2.1. The many-to-one lemma

The well-known many-to-one lemma is a powerful tool in the study of branching random walks; see Shi [Reference Shi29] and the references therein. In this paper, we need a time-inhomogeneous version of this lemma. For all $n\geq1$ , we define the probability measure $\mu_{n}$ on $\mathbb{R}$ by

$$\mu_{n}\left(B\right)\,:\!=\,\mathbb{E}_{\xi}\left[\sum_{x\in L_{n}}\mathbf{1}_{\left\{x\in B\right\}}\mathrm{e}^{-x}\right], \quad \text{for all}\,\ B\in\mathcal{B}(\mathbb{R}).$$

Note that $\mu_{n}\left(B\right)$ is a random variable depending on $\xi$ . Up to a possible enlargement of the probability space, we define a sequence $\left(X_{n}, n\geq1\right)$ of independent random variables, where $X_{n}$ has law $\mu_{n}$ . Let $S_{n}\,:\!=\,S_{0}+\sum_{i=1}^{n}X_{i}$ . The process $\left(S_{n},n\geq0\right)$ is a random walk in a time-dependent random environment. For convenience, $\mathbb{P}_{\xi}$ also stands for the joint law of the BRWRE and the RWRE, given the environment $\xi$ . If we emphasize that the process starts from $a\in\mathbb{R}$ , this law will be denoted by $\mathbb{P}_{\xi,a}$ and $\mathbb{P}_{\xi}\,:\!=\,\mathbb{P}_{\xi,0}$ . The following time-inhomogeneous many-to-one lemma can be found in Lemma 2.2 of Mallein [Reference Mallein25].

Lemma 1. (Many-to-one) For all $n\geq1$ and any measurable function $f:\mathbb{R}^{n}\to\mathbb{R}_{+}$ , we have

(2.1) \begin{equation} \mathbb{E}_{\xi, a}\left[\sum_{|u|=n}f\left(V(u_{1}), \cdots, V(u_{n})\right)\right]=\mathbb{E}_{\xi, a}\left[\mathrm{e}^{S_{n}-a}f\left(S_{1}, \cdots, S_{n}\right)\right], \quad \mathrm{P}\text{-}\mathit{a.s.,} \end{equation}

with $\mathbb{P}_{\xi, a}\left(S_{0}=a\right)=1$ , P-a.s.

2.2. Quenched harmonic function

In this subsection, we introduce the quenched harmonic function which will be used to construct the random walk conditioned to stay in a given interval.

It follows from (1.1), (1.2), (1.3) and (2.1) that

(2.2) \begin{equation} \begin{aligned} &\mathbb{P}_{\xi}\left(S_{1} \gt 0\right)=\mathbb{E}_{\xi}\left[\sum_{|u|=1}\mathbf{1}_{\left\{V(u) \gt 0\right\}}\mathrm{e}^{-V(u)}\right] \gt 0, \quad \mathrm{P}\text{-a.s.},\\ &\mathbb{E}_{\xi}(S_{1})=\mathbb{E}_{\xi}\left[\sum_{|u|=1}V(u)\mathrm{e}^{-V(u)}\right]=0,\quad \mathrm{P}\text{-a.s.},\\ &\mathbb{E}(S^{2+\delta}_{1})=\mathbb{E}\left[\sum_{|u|=1}V(u)^{2+\delta}\mathrm{e}^{-V(u)}\right] \lt \infty. \end{aligned}\end{equation}

Under (2.2), we can formulate the quenched harmonic function as follows.

Let $y\geq0$ , and denote by $\tau_{y}$ the first time when $\{S_n\}$ enters the interval $({-}\infty, -y)$ :

$$\tau_{y}\,:\!=\,\inf\left\{n\geq1: y+S_n \lt 0\right\}.$$

Define

$$U_n(\xi,y)\,:\!=\,\mathbb{E}_\xi\left((y+S_n)\mathbf{1}_{\left\{\tau_{y} \gt n\right\}}\right).$$

Let $\theta$ be the shift operator, i.e. $\theta\xi\,:\!=\,(\xi_2,\xi_3,\cdots)$ . For $n\geq1$ , $\theta^n\xi\,:\!=\,\theta(\theta^{n-1}\xi)$ , with the convention that $\theta^0\xi\,:\!=\,\xi$ .

The following proposition (see Hong and Liang [Reference Hong and Liang15]) proves the existence and asymptotic behavior of a positive quenched harmonic function.

2.3. Random walk conditioned to stay above a line

2.3.1. Quenched probability $\mathbb{P}^{+,(\beta)}_{\xi}\left(x;\mathrm{d}y\right)$ and annealed renewal measure $\mathcal{R}(\mathrm{d}y)$

For any fixed $\beta\geq0$ , we introduce the quenched random walk conditioned to stay above $-\beta$ for almost all $\xi$ , denoted by $\zeta^{(\beta)}_{n}$ , in the sense of Doob’s h-transform. By Proposition 1(iii), for any $n\geq1$ and $B\in\mathcal{B}(\mathbb{R})$ , we can define the law of $\zeta^{(\beta)}_{n}$ by

(2.4) \begin{equation} \mathbb{P}_{\xi,a}(\zeta^{(\beta)}_{0}=a)\,:\!=\,1, \quad \mathbb{P}_{\xi,a}(\zeta^{(\beta)}_{n}\in B)\,:\!=\,\frac{\mathbb{E}_{\xi,a}\left(U(\theta^n \xi, S_n+\beta)\mathbf{1}_{\{\tau_{\beta} \gt n\}}\mathbf{1}_{\left\{S_{n}\in B\right\}}\right)}{U(\xi,a+\beta)}.\end{equation}

The process $\big(\zeta^{(\beta)}_{n},n\geq0\big)$ is called a random walk in time-inhomogeneous random environment conditioned to stay above $-\beta$ . In fact, $\big(\zeta^{(\beta)}_{n},n\geq0\big)$ is a Markov chain with state space $[{-}\beta,\infty)$ and a transition kernel that is given by

$$\mathbb{P}^{+,(\beta)}_{\xi}\left(x;\mathrm{d}y\right)\,:\!=\,\frac{U(\theta\xi,y+\beta)\mathbf{1}_{\left\{y\geq -\beta\right\}}}{U(\xi,x+\beta)}\mathbb{P}_{\xi,x}\left(S_{1}\in \mathrm{d}y\right).$$

On the other hand, $\left(S_{n},n\geq0\right)$ is simply a random walk under the annealed law $\mathbb{P}$ , thanks to the i.i.d. nature of the environment. Recall that given the environment $\xi$ , $X_{n}$ has the law $\mu_{n}$ . Let $\mathrm{E}\left(\mu_{n}\right)$ be the annealed probability measure corresponding to averaging $\mu_{n}$ over $\xi$ , and let $\mu^{\infty}\,:\!=\,\prod_{n=1}^{\infty}\mathrm{E}\left(\mu_{n}\right)$ be the product probability measure; denote by $\mathbb{E}_{\mu^{\infty}}$ the corresponding expectation. Then $\left(S_{n},n\geq0\right)$ is a usual random walk under $\mu^{\infty}$ . When we are considering the annealed random walk (that is, $\left(S_{n},n\geq0\right)$ under $\mathbb{P}_{a}$ ), we shall identify the law $\mathbb{P}_{a}$ with the law $\mu^{\infty}_{a}$ .

In what follows, we state the usual construction of a classical random walk conditioned to stay above a given value, which is indicated in Remark 1. Define the strict descending ladder epochs of the random walk $(S_{n},n\geq0)$ as

$$\gamma_{0}\,:\!=\,0,\quad\quad \gamma_{k+1}\,:\!=\,\inf\left\{n \gt \gamma_{k}:S_{n}\lt S_{\gamma_{k}}\right\},\quad k\geq0.$$

Let $R^{-}$ be the function associated with $(S_{n},n\geq0)$ that is defined by

$$R^{-}(0)\,:\!=\,1, \quad\quad R^{-}(x)\,:\!=\,\sum_{k=0}^{\infty}\mu^{\infty}(S_{\gamma_{k}}\geq -x),\quad x \gt 0.$$

Then $R^{-}(x)$ is a renewal function of the ladder heights $({-}S_{\gamma_{k}})$ . Let $\mathcal{R}^{-}(\mathrm{d}x)$ be the corresponding renewal measure. By the renewal theorem (cf. [Reference Feller11, Chapter XI, Section 1]), in our setting we have

\begin{equation*} \lim_{x\to \infty}\frac{R^{-}(x)}{x}=c_{0}\in(0,\infty).\end{equation*}

The function $R^{-}(x)$ satisfies (cf. [Reference Tanaka30, Lemma 1])

(2.5) \begin{equation} \mu^{\infty}\left[R^{-}(x+X_{1})\mathbf{1}_{\left\{x+X_{1}\geq 0\right\}}\right]=R^{-}(x),\quad \text{for}\ x\geq0.\end{equation}

From (2.5), it follows that $(R^{-}(S_{n}+\beta)\mathbf{1}_{\left\{\underline{S}_{n}\geq -\beta\right\}},n\geq1)$ is a martingale under $\mu^{\infty}$ , where $\underline{S}_{n}\,:\!=\,\min\left\{S_{0},{\cdots},S_{n}\right\}$ . Thus, we can construct the random walk conditioned to stay above $-\beta$ , denoted by $\eta^{(\beta)}_{n}$ ; that is, for any $B\in\mathcal{B}(\mathbb{R})$ ,

\begin{equation*} \mu^{\infty}_{a}(\eta^{(\beta)}_{n}\in B)\,:\!=\,\frac{\mu^{\infty}_{a}[R^{-}(S_{n}+\beta)\mathbf{1}_{\left\{\underline{S}_{n}\geq -\beta\right\}}\mathbf{1}_{\left\{S_{n}\in B\right\}}]}{R^{-}(a+\beta)}.\end{equation*}

Similarly, define the weak ascending ladder epochs of the random walk $(S_{n},n\geq0)$ as

$$\Gamma_{0}\,:\!=\,0, \quad \Gamma_{k+1}\,:\!=\,\inf\left\{n \gt \Gamma_{k}\,:\,S_{n}\geq S_{\Gamma_{k}}\right\},\quad k\geq0.$$

Let R(x) be the renewal function associated with the weak ascending ladder height process $\left(S_{\Gamma_{n}},n\geq1\right)$ , i.e.

$$R(0)\,:\!=\,1, \quad R(x)\,:\!=\,\sum_{n=1}^{\infty}\mu^{\infty}(S_{\Gamma_{n}}\lt x), \quad x \gt 0,$$

and denote by $\mathcal{R}(\mathrm{d}x)$ the corresponding renewal measure. Define $R^{(\beta)}(x)\,:\!=\,R(x+\beta)$ with the corresponding measure $\mathcal{R}^{(\beta)}(\mathrm{d}x)$ . By the renewal theorem again, there exist $c_{1},c_{2} \gt 0$ such that for any non-negative measurable function f,

(2.6) \begin{equation} c_{1}\int_{0}^{\infty}f(x-\beta)\,\mathrm{d}x\leq \int_{-\beta}^{\infty}f(x)\mathcal{R}^{(\beta)}(\mathrm{d}x)\leq c_{2}\int_{0}^{\infty}f(x-\beta)\,\mathrm{d}x.\end{equation}

We give the following lemma, which allows us to express the expectation of the series of the form $\sum_{n=1}^{\infty}\frac{G(\theta^n\xi,\zeta^{(\beta)}_{n})}{U(\theta^n\xi,\zeta^{(\beta)}_{n}+\beta)}$ (where $G(\theta^n\xi,x)$ is a non-negative measurable function depending on the n-step shifted environment $\theta^n\xi$ ) under the annealed probability $\mathbb{P}$ in the form of the integral with respect to $\mathcal{R}^{(\beta)}(\mathrm{d}x)$ .

Lemma 2. Let $\zeta^{(\beta)}_{n}$ be defined as in (2.4), and for almost all $\xi$ , let $G(\xi,{\cdot}):[{-}\beta,\infty)\to[0,\infty)$ be a measurable function, with $G\left(x\right)\,:\!=\,\mathrm{E}[G(\xi,x)] \lt \infty$ for all $x\geq -\beta$ . Then

\begin{equation*} \mathbb{E}\left[\sum_{n=1}^{\infty}\frac{G\big(\theta^n\xi,\zeta^{(\beta)}_{n}\big)}{U(\theta^n\xi,\zeta^{(\beta)}_{n}+\beta)}U(\xi,\beta)\right]=\int_{-\beta}^{\infty}G(x)\mathcal{R}^{(\beta)}(\mathrm{d}x). \end{equation*}

Proof. By the duality principle for classical random walks, following the arguments of Sections 2 and 6 of [Reference Biggins6], for any non-negative measurable function f we get

(2.7) \begin{equation} \sum_{n=1}^{\infty}\mathbb{E}_{\mu^{\infty}}\left(f\left(S_{n}\right)\mathbf{1}_{\left\{\underline{S}_{n}\geq -\beta\right\}}\right)=\int_{-\beta}^{\infty}f(x)\mathcal{R}^{(\beta)}(\mathrm{d}x). \end{equation}

Then, by the definition of $\zeta^{(\beta)}_{n}$ , we obtain

\begin{equation*} \begin{aligned} \mathbb{E}\left[\sum_{n=1}^{\infty}\frac{G\big(\theta^n\xi,\zeta^{(\beta)}_{n}\big)}{U(\theta^n\xi,\zeta^{(\beta)}_{n}+\beta)}U(\xi,\beta)\right]=&\mathrm{E}\left[\sum_{n=1}^{\infty}\mathbb{E}_{\xi}\left(\frac{G\big(\theta^n\xi,\zeta^{(\beta)}_{n}\big)}{U\big(\theta^n\xi,\zeta^{(\beta)}_{n}+\beta\big)}\right)U(\xi,\beta)\right]\\ = &\mathrm{E}\left[\sum_{n=1}^{\infty}\mathbb{E}_{\xi}\!\left(\!\frac{U(\theta^n\xi,S_n+\beta)\mathbf{1}_{\{\tau_{\beta} \gt n\}}G(\theta^n\xi,S_{n})}{U(\theta^n\xi,S_{n}+\beta)U(\xi,\beta)}\!\right)\!U(\xi,\beta)\right]\end{aligned} \end{equation*}

\begin{equation*} \begin{aligned} & = \mathrm{E}\left[\sum_{n=1}^{\infty}\mathbb{E}_{\xi}\left(\mathbf{1}_{\{\tau_{\beta} \gt n\}}G\left(\theta^n\xi,S_{n}\right)\right)\right]\\=&\mathrm{E}\left[\sum_{n=1}^{\infty}\int_{-\beta}^{\infty}G\left(\theta^n\xi,x\right)\mathbb{P}_{\xi}\left(S_{n}\in \mathrm{d}x,\tau_{\beta} \gt n\right)\right]. \end{aligned} \end{equation*}

Because of the i.i.d. random environment, for each fixed x, $G(\theta^n\xi,x)$ forms a stationary and ergodic sequence (see e.g. [Reference Kallenberg18, Lemmas 10.1 and 10.5]), $\mathrm{E}\left[G(\theta^n\xi,x)\right]=G(x)$ , and by the independence, we deduce

\begin{equation*} \begin{aligned} \mathrm{E}\left[\sum_{n=1}^{\infty}\int_{-\beta}^{\infty}G\left(\theta^n\xi,x\right)\mathbb{P}_{\xi}\left(S_{n}\in\mathrm{d}x,\tau_{\beta} \gt n\right)\right]=&\sum_{n=1}^{\infty}\int_{-\beta}^{\infty}\mathrm{E}\left[G\left(\theta^n\xi,x\right)\mathbb{P}_{\xi}\left(S_{n}\in \mathrm{d}x,\tau_{\beta} \gt n\right)\right]\\ =&\int_{-\beta}^{\infty}G(x)\sum_{n=1}^{\infty}\mathbb{P}\left(S_{n}\in \mathrm{d}x,\tau_{\beta} \gt n\right)\\ =&\int_{-\beta}^{\infty}G(x)\sum_{n=1}^{\infty}\mu^{\infty}\left(S_{n}\in \mathrm{d}x,\underline{S}_{n}\geq -\beta\right)\\ =&\int_{-\beta}^{\infty}G(x)\mathcal{R}^{(\beta)}(\mathrm{d}x), \end{aligned} \end{equation*}

where the last equality follows from (2.7). This yields the lemma.

2.3.2. Quenched Tanaka’s decomposition

Tanaka’s decomposition is a fundamental tool for investigating the behavior of the random walk conditioned to stay non-negative; see [Reference Afanasyev, Geiger, Kersting and Vatutin1, Reference Biggins6, Reference Tanaka30], for example. With the above preparations in hand, we can now specify a quenched version of Tanaka’s decomposition for the RWRE conditioned to stay non-negative. We will then proceed to discuss the relationship between two kinds of probability measures. For simplicity, we will restrict our analysis to the case where $\beta=0$ and write $\zeta_{n}\,:\!=\,\zeta^{(0)}_{n}$ . Let $\nu$ denote the time at which the first prospective minimal value of the process $(\zeta_{n},n\geqslant0)$ occurs, i.e.

(2.8) \begin{equation} \nu\,:\!=\,\inf\left\{m\geq1\,:\,\zeta_{m+n}\geq \zeta_{m},\ \text{for all}\,\ n\geq0\right\}.\end{equation}

Write $\zeta^{\nu}_{k}\,:\!=\,\zeta_{\nu+k}-\zeta_{\nu}$ , $k\geq1$ .

Proposition 2. (Quenched Tanaka’s decomposition.) For almost all $\xi$ , we have the following:

1. (i) $\zeta_{n}\to\infty$ $\mathbb{P}_{\xi}$ -a.s. as $n\to\infty$ ;

2. (ii) $\left(\nu,\zeta_{1},\cdots,\zeta_{\nu}\right)$ and $\left(\zeta^{\nu}_{1},\zeta^{\nu}_{2},\cdots\right)$ are independent with respect to $\mathbb{P}_{\xi}$ ;

3. (iii) $U(\xi,0)\mathbb{P}_{\xi}\left(\nu=k,\zeta_{\nu}\in \mathrm{d}x\right)=U\left(\theta^k\xi,0\right)\mathbb{P}_{\xi}\left(S_{k} \lt S_{k-1},\cdots,S_{k} \lt S_{1},S_{k}\in \mathrm{d}x\right)$ for all $k\geq1$ .

Proof. (i) We claim that $U(y)\,:\!=\,\mathrm{E}\left[U(\xi,y)\right] \lt \infty$ for any $y\geq0$ . In fact, by the definition of $U(\xi,y)$ , we have

\begin{equation*} \begin{aligned} U(0)=\mathrm{E}\left[\mathbb{E}_{\xi}\left({-}S_{\tau_{0}}\right)\right]=&\mathrm{E}\left[\sum_{n=1}^{\infty}\mathbb{E}_{\xi}\left({-}S_{n},\tau_{0}=n\right)\right]\\ =&\mathrm{E}\left[\sum_{n=1}^{\infty}\mathbb{E}_{\xi}\left({-}S_{n},S_{1}\geq 0,\cdots,S_{n-1}\geq 0,S_{n} \lt 0\right)\right]\\ =&\sum_{n=1}^{\infty}\mathbb{E}_{\mu^{\infty}}\left({-}S_{n},S_{1}\geq 0,\cdots,S_{n-1}\geq 0,S_{n} \lt 0\right)\\ =&\sum_{n=1}^{\infty}\mathbb{E}_{\mu^{\infty}}\left({-}S_{n},\gamma_{1}=n\right)\\ =&-\mathbb{E}_{\mu^{\infty}}\left(S_{\gamma_{1}}\right) \lt \infty, \end{aligned} \end{equation*}

where the validity of the finiteness of $\mathbb{E}_{\mu^{\infty}}\left(S_{\gamma_{1}}\right)$ is proved by Theorem 1 in Chapter XVIII.5 of [Reference Feller11]. By Proposition 1(ii), U(y) satisfies the annealed harmonic property:

(2.9) \begin{equation} \mathbb{E}\left[U\left(y+S_{1}\right)\mathbf{1}_{\left\{y+S_{1}\geq 0\right\}}\right]=U(y),\quad y\geq0. \end{equation}

Since $\mathbb{P}\left(S_{1} \gt y_{0}\right) \gt 0$ for some $y_{0} \gt 0$ , by (2.9) with $y=0$ we have $U(y_{1}) \lt \infty$ for some $y_{1} \gt y_{0}$ . Again applying (2.9) with $y=y_{1}$ , we have $U(y_{2}) \lt \infty$ for some $y_{2} \gt y_{1}+y_{0}$ . Repeating this argument, we deduce that there exists a sequence $\left(y_{n},n\geq1\right)$ such that $U\left(y_{n}\right) \lt \infty$ for all n. By the monotonicity of U(y), we conclude that $U(y) \lt \infty$ for all $y\geq0$ .

Since $U(\xi,y)$ is positive and non-decreasing in y, for any $y \gt 0$ , by the definition of $\zeta_{n}$ , we have

\begin{equation*} \begin{aligned} \mathrm{E}\left[U\left(\xi,0\right)\sum_{n=1}^{\infty}\mathbb{P}_{\xi}\left(\zeta_{n} \lt y\right)\right]=&\mathrm{E}\left[\sum_{n=1}^{\infty}\frac{\mathbb{E}_{\xi}\left(U(\theta^n\xi,S_n)\mathbf{1}_{\{\tau_{0} \gt n\}}\mathbf{1}_{\left\{S_{n} \lt y\right\}}\right)}{U(\xi,0)}U\left(\xi,0\right)\right]\\ \leq&\sum_{n=1}^{\infty}\mathrm{E}\left[U(\theta^n\xi,y)\mathbb{E}_{\xi}\left(\mathbf{1}_{\{\tau_{0} \gt n\}}\mathbf{1}_{\left\{S_{n} \lt y\right\}}\right)\right]. \end{aligned} \end{equation*}

Following the same argument as in the proof of Lemma 2, we have

\begin{equation*} \begin{aligned} \sum_{n=1}^{\infty}\mathrm{E}\left[U(\theta^n\xi,y)\mathbb{E}_{\xi}\left(\mathbf{1}_{\{\tau_{0} \gt n\}}\mathbf{1}_{\left\{S_{n} \lt y\right\}}\right)\right]=&\sum_{n=1}^{\infty}U(y)\mathbb{P}\left(S_{n} \lt y,\tau_{0} \gt n\right)\\ =&U(y)\sum_{n=1}^{\infty}\mu^{\infty}\left(S_{n} \lt y,\underline{S}_{n}\geq0\right)\\ =&U(y)\int_{0}^{y}\mathcal{R}^{(0)}(\mathrm{d}x) \lt \infty. \end{aligned} \end{equation*}

Thus, for almost all $\xi$ , $\sum_{n=1}^{\infty}\mathbb{P}_{\xi}\left(\zeta_{n} \lt y\right) \lt \infty$ . The Borel–Cantelli lemma yields that, for almost all $\xi$ , $\zeta_{n}\to\infty$ $\mathbb{P}_{\xi}$ -a.s. as $n\to\infty$ .

(ii) Let

$$H\left(\xi,x,z\right)\,:\!=\,\frac{U\left(\xi,x-z\right)}{U\left(\xi,x\right)}, \quad x\geq z\geq0, \quad\quad H\left(\xi,x,z\right)\,:\!=\,0, \quad x \lt z,$$

and

\begin{align*} \mathbb{P}^+_{\theta^{j-1}\xi}\left(y_{j-1};\mathrm{d}y_{j}\right)\,:\!=\,&\mathbb{P}_{\xi}\left.\left(\zeta_{j}\in \mathrm{d}y_{j}\,\right| \zeta_{j-1}=y_{j-1}\right)\\ =&\frac{U(\theta^j\xi,y_{j})\mathbf{1}_{\left\{y_{j}\geq 0\right\}}}{U(\theta^{j-1}\xi,y_{j-1})}\mathbb{P}_{\theta^{j-1}\xi,y_{j-1}}\left(S_{1}\in \mathrm{d}y_{j}\right), \quad j\geq 1 .\end{align*}

Then, by Proposition 1(ii), $H\left(\xi,{\cdot},z\right)$ is quenched harmonic with respect to the transition kernel $\mathbb{P}^+_{\xi}$ in the following sense:

(2.10) \begin{equation} \int H\left(\theta\xi,y,z\right)\mathbb{P}^+_{\xi}\left(x;\mathrm{d}y\right)=H\left(\xi,x,z\right), \quad x\geq z\geq0. \end{equation}

Define $\hat{\tau}_{z}\,:\!=\,\inf\left\{n \gt 0:\zeta_{n} \lt z\right\}$ . Since $\hat{\tau}_{z}$ is a stopping time, it follows from (2.10) that the process $\left(H\left(\theta^{n\wedge\hat{\tau}_{z}}\xi,\zeta_{n\wedge\hat{\tau}_{z}},z\right),n\geq 0\right)$ is a martingale under $\mathbb{P}_{\xi}$ . Thus, for all $n\geq 0$ , we have

$$\mathbb{E}_{\xi,x}\left[H\left(\theta^{n\wedge\hat{\tau}_{z}}\xi,\zeta_{n\wedge\hat{\tau}_{z}},z\right) \right]=H\left(\xi,x,z\right).$$

Note that for almost all $\xi$ , $0\leq H\left(\xi,x,z\right)\leq1$ , $H\left(\xi,x,z\right)\,:\!=\,0$ for $x \lt z$ , and $H\left(\theta^n\xi,\zeta_{n},z\right)\to1$ as $n\to\infty$ by (2.3) and Part (i). It follows that

$$\lim_{n\to\infty}H\left(\theta^{n\wedge\hat{\tau}_{z}}\xi,\zeta_{n\wedge\hat{\tau}_{z}},z\right)=\lim_{n\to\infty}H\left(\theta^{n}\xi,\zeta_{n},z\right)\mathbf{1}_{\left\{\hat{\tau}_{z} \gt n\right\}}=\mathbf{1}_{\left\{\zeta_{n}\geq z\ \text{for all}\ n \gt 0\right\}}.$$

By the dominated convergence theorem, we get

$$\mathbb{P}_{\xi,x}\left(\zeta_{n}\geq z\,\ \text{for all}\,\ n \gt 0\right)=\mathbb{E}_{\xi,x}\left[\lim_{n\to\infty}H\left(\theta^{n\wedge\hat{\tau}_{z}}\xi,\zeta_{n\wedge\hat{\tau}_{z}},z\right) \right]=H\left(\xi,x,z\right).$$

This tells us that $H\left(\xi,x,z\right)$ is the probability that, starting at x, the process $\left(\zeta_{n},n\geq0\right)$ never hits $\left({-}\infty,z\right)$ .

For any $x_{1},\cdots,x_{k},y_{1},\cdots,y_{m}\geq 0$ with $x_{0}=y_{0}=0$ , we have

\begin{equation*} \begin{aligned} &\prod_{j=1}^{m}\mathbb{P}^+_{\theta^{k+j-1}\xi}\left(y_{j-1}+x_{k};\mathrm{d}y_{j}+x_{k}\right)H\left(\theta^{k+m}\xi,y_{m}+x_{k},x_{k}\right)\\ =&\prod_{j=1}^{m}\mathbb{P}^+_{\theta^{k+j-1}\xi}\left(y_{j-1}+x_{k};\mathrm{d}y_{j}+x_{k}\right)\frac{U\left(\theta^{k+m}\xi,y_{m}\right)}{U\left(\theta^{k+m}\xi,y_{m}+x_{k}\right)}\\ =&\prod_{j=1}^{m}\mathbb{P}_{\theta^{k+j-1}\xi}\left(y_{j-1}+x_{k};\mathrm{d}y_{j}+x_{k}\right)\frac{U\left(\theta^{k+j}\xi,y_{j}+x_{k}\right)}{U\left(\theta^{k+j-1}\xi,y_{j-1}+x_{k}\right)}\frac{U\left(\theta^{k+m}\xi,y_{m}\right)}{U\left(\theta^{k+m}\xi,y_{m}+x_{k}\right)}\\ = & \prod_{j=1}^{m}\mathbb{P}_{\theta^{k+j-1}\xi}\left(y_{j-1};\mathrm{d}y_{j}\right)\frac{U\left(\theta^{k+m}\xi,y_{m}\right)}{U\left(\theta^{k}\xi,x_{k}\right)} \end{aligned} \end{equation*}

\begin{equation*} \begin{aligned} =&\prod_{j=1}^{m}\mathbb{P}_{\theta^{k+j-1}\xi}\left(y_{j-1};\mathrm{d}y_{j}\right)\frac{U\left(\theta^{k+j}\xi,y_{j}\right)}{U\left(\theta^{k+j-1}\xi,y_{j-1}\right)}\frac{U\left(\theta^{k}\xi,0\right)}{U\left(\theta^{k}\xi,x_{k}\right)}\\ =&\prod_{j=1}^{m}\mathbb{P}^+_{\theta^{k+j-1}\xi}\left(y_{j-1};\mathrm{d}y_{j}\right)H\left(\theta^{k}\xi,x_{k},x_{k}\right). \end{aligned} \end{equation*}

As a result,

\begin{align*} & \mathbb{P}_{\xi}\left(\nu=k,\zeta_{1}\in \mathrm{d}x_{1},\cdots,\zeta_{k}\in \mathrm{d}x_{k},\zeta^{\nu}_{1}\in \mathrm{d}y_{1},\cdots,\zeta^{\nu}_{m}\in \mathrm{d}y_{m}\right)\\ = & \mathbf{1}_{\left\{x_{1},\cdots,x_{k-1} \gt x_{k}\right\}}\mathbf{1}_{\left\{y_{1},\cdots,y_{m}\geq0\right\}}\prod_{i=1}^{k}\mathbb{P}^+_{\theta^i\xi}\left(x_{i-1};\mathrm{d}x_{i}\right)\\ & \times\prod_{j=1}^{m}\mathbb{P}^+_{\theta^{k+j-1}\xi}\left(y_{j-1}+x_{k};\mathrm{d}y_{j}+x_{k}\right)H\left(\theta^{k+m}\xi,y_{m}+x_{k},x_{k}\right)\\ =& \mathbf{1}_{\left\{x_{1},\cdots,x_{k-1} \gt x_{k}\right\}}\prod_{i=1}^{k}\mathbb{P}^+_{\theta^i\xi}\left(x_{i-1};\mathrm{d}x_{i}\right)H\left(\theta^{k}\xi,x_{k},x_{k}\right)\prod_{j=1}^{m}\mathbb{P}^+_{\theta^{k+j-1}\xi}\left(y_{j-1};\mathrm{d}y_{j}\right)\\ =& \mathbb{P}_{\xi}\left(\nu=k,\zeta_{1}\in \mathrm{d}x_{1},\cdots,\zeta_{k}\in \mathrm{d}x_{k}\right)\mathbb{P}_{\xi}\left(\zeta^{\nu}_{1}\in \mathrm{d}y_{1},\cdots,\zeta^{\nu}_{m}\in \mathrm{d}y_{m}\right), \end{align*}

which proves that $\left(\nu,\zeta_{1},\cdots,\zeta_{\nu}\right)$ and $\left(\zeta^{\nu}_{1},\zeta^{\nu}_{2},\cdots\right)$ are independent with respect to $\mathbb{P}_{\xi}$ .

(iii) For all $k\geq1$ and $x\geq 0$ , we have

\begin{equation*} \begin{aligned} \mathbb{P}_{\xi}\left(\nu=k,\zeta_{\nu}\in \mathrm{d}x\right)=&\mathbb{P}_{\xi}\left(\zeta_{k} \lt \zeta_{k-1},\cdots,\zeta_{k} \lt \zeta_{1},\zeta_{k}\in \mathrm{d}x\right)H\left(\theta^{k}\xi,x,x\right)\\ =&\mathbb{P}_{\xi}\left(S_{k} \lt S_{k-1},\cdots,S_{k} \lt S_{1},S_{k}\in \mathrm{d}x\right)\frac{U\left(\theta^k\xi,x\right)}{U(\xi,0)}\frac{U\left(\theta^k\xi,0\right)}{U(\theta^k\xi,x)}\\ =&\mathbb{P}_{\xi}\left(S_{k} \lt S_{k-1},\cdots,S_{k}\lt S_{1},S_{k}\in \mathrm{d}x\right)\frac{U\left(\theta^k\xi,0\right)}{U(\xi,0)}, \end{aligned} \end{equation*}

as desired.

Define $\tilde{\mathbb{P}}({\cdot})\,:\!=\,\int_{\Omega}\frac{U(\xi,0)}{U(0)}\mathbb{P}_{\xi}({\cdot})\,\mathrm{d}\mathrm{P}$ and denote by $\tilde{\mathbb{E}}$ the corresponding expectation. We show that the excursion $\left(\zeta_{1},\cdots,\zeta_{\nu}\right)$ under the annealed probability $\tilde{\mathbb{P}}$ is distributed as $\left(S_{\Gamma_1}-S_{\Gamma_1-1},\cdots,S_{\Gamma_1}\right)$ under $\mu^{\infty}$ .

Proposition 3. (Annealed excursion distribution.) If $\nu$ is the time of the first prospective minimal value of the process $(\zeta_{n},n\geqslant0)$ defined as in (2.8), then we have the following:

1. (i) $\tilde{\mathbb{P}}\left(\zeta_{\nu}\in \mathrm{d}x\right)=\mu^{\infty}\left(S_{\Gamma_1}\in \mathrm{d}x\right)$ ;

2. (ii) $\tilde{\mathbb{E}}\left[f\left(\nu,\zeta_{1},\cdots,\zeta_{\nu}\right)\right]=\mathbb{E}_{\mu^{\infty}}\left[f\left(\Gamma_1,S_{\Gamma_1}-S_{\Gamma_1-1},\cdots,S_{\Gamma_1}\right)\right]$ for any measurable function f.

Proof. (i) Proposition 2(iii) yields

\begin{align*} \mathrm{E}\left[U(\xi,0)\mathbb{P}_{\xi}\left(\nu=k,\zeta_{\nu}\in \mathrm{d}x\right)\right] = & \mathrm{E}\left[\mathbb{P}_{\xi}\left(S_{k} \lt S_{k-1},\cdots,S_{k} \lt S_{1},S_{k}\in \mathrm{d}x\right)U\left(\theta^k\xi,0\right)\right]\\ =&U\left(0\right)\mu^{\infty}\left(S_{k}-S_{k-1} \lt 0,\cdots,S_{k}-S_{1} \lt 0,S_{k}\in \mathrm{d}x\right) \end{align*}

\begin{align*} =&U\left(0\right)\mu^{\infty}\left(S_{1} \lt 0,\cdots,S_{k-1} \lt 0,S_{k}\in \mathrm{d}x\right)\\ =&U\left(0\right)\mu^{\infty}\left(\Gamma_1=k,S_{\Gamma_1}\in \mathrm{d}x\right). \end{align*}

Dividing by U(0) and summing over k, we have $\tilde{\mathbb{P}}\left(\zeta_{\nu}\in \mathrm{d}x\right)=\mu^{\infty}\left(S_{\Gamma_1}\in \mathrm{d}x\right)$ .

(ii) Similarly to the proof of Proposition 2(iii), we get

\begin{equation*} \begin{aligned} &\tilde{\mathbb{E}}\left[f\left(\nu,\zeta_{1},\cdots,\zeta_{\nu}\right)\right]\\ =&\,\mathrm{E}\left[\int f\left(k,x_{1},\cdots,x_{k}\right)\frac{U(\xi,0)}{U(0)}\mathbb{P}_{\xi}\left(\nu=k,\zeta_{1}\in \mathrm{d}x_{1},\cdots,\zeta_{k}\in \mathrm{d}x_{k}\right)\right]\\ =&\,\mathrm{E}\left[\int f\left(k,x_{1},\cdots,x_{k}\right)\frac{U\left(\theta^k\xi,0\right)}{U(0)}\mathbb{P}_{\xi}\left(S_{k} \lt S_{k-1},\cdots,S_{k} \lt S_{1}, S_{1}\in \mathrm{d}x_{1},\cdots,S_{k}\in \mathrm{d}x_{k}\right)\right]\\ =&\int f\left(k,x_{1},\cdots,x_{k}\right)\mu^{\infty}\left(S_{k} \lt S_{k-1},\cdots,S_{k} \lt S_{1}, S_{1}\in \mathrm{d}x_{1},\cdots,S_{k}\in \mathrm{d}x_{k}\right)\\ =&\int f\left(k,x_{1},\cdots,x_{k}\right)\mu^{\infty}\left(\tilde{S}_{1} \lt 0,\cdots,\tilde{S}_{k-1} \lt 0, \tilde{S}_{k}-\tilde{S}_{k-1}\in \mathrm{d}x_{1},\cdots,\tilde{S}_{k}\in \mathrm{d}x_{k}\right)\\ =&\, \mathbb{E}_{\mu^{\infty}}\left[f\left(\Gamma_1,S_{\Gamma_1}-S_{\Gamma_1-1},\cdots,S_{\Gamma_1}\right)\right], \end{aligned} \end{equation*}

where $\tilde{S}_{j}\,:\!=\,S_{k}-S_{k-j}$ , $j\leq k$ , and the last equality follows from the duality property for the random walk under $\mu^{\infty}$ .

2.3.3. Application of Tanaka’s decomposition

The following proposition gives an integral criterion for the almost sure divergence of the infinite series $\sum_{n=1}^{\infty}U(\xi,\beta)F\big(\zeta^{(\beta)}_{n}\big)$ .

Proposition 4. Let $\zeta^{(\beta)}_{n}$ be defined as in (2.4), and let $F:[{-}\beta,\infty)\to[0,\infty)$ be a non-increasing measurable function. Then

\begin{equation*} \begin{aligned} \int_{-\beta}^{\infty}F(x)(x+\beta)\, \mathrm{d}x=\infty \quad &\Longleftrightarrow \quad \sum_{n=1}^{\infty}\frac{U(\xi,\beta)F\big(\zeta^{(\beta)}_{n}\big)\big(\zeta^{(\beta)}_{n}+\beta\big)}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}=\infty,\quad \mathbb{P}\text{-a.s.}\\ \quad &\Longleftrightarrow \quad \sum_{n=1}^{\infty}U(\xi,\beta)F\big(\zeta^{(\beta)}_{n}\big)=\infty,\quad \mathbb{P}\text{-a.s.} \end{aligned} \end{equation*}

Proof. For the second equivalence, note that for almost all $\xi$ , $\zeta^{(\beta)}_{n}+\beta\to\infty$ $\mathbb{P}_{\xi}$ -a.s. as $n\to\infty$ (by Proposition 2(i)), and by (2.3), we have $U\big(\theta^n\xi,\zeta^{(\beta)}_{n}+\beta\big)\sim\zeta^{(\beta)}_{n}+\beta$ as $n\to\infty$ ; hence,

$$\sum_{n=1}^{\infty}\frac{U(\xi,\beta)F\big(\zeta^{(\beta)}_{n}\big)\big(\zeta^{(\beta)}_{n}+\beta\big)}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}=\infty\quad \Longleftrightarrow\quad \sum_{n=1}^{\infty}U(\xi,\beta)F\big(\zeta^{(\beta)}_{n}\big)=\infty, \quad \mathbb{P}\text{-a.s.}$$

By Lemma 2, we have

$$\mathbb{E}\left[\sum_{n=1}^{\infty}\frac{U(\xi,\beta)F\big(\zeta^{(\beta)}_{n}\big)\big(\zeta^{(\beta)}_{n}+\beta\big)}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\right]=\int_{-\beta}^{\infty}F(x)(x+\beta)\mathcal{R}^{(\beta)}(\mathrm{d}x).$$

It follows from (2.6) that

$$\sum_{n=1}^{\infty}\frac{U(\xi,\beta)F\big(\zeta^{(\beta)}_{n}\big)\big(\zeta^{(\beta)}_{n}+\beta\big)}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}=\infty,\quad \mathbb{P}\text{-a.s.} \quad \Longrightarrow \quad \int_{-\beta}^{\infty}F(x)(x+\beta)\, \mathrm{d}x=\infty.$$

Thus we need only to prove that

$$\int_{-\beta}^{\infty}F(x)(x+\beta)\, \mathrm{d}x=\infty \quad \Longrightarrow \quad \sum_{n=1}^{\infty}U(\xi,\beta)F\big(\zeta^{(\beta)}_{n}\big)=\infty,\quad \mathbb{P}\text{-a.s.}$$

For simplicity, we only consider $\beta=0$ and write $\zeta_{n}\,:\!=\,\zeta^{(0)}_{n}$ , since the case of $\beta \gt 0$ is similar to this case. Note that

$$\sum_{n=1}^{\infty}U\left(\xi,0\right)F\left(\zeta_{n}\right)=\infty,\quad \mathbb{P}\text{-a.s.}\quad \Longleftrightarrow\quad \sum_{n=1}^{\infty}F\left(\zeta_{n}\right)=\infty,\quad \tilde{\mathbb{P}}\text{-a.s.}$$

Therefore it remains to show that

(2.11) \begin{equation} \int_{0}^{\infty}F(x)x\, \mathrm{d}x=\infty \quad \Longrightarrow\quad \sum_{n=1}^{\infty}F\left(\zeta_{n}\right)=\infty,\quad \tilde{\mathbb{P}}\text{-a.s.} \end{equation}

To prove (2.11), it suffices to check that

$$\tilde{\mathbb{P}}\left(\sum_{n=1}^{\infty}F\left(\zeta_n\right)=\infty\right) \lt 1 \quad \Longrightarrow\quad \int_0^{\infty} F(x)x \,\mathrm{d}x \lt \infty.$$

We assume that $\tilde{\mathbb{P}}\left(\sum_{n=1}^{\infty}F\left(\zeta_n\right)=\infty\right) \lt 1$ , that is, $\tilde{\mathbb{P}}\left(\sum_{n=1}^{\infty}F\left(\zeta_n\right) \lt \infty\right) \gt 0$ .

We first use Tanaka’s decomposition (Propositions 2 and 3) to reconstruct the process $\left(\zeta_{n},n\geq0\right)$ . Recall that

$$\nu\,:\!=\,\inf\left\{m\geq1:\zeta_{m+n}\geq \zeta_{m}, \,\text{for all}\,\ n\geq0\right\}.$$

We have an excursion $\left(\zeta_j, 0\leq j \leq \nu\right)$ , which is denoted by $\omega=(\omega(j), 0\leq j \leq \nu)$ . Let $\left\{\omega_k=\left(\omega_k(j), 0 \leq j \leq \nu_k\right), k \geq 1\right\}$ be a sequence of independent copies of $\omega$ under $\tilde{\mathbb{P}}$ . Define

$$V_0\,:\!=\,0,\quad\quad V_k\,:\!=\,\nu_1+\cdots+\nu_k,\quad \text{for all}\,\ k \geq 1.$$

The process

$$\zeta_0=0,\quad\quad \zeta_n=\sum_{i=1}^{k}\omega_i\left(\nu_i\right)+\omega_{k+1}\left(n-V_{k}\right), \quad \text{for}\ V_k \lt n \leq V_{k+1},$$

is what we need. Then,

\begin{equation*} \begin{aligned} \sum_{n=1}^{\infty}F\left(\zeta_{n}\right)&=\sum_{k=1}^{\infty}\sum_{n=V_{k-1}+1}^{V_k}F\left(\sum_{i=1}^{k-1}\omega_i\left(\nu_i\right)+\omega_{k}\left(n-V_{k-1}\right)\right)\\ &=\sum_{k=1}^{\infty}\sum_{j=1}^{\nu_k}F\left(\sum_{i=1}^{k}\omega_i\left(\nu_i\right)-\left(\omega_{k}\left(\nu_k\right)-\omega_{k}\left(j\right)\right)\right). \end{aligned} \end{equation*}

Hence, by hypothesis,

$$\tilde{\mathbb{P}}\left(\sum_{n=1}^{\infty}F\left(\zeta_{n}\right) \lt \infty\right)=\tilde{\mathbb{P}}\left(\sum_{k=1}^{\infty}\sum_{j=1}^{\nu_k}F\left(\sum_{i=1}^{k}\omega_i\left(\nu_i\right)-\left(\omega_{k}\left(\nu_k\right)-\omega_{k}\left(j\right)\right)\right) \lt \infty\right) \gt 0.$$

On the other hand, by Proposition 3(i), it follows from the strong law of large numbers that

$$\lim _{k \rightarrow \infty} \frac{\sum_{i=1}^{k}\omega_i\left(\nu_i\right)}{k}=C,\quad \tilde{\mathbb{P}}\text{-a.s.,}$$

where $C\,:\!=\,\mathbb{E}_{\mu^\infty}\left(S_{\Gamma_1}\right) \lt \infty$ is due to [Reference Feller11, Chapter XVIII, Section 5, Theorem 1]. Let $\epsilon \gt 0$ and $A\,:\!=\,(C+\epsilon)\vee 1$ ; then, for all sufficiently large k, $\sum_{i=1}^{k}\omega_i\left(\nu_i\right)\leq Ak$ . Since F is non-increasing, we obtain

\begin{equation*} \begin{aligned} &\tilde{\mathbb{P}}\left(\sum_{k=1}^{\infty}\sum_{j=1}^{\nu_k}F\left(Ak-\left(\omega_{k}\left(\nu_k\right)-\omega_{k}\left(j\right)\right)\right) \lt \infty\right)\\ \geq\ &\tilde{\mathbb{P}}\left(\sum_{k=1}^{\infty}\sum_{j=1}^{\nu_k}F\left(\sum_{i=1}^{k}\omega_i\left(\nu_i\right)-\left(\omega_{k}\left(\nu_k\right)-\omega_{k}\left(j\right)\right)\right) \lt \infty\right) \gt 0. \end{aligned} \end{equation*}

Let

$$\chi_k(\nu,\omega,F)\,:\!=\,\sum_{j=1}^{\nu_k} F\left(Ak-\left(\omega_k(\nu_{k})-\omega_k(j)\right)\right);$$

then $\tilde{\mathbb{P}}\left(\sum_{k=1}^{\infty}\chi_k(\nu,\omega,F) \lt \infty\right) \gt 0$ . Note that the independence of the sequence $\left\{\omega_k, k \geq 1\right\}$ yields the independence of the sequence $\left\{\chi_k\left(\nu,\omega,F\right), k \geq 1\right\}$ . By Kolmogorov’s 0–1 law, it follows that

(2.12) \begin{equation} \tilde{\mathbb{P}}\left(\sum_{k=1}^{\infty}\chi_k\left(\nu,\omega,F\right) \lt \infty\right)=1. \end{equation}

From now on, we proceed in the same way as [Reference Chen9]. Let $E_M\,:\!=\,\left\{\sum_{k=1}^{\infty}\chi_k\left(\nu,\omega,F\right) \lt M\right\}$ for any $M \gt 0$ . Either $\mathbb{P}\left(E_{M_0}\right)=1$ for some $M_0 \lt \infty$ , or $\tilde{\mathbb{P}}\left(E_M\right) \lt 1$ for all $M \in(0, \infty)$ . For the first case—that is, if there exists some $M_0 \lt \infty$ such that $\mathbb{P}\left(E_{M_0}\right)=1$ —we have

$$ \begin{aligned} M_0 &\geq \tilde{\mathbb{E}}\left(\sum_{k=1}^{\infty}\chi_k\left(\nu,\omega,F\right)\right)\\ & =\ {\mathbb{E}}\left(\sum_{k=1}^{\infty}\sum_{j=1}^{\nu_k}F\left(Ak-\left(\omega_k(\nu_{k})-\omega_k(j)\right)\right)\right)\\ &=\sum_{k=1}^{\infty}\tilde{\mathbb{E}}\left(\sum_{j=1}^{\nu} F\left(Ak-\left(\omega(\nu)-\omega(j)\right)\right)\right)\\ &=\sum_{k=1}^{\infty}\mu^{\infty}\left(\sum_{j=0}^{\Gamma_1-1} F\left(Ak-S_{j}\right)\right), \end{aligned} $$

where the last equality follows from Proposition 3(ii). Similarly to (2.7), we have

$$ \begin{aligned} \mathbb{E}_{\mu^{\infty}}\left(\sum_{j=0}^{\Gamma_1-1} F\left(Ak-S_{j}\right)\right)=&\,\mathbb{E}_{\mu^{\infty}}\left(\sum_{n=0}^{\infty}F\left(Ak-S_{n}\right)\mathbf{1}_{\left\{n \lt \Gamma_1\right\}}\right)\\ =&\,\mathbb{E}_{\mu^{\infty}}\left(F\left(Ak-S_{n}\right)\mathbf{1}_{\left\{\max_{j\leq n}S_{j} \lt 0\right\}}\right)\\ =&\int_0^{\infty}F(Ak+x)\mathcal{R}^{-}(\mathrm{d}x), \end{aligned} $$

where $\mathcal{R}^{-}(\mathrm{d}x)$ is the renewal measure of $R^{-}(x)$ , i.e. the renewal measure associated with the strict descending ladder height process. Thus, $\sum_{k=1}^{\infty}\int_0^{\infty}F(Ak+x)\mathcal{R}^{-}(\mathrm{d}x) \lt \infty$ , which implies that

$$\int_0^{\infty}F(x)x \,\mathrm{d}x \lt \infty.$$

For the second case, $\tilde{\mathbb{P}}\left(E_M\right) \lt 1$ for all $M \in(0, \infty)$ , so $\lim _{M \to \infty} \tilde{\mathbb{P}}\left(E_M\right)=1$ by (2.12). Let

$$\Lambda_{k,l}(\nu,\omega)\,:\!=\,\sum_{j=1}^{\nu_k}\mathbf{1}_{\left\{A(l-1)\leq-\left(\omega_k(\nu_{k})-\omega_k(j)\right) \lt Al\right\}}, \quad \text{for all}\,\ k\geq1,\ l\geq1.$$

Note that, for any $k \geq 1$ ,

$$ \begin{aligned} \chi_k\left(\nu,\omega,F\right)&=\sum_{j=1}^{\nu_k} F\left(Ak-\left(\omega_k(\nu_{k})-\omega_k(j)\right)\right) \sum_{l=1}^{\infty} \mathbf{1}_{\left\{A(l-1) \leq-\left(\omega_k(\nu_{k})-\omega_k(j)\right) \lt Al\right\}}\\ &=\sum_{l=1}^{\infty} \sum_{j=1}^{\nu_k} F\left(Ak-\left(\omega_k(\nu_{k})-\omega_k(j)\right)\right) \mathbf{1}_{\left\{A(l-1) \leq-\left(\omega_k(\nu_{k})-\omega_k(j)\right) \lt Al\right\}}\\ &\geq \sum_{l=1}^{\infty} F(Ak+Al) \Lambda_{k,l}(\nu,\omega), \end{aligned} $$

where the last inequality holds because F is non-increasing. Thus, we have

$$ \begin{aligned} \sum_{k=1}^{\infty} \chi_k\left(\nu,\omega,F\right) &\geq \sum_{k=1}^{\infty}\sum_{l=1}^{\infty}F(Ak+Al) \Lambda_{k,l}(\nu,\omega)\\ &=\sum_{m=1}^{\infty} F(Am+A) \sum_{k=1}^{m} \Lambda_{k,m+1-k}(\nu,\omega)\\ &=\sum_{m=1}^{\infty} F(Am+A) m Y_m, \end{aligned} $$

where $Y_m\,:\!=\,\sum_{k=1}^m \Lambda_{k,m+1-k}(\nu,\omega) / m$ for all $m \geq 1$ . Note that $\left(\Lambda_{k,{\cdot}}(\nu,\omega), k\geq1\right)$ are i.i.d. under $\tilde{\mathbb{P}}$ , and for all $l\geq1$ , $\Lambda_{1,l}(\nu,\omega)$ has the same law as $\sum_{j=0}^{\Gamma_{1}-1}\mathbf{1}_{\left\{A(l-1)\leq -S_j \lt Al\right\}}$ under $\mu^{\infty}$ . Following the same first- and second-moment argument for $Y_m$ as [Reference Chen9], we obtain that there exists a sufficiently large number $M \gt 0$ so that, for any $m \geq 1$ ,

$$ C_2 \geq \tilde{\mathbb{E}}\left(Y_m \mathbf{1}_{E_M}\right) \geq C_1 \gt 0, $$

where $C_1, C_2$ are positive constants. Therefore, we have

$$ \begin{aligned} M & \geq \tilde{\mathbb{E}}\left(\sum_{k=1}^{\infty} \chi_k\left(\nu,\omega,F\right) \mathbf{1}_{E_M}\right)\\ & \geq \tilde{\mathbb{E}}\left(\sum_{m=1}^{\infty} F(Am+A) m Y_m \mathbf{1}_{E_M}\right)\\ & = \sum_{m=1}^{\infty} F(Am+A) m \tilde{\mathbb{E}}\left(Y_m \mathbf{1}_{E_M}\right)\\ & \geq \sum_{m=1}^{\infty} F(Am+A) m C_1. \end{aligned} $$

This yields

$$\sum_{m=1}^{\infty} F(A m+A) m \leq \frac{M}{C_1} \lt \infty,$$

which implies that $\int_0^{\infty} F(y) y \, \mathrm{d} y \lt \infty$ and completes the proof of (2.11).

3.1. Truncated martingales and change of probabilities

To investigate the limit of the derivative martingale, we introduce a non-negative process with a barrier.

Let $\beta\geq0$ and $V\left(\varnothing\right)=a\geq0$ . We define

$$D^{(\beta)}_{n}\,:\!=\,\sum_{|u|=n}U\left(\theta^{n}\xi,V(u)+\beta\right)\mathrm{e}^{-V(u)}\mathbf{1}_{\left\{\min_{0\leq k\leq n}V(u_{k})\geq-\beta\right\}}, \quad n\geq1,$$

and $D^{(\beta)}_{0}\,:\!=\,U\left(\xi,a+\beta\right)\mathrm{e}^{-a}$ .

The lemma presented below establishes a link between the limits of the truncated martingales and the derivative martingale. Therefore, we can examine the non-triviality of the limit of the derivative martingale by the mean convergence of the truncated martingales.

Since $\big(D^{(\beta)}_{n},n\geq0\big)$ is a non-negative martingale with $\mathbb{E}_{\xi,a}\big(D^{(\beta)}_{n}\big)=U(\xi,a+\beta)\mathrm{e}^{-a}$ , it follows from Kolmogorov’s extension theorem that there exists a unique probability measure $\mathbb{Q}^{(\beta)}_{\xi,a}$ on $\mathcal{F}_{\infty}\,:\!=\,\vee_{n\geq0}\mathcal{F}_{n}$ such that, for all $n\geq1$ ,

(3.1) \begin{equation} \frac{\mathrm{d}\mathbb{Q}^{(\beta)}_{\xi,a}}{\mathrm{d}\mathbb{P}_{\xi,a}}\bigg|_{\mathcal{F}_{n}}\,:\!=\,\frac{D^{(\beta)}_{n}}{U\left(\xi,a+\beta\right)\mathrm{e}^{-a}}.\end{equation}

An intuitive description of the new probability measure is presented in the next subsection.

3.2. Spinal decomposition of the time-inhomogeneous branching random walk

This subsection is devoted to giving a time-inhomogeneous version of the spinal decomposition of the branching random walk. The spinal decomposition method was first introduced by Lyons, Pemantle and Peres [Reference Lyons, Pemantle and Peres24] to investigate Galton–Watson processes. Lyons [Reference Lyons23] subsequently applied this approach to study the additive martingale for the branching random walk. Later, Biggins and Kyprianou [Reference Biggins and Kyprianou7] expanded on this work, adapting it to general martingales based on additive functional of multitype branching.

The spinal decomposition provides an alternative explanation for the distribution of a branching random walk that is biased by a non-negative martingale as a branching random walk with a special infinite ray called the ‘spine’. For clarity, we first give the main steps and then give a detailed description below. Firstly, we define the branching random walk with a random infinite ray $w^{(\beta)}=\big(w^{(\beta)}_{n},n\geq0\big)$ : $w^{(\beta)}_{0}\,:\!=\,\varnothing$ , and $w^{(\beta)}_{n}$ is a child of $w^{(\beta)}_{n-1}$ with $|w^{(\beta)}_{n}|=n$ for each $n\geq1$ . Secondly, we construct a new probability measure via a non-negative martingale. The special individual (spine) reproduces according to the new probability measure, while the other normal particles behave as before. Lastly, we identify this new process as the branching random walk under the new probability measure. The spine approach helps us tackle difficult moment calculations for branching random walks.

Now we introduce the time-inhomogeneous branching random walk with a spine. The process starts with a single particle $w^{(\beta)}_{0}$ at position $V(w^{(\beta)}_{0})=a$ . It dies at time 1 and gives birth to children distributed as $\hat{L}^{(\beta)}_{1,a}$ , whose distribution is the law of $L_{1}$ under $\mathbb{Q}^{(\beta)}_{\xi,a}$ . The particle u is chosen as the spine element $w^{(\beta)}_{1}$ among the children of $w^{(\beta)}_{0}$ with probability proportional to $U\big(\theta\xi,V(u)\big)\mathrm{e}^{-V(u)}\mathbf{1}_{\left\{V(u)\geq-\beta\right\}}$ , while all other children are normal particles. For any $n\geq1$ , each particle alive at generation n dies at time $n+1$ and gives birth independently to children. The children of the normal particle z are distributed as $L_{n+1,V(z)}$ (i.e. $L_{n+1}$ under $\mathbb{P}_{\xi,V(z)}$ ), while the spine element $w^{(\beta)}_{n}$ reproduces according to the point process $\hat{L}^{(\beta)}_{n+1,V(w^{(\beta)}_{n})}$ , which is distributed as $L_{n+1}$ under $\mathbb{Q}^{(\beta)}_{\xi,V(w^{(\beta)}_{n})}$ , and the particle v is chosen as the spine element $w^{(\beta)}_{n+1}$ among the children of $w^{(\beta)}_{n}$ with probability proportional to $U\big(\theta^{n+1}\xi,V(v)+\beta\big)\mathrm{e}^{-V(v)}\mathbf{1}_{\left\{\min_{0\leq k\leq n+1}V(v_{k})\geq-\beta\right\}}$ ; all other children are normal particles. The process continues as described as above. We continue to use $\mathbb{T}$ to denote the genealogical tree. Let $\hat{\mathbb{P}}^{(\beta)}_{\xi,a}$ denote the law of the new process, which is a probability measure on the product of the space of all marked trees and the space of all infinite spines.

The following spinal decomposition consists of an alternative construction of the law $\mathbb{Q}^{(\beta)}_{\xi,a}$ as the projection of the law $\hat{\mathbb{P}}^{(\beta)}_{\xi,a}$ on the space of all marked trees. By an abuse of notation, the marginal law of $\hat{\mathbb{P}}^{(\beta)}_{\xi,a}$ on the space of marked trees is also denoted by $\hat{\mathbb{P}}^{(\beta)}_{\xi,a}$ . This alternative construction allows us to study the mean convergence of the corresponding martingale in Section 4.

Using Proposition 5, we will identify the branching random walk under $\mathbb{Q}^{(\beta)}_{\xi,a}$ with $\hat{\mathbb{P}}^{(\beta)}_{\xi,a}$ in the following.

Proposition 6. (Law of the spine.) Let the spine $w^{(\beta)}=\big(w^{(\beta)}_{n}\big)$ and probability measure $\mathbb{Q}^{(\beta)}_{\xi,a}$ be defined as above. We have the following:

1. (i) For any n and any vertex $v\in\mathbb{T}$ with $|v|=n$ , we have

   $$\mathbb{Q}^{(\beta)}_{\xi,a}\big(w^{(\beta)}_{n}=v\mid \mathcal{F}_{n}\big)=\frac{U\left(\theta^{n}\xi,V(v)+\beta\right)\mathrm{e}^{-V(v)}\mathbf{1}_{\left\{\min_{0\leq k\leq n}V(v_{k})\geq-\beta\right\}}}{D^{(\beta)}_{n}}.$$

2. (ii) The process $\big(V(w^{(\beta)}_{n}),n\geq0\big)$ under $\mathbb{Q}^{(\beta)}_{\xi,a}$ is distributed as the random walk $\left(S_{n},n\geq0\right)$ under $\mathbb{P}_{\xi,a}$ conditioned to stay in $[{-}\beta,\infty)$ . Equivalently, for all n and any measurable function $f:\mathbb{R}^{n+1}\to\mathbb{R}_{+}$ , we have P-a.s.

(3.2) \begin{equation} \mathbb{E}_{\mathbb{Q}^{(\beta)}_{\xi,a}}\left[f\big(V(w^{(\beta)}_{0}), \cdots, V(w^{(\beta)}_{n})\big)\right]=\mathbb{E}_{\xi, a}\left[f\left(S_{0}, \cdots, S_{n}\right)\frac{U(\theta^{n}\xi,S_{n}+\beta)}{U(\xi,a+\beta)}\mathbf{1}_{\left\{\min_{0\leq k\leq n}S_{k}\geq-\beta\right\}}\right].\qquad\quad \end{equation}

Note that $\mathbb{Q}^{(\beta)}_{\xi,a}\big(D^{(\beta)}_{n} \gt 0\big)=\mathbb{E}_{\xi,a}\Big(\frac{D^{(\beta)}_{n}}{U(\xi,a+\beta)\mathrm{e}^{-a}}\Big)=1$ ; the right-hand side of the identity in Proposition 6(i) is $\mathbb{Q}^{(\beta)}_{\xi,a}\text{-a.s.}$ well-defined. For (ii), by (2.4) and (3.2), we obtain the following identity:

\begin{equation*} \mathbb{E}_{\mathbb{Q}^{(\beta)}_{\xi,a}}\left[f\big(V(w^{(\beta)}_{0}), \cdots, V(w^{(\beta)}_{n})\big)\right]=\mathbb{E}_{\xi, a}\left[f\big(\zeta^{(\beta)}_{0}, \cdots, \zeta^{(\beta)}_{n}\big)\right].\end{equation*}

4.1. The sufficient condition

In this subsection, we show that for all $\beta\geq0$ , $D^{(\beta)}_{n}$ converges in $L^{1}\left(\mathbb{P}_{\xi}\right)$ to $D^{(\beta)}_{\infty}$ for almost all $\xi$ under the assumption (1.4).

Proof. According to Proposition 7(i), it suffices to prove that

(4.2) \begin{equation} \sum_{n=1}^{\infty}\mathbb{E}_{\xi,\zeta^{(\beta)}_{n}}\left[\tilde{X}\left(\big(U(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta)\mathrm{e}^{-\zeta^{(\beta)}_{n}}\tilde{X}\big) \wedge 1\right)\right]U(\xi,\beta) \lt \infty, \quad \mathbb{P}\text{-a.s.} \end{equation}

By (2.3) and the fact that, for almost all $\xi$ , $\zeta^{(\beta)}_{n}+\beta\to\infty$ $\mathbb{P}_{\xi}$ -a.s. as $n\to\infty$ , we have, as $n\to\infty$ ,

\begin{equation*} \begin{aligned} &\frac{\sum_{|u|=1}U\big(\theta^{n+1}\xi,\zeta^{(\beta)}_{n}+\Delta V(u)+\beta\big)\mathrm{e}^{-\Delta V(u)}\mathbf{1}_{\left\{\zeta^{(\beta)}_{n}+\Delta V(u)\geq -\beta\right\}}}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\\ \sim&\frac{\sum_{|u|=1}\big(\zeta^{(\beta)}_{n}+\Delta V(u)+\beta\big)\mathrm{e}^{-\Delta V(u)}\mathbf{1}_{\left\{\zeta^{(\beta)}_{n}+\Delta V(u)\geq -\beta\right\}}}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}. \end{aligned} \end{equation*}

Since $\big(\zeta^{(\beta)}_{n}+\Delta V(u)+\beta\big)\mathbf{1}_{\left\{\zeta^{(\beta)}_{n}+\Delta V(u)\geq -\beta\right\}}\leq \zeta^{(\beta)}_{n}+\beta+\Delta V(u)\mathbf{1}_{\left\{\Delta V(u)\geq 0\right\}}$ , we obtain

\begin{align*} &\frac{\sum_{|u|=1}\big(\zeta^{(\beta)}_{n}+\Delta V(u)+\beta\big)\mathrm{e}^{-\Delta V(u)}\mathbf{1}_{\left\{\zeta^{(\beta)}_{n}+\Delta V(u)\geq -\beta\right\}}}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\\ \leq&\frac{\sum_{|u|=1}\big(\zeta^{(\beta)}_{n}+\beta\big)\mathrm{e}^{-\Delta V(u)}}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}+\frac{\sum_{|u|=1}\Delta V(u)\mathrm{e}^{-\Delta V(u)}\mathbf{1}_{\left\{\Delta V(u)\geq 0\right\}}}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)} \end{align*}

\begin{align*} =&:\frac{\big(\zeta^{(\beta)}_{n}+\beta\big)\tilde{Y}(\xi_{n+1})}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}+\frac{\tilde{Z}(\xi_{n+1})}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\\ \leq&2\max\left\{\frac{\big(\zeta^{(\beta)}_{n}+\beta\big)\tilde{Y}(\xi_{n+1})}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)},\frac{\tilde{Z}(\xi_{n+1})}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\right\}, \end{align*}

where $\big(\tilde{Y}(\xi_{n+1}),\tilde{Z}(\xi_{n+1})\big)$ is independent of $\zeta^{(\beta)}_{n}$ under $\mathbb{P}_{\xi}$ .

Therefore, we only need to show that

(4.3) \begin{equation} \begin{cases} \mathbb{E}\left\{\sum_{n=1}^{\infty}\mathbb{E}_{\xi,\zeta^{(\beta)}_{n}}\left[\frac{\big(\zeta^{(\beta)}_{n}+\beta\big)\tilde{Y}(\xi_{n+1})}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\left(\big(\mathrm{e}^{-\zeta^{(\beta)}_{n}}\big(\zeta^{(\beta)}_{n}+\beta\big)\tilde{Y}(\xi_{n+1})\big) \wedge 1\right)\right]U(\xi,\beta)\right\} \lt \infty,\\ \\ \mathbb{E}\left\{\sum_{n=1}^{\infty}\mathbb{E}_{\xi,\zeta^{(\beta)}_{n}}\left[\frac{\tilde{Z}(\xi_{n+1})}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\left(\big(\mathrm{e}^{-\zeta^{(\beta)}_{n}}\tilde{Z}(\xi_{n+1})\big) \wedge 1\right)\right]U(\xi,\beta)\right\} \lt \infty, \end{cases} \end{equation}

which implies (4.2).

To prove the first term of (4.3), by the inequality $\mathrm{e}^{x/2}\geq x$ for all $x \gt 0$ , we have

\begin{equation*} \begin{aligned} &\mathbb{E}\left\{\sum_{n=1}^{\infty}\mathbb{E}_{\xi,\zeta^{(\beta)}_{n}}\left[\frac{\big(\zeta^{(\beta)}_{n}+\beta\big)\tilde{Y}(\xi_{n+1})}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\left(\big(\mathrm{e}^{-\zeta^{(\beta)}_{n}}\big(\zeta^{(\beta)}_{n}+\beta\big)\tilde{Y}(\xi_{n+1})\big) \wedge1\right)\right]U(\xi,\beta)\right\}\\ \leq&\mathbb{E}\left\{\sum_{n=1}^{\infty}\mathbb{E}_{\xi,\zeta^{(\beta)}_{n}}\left[\frac{\big(\zeta^{(\beta)}_{n}+\beta\big)\tilde{Y}(\xi_{n+1})}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\left(\big(\mathrm{e}^{-\zeta^{(\beta)}_{n}/2+\beta/2}\tilde{Y}(\xi_{n+1})\big) \wedge1\right)\right]U(\xi,\beta)\right\}\\ \leq&\mathbb{E}\left\{\sum_{n=1}^{\infty}\mathbb{E}_{\xi}\left[\frac{\big(\zeta^{(\beta)}_{n}+\beta\big)\mathrm{e}^{-\zeta^{(\beta)}_{n}/2+\beta/2}\tilde{Y}^{2}(\xi_{n+1})}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\mathbf{1}_{\left\{\zeta^{(\beta)}_{n}\geq2\log\tilde{Y}(\xi_{n+1})+\beta\right\}}\bigg | \zeta^{(\beta)}_{n}\right]U(\xi,\beta)\right\}\\ &+\mathbb{E}\left\{\sum_{n=1}^{\infty}\mathbb{E}_{\xi}\left[\frac{\big(\zeta^{(\beta)}_{n}+\beta\big)\tilde{Y}(\xi_{n+1})}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\mathbf{1}_{\left\{\zeta^{(\beta)}_{n} \lt 2\log\tilde{Y}(\xi_{n+1})+\beta\right\}}\bigg | \zeta^{(\beta)}_{n}\right]U(\xi,\beta)\right\}. \end{aligned} \end{equation*}

Hence, it follows by Lemma 2 and (2.6) that

\begin{align*} &\mathbb{E}\left\{\sum_{n=1}^{\infty}\mathbb{E}_{\xi,\zeta^{(\beta)}_{n}}\left[\frac{\big(\zeta^{(\beta)}_{n}+\beta\big)\tilde{Y}(\xi_{n+1})}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\left(\big(\mathrm{e}^{-\zeta^{(\beta)}_{n}}\big(\zeta^{(\beta)}_{n}+\beta\big)\tilde{Y}(\xi_{n+1})\big) \wedge1\right)\right]U(\xi,\beta)\right\}\\ \leq&\int_{-\beta}^{\infty}\mathbb{E}\left[\left(x+\beta\right)\mathrm{e}^{-x/2+\beta/2}\tilde{Y}^{2}(\xi_{n+1})\mathbf{1}_{\left\{x\geq2\log\tilde{Y}(\xi_{n+1})+\beta\right\}}\right]\mathcal{R}^{(\beta)}(\mathrm{d}x)\\ &+\int_{-\beta}^{\infty}\mathbb{E}\left[\left(x+\beta\right)\tilde{Y}(\xi_{n+1})\mathbf{1}_{\left\{x \lt 2\log\tilde{Y}(\xi_{n+1})+\beta\right\}}\right]\mathcal{R}^{(\beta)}(\mathrm{d}x)\\ \leq&c_{2}\int_{0}^{\infty}\mathbb{E}\left[x\mathrm{e}^{-x/2+\beta}\tilde{Y}^{2}(\xi_{n+1})\mathbf{1}_{\left\{x\geq2\log\tilde{Y}(\xi_{n+1})+2\beta\right\}}\right]\mathrm{d}x\\ &+c_{2}\int_{0}^{\infty}\mathbb{E}\left[x\tilde{Y}(\xi_{n+1})\mathbf{1}_{\left\{x \lt 2\log\tilde{Y}(\xi_{n+1})+2\beta\right\}}\right]\mathrm{d}x \end{align*}

\begin{align*}=&c_{2}\mathrm{e}^{\beta}\mathbb{E}\left[\tilde{Y}^{2}(\xi_{n+1})\int_{2\left(\log\tilde{Y}(\xi_{n+1})+\beta\right)_{+}}^{\infty}x\mathrm{e}^{-x/2}\,\mathrm{d}x\right]+c_{2}\mathbb{E}\left[\tilde{Y}(\xi_{n+1})\int_{0}^{2\left(\log\tilde{Y}(\xi_{n+1})+\beta\right)_{+}}x\,\mathrm{d}x\right]\\ \leq&c_{1}(\beta)\mathbb{E}\big(\tilde{Y}(\xi_{n+1})\log_{+}\tilde{Y}(\xi_{n+1})\big)+c_{2}(\beta)\mathbb{E}\big(\tilde{Y}(\xi_{n+1})\log^2_{+}\tilde{Y}(\xi_{n+1})\big)\\ =&c_{1}(\beta)\mathbb{E}(Y\log_{+}Y)+c_{2}(\beta)\mathbb{E}(Y\log^2_{+}Y) \lt \infty. \end{align*}

For the second term of (4.3), by the same argument, we have

\begin{equation*} \begin{aligned} &\mathbb{E}\left\{\sum_{n=1}^{\infty}\mathbb{E}_{\xi,\zeta^{(\beta)}_{n}}\left[\frac{\tilde{Z}(\xi_{n+1})}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\left(\big(\mathrm{e}^{-\zeta^{(\beta)}_{n}}\tilde{Z}(\xi_{n+1})\big) \wedge 1\right)\right]U(\xi,\beta)\right\}\\ \leq&c_{3}(\beta)\mathbb{E}\big(\tilde{Z}(\xi_{n+1})\log_{+}\tilde{Z}(\xi_{n+1})\big)\\ =&c_{3}(\beta)\mathbb{E}(Z\log_{+}Z) \lt \infty. \end{aligned} \end{equation*}

This completes the proof of (4.3), and so Lemma 5 is now proved.

Proof of the sufficient condition of Theorem 1(ii). Assume that (1.4) holds. By Lemma 5, for all $\beta\geq0$ , $D^{(\beta)}_{n}$ is $L^{1}\left(\mathbb{P}_{\xi}\right)$ -convergent for almost all $\xi$ . Therefore, in view of Lemma 4(ii), we prove that $D_{\infty}$ is non-degenerate for almost all $\xi$ , which completes the proof of sufficiency.

4.2. The necessary condition

In this subsection, we show that $D^{(\beta)}_{\infty}=0$ $\mathbb{P}\text{-a.s.}$ for all $\beta\geq0$ when (1.4) does not hold.

Proof. According to Proposition 7(ii), it suffices to prove that, for any $c\geq1$ ,

(4.4) \begin{align} & \mathbb{E}\left[Y\log^2_{+}Y+Z\log_{+}Z\right]=\infty\nonumber\\ \Longrightarrow\ & \sum_{n=1}^{\infty}\mathbb{E}_{\zeta^{(\beta)}_{n}}\left[\tilde{X}\mathbf{1}_{\left\{U(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta)\mathrm{e}^{-\zeta^{(\beta)}_{n}}\tilde{X}\geq c\right\}}\right]U(\xi,\beta)=\infty, \quad \mathbb{P}\text{-a.s.} \end{align}

Following the idea of Chen [Reference Chen9], we divide the assumption on the left-hand side of (4.4) into three cases as follows:

(4.5) \begin{equation} \begin{cases} (a)\,\, \mathbb{E}\left[Y\log^2_{+}Y\right]=\infty,\quad \mathbb{E}\left[Y\log_{+}Y\right] \lt \infty,\\ (b)\,\ \mathbb{E}\left[Y\log_{+}Y\right]=\infty,\\ (c)\,\ \mathbb{E}\left[Z\log_{+}Z\right]=\infty. \end{cases} \end{equation}

Note that under $\mathbb{P}_{\xi,\zeta^{(\beta)}_{n}}$ , $\left(\Delta V(u)\,:\,|u|=1\right)$ is distributed as $L_{n+1}$ . For any $x\in\mathbb{R}$ , we define

$$Y_{+}\left(\xi_{n+1},x\right)\,:\!=\,\sum_{|u|=1}\mathrm{e}^{-\Delta V(u)}\mathbf{1}_{\left\{\Delta V(u)\geq -x\right\}},$$

$$Y_{-}\left(\xi_{n+1},x\right)\,:\!=\,\sum_{|u|=1}\mathrm{e}^{-\Delta V(u)}\mathbf{1}_{\left\{\Delta V(u) \lt -x\right\}}.$$

Observe that $\tilde{Y}\left(\xi_{n+1}\right)=Y_{+}\left(\xi_{n+1},x\right)+Y_{-}\left(\xi_{n+1},x\right)$ , and $\left(Y_{+}\left(\xi_{n+1},x\right),Y_{-}\left(\xi_{n+1},x\right),x\in\mathbb{R}\right)$ is independent of $\zeta^{(\beta)}_{n}$ under $\mathbb{P}_{\xi}$ .

Firstly, we give the proof of (4.4) under the assumption (a) of (4.5). By (4.1) and Proposition 1(iv), under $\mathbb{P}_{\xi,\zeta^{(\beta)}_{n}}$ , we have

\begin{equation*} \begin{aligned} \tilde{X}=&\frac{\sum_{|u|=1}U\big(\theta^{n+1}\xi,\zeta^{(\beta)}_{n}+\Delta V(u)+\beta\big)\mathrm{e}^{-\Delta V(u)}\mathbf{1}_{\left\{\zeta^{(\beta)}_{n}+\Delta V(u)+\beta\geq 0\right\}}}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\\ \geq&\frac{\sum_{|u|=1}\big(\zeta^{(\beta)}_{n}+\Delta V(u)+\beta\big)\mathrm{e}^{-\Delta V(u)}\mathbf{1}_{\left\{\Delta V(u)\geq -(\zeta^{(\beta)}_{n}+\beta)/2\right\}}}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\\ \geq&\frac{\sum_{|u|=1}\big(\zeta^{(\beta)}_{n}+\beta\big)\mathrm{e}^{-\Delta V(u)}\mathbf{1}_{\left\{\Delta V(u)\geq -(\zeta^{(\beta)}_{n}+\beta)/2\right\}}}{2U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\\ =&\frac{\zeta^{(\beta)}_{n}+\beta}{2U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}Y_{+}\bigg(\xi_{n+1},\frac{\zeta^{(\beta)}_{n}+\beta}{2}\bigg). \end{aligned} \end{equation*}

Thus we only need to show that, for any fixed $c\geq1$ , $\mathbb{P}\text{-a.s.}$ ,

\begin{equation*} \sum_{n=1}^{\infty}\mathbb{E}_{\zeta^{(\beta)}_{n}}\left[\frac{\big(\zeta^{(\beta)}_{n}+\beta\big)Y_{+}\big(\xi_{n+1},(\zeta^{(\beta)}_{n}+\beta)/2\big)}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\mathbf{1}_{\left\{\mathrm{e}^{-\zeta^{(\beta)}_{n}}(\zeta^{(\beta)}_{n}+\beta)Y_{+}\big(\xi_{n+1},\frac{\zeta^{(\beta)}_{n}+\beta}{2}\big)\geq c\right\}}\right]U(\xi,\beta)=\infty. \end{equation*}

Since, for almost all $\xi$ , $\zeta^{(\beta)}_{n}+\beta\to\infty$ $\mathbb{P}_{\xi}$ -a.s. as $n\to\infty$ , and by (2.3), it suffices to prove that

\begin{equation*} \sum_{n=1}^{\infty}\mathbb{E}\left[Y_{+}\bigg(\xi_{n+1},\frac{\zeta^{(\beta)}_{n}+\beta}{2}\bigg)\mathbf{1}_{\left\{\log Y_{+}\big(\xi_{n+1},\frac{\zeta^{(\beta)}_{n}+\beta}{2}\big)\geq\zeta^{(\beta)}_{n}\right\}}\ \Big|\ \zeta^{(\beta)}_{n}\right]U(\xi,\beta)=\infty,\quad \mathbb{P}\text{-a.s.}, \end{equation*}

which we can write as

(4.6) \begin{equation} \sum_{n=1}^{\infty}F\bigg(\frac{\zeta^{(\beta)}_{n}+\beta}{2},\zeta^{(\beta)}_{n}\bigg)U(\xi,\beta)=\infty, \quad \mathbb{P}\text{-a.s.}, \end{equation}

where $F\left(x,y\right)\,:\!=\,\mathrm{E}\left[F\left(\xi_{n+1},x,y\right)\right]$ and $F\left(\xi_{n+1},x,y\right)\,:\!=\,\mathbb{E}_{\xi}\left[Y_{+}\left(\xi_{n+1},x\right)\right.$ $\left.\mathbf{1}_{\left\{\log Y_{+}\left(\xi_{n+1},x\right)\geq y\right\}}\right]$ , $x,y\in\mathbb{R}$ .

Let

$$F_{1}\left(\xi_{n+1},y\right)\,:\!=\,\mathbb{E}_{\xi}\left[\tilde{Y}\left(\xi_{n+1}\right)\mathbf{1}_{\left\{\log \tilde{Y}\left(\xi_{n+1}\right)\geq y\right\}}\right], \quad y\in\mathbb{R},$$

and $F_{1}\left(y\right)\,:\!=\,\mathrm{E}\left[F_{1}\left(\xi_{n+1},y\right)\right]$ . Note that, by the assumption (1.2),

$$0\leq F\left(\xi_{n+1},x,y\right)\leq F_{1}\left(\xi_{n+1},y\right)\leq \mathbb{E}_{\xi}\big(\tilde{Y}(\xi_{n+1})\big)=1, \quad \mathrm{P}\text{-a.s.}$$

It follows that $F_{1}\left(y\right)$ is a non-negative, non-increasing function. By the assumption (a) of (4.5), we obtain

\begin{align*} \int_{-\beta}^{\infty}F_{1}\left(y\right)(y+\beta)\,\mathrm{d}y&=\int_{-\beta}^{\infty}\mathbb{E}\left[\tilde{Y}\left(\xi_{n+1}\right)\mathbf{1}_{\left\{\log \tilde{Y}\left(\xi_{n+1}\right)\geq y\right\}}\right](y+\beta)\,\mathrm{d}y\\ &=\mathbb{E}\left[\tilde{Y}\left(\xi_{n+1}\right)\int_{-\beta}^{\log_{+}\tilde{Y}\left(\xi_{n+1}\right)}(y+\beta)\,\mathrm{d}y\right] \end{align*}

\begin{align*} &=\frac{\mathbb{E}\left[\tilde{Y}\left(\xi_{n+1}\right)\left(\log_{+}\tilde{Y}\left(\xi_{n+1}\right)+\beta\right)^2\right]}{2}\\ &=\frac{\mathbb{E}\left[Y\left(\log_{+}Y+\beta\right)^2\right]}{2}=\infty.\end{align*}

By Proposition 4, we have

(4.7) \begin{equation} \sum_{n=1}^{\infty}F_{1}\left(\zeta^{(\beta)}_{n}\right)U(\xi,\beta)=\infty, \quad \mathbb{P}\text{-a.s.} \end{equation}

It remains to prove that

\begin{equation*} \sum_{n=1}^{\infty}\left[F_{1}\left(\zeta^{(\beta)}_{n}\right)-F\bigg(\frac{\zeta^{(\beta)}_{n}+\beta}{2},\zeta^{(\beta)}_{n}\bigg)\right]U(\xi,\beta) \lt \infty, \quad \mathbb{P}\text{-a.s.}, \end{equation*}

which, together with (4.7), implies (4.6). Equivalently, we only need to prove that

(4.8) \begin{equation} \sum_{n=1}^{\infty}\frac{\zeta^{(\beta)}_{n}+\beta}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\left[F_{1}\left(\zeta^{(\beta)}_{n}\right)-F\bigg(\frac{\zeta^{(\beta)}_{n}+\beta}{2},\zeta^{(\beta)}_{n}\bigg)\right]U(\xi,\beta) \lt \infty, \quad \mathbb{P}\text{-a.s.} \end{equation}

We begin our proof by giving an upper bound for $F_{1}\left(\xi_{n+1},y\right)-F\left(\xi_{n+1},x,y\right)$ . For any $x,y\in\mathbb{R}$ , we obtain

\begin{equation*} \begin{aligned} &F_{1}\left(\xi_{n+1},y\right)-F\left(\xi_{n+1},x,y\right)\\ =&\mathbb{E}_{\xi}\left[\tilde{Y}\left(\xi_{n+1}\right)\mathbf{1}_{\left\{\log \tilde{Y}\left(\xi_{n+1}\right)\geq y\right\}}\right]-\mathbb{E}_{\xi}\left[Y_{+}\left(\xi_{n+1},x\right)\mathbf{1}_{\left\{\log Y_{+}\left(\xi_{n+1},x\right)\geq y\right\}}\right]\\ =&\mathbb{E}_{\xi}\left[\tilde{Y}\left(\xi_{n+1}\right)\mathbf{1}_{\left\{\log \tilde{Y}\left(\xi_{n+1}\right)\geq y \gt \log Y_{+}\left(\xi_{n+1},x\right)\right\}}\right]+\mathbb{E}_{\xi}\left[\tilde{Y}\left(\xi_{n+1}\right)\mathbf{1}_{\left\{\log Y_{+}\left(\xi_{n+1},x\right)\geq y\right\}}\right]\\ &-\mathbb{E}_{\xi}\left[Y_{+}\left(\xi_{n+1},x\right)\mathbf{1}_{\left\{\log Y_{+}\left(\xi_{n+1},x\right)\geq y\right\}}\right]\\ =&\mathbb{E}_{\xi}\left[\tilde{Y}\left(\xi_{n+1}\right)\mathbf{1}_{\left\{\log \tilde{Y}\left(\xi_{n+1}\right)\geq y \gt \log Y_{+}\left(\xi_{n+1},x\right)\right\}}\right]+\mathbb{E}_{\xi}\left[Y_{-}\left(\xi_{n+1},x\right)\mathbf{1}_{\left\{\log Y_{+}\left(\xi_{n+1},x\right)\geq y\right\}}\right]\\ \leq&\mathbb{E}_{\xi}\left[2Y_{-}\left(\xi_{n+1},x\right)\mathbf{1}_{\left\{\log \tilde{Y}\left(\xi_{n+1}\right)\geq y \gt \log Y_{+}\left(\xi_{n+1},x\right),\ Y_{-}\left(\xi_{n+1},x\right)\geq Y_{+}\left(\xi_{n+1},x\right)\right\}}\right]\\ &+\mathbb{E}_{\xi}\left[\tilde{Y}\left(\xi_{n+1}\right)\mathbf{1}_{\left\{\log \tilde{Y}\left(\xi_{n+1}\right)\geq y \gt \log Y_{+}\left(\xi_{n+1},x\right),\ Y_{-}\left(\xi_{n+1},x\right) \lt Y_{+}\left(\xi_{n+1},x\right)\right\}}\right]\\ &+\mathbb{E}_{\xi}\left[Y_{-}\left(\xi_{n+1},x\right)\mathbf{1}_{\left\{\log Y_{+}\left(\xi_{n+1},x\right)\geq y\right\}}\right]\\ \leq&3\mathbb{E}_{\xi}\left[Y_{-}\left(\xi_{n+1},x\right)\right]+\mathbb{E}_{\xi}\left[\tilde{Y}\left(\xi_{n+1}\right)\mathbf{1}_{\left\{\log \tilde{Y}\left(\xi_{n+1}\right)\geq y \gt \log Y_{+}\left(\xi_{n+1},x\right),\ Y_{-}\left(\xi_{n+1},x\right) \lt Y_{+}\left(\xi_{n+1},x\right)\right\}}\right]\\ \leq&3\mathbb{E}_{\xi}\left[Y_{-}\left(\xi_{n+1},x\right)\right]+\mathbb{E}_{\xi}\left[\tilde{Y}\left(\xi_{n+1}\right)\mathbf{1}_{\left\{\log \tilde{Y}\left(\xi_{n+1}\right)\geq y \gt \log \left(\tilde{Y}\left(\xi_{n+1}\right)/2\right)\right\}}\right]\\ =&:3A_{1}\left(\xi_{n+1},x\right)+A_{2}\left(\xi_{n+1},y\right), \end{aligned} \end{equation*}

where the first and last inequalities follow from the fact that $\tilde{Y}\left(\xi_{n+1}\right)\leq2\max\left\{Y_{+}\left(\xi_{n+1},x\right),Y_{-}\left(\xi_{n+1},x\right)\right\}$ . Therefore,

\begin{equation*} \begin{aligned} &\sum_{n=1}^{\infty}\frac{\zeta^{(\beta)}_{n}+\beta}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\left[F_{1}\big(\zeta^{(\beta)}_{n}\big)-F\bigg(\frac{\zeta^{(\beta)}_{n}+\beta}{2},\zeta^{(\beta)}_{n}\bigg)\right]U(\xi,\beta)\\ \leq&\sum_{n=1}^{\infty}\frac{\zeta^{(\beta)}_{n}+\beta}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\mathrm{E}\left[3A_{1}\bigg(\xi_{n+1},\frac{\zeta^{(\beta)}_{n}+\beta}{2}\bigg)+A_{2}\big(\xi_{n+1},\zeta^{(\beta)}_{n}\big)\right]U(\xi,\beta). \end{aligned} \end{equation*}

Taking the expectation on both sides, we have

\begin{equation*} \begin{aligned} &\mathbb{E}\left\{\sum_{n=1}^{\infty}\frac{\zeta^{(\beta)}_{n}+\beta}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\left[F_{1}\big(\zeta^{(\beta)}_{n}\big)-F\bigg(\frac{\zeta^{(\beta)}_{n}+\beta}{2},\zeta^{(\beta)}_{n}\bigg)\right]U(\xi,\beta)\right\}\\ \leq&\mathbb{E}\left\{\sum_{n=1}^{\infty}\frac{\zeta^{(\beta)}_{n}+\beta}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\mathrm{E}\left[3A_{1}\bigg(\xi_{n+1},\frac{\zeta^{(\beta)}_{n}+\beta}{2}\bigg)+A_{2}\big(\xi_{n+1},\zeta^{(\beta)}_{n}\big)\right]U(\xi,\beta)\right\}. \end{aligned} \end{equation*}

Then, using Lemma 2, we deduce that

\begin{equation*} \begin{aligned} &\mathbb{E}\left\{\sum_{n=1}^{\infty}\frac{\zeta^{(\beta)}_{n}+\beta}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\left[F_{1}\big(\zeta^{(\beta)}_{n}\big)-F\bigg(\frac{\zeta^{(\beta)}_{n}+\beta}{2},\zeta^{(\beta)}_{n}\bigg)\right]U(\xi,\beta)\right\}\\ \leq&\int_{-\beta}^{\infty}3(x+\beta)\mathrm{E}\left[A_{1}\bigg(\xi_{n+1},\frac{x+\beta}{2}\bigg)\right]\mathcal{R}^{(\beta)}(\mathrm{d}x)+\int_{-\beta}^{\infty}(x+\beta)\mathrm{E}\left[A_{2}\left(\xi_{n+1},x\right)\right]\mathcal{R}^{(\beta)}(\mathrm{d}x). \end{aligned} \end{equation*}

Our aim is to prove that the values of the two integrals in the last equality are finite, which completes the proof of (4.8). For the first term, by the many-to-one lemma, we have

(4.9) \begin{equation} \mathrm{E}\left[A_{1}\left(\xi_{n+1},x\right)\right]=\mathbb{E}\left[Y_{-}\left(\xi_{n+1},x\right)\right]=\mathbb{E}\left[\sum_{|u|=1}\mathrm{e}^{-\Delta V(u)}\mathbf{1}_{\left\{\Delta V(u) \lt -x\right\}}\right]=\mathbb{P}\left(S_{1} \lt -x\right). \end{equation}

Thus, from (4.9), (2.6), and the assumption (1.3), we obtain

\begin{equation*} \begin{aligned} \int_{-\beta}^{\infty}(x+\beta)\mathrm{E}\left[A_{1}\left(\xi_{n+1},\frac{x+\beta}{2}\right)\right]\mathcal{R}^{(\beta)}(\mathrm{d}x)=&\int_{-\beta}^{\infty}(x+\beta)\mathbb{P}\left(S_{1} \lt -\frac{x+\beta}{2}\right)\mathcal{R}^{(\beta)}(\mathrm{d}x)\\ \leq&c_{2}\int_{0}^{\infty}x\mathbb{P}\left(S_{1} \lt -\frac{x}{2}\right)\mathrm{d}x\\ =&c_{2}\mathbb{E}\left[\int_{0}^{2({-}S_{1})_{+}}x\,\mathrm{d}x\right]\\ =&2c_{2}\mathbb{E}\left[\left(({-}S_{1})_{+}\right)^{2}\right] \lt \infty. \end{aligned} \end{equation*}

For the second term, by (2.6) and the assumption (a) of (4.5), we have

\begin{equation*} \begin{aligned} &\int_{-\beta}^{\infty}(x+\beta)\mathrm{E}\left[A_{2}\left(\xi_{n+1},x\right)\right]\mathcal{R}^{(\beta)}(\mathrm{d}x)\\ =&\int_{-\beta}^{\infty}(x+\beta)\mathbb{E}\left[\tilde{Y}\left(\xi_{n+1}\right)\mathbf{1}_{\left\{\log \tilde{Y}\left(\xi_{n+1}\right)\geq x \gt \log \left(\tilde{Y}\left(\xi_{n+1}\right)/2\right)\right\}}\right]\mathcal{R}^{(\beta)}(\mathrm{d}x)\\ \leq&c_{2}\int_{0}^{\infty}x\mathbb{E}\left[\tilde{Y}\left(\xi_{n+1}\right)\mathbf{1}_{\left\{\log \tilde{Y}\left(\xi_{n+1}\right)+\beta \geq x \gt \log \left(\tilde{Y}\left(\xi_{n+1}\right)/2\right)+\beta\right\}}\right]\mathrm{d}x\\ =&c_{2}\mathbb{E}\left[\tilde{Y}\left(\xi_{n+1}\right)\int_{\left(\log\left(\tilde{Y}(\xi_{n+1})/2\right)+\beta\right)_{+}}^{\left(\log\tilde{Y}(\xi_{n+1})+\beta\right)_{+}}x\,\mathrm{d}x\right]\\ \leq&c_{4}(\beta)\mathbb{E}\left[Y\log_{+}Y\right] \lt \infty. \end{aligned} \end{equation*}

We conclude that (4.4) holds for the first case in (4.5).

Secondly, we give the proof of (4.4) under the assumption (b) of (4.5). By (4.1) and Proposition 1 (iv), under $\mathbb{P}_{\xi,\zeta^{(\beta)}_{n}}$ , we get

\begin{equation*} \begin{aligned} \tilde{X}=&\frac{\sum_{|u|=1}U\big(\theta^{n+1}\xi,\zeta^{(\beta)}_{n}+\Delta V(u)+\beta\big)\mathrm{e}^{-\Delta V(u)}\mathbf{1}_{\left\{\zeta^{(\beta)}_{n}+\Delta V(u)+\beta\geq 0\right\}}}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\\ \geq&\frac{\sum_{|u|=1}\big(\zeta^{(\beta)}_{n}+\Delta V(u)+\beta\big)\mathrm{e}^{-\Delta V(u)}\mathbf{1}_{\left\{\zeta^{(\beta)}_{n}+\Delta V(u)+\beta\geq 1\right\}}}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\\ \geq&\frac{\sum_{|u|=1}\mathrm{e}^{-\Delta V(u)}\mathbf{1}_{\left\{\Delta V(u)\geq -\left(\zeta^{(\beta)}_{n}+\beta-1\right)\right\}}}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\\ =&\frac{Y_{+}\big(\xi_{n+1},\zeta^{(\beta)}_{n}+\beta-1\big)}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}. \end{aligned} \end{equation*}

Hence, it suffices to prove that, for any fixed $c\geq1$ ,

\begin{equation*} \sum_{n=1}^{\infty}\mathbb{E}_{\zeta^{(\beta)}_{n}}\left[\frac{Y_{+}\big(\xi_{n+1},\zeta^{(\beta)}_{n}+\beta-1\big)}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\mathbf{1}_{\left\{\mathrm{e}^{-\zeta^{(\beta)}_{n}}Y_{+}(\xi_{n+1},\zeta^{(\beta)}_{n}+\beta-1)\geq c\right\}}\right]U(\xi,\beta)=\infty, \quad \mathbb{P}\text{-a.s.} \end{equation*}

Since, for almost all $\xi$ , $\zeta^{(\beta)}_{n}+\beta\to\infty$ $\mathbb{P}_{\xi}$ -a.s. as $n\to\infty$ , and by (2.3), this is equivalent to

\begin{equation*} \sum_{n=1}^{\infty}\mathbb{E}\left[\frac{Y_{+}\big(\xi_{n+1},\zeta^{(\beta)}_{n}+\beta-1\big)}{\zeta^{(\beta)}_{n}+\beta+1}\mathbf{1}_{\left\{\log Y_{+}(\xi_{n+1},\zeta^{(\beta)}_{n}+\beta-1)\geq\log c+\zeta^{(\beta)}_{n}\right\}}\ \Big|\ \zeta^{(\beta)}_{n}\right]U(\xi,\beta)=\infty, \quad \mathbb{P}\text{-a.s.}, \end{equation*}

which we can write as

(4.10) \begin{equation} \sum_{n=1}^{\infty}\frac{F\big(\zeta^{(\beta)}_{n}+\beta-1,\log c+\zeta^{(\beta)}_{n}\big)}{\zeta^{(\beta)}_{n}+\beta+1}U(\xi,\beta)=\infty, \quad \mathbb{P}\text{-a.s.}, \end{equation}

recalling that $F\left(x,\log c+y\right)=\mathbb{E}\left[Y_{+}\left(\xi_{n+1},x\right)\mathbf{1}_{\left\{\log Y_{+}\left(\xi_{n+1},x\right)\geq \log c+y\right\}}\right]$ , $x,y\in\mathbb{R}$ .

Let

$$F_{2}\left(\xi_{n+1},y\right)\,:\!=\,\frac{\mathbb{E}_{\xi}\left[\tilde{Y}\left(\xi_{n+1}\right)\mathbf{1}_{\left\{\log \tilde{Y}\left(\xi_{n+1}\right)\geq \log c+y\right\}}\right]}{y+\beta+1}=\frac{F_{1}\left(\xi_{n+1},\log c+y\right)}{y+\beta+1}, \quad y\geq -\beta,$$

and let $F_{2}\left(y\right)\,:\!=\,\mathrm{E}\left[F_{2}\left(\xi_{n+1},y\right)\right]$ . Clearly $F_{2}\left(y\right)$ is non-increasing and

$$0\leq F_{2}\left(y\right)=\mathrm{E}\left[\frac{F_{1}\left(\xi_{n+1},\log c+y\right)}{y+\beta+1}\right]=\frac{F_{1}\left(\log c+y\right)}{y+\beta+1}\leq1.$$

By the assumption (b) of (4.5), we have

\begin{align*} \int_{-\beta}^{\infty}F_{2}\left(y\right)(y+\beta)\,\mathrm{d}y&=\int_{-\beta}^{\infty}\frac{F_{1}\left(\log c+y\right)}{y+\beta+1}(y+\beta)\,\mathrm{d}y\\ &\geq \int_{1}^{\infty}\frac{F_{1}\left(\log c+y\right)}{y+\beta+1}(y+\beta)\,\mathrm{d}y\\ &\geq \frac{1}{2}\int_{1}^{\infty}F_{1}\left(\log c+y\right)\,\mathrm{d}y \end{align*}

\begin{align*} &=\frac{1}{2}\int_{1}^{\infty}\mathbb{E}\left[\tilde{Y}\left(\xi_{n+1}\right)\mathbf{1}_{\left\{\log \tilde{Y}\left(\xi_{n+1}\right)\geq \log c+y\right\}}\right]\,\mathrm{d}y\\ &=\frac{1}{2}\mathbb{E}\left[\tilde{Y}\left(\xi_{n+1}\right)\big(\log(\tilde{Y}\left(\xi_{n+1}\right)/c)-1\big)_{+}\right]=\infty, \end{align*}

which implies $\int_{-\beta}^{\infty}F_{2}\left(y\right)(y+\beta)\,\mathrm{d}y=\infty$ . By Proposition 4, we get

(4.11) \begin{equation} \sum_{n=1}^{\infty}\frac{F_{1}\big(\log c+\zeta^{(\beta)}_{n}\big)}{\zeta^{(\beta)}_{n}+\beta+1}U(\xi,\beta)=\infty, \quad \mathbb{P}\text{-a.s.} \end{equation}

It remains to prove that

\begin{equation*} \sum_{n=1}^{\infty}\frac{F_{1}\big(\log c+\zeta^{(\beta)}_{n}\big)-F\big(\zeta^{(\beta)}_{n}+\beta-1,\log c+\zeta^{(\beta)}_{n}\big)}{\zeta^{(\beta)}_{n}+\beta+1}U(\xi,\beta) \lt \infty, \quad \mathbb{P}\text{-a.s.}, \end{equation*}

which, combined with (4.11), implies (4.10). Equivalently, we only need to prove that

(4.12) \begin{equation} \sum_{n=1}^{\infty}\frac{F_{1}\big(\log c+\zeta^{(\beta)}_{n}\big)-F\big(\zeta^{(\beta)}_{n}+\beta-1,\log c+\zeta^{(\beta)}_{n}\big)}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}U(\xi,\beta) \lt \infty, \quad \mathbb{P}\text{-a.s.} \end{equation}

By the same argument as in the proof of the first part, we have

\begin{equation*} \begin{aligned} &\sum_{n=1}^{\infty}\frac{F_{1}\big(\log c+\zeta^{(\beta)}_{n}\big)-F\big(\zeta^{(\beta)}_{n}+\beta-1,\log c+\zeta^{(\beta)}_{n}\big)}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}U(\xi,\beta)\\ \leq&\sum_{n=1}^{\infty}\frac{\mathrm{E}\left[3A_{1}\big(\xi_{n+1},\zeta^{(\beta)}_{n}+\beta-1\big)+A_{2}\big(\xi_{n+1},\log c+\zeta^{(\beta)}_{n}\big)\right]}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}U(\xi,\beta). \end{aligned} \end{equation*}

Taking the expectation on both sides, we get

\begin{equation*} \begin{aligned} &\mathbb{E}\left\{\sum_{n=1}^{\infty}\frac{F_{1}\big(\log c+\zeta^{(\beta)}_{n}\big)-F\big(\zeta^{(\beta)}_{n}+\beta-1,\log c+\zeta^{(\beta)}_{n}\big)}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}U(\xi,\beta)\right\}\\ \leq&\mathbb{E}\left\{\sum_{n=1}^{\infty}\frac{\mathrm{E}\left[3A_{1}\big(\xi_{n+1},\zeta^{(\beta)}_{n}+\beta-1\big)+A_{2}\big(\xi_{n+1},\log c+\zeta^{(\beta)}_{n}\big)\right]}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}U(\xi,\beta)\right\}. \end{aligned} \end{equation*}

Then, by Lemma 2, we have

\begin{equation*} \begin{aligned} &\mathbb{E}\left\{\sum_{n=1}^{\infty}\frac{F_{1}\big(\xi_{n+1},\log c+\zeta^{(\beta)}_{n}\big)-F\big(\xi_{n+1},\zeta^{(\beta)}_{n}+\beta-1,\log c+\zeta^{(\beta)}_{n}\big)}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}U(\xi,\beta)\right\}\\ \leq&\int_{-\beta}^{\infty}\left[3\mathrm{E}\left(A_{1}\left(\xi_{n+1},x+\beta-1\right)\right)+\mathrm{E}\left(A_{2}\left(\xi_{n+1},\log c+x\right)\right)\right]\mathcal{R}^{(\beta)}(\mathrm{d}x). \end{aligned} \end{equation*}

We now turn to proving the finiteness of the above two integrals, which completes the proof of (4.12). For the first integral, by (4.9) and (2.6), we obtain

\begin{align*} \int_{-\beta}^{\infty}\mathrm{E}\left[A_{1}\left(\xi_{n+1},x+\beta-1\right)\right]\mathcal{R}^{(\beta)}(\mathrm{d}x)=&\int_{-\beta}^{\infty}\mathbb{P}\left(S_{1} \lt -(x+\beta-1)\right)\mathcal{R}^{(\beta)}(\mathrm{d}x)\\ \leq&c_{2}\int_{0}^{\infty}\mathbb{P}\left(S_{1} \lt -(x-1)\right)\,\mathrm{d}x \end{align*}

\begin{align*} =&c_{2}\mathbb{E}\left[\int_{0}^{({-}S_{1}+1)_{+}}\,\mathrm{d}x\right]\\ =&c_{2}\mathbb{E}\left[({-}S_{1}+1)_{+}\right] \lt \infty. \end{align*}

For the second integral, by (2.6), we get

\begin{equation*} \begin{aligned} &\int_{-\beta}^{\infty}\mathrm{E}\left[A_{2}\left(\xi_{n+1},\log c+x\right)\right]\mathcal{R}^{(\beta)}(\mathrm{d}x)\\ =&\int_{-\beta}^{\infty}\mathbb{E}\left[\tilde{Y}\left(\xi_{n+1}\right)\mathbf{1}_{\left\{\log \tilde{Y}\left(\xi_{n+1}\right)\geq \log c+x \gt \log \left(\tilde{Y}\left(\xi_{n+1}\right)/2\right)\right\}}\right]\mathcal{R}^{(\beta)}(\mathrm{d}x)\\ \leq&c_{2}\int_{0}^{\infty}\mathbb{E}\left[\tilde{Y}\left(\xi_{n+1}\right)\mathbf{1}_{\left\{\log \tilde{Y}\left(\xi_{n+1}\right)+\beta\geq \log c+x \gt \log \left(\tilde{Y}\left(\xi_{n+1}\right)/2\right)+\beta\right\}}\right]\mathrm{d}x\\ =&c_{2}\mathbb{E}\left[\tilde{Y}\left(\xi_{n+1}\right)\int_{\left(\log\left(\tilde{Y}(\xi_{n+1})/2c\right)+\beta\right)_{+}}^{\left(\log\left(\tilde{Y}(\xi_{n+1})/c\right)+\beta\right)_{+}}\,\mathrm{d}x\right]\\ \leq&c_{5}(\beta)\mathbb{E}\left[\tilde{Y}\left(\xi_{n+1}\right)\right] \lt \infty. \end{aligned} \end{equation*}

We obtain that (4.4) holds for the second case in (4.5).

Finally, we give the proof of (4.4) under the assumption (c) of (4.5). By (4.1) and Proposition 1(iv), under $\mathbb{P}_{\xi,\zeta^{(\beta)}_{n}}$ , we get

\begin{equation*} \begin{aligned} \tilde{X}=&\frac{\sum_{|u|=1}U\big(\theta^{n+1}\xi,\zeta^{(\beta)}_{n}+\Delta V(u)+\beta\big)\mathrm{e}^{-\Delta V(u)}\mathbf{1}_{\left\{\zeta^{(\beta)}_{n}+\Delta V(u)+\beta\geq 0\right\}}}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\\ \geq&\frac{\sum_{|u|=1}\big(\zeta^{(\beta)}_{n}+\Delta V(u)+\beta\big)\mathrm{e}^{-\Delta V(u)}\mathbf{1}_{\left\{\Delta V(u)\geq 0\right\}}}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\\ \geq&\frac{\sum_{|u|=1}\Delta V(u)\mathrm{e}^{-\Delta V(u)}\mathbf{1}_{\left\{\Delta V(u)\geq 0\right\}}}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\\ =&\frac{\tilde{Z}\left(\xi_{n+1}\right)}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}. \end{aligned} \end{equation*}

Hence we just need to prove that, for any fixed $c\geq1$ ,

\begin{equation*} \sum_{n=1}^{\infty}\mathbb{E}\left[\frac{\tilde{Z}\left(\xi_{n+1}\right)}{U\big(\theta^{n}\xi,\zeta^{(\beta)}_{n}+\beta\big)}\mathbf{1}_{\left\{\mathrm{e}^{-\zeta^{(\beta)}_{n}}\tilde{Z}\left(\xi_{n+1}\right)\geq c\right\}}\ \bigg|\ \zeta^{(\beta)}_{n}\right]U(\xi,\beta)=\infty, \quad \mathbb{P}\text{-a.s.} \end{equation*}

Since, for almost all $\xi$ , $\zeta^{(\beta)}_{n}+\beta\to\infty$ $\mathbb{P}_{\xi}$ -a.s. as $n\to\infty$ , and by (2.3), this is equivalent to

\begin{equation*} \sum_{n=1}^{\infty}\mathbb{E}\left[\frac{\tilde{Z}\left(\xi_{n+1}\right)}{\zeta^{(\beta)}_{n}+\beta+1}\mathbf{1}_{\left\{\log \tilde{Z}\left(\xi_{n+1}\right)\geq \log c+\zeta^{(\beta)}_{n}\right\}}\ \bigg|\ \zeta^{(\beta)}_{n}\right]U(\xi,\beta)=\infty, \quad \mathbb{P}\text{-a.s.}, \end{equation*}

which can be written as

(4.13) \begin{equation} \sum_{n=1}^{\infty}F_{3}\big(\zeta^{(\beta)}_{n}\big)U(\xi,\beta)=\infty, \quad \mathbb{P}\text{-a.s.}, \end{equation}

where

$$F_{3}\left(x\right)\,:\!=\,\frac{\mathbb{E}\left[\tilde{Z}\left(\xi_{n+1}\right)\mathbf{1}_{\left\{\log \tilde{Z}\left(\xi_{n+1}\right)\geq \log c+x\right\}}\right]}{x+\beta+1}, \quad x\geq -\beta.$$

It is obvious that $F_{3}(x)$ is non-increasing and

$$0\leq F_{3}\left(x\right)=\frac{\mathbb{E}\left[\tilde{Z}\left(\xi_{n+1}\right)\mathbf{1}_{\left\{\log \tilde{Z}\left(\xi_{n+1}\right)\geq \log c+x\right\}}\right]}{x+\beta+1}\leq\mathbb{E}(Z)=\mathbb{E}\left[(S_{1})_{+}\right] \lt \infty.$$

Observe that, by the assumption (c) of (4.5),

\begin{equation*} \begin{aligned} \int_{-\beta}^{\infty}F_{3}\left(x\right)(x+\beta)\,\mathrm{d}x&\geq\int_{1}^{\infty}\frac{\mathbb{E}\left[\tilde{Z}\left(\xi_{n+1}\right)\mathbf{1}_{\left\{\log \tilde{Z}\left(\xi_{n+1}\right)\geq\log c+x\right\}}\right]}{x+\beta+1}(x+\beta)\,\mathrm{d}x\\ &\geq\frac{1}{2}\int_{1}^{\infty}\mathbb{E}\left[\tilde{Z}\left(\xi_{n+1}\right)\mathbf{1}_{\left\{\log \tilde{Z}\left(\xi_{n+1}\right)\geq \log c+x\right\}}\right]\,\mathrm{d}x\\ &=\frac{1}{2}\mathbb{E}\left[\tilde{Z}\left(\xi_{n+1}\right)\big(\log(\tilde{Z}\left(\xi_{n+1}\right)/c)-1\big)_{+}\right]=\infty. \end{aligned} \end{equation*}

By Proposition 4, it follows that (4.13) is valid. Therefore, we conclude that (4.4) holds for the third case in (4.5).

Proof of the necessary condition of Theorem 1(ii). By Lemma 6, for almost all $\xi$ , $D^{(\beta)}_{\infty}=0$ $\mathbb{P}_{\xi}$ -a.s. for all $\beta\geq0$ when (1.4) does not hold. Therefore, using Lemma 4(iii), we obtain that $D_{\infty}$ is degenerate for almost all $\xi$ , which completes the proof of necessity.