^{1-w}S_t + t^{1-w}\lambda_t\), and hence the thesis follows by the fact that \(t^{1-w}\lambda_t=\alpha t^{\beta-w}+O(t^{-(1+w-\beta)})= \alpha t^{\beta-w}+o(t^{-w/2})\). In addition, note that, from ?? , for \(w/2<\beta<w\) we have \[\begin{align} {t}^{w-\beta}(t^{1-w}Z_t -Z^*_\infty) +\tfrac{\alpha\beta}{w-\beta} &= \tfrac{1}{t^{\beta-w/2}}\, t^{w/2}\, t^{\beta-w} \,[t^{w-\beta}(t^{1-w}Z_t-Z^*_\infty)+\tfrac{\alpha\beta}{w-\beta} ]\\ &= \tfrac{1}{t^{\beta-w/2}}\, t^{w/2}\,[(t^{1-w}Z_t-Z^*_\infty)+\tfrac{\alpha\beta}{w-\beta} t^{\beta-w} ]\stackrel{P}\rightarrow 0. \qedhere \end{align}\] ◻

Proof of Corollary 4. From from ?? , we can derive the corresponding result for \(\overline{T}_t\). Indeed, by 4 , we have \(\overline{T}_t=(1+\tfrac{\theta}{t})S_t/w\) and, in addition, we have \(\lambda_t=\alpha/(t+1)^{1-\beta}=\alpha/t^{1-\beta}+(1-\beta)O(1/t^{2-\beta})\), so that we obtain \[\begin{align} &t^{w/2}(t^{1-w}\overline{T}_t+\tfrac{\alpha}{(w-\beta)}t^{-(w-\beta)}-\tfrac{1}{w}Z^*_\infty)=\\ &t^{w/2}(t^{1-w}\overline{T}_t+\tfrac{\alpha\beta}{w(w-\beta)}t^{-(w-\beta)}+\tfrac{t^{1-w}}{w}\tfrac{\alpha}{t^{1-\beta}}-\tfrac{1}{w}Z^*_\infty) =\\ &t^{w/2}(t^{1-w}\overline{T}_t+\tfrac{\alpha\beta}{w(w-\beta)}t^{-(w-\beta)}+\tfrac{t^{1-w}}{w}\lambda_t-\tfrac{1}{w}Z^*_\infty\Big)+O(1/t^{1+w/2-\beta})=\\ &\tfrac{t^{w/2}}{w}(t^{1-w}Z_t+\tfrac{\alpha\beta}{(w-\beta)}t^{\beta-w}-Z^*_\infty)+ \tfrac{\theta}{w} \tfrac{S_t}{t^{w/2}}+O(\tfrac{1}{t^{1+w/2-\beta}})\stackrel{stably}\longrightarrow {\mathcal{N}}(0, \Sigma_\infty/w) \end{align}\] and this convergence also holds true in the sense of the a.s. conditional convergence with respect to \(\mathcal{F}\). Consequently, when \((w-\beta)<w/2\), the quantity \[\begin{align} t^{w-\beta}(t^{1-w}\overline{T}_t-\tfrac{1}{w}Z^*_\infty)+\tfrac{\alpha}{(w-\beta)}= \tfrac{t^{w-\beta}}{t^{w/2}} t^{w/2} (t^{1-w}\overline{T}_t+\tfrac{\alpha}{(w-\beta)}t^{-(w-\beta)}-\tfrac{1}{w}Z^*_\infty) \end{align}\] converges in probability toward zero.
Now, remember that \(\overline{P}_t=S_t/D_t=(Z_t-\lambda_t)/D_t\) so that, using the notation and the CLT for \((D_t)_t\) provided in Remark 1, we can write \[\begin{align} \overline{P}_t&= \tfrac{1}{\lambda(\beta)a_t(\beta)} S_t ( 1 - \tfrac{D_t-\lambda(\beta)a_t(\beta)}{D_t} ) \\ &=\tfrac{1}{\lambda(\beta)a_t(\beta)} S_t \Big[1-\tfrac{1}{\tfrac{D_t}{a_t(\beta)\lambda(\beta)}}\tfrac{1}{\sqrt{a_t(\beta)}} \cdot\tfrac{\sqrt{a_t(\beta)}}{\lambda(\beta)}\big(\tfrac{D_t}{a_t(\beta)}-\lambda(\beta)\big)\Big]\\ &=\tfrac{1}{\lambda(\beta)a_t(\beta)} S_t [1+\tfrac{1}{\sqrt{a_t(\beta)}} W_t]\,, \end{align}\] where \(W_t\stackrel{stably}\longrightarrow {\mathcal{N}}(0,1/\lambda(\beta))\). Hence, since \(\lambda_t=\alpha/t^{1-\beta}+(1-\beta)O(1/t^{2-\beta})\), we get \[\begin{align} &t^{1-w}\lambda(\beta)a_t(\beta)\overline{P}_t+\tfrac{\alpha w}{w-\beta}t^{-(w-\beta)}-Z^*_\infty =\\ &t^{1-w}\lambda(\beta)a_t(\beta)\overline{P}_t + \alpha t^{-(w-\beta)} +\tfrac{\alpha\beta}{w-\beta}t^{-(w-\beta)} -Z^*_\infty =\\ &\left(t^{1-w}Z_t+\tfrac{\alpha\beta}{w-\beta}t^{-(w-\beta)}-Z_\infty^*\right)+O(\tfrac{1}{t^{1+w-\beta}})+\tfrac{t^{1-w}S_t}{\sqrt{a_t(\beta)}}W_t\,. \end{align}\] Hence, taking into account ?? , since \(\sqrt{a_t(\beta)}=o(t^{w/2})\), we obtain ?? . As a consequence, when \((w-\beta)<\beta/2\), we have \(\sqrt{a_t(\beta)}=t^{\beta/2}\) and \[\begin{align} &t^{w-\beta}\left(t^{1-w}\lambda(\beta)a_t(\beta)\overline{P}_t-Z^*_\infty\right) +\tfrac{\alpha w}{w-\beta}=\\ &\tfrac{t^{w-\beta}}{\sqrt{a_t(\beta)}} \sqrt{a_t(\beta)}\Bigl(t^{1-w}\lambda(\beta)a_t(\beta)\overline{P}_t+\tfrac{\alpha w}{w-\beta}t^{-(w-\beta)}-Z^*_\infty\Bigr) \stackrel{P}\longrightarrow 0\,. \end{align}\] Finally, recalling that \(\overline{K}_t=(\theta+t)\overline{P}_t/w\) (see 5 ), we have \[\begin{gather} w\lambda(\beta)\tfrac{a_t(\beta)}{t^w}\overline{K}_t+\tfrac{\alpha w}{w-\beta}t^{-(w-\beta)}-Z^*_\infty \\ = (t^{1-w}\lambda(\beta)a_t(\beta)\overline{P}_t+\tfrac{\alpha w}{w-\beta}t^{-(w-\beta)}-Z^*_\infty )+\theta\lambda(\beta)\tfrac{a_t(\beta)}{t^w}\overline{P}_t \end{gather}\] so that, from ?? , we obtain ?? and, as a consequence, when \((w-\beta)<\beta/2\), we have (with the same arguments used above) \[t^{w-\beta}\left( \lambda(\beta)\tfrac{a_t(\beta)}{t^w}\overline{K}_t-\tfrac{Z^*_\infty}{w} \right)+\tfrac{\alpha}{w-\beta} \stackrel{P}\longrightarrow 0\,. \qedhere\] ◻

Proof. We split the proof in some steps.
First notice that, for any \(z>0\), \[\begin{align} E\bigl[e^{z\,T_{t+1}}\mid\mathcal{F}_t\bigr]& = E\bigl[e^{z\,\sum_{j\in\mathcal{O}_t}X_{t+1,j}+z\,N_{t+1}}\mid\mathcal{F}_t\bigr] = E\bigl[e^{z\,N_{t+1}}\mid\mathcal{F}_t\bigr] \prod_{j\in\mathcal{O}_t}E\bigl[e^{z\,X_{t+1,j}}\mid\mathcal{F}_t\bigr]\\ &=e^{(e^z-1)\,\lambda_t}\prod_{j\in\mathcal{O}_t} \bigl( e^z P_{t,j} + 1-P_{t,j} \bigr)\,. \end{align}\] Then, denoting \(g(z)=e^z-1\), since \(g(z)\geq 0\) and \(1+x\leq e^x\) for \(x\geq 0\) and using 4 we have \[\label{eq-proof-1} \begin{align} E\bigl[e^{z\,T_{t+1}}\mid\mathcal{F}_t\bigr] &=e^{g(z)\,\lambda_t}\,\prod_{j\in {\mathcal{O}}_t}\bigl(1+g(z)\,P_{t,j}\bigr) \leq\exp{\Bigl(g(z)\,\lambda_t+g(z)\sum_{j\in {\mathcal{O}}_t}P_{t,j}\Bigr)} \\ &= \exp \Bigl( g(z)\,\lambda_t + g(z)\, \tfrac{w\sum_{n=1}^t T_n}{\theta+t} \Bigr) = \exp \Bigl( g(z)\,\lambda_t + g(z)\,w\,\tfrac{t}{\theta+t}\overline{T}_t \Bigr) \quad\text{a.s.}\,. \end{align}\tag{37}\] Define \(\overline{T}_t^*=r_t\overline{T}_t\) and note that \[\label{eq:u95v} \begin{align} \overline{T}_{t+1}^*&= r_{t+1}\tfrac{\sum_{n=1}^t T_n + T_{t+1}}{t+1} = \tfrac{r_{t+1}}{r_t}\tfrac{t}{t+1}\overline{T}_t^* + T_{t+1}\tfrac{r_{t+1}}{t+1} = (1-v_t)\,\overline{T}_t^*+ u_t\, T_{t+1}\,, \end{align}\tag{38}\] where by 20 and 22 \[u_t=\tfrac{r_{t+1}}{t+1}\rightarrow 0 \qquad\text{and}\qquad v_t=1-\tfrac{u_{t}}{u_{t-1}}\geq 0\,.\] Hence, by 37 and 38 , we get \[\label{eq-proof-2} \begin{align} E\bigl[& e^{z\,\overline{T}^*_{t+1}}\bigr] = E\Bigl[\,\exp{\Bigl(z\,\overline{T}^*_t\,(1-v_t)\Bigr)} \,E\bigl[e^{z\,u_t\,T_{t+1}}\,\mid\,\mathcal{F}_t\bigr]\,\Bigr] \\ & \leq \exp{\Bigl(\lambda_t\,g(z\,u_t)\Bigr)} E\Bigl[\,\exp{\Bigl( g(z\,u_t)w\overline{T}_t+z\,(1-v_t)\,\overline{T}^*_t\Bigr) } \Bigr] \\ &=\exp{\Bigl(\lambda_t\,g(z\,u_t)\Bigr)} E\Bigl[\,\exp{\Bigl( g(z\,u_t)\tfrac{w}{r_t}\overline{T}^*_t+z\,(1-v_t)\,\overline{T}^*_t\Bigr) } \Bigr]\,. \end{align}\tag{39}\] Since \(v_t\geq 0\) and \(w\leq 1\), it follows \[\begin{gather} E\bigl[ e^{z\,\overline{T}^*_{t+1}}\bigr] \leq \exp{\Bigl(\lambda_t\,g(zu_t)\Bigr)}\,\, E\Bigl[\exp{\Bigl(\Big(z+\tfrac{g(zu_t)}{r_t}\Big)\,\overline{T}^*_t\Bigr)}\Bigr]. \end{gather}\] Iterating this last inequality, we obtain \[\begin{gather} E\bigl[ e^{z\,\overline{T}^*_{t+1}}\bigr]\leq a_t(z)\,E\bigl[e^{b_t(z)\, \overline{T}^*_1}\bigr] \end{gather}\] for suitable deterministic quantities \(a_t(z)\) and \(b_t(z)\). Since we have \(\overline{T}^*_1=r_1T_1\) with \(T_1\stackrel{d}=\,\)Poi\((\alpha)\), then \(E\bigl[e^{b_t(z)r_1\,T_1}\bigr]<+\infty\) so that \(E\bigl[e^{z\,\overline{T}^*_t}\bigr]<+\infty\) for all fixed \(t\geq 1\) and \(z>0\). This also implies \(E[e^{z'\,\overline{T}_t}]<+\infty\) for all fixed \(t\geq 1\) and \(z'>0\) (take in the above relation \(z=z'/r_t\)), which by 37 also entails \(E[e^{zT_t}]<+\infty\) for all fixed \(t\geq 1\) and \(z>0\).
Step 3): Observe now that \(g(x)\leq 2\,x\) and \(g(x)-x\leq x^2\) for \(x\in [0,\,1/2]\). Moreover, since \(u_t\to0\), we know \(z u_t\leq u_t\leq 1/2\) for \(z\in (0,1]\) and \(t\) large enough. These facts are used for \(z\in(0,1]\) to write \(g(z u_t)\leq 2 z u_t\) and \(g(z u_t)\leq (z u_t)+(z u_t)^2\) when \(t\) is large enough. Hence, by 39 , we get for \(z\in (0,1]\) and \(t\) large enough: \[\begin{align} E\bigl[e^{z\,\overline{T}^*_{t+1}}\bigr] &\leq \exp{\Bigl(2z\,\lambda_t\,u_t\Bigr)}\, E\Bigl[\exp \Bigl( (z\,u_t)^2 \tfrac{w}{r_t}\overline{T}^*_t +\Big(1-v_t+u_t\tfrac{w}{r_t}\Big)z\overline{T}^*_t \Bigr) \Bigr]\\ &\leq \exp{\Bigl(2z\,\lambda_t\,u_t\Bigr)}\, E\Bigl[\Big(1-v_t+(u_t+u_t^2)\tfrac{w}{r_t}\Big)z\overline{T}^*_t \Bigr) \Bigr],\\ \end{align}\] where we have used that \(z^2\leq z\) as \(z\in (0,1]\). Recalling that \(v_t=1-u_{t}/u_{t-1}\) and \(u_{t-1}=r_{t}/t\), we have \[\begin{align} \pi_t&= 1-v_t+(u_t+u_t^2)\tfrac{w}{r_t} = \tfrac{u_{t}}{u_{t-1}} + (1+u_t)\tfrac{u_{t}}{u_{t-1}}\tfrac{w}{t} \\ &= \tfrac{u_{t}}{u_{t-1}}\Big(1+ \tfrac{w}{t} +\tfrac{w}{t} u_t \Big) = \tfrac{u_{t}}{u_{t-1}}\Big(1+ \tfrac{w}{t}\Big)\Big(1 +\tfrac{w/t}{1+w/t} u_t \Big)\\ &= \tfrac{u_{t}}{u_{t-1}}\tfrac{\widetilde{\zeta}_t(w)}{\widetilde{\zeta}_{t-1}(w)}\Big(1 +\tfrac{w}{t+w} u_t \Big)\,, \end{align}\] where \(\widetilde{\zeta}_t(x)\) is defined as \(\zeta_t(x)\) in 17 with \(\theta=0\). This leads for \(z\in (0,1]\) to \[\label{eq:sum95prod950} E\bigl[e^{z\,\overline{T}^*_{t+1}}\bigr]\leq \exp{\Bigl(2z\,\lambda_t\,u_t\Bigr)}\, E\Bigl[\exp \Bigl( z\,\pi_t\,\overline{T}^*_t \Bigr) \Bigr].\tag{40}\] Now, note that for any \(n<t\) \[\label{eq:prod95telescopic} \prod_{j=n+1}^t \pi_j = \prod_{j=n+1}^t \tfrac{u_{j}}{u_{j-1}}\tfrac{\widetilde{\zeta}_j(w)}{\widetilde{\zeta}_{j-1}(w)}\big(1 +\tfrac{w}{j+w} u_j \big)= \tfrac{u_{t}}{u_{n}}\tfrac{\widetilde{\zeta}_t(w)}{\widetilde{\zeta}_{n}(w)}\prod_{j=n+1}^t\Big(1 +\tfrac{w}{j+w} u_j \Big)\tag{41}\] and since

• \(\limsup_{t\rightarrow\infty}u_t\widetilde{\zeta}_t(w)\leq 1\) and

• \(C=\prod_{t\geq 1} \Big(1 +\tfrac{w}{t+w} u_t \Big)<+\infty\) as \(\sum_{t\geq 1} \tfrac{w}{t+w} u_t<+\infty\)

we can choose \(z_0\in(0,1)\) and \(t_0\) large enough such that \[\sup_{t\geq t_0}\Big(\prod_{j=t_0}^t \pi_j\Big)z_0\leq 1\,.\] This ensures that we can iterate the inequality 40 , so obtaining \[\begin{align} \label{eq:sum95prod951}E\bigl[e^{ z_0\,\overline{T}^*_{t+1}}\bigr]& \leq \exp{\Bigl( 2z_0\,\sum_{n= t_0}^t \lambda_n\, u_n \prod_{j=n+1}^t \pi_j\Bigr)} E\bigl[e^{ (\prod_{j=t_0}^{t} \pi_j)z_0\,\overline{T}_{t_0}^*}\bigr] \leq \exp{\Bigl( 2z_0\,\sum_{n= t_0}^t \lambda_n\, u_n \prod_{j=n+1}^t \pi_j\Bigr)} E\bigl[e^{\overline{T}_{t_0}^*}\bigr]\end{align}\tag{42}\] for each \(t\geq t_0\). Then, the second term in the right-hand product in 42 is always bounded by Step 2) and so we can now focus on the first term, which by 41 becomes less or equal than \[\exp{\Bigl(2z_0\,C\,u_t \widetilde{\zeta}_t(w) \sum_{n= t_0}^t \tfrac{\lambda_n}{\widetilde{\zeta}_{n}(w)}\Bigr)}\sim \exp{\Bigl( 2z_0\,C\,\tfrac{r_t}{t^{1-w}}\sum_{n= t_0}^t \tfrac{\alpha}{n^{1+w-\beta}}\Bigr)}\,,\] that is finite in all cases as

• in the case \(w>\beta\) we have \(\tfrac{r_t}{t^{1-w}}\sim 1\) and \(\sum_{n= t_0}^t \tfrac{\alpha}{n^{1+w-\beta}}<+\infty\);

• in the case \(w=\beta\) we have \(\tfrac{r_t}{t^{1-w}}\sim 1/\ln(t)\) and \(\sum_{n= t_0}^t \tfrac{\alpha}{n^{1+w-\beta}}=O(\ln(t))\);

• in the case \(w<\beta\) we have \(\tfrac{r_t}{t^{1-w}}\sim 1/t^{\beta-w}\) and \(\sum_{n= t_0}^t \tfrac{\alpha}{n^{1+w-\beta}}=O( t^{\beta-w})\).

Therefore from 42 we can conclude that, for any \(z\leq z_0\), \(\sup_{t\geq 1} E[e^{z\,\overline{T}_t^*}]\leq \sup_{t\geq 1} E[e^{z_0\,\overline{T}_t^*}]<+\infty\).
Finally, since by 4 we have \(r_t Z_t=w\tfrac{t}{t+\theta}\overline{T}_t^*+r_t\lambda_t\leq \overline{T}_t^*+c\) for some suitable \(c>0\), we also obtain \(\sup_{t\geq 1} E[e^{z\,r_t Z_t}]<+\infty\) for any \(z\leq z_0\).
The \(L^p\)-boundness of the sequences \((r_t\,Z_t)_t\) and \((r_t\,\overline{T}_t)_t\) follows immediately, as for non-negative random variables we have \[\begin{gather} \sup_t E( e^{z_0 r_t Z_t}) <\infty \iff \sup_t \sum_{p\geq 0} \tfrac{z_0^p E[\,(r_t Z_t)^p\,]}{p!} < +\infty \Longrightarrow \sup_t E[\,(r_t Z_t)^p\,] < +\infty\; \forall p\geq 1. \end{gather}\] and the same for \((r_t\,\overline{T}_t)_t\). For what concerns \(({T}_t)_t\), from 37 , 4 , 7 and the fact that \(g(z)\leq 2z\) when \(z\leq 1/2\), we obtain for any \(t\geq 0\) and \(z\leq 1/2\) \[\label{eq-lemma-new} E\bigl[e^{z\,T_{t+1}}\mid\mathcal{F}_t\bigr]\leq \exp ( g(z)\, (\lambda_t + S_t)) = \exp ( g(z)\, Z_t) \leq \exp ( 2z\, Z_t)\,.\tag{43}\] Now, note that \(r_t \equiv 1\) when \(w<\beta=1\) or \(\beta<w=1\), and in this case the two expressions in ?? coincide. More generally, we can consider the following cases.
Case \(w=1\) or \(\beta=1\): we have \(r_t \leq 1\) and so, by 43 , for each \(t\geq 0\), we have \[\begin{align} E[ e^{\tfrac{z_0}{2} r_t\,T_{t+1}}] &\leq E[e^{z_0\, r_t Z_t}]\,. \end{align}\] Then, by ?? , we get \(\sup_t E[e^{\tfrac{z_0}{2} r_t\,T_{t+1}}] <+\infty\), that implies as above \(\sup_t E[\,(r_tT_{t+1})^p\,] < +\infty\) for all \(p\geq 1\).
Case \(w<1\) or \(\beta<1\): we have \(r_t \geq 1\) for \(t\) large enough, say \(t\geq \bar{t}\). Hence, by 43 , for \(t\geq \bar{t}\), we can write
\[\begin{align} r_t(E[e^{\tfrac{z_0}{2}\,T_{t+1}}]-1)&\leq r_t ( E[e^{z_0\, Z_t}] -1) =r_t ( E[e^{z_0\, \tfrac{ Z_t^*}{r_t}}] -1) \\ &= \sum_{p\geq 1} \tfrac{z_0^p E[(Z^*_t)^p]}{(r_t)^{p-1}\,p!} \leq \sum_{p\geq 1} \tfrac{z_0^pE[(Z^*_t)^p]}{p!} = (E[e^{z_0\, Z^*_t}] -1)\,. \end{align}\] Hence, we can conclude that \(\sup_{t\geq \bar{t}} \,r_t(E[e^{\tfrac{z_0}{2}\,T_{t+1}}]-1) <+\infty\), that reads \[\sup_{t\geq \bar{t}}\, \sum_{p\geq 1} \tfrac{(z_0)^p E[r_t(T_{t+1})^p]}{2^p \, p!} < +\infty\,,\] and so we can get \(\sup_{t\geq\bar{t}} \, E[\,r_t(T_{t+1})^p\,] < +\infty\), for all \(p\geq 1\). Since, by Step 2), for all \(p\geq 1\) and each fixed \(t\) we have \(E[(T_{t+1})^p]<+\infty\), and so \(E[r_t (T_{t+1})^p]<+\infty\), we can also affirm that \(\sup_{t} \, E[\,r_t(T_{t+1})^p\,] < +\infty\) for all \(p\geq 1\). ◻

Proof. From 6 , since \(D_t \overline{P}_t^2=\overline{P}_t S_t\) we get \[\begin{align} R_t&=\sum_{j\in {\mathcal{O}}_t} P_{t,j}^2 =\sum_{j\in {\mathcal{O}}_t} \big(w(1-\iota)\tfrac{K_{t,j}}{(\theta+t)}+\iota\overline{P}_t\big)^2 =(w(1-\iota))^2 \widetilde{R}_t+ \iota^2 D_t \overline{P}_t^2 + 2 \iota(1-\iota) \overline{P}_t S_t\\ &= (w(1-\iota))^2 \widetilde{R}_t+ \iota(2-\iota)\overline{P}_t S_t \,, \end{align}\] where, by 4 , \[0 \leq \widetilde{R}_t=\tfrac{\sum_{j\in {\mathcal{O}}_t}K_{t,j}^2}{(\theta+t)^2}\leq \left(\tfrac{\sum_{j\in {\mathcal{O}}_t}K_{t,j}}{\theta+t}\right)^2= S_t^2/w^2\,.\] Since \(\overline{P}_t\to 0\) a.s.and \(r_t S_t = O(1)\) from 4 and Table 1, then \(r_t R_t \to 0\) a.s.when \(\iota=1\), otherwise if and only if \(r_t \widetilde{R}_{t} \to 0\) a.s.and this is always the case when \(\max\{w,\beta\}<1\), because \(r_t \widetilde{R}_{t} \leq r_t S_t^2/w^2\to 0\) a.s..

From now on, we assume \(\iota<1\) and \(\max\{w,\beta\}=1\) (and so \(r_t=1\) when \(\beta\neq w\) and \(r_t\to 0\) when \(\beta=w\)) and we are going to prove that \(r_t \widetilde{R}_{t}\to 0\) a.s.unless \(\iota=0\), \(w=1\) and \(\beta<1\). We observe that \[\begin{align} \widetilde{R}_{t+1}= \sum_{j\in {\mathcal{O}}_t}\big(\tfrac{K_{t+1,j}}{\theta+t+1}\big)^2+ \sum_{j\in {\mathcal{O}}_{t+1}\setminus{\mathcal{O}}_t} \big(\tfrac{1}{\theta+t}\big)^2 =\sum_{j\in {\mathcal{O}}_t}\big(\tfrac{K_{t,j}+\Delta K_{t+1,j}}{\theta+t+1}\big)^2 + \tfrac{N_{t+1}}{(\theta+t+1)^2} \end{align}\] and so, since \((\Delta K_{t+1,j})^2=\Delta K_{t+1,j}\), we have \[\label{eq:R42pred} \begin{align} E[\widetilde{R}_{t+1}\mid{\mathcal{F}}_t] &=\sum_{j\in {\mathcal{O}}_t} \tfrac{K_{t,j}^2+P_{t,j}+2K_{t,j}P_{t,j}}{(\theta+t+1)^2} + \tfrac{\lambda_t}{(\theta+t+1)^2} \,, \end{align}\tag{44}\] while, the martingale difference reads \[\widetilde{R}_{t+1}-E[\widetilde{R}_{t+1}\mid{\mathcal{F}}_t] = \tfrac{1}{\theta+t+1}\Big( \sum_{j\in {\mathcal{O}}_t} \big(\Delta K_{t+1,j}-P_{t,j}\big)\tfrac{1+2K_{t,j}}{\theta+t+1} +\tfrac{N_{t+1}-\lambda_t}{(\theta+t+1)}\Big) = \tfrac{\Delta M^{(\widetilde{R})}_{t+1}}{\theta+t+1}\,.\] Since the random variables \((\Delta K_{t+1,j})_{j\in\mathcal{O}_t}\) are, conditionally on \(\mathcal{F}_t\), independent among each other, and \(N_{t+1}\) is independent of them and \({\mathcal{F}}_t\) (see the description of the model dynamics), we have \[\label{eq:R42martComp} \begin{align} E[(\Delta M^{(\widetilde{R})}_{t+1})^2\mid{\mathcal{F}}_t] &=\sum_{j\in {\mathcal{O}}_t}P_{t,j}(1-P_{t,j})\Big(\tfrac{1+2K_{t,j}}{\theta+t+1}\Big)^2 +\tfrac{\lambda_t}{(\theta+t+1)^2}\,. \end{align}\tag{45}\] Now, we go on with the computations for \(E[\widetilde{R}_{t+1}\mid {\mathcal{F}}_t]\). From 44 , recalling that \(\widetilde{R}_t=\sum_{j\in{\mathcal{O}}_t} K_{t,j}^2/(\theta+t)^2\) and using 4 and 6 , we get \[\begin{align} E[\widetilde{R}_{t+1}\mid{\mathcal{F}}_t] &=\widetilde{R}_t+ \sum_{j\in {\mathcal{O}}_t}K_{t,j}^2\left(\tfrac{1}{(\theta+t+1)^2}-\tfrac{1}{(\theta+t)^2}\right)+\tfrac{S_t}{(\theta+t+1)^2} + \tfrac{2}{(\theta+t+1)^2}\sum_{j\in{\mathcal{O}}_t} K_{t,j}\left((1-\iota)w\tfrac{K_{t,j}}{\theta+t}+\iota \overline{P}_t\right) +\tfrac{\lambda_t}{(\theta+t+1)^2} \\ &=\widetilde{R}_t+ \sum_{j\in {\mathcal{O}}_t}K_{t,j}^2\left(\tfrac{1}{(\theta+t+1)^2}-\tfrac{1}{(\theta+t)^2} + \tfrac{2(1-\iota)w}{(\theta+t)(\theta+t+1)^2}\right) +\tfrac{S_t}{(\theta+t+1)^2}+ \tfrac{2\iota}{(\theta+t+1)^2}\overline{P}_t\sum_{j\in{\mathcal{O}}_t} K_{t,j} +\tfrac{\lambda_t}{(\theta+t+1)^2} \\ &=\widetilde{R}_t-\tfrac{2(\iota+1-w)}{\theta+t+1}\widetilde{R}_t + \tfrac{2\iota}{(\theta+t+1)^2}\overline{P}_t\sum_{j\in{\mathcal{O}}_t} K_{t,j} + \rho_t \,. \end{align}\] with \(\rho_t\) being \({\mathcal{F}}_t\)-measurable and such that \(\sum_t|\rho_t|<+\infty\) since \(\rho_t=O((\widetilde{R}_t+S_t+1)/t^2)=O((S_t+1)/t^2)\). Using 5 , we finally obtain \[\begin{align} E[\widetilde{R}_{t+1}\mid{\mathcal{F}}_t] &= \widetilde{R}_t-\tfrac{2(\iota+1-w)}{\theta+t+1}R_t^*+ \tfrac{2\iota (\theta+t)D_t}{(\theta+t+1)^2{w}}\overline{P}_t^2 +\rho_t \end{align}\] and hence \[\label{eq:R4295in95proof} \widetilde{R}_{t+1} = \widetilde{R}_t + \tfrac{1}{\theta+t+1}\Big( \tfrac{2\iota (\theta+t)D_t\overline{P}_t^2}{(\theta+t+1)w} -2(\iota+1-w) \widetilde{R}_t\Big) + \tfrac{\Delta M^{(\widetilde{R})}_{t+1}}{\theta+t+1} +\rho_t \,.\tag{46}\] Now, we consider two cases.
Case \(\iota > 0\) or \(w<1\) (and \(\beta=1\)): in this case we can apply Theorem 11 to 46 with \(X_t=\widetilde{R}_t\), \(A_t= \tfrac{2\iota(\theta+t) D_t\overline{P}^2_t}{(\theta+t+1)w}\), \(b=2(\iota+1-w)>0\), \(\eta_t=(\theta+t+1)^{-1}\) and \(\Delta M_{t+1}=\Delta M^{(\widetilde{R})}_{t+1}\) (and in the present case the reminder term \(\rho_{t+1}\) in Theorem 11 coincides here with \(\rho_t\) and so it is \({\mathcal{F}}_t\)-measurable). Indeed, by 45 , we have \[\begin{align} \sum_t \eta_t^2 E[(\Delta M_{t+1})^2\mid{\mathcal{F}}_t] & = \sum_t \tfrac{1}{(\theta+t+1)^2}\bigg( \sum_{j\in {\mathcal{O}}_t}P_{t,j}(1-P_{t,j})\Big(\tfrac{1+2K_{t,j}}{\theta+t+1}\Big)^2+\tfrac{\lambda_t}{(\theta+t+1)^2}\bigg) \\ & \leq \sum_t \tfrac{1}{(\theta+t+1)^2} \bigg(\sum_{j\in {\mathcal{O}}_t} P_{t,j} \Big(\tfrac{1+2t}{\theta+t+1}\Big)^2 +\tfrac{\lambda_t}{(\theta+t+1)^2}\bigg) \\ & \leq \sum_t \Big(\tfrac{4S_{t}}{(\theta+t+1)^2} + \tfrac{\lambda_t}{(\theta+t+1)^4} \Big)<+\infty\,. \end{align}\] In addition, we have \[A_t = \tfrac{2\iota}{w}\tfrac{(\theta+t)}{(\theta+t+1)}D_t\overline{P}^2_t \stackrel{a.s.}\longrightarrow 0\,.\] Hence, by Theorem 11, we obtain \(\widetilde{R}_t\to 0\) a.s.and, since in this case \(r_t=1\), this means \(r_t\widetilde{R}_t\to 0\) a.s..
Case \(\iota = 0\) and \(w=1\) (and \(\beta<1\) or \(=1\)): in this case \(R_t=\widetilde{R}_t\) and \(\widetilde{R}_t\) is a non-negative almost super-martingale converging a.s. to a non-negative real random variable \(\widetilde{R}_\infty\) (see Theorem 22). Then, when \(w=\beta=1\), since \(r_t\to 0\) in this case, we trivially get \(r_tR_t=r_t \widetilde{R}_t\to 0\) a.s..
Let us focus on the remaining sub-case, that is when \(\iota=0\), \(w=1\) and \(\beta<1\) (and so \(r_t=1\)). First, since \(P_{t,j}^2\leq P_{t,j}\), we have \(\widetilde{R}_t=R_t\leq S_t=(Z_t-\lambda_t)\to Z_\infty^*\) and so we obtain \(0\leq\widetilde{R}_\infty\leq Z^*_\infty\). Now, let \(\tau_1=\inf\{t\colon N_t>0\}\) and note that in this case 1 reads \[P_{\tau_1+n}(1) = \tfrac{K_{\tau_1+n}}{\tau_1+n+\theta}\,,\] that evolves, conditionally on the value taken by \(\tau_1\) (note that \(\tau_1<+\infty\) a.s.), as the proportion of black balls in a Pòlya urn that starts with \(1\) black ball and \((\tau_1+\theta)\) white balls. Then, conditionally on the value taken by \(\tau_1\), we have \(P_{\tau_1+n}(1)\stackrel{a.s.}\to P_\infty(1) \stackrel{d}= Beta(1,\tau_1+\theta)\), so that we have \(P_\infty(1)(1-P_\infty(1))>0\). Note that, since \(\beta<w=1\) and \(\iota=0\), we have \[\widetilde{R}_\infty = P_\infty(1)^2 + \text{a.s.-}\lim_t \sum_{j>1,j\in{\mathcal{O}}_t} P_{t,j}^2\] and \[Z^*_\infty = S^*_\infty= P_\infty(1) + \text{a.s.-}\lim_t \sum_{j>1,j\in{\mathcal{O}}_t} P_{t,j}\] and hence \[0 <P_\infty(1)^2\leq \widetilde{R}_\infty < \widetilde{R}_\infty + P_\infty(1)(1-P_\infty(1))\leq Z^*_\infty.\] Hence, we can conclude because in this last sub-case \(r_t=1\) and so \(r_t\widetilde{R}_t=\widetilde{R}_t\). ◻

Proof. First, recall from Section 3.1 that \(D_t=\sum_{n=1}^t N_{n}\stackrel{d}=\)Poi\((\Lambda_t)\) with \(\Lambda_t=\sum_{n=1}^t \lambda_{n-1}\sim (\alpha/\beta) t^\beta\) when \(\beta>0\). Then, consider the sequence \((r_t)\) as defined in ?? with \(\delta=w\) e \(\rho=\beta\). Since \(E[\Delta K_{t+1,j}]=w(1-\iota) E[B_{t,j}]+ \iota E[\overline{P}_t]\), the proof will be divided in the proof of the following two facts:

• \(E[t^{1-w+w\iota}\overline{P}_t]=O(t^{-(\min\{\beta,w\}-w\iota)}\ln(t))\to 0\),

• \(\sup_{t\geq1} t^{1-w+w\iota} E[B_{t,j}] < +\infty\).

Proof of (1): Consider \[E[t^{1-w+w\iota}\overline{P}_t] = \tfrac{t^{1-w}}{r_t} E\Big[\tfrac{r_t S_t}{t^{-w\iota}D_t}\Big]\,.\] Since \(D_t\stackrel{d}=\)Poi\((\Lambda_t)\), by Chernoff’s bound (see [39]), we have \[P\big(D_t < (1-\epsilon)\Lambda_t\big) \leq e^{-\tfrac{\epsilon^2}{2}\Lambda_t}\] and so, using also Holder’s inequality, we get \[\begin{align} E\Big[\tfrac{r_t S_t}{t^{-w\iota}D_t}\Big] &= E\Big[\tfrac{r_tS_t}{t^{-w\iota}D_t}\mathbb{1}_{\{D_t \geq (1-\epsilon)\Lambda_t\}}\Big] + E\Big[\tfrac{r_tS_t}{t^{-w\iota}D_t}\mathbb{1}_{\{D_t < (1-\epsilon)\Lambda_t\}}\Big]\\ &\leq E\Big[\tfrac{r_tS_t}{t^{-w\iota}(1-\epsilon)\Lambda_t}\Big] + E[(r_tS_t)^{2}]^{\frac{1}{2}}\,t^{w\iota} P\big(D_t < (1-\epsilon)\Lambda_t\big)^{\frac{1}{2}}\\ &\leq E[r_tS_t] \tfrac{t^{w\iota}}{(1-\epsilon)\Lambda_t} + E[(r_tS_t)^{2}]^{\frac{1}{2}} \, t^{w\iota}e^{-\tfrac{\epsilon^2}{4}\Lambda_t}\,. \end{align}\] Now, we observe that, by Lemma 16 and the fact that \(0\leq S_t=Z_t-\lambda_t\leq Z_t\), we have \(\sup_{t} E[(r_tS_t)^p]\leq \sup_{t} E[(r_tZ_t)^p] <+\infty\) for \(p=1,2\) and so the above inequality implies that \[E[t^{1-w+w\iota}\overline{P}_t]=\tfrac{t^{1-w}}{r_t}O\big(\tfrac{t^{w\iota}}{\Lambda_t}\big)= O\big(t^{-(\min\{\beta,w\}-w\iota)}\ln(t)\big)\rightarrow 0\,.\] Proof of (2): Setting \(m_{t}=E[B_{t}]\), by 12 we have \[\begin{align} m_{t+1}= \big(1-\tfrac{1}{\theta+t+1}\big) m_t +\tfrac{1}{\theta+t+1} E[P_{t,j}] =\Big(1-\tfrac{1-w(1-\iota)}{\theta+t+1}\Big) m_t +\tfrac{\iota}{\theta+t+1} E[\overline{P_t}]\,. \end{align}\] Then, setting \(m^{*}_t=\zeta_t(1-w+w\iota)m_t\) and using 20 , we get \[\begin{align} m^{*}_{t+1}&=\Big(1-\tfrac{(1-w(1-\iota))^2}{(\theta+t+1)^2}\Big) m^{*}_t +\tfrac{\iota}{\theta+t+1}E[t^{1-w+w\iota}\overline{P}_t]\\ &\leq m^{*}_t+\tfrac{\iota}{\theta+t+1}E[t^{1-w+w\iota}\overline{P}_t]\,, \end{align}\] which implies that \((m^*_t)_t\) is bounded because, by fact \((1)\), we have \[\sum_tE[t^{1-w+w\iota}\overline{P}_t]/(\theta+t+1)=\sum_t O(1/t^{1+\min\{\beta,w\}-w\iota}) < +\infty\,. \qedhere\] ◻

Proof. Recall that \(D_t - \Lambda_t = \sum_{n=1}^t (N_n -\lambda_{n-1})\), where \((N_n -\lambda_{n-1})_n\) is a sequence of zero-mean independent random variables and \(Var[D_t] = \sum_{n=1}^t Var[N_n] = \sum_{n=1}^t\lambda_{n-1} = \Lambda_t\). Set \(c_n = \sqrt{\tfrac{\Lambda_n}{{\ln (\Lambda_n)}}}\). For \(n\) large enough, we have \(\lambda_{n-1} < c_n\) and consequently \(P(N_n-\lambda_{n-1} \leq - c_n) = P(N_n\leq \lambda_{n-1}-c_n)=0\). Then, by the bounds for the tail probabilities of a Poisson random variable (e.g. [40]), we obtain for \(n\) large enough \[\begin{align} P(|N_n-\lambda_{n-1}|\geq c_n)&= P(N_n \geq \lambda_{n-1} + c_n) \leq e^{-\lambda_{n-1}} \tfrac{ (e\lambda_{n-1})^{\lambda_{n-1}+c_n} }{(\lambda_{n-1}+c_n)^{\lambda_{n-1}+c_n}} \\ &= \big(\tfrac{e\lambda_{n-1}}{c_n}\big)^{c_n} \big(\tfrac{\lambda_{n-1}}{c_n}\big)^{\lambda_{n-1}} \tfrac{ 1 }{(\tfrac{\lambda_{n-1}}{c_n}+1)^{c_n}(\tfrac{\lambda_{n-1}}{c_n}+1)^{\lambda_{n-1}}} \\ &\leq C \big(\tfrac{e\lambda_{n-1}}{c_n}\big)^{c_n}\,, \end{align}\] where the last inequality is due to the fact that \(\lambda_{n-1} = o(c_n)\). Moreover, we have \(\sum_n \big(\tfrac{e\lambda_{n-1}}{c_n}\big)^{c_n} < +\infty\). In fact, for \(\beta=0\), we have \(\lambda_{n-1} = \alpha / n\) and hence \(\big(\tfrac{e\lambda_{n-1}}{c_n}\big)^{c_n} = \big(\tfrac{e\alpha}{nc_n}\big)^{c_n} = o(n^{-2})\) and for \(\beta>0\), we have \(\lambda_{n-1}=\alpha/n^{1-\beta}\) and so \(c_n \geq n^{\beta/(2+\epsilon)}\) eventually, which, together with \(\lambda_{n-1}/c_n\to 0\), implies \(\big(\tfrac{e\lambda_{n-1}}{c_n}\big)^{c_n}\leq 2^{- n^{\beta/(2+\epsilon)}}\) eventually. By Borel-Cantelli Lemma, we finally obtain \[P\Big(\limsup_n\{|N_n-\lambda_{n-1}|\geq c_n\}\Big)=0\] and so \[|N_n-\lambda_{n-1}| = O (c_n) = o \left(\sqrt{\tfrac{\Lambda_n}{\ln\ln\Lambda_n}}\right)\qquad a.s.\,.\] Therefore, ?? is a consequence of Kolmogorov’s law of the iterated logarithm [41] applied to the sequence \((N_n -\lambda_{n-1})_n\) (see [42]). ◻

Proof. Since Cramér-Wold device and [31], it is enough to prove that, for each \(a_1,a_2\in\mathbb{R}\), we have \[E[e^{i(a_1X_t+a_2Y_t)}\mid{\mathcal{F}}_t]\stackrel{a.s.}\longrightarrow \left(\textstyle{\int} e^{ia_1x}K(\cdot)(dx)\right)e^{ia_2Y}\,.\] We first prove the above convergence under the additional assumption that \(Y\) is integrable and then we will show how to get the result in the general case. Suppose \(Y\) to be integrable and set \(Z_{t}=E[Y\mid{\mathcal{F}}_{t}]\) so that we have \(Z_{t}\stackrel{a.s.}\to Y\) (by the martingale convergence theorem for martingales bounded in \(L^1\)). Moreover, we have \[\begin{align} |D_t|&=\big|E[e^{i(a_1X_t+a_2Y_t)}\mid{\mathcal{F}}_t]- \left(\textstyle{\int} e^{ia_1x}K(\cdot)(dx)\right)e^{ia_2Y}\big|\\ &\leq |T_{1,t}|+|T_{2,t}|\,, \end{align}\] where

• \(|T_{1,t}|=|E[e^{ia_1X_t}(e^{ia_2Y_t}-e^{ia_2Z_t})\mid {\mathcal{F}}_t]|\), which is smaller than or equal to \(E[\,|e^{ia_2Y_t}-e^{ia_2Z_t}|\mid {\mathcal{F}}_t]\), that converges a.s.to zero, because of [31] (see also [52]) and the a.s. convergence of both \(Y_t\) and \(Z_t\) to the same limit \(Y\);

• \(|T_{2,t}|=|E[e^{i(a_1X_t+a_2Z_{t})}\mid{\mathcal{F}}_t]- \left(\int e^{ia_1x}K(\cdot)(dx)\right)e^{ia_2Y}|\), which is equal to the difference \[|e^{ia_2Z_{t}}E[e^{ia_1X_t}\mid{\mathcal{F}}_t]- \left(\textstyle{\int} e^{ia_1x}K(\cdot)(dx)\right)e^{ia_2Y}|\] and converges a.s. to zero by the assumption on \(X_t\) and the a.s. converge of \(Z_t\) to \(Y\).

Therefore, we can conclude that \(D_t\stackrel{a.s.}\to 0\) and so the result is proven when \(Y\) is integrable.
If \(Y\) is not integrable, then, for any \(N\in \mathbb{N}\), we can define \(Y^{(N)}_t=(-N)\vee Y_t\wedge N\) and \(Y^{(N)}=(-N)\vee Y\wedge N\) so that \(|Y^{(N)}_t|\leq N\), \(|Y^{(N)}|\leq N\), \(Y^{(N)}_t\stackrel{a.s.}\to Y^{(N)}\) as \(t\to +\infty\), \(Y^{(N)}_t\stackrel{a.s.}\to Y_t\) and \(Y^{(N)}\stackrel{a.s.}\to Y\) as \(N\to +\infty\). For each fixed \(N\), applying the result to \(X_t\longrightarrow K\) and \(Y^{(N)}_t\stackrel{a.s.}\longrightarrow Y^{(N)}\) (which is trivially integrable), we find for \(t\to +\infty\) \[D^{(N)}_t= E[ e^{ i (a_1X_t+a_2Y^{(N)}_t)} \mid{\mathcal{F}}_t]- \left(\textstyle{\int} e^{ia_1x}K(\cdot)(dx)\right)e^{ia_2 Y^{(N)}}\stackrel{a.s.}\longrightarrow 0\,.\] Moreover, since [31] and the a.s.convergence of the difference \(|e^{ia_2 Y_t}-e^{ia_2Y^{(N)}_t}|\) to \(|e^{ia_2 Y}-e^{ia_2Y^{(N)}}|\) for \(t\to +\infty\) (and each fixed \(N\)), we have \(E[\,|e^{ia_2 Y_t}-e^{ia_2Y^{(N)}_t}|\,|{\mathcal{F}}_t]\) to \(|e^{ia_2 Y}-e^{ia_2Y^{(N)}}|\) for \(t\to +\infty\) (and each fixed \(N\)) and we get \[\limsup_t |D_t-D_t^{(N)}|\leq 2|e^{ia_2Y}-e^{ia_2 Y^{(N)}}|\stackrel{a.s.}\longrightarrow 0\quadasN\to +\infty\,.\] Hence, \(D_t\stackrel{a.s.}\to 0\) and the proof is so concluded. ◻