Proof. Due to the Markov property of \((M_i)_{i\in\mathbb{N}}\) with transition kernel \(p\): \[\begin{align} \lim_{n\to\infty}\frac{1}{c(\theta,a)\lambda^{n}} \, p_n^{a,x}(\theta) &=\lim_{n\to\infty}\frac{1}{c(\theta,a)\lambda^{n}}\,\mathbb{P}_x(M_1\in S,\dots,M_n\in S) \\&=\lim_{n\to\infty}\frac{1}{c(\theta,a)\lambda^{n}}\int\mathbb{P}_y(M_1\in S,\dots M_{n-1}\in S)\mathbb{1}_{\{(y,x)\in S\}}p(x,\mathrm{d}y) \\&=\lim_{n\to\infty}\int_{\{y>\theta x\}\cap[-a,1]}\frac{\mathbb{P}_y(M_1\in S,\dots M_{n-1}\in S)}{c(\theta,a)\lambda^{n}}\,\frac{1}{a+1}\,\mathrm{d}y \\&\overset{ }{=}\int_{\{y>\theta x\}\cap[-a,1]}\lim_{n\to\infty}\frac{\mathbb{P}_y(M_1\in S,\dots M_{n-1}\in S)}{c(\theta,a)\lambda^{n}}\frac{1}{a+1}\,\mathrm{d}y \\&=\int_{\{y>\theta x\}\cap[-a,1]}\lim_{n\to\infty}\frac{p_{n-1}^{a,y}(\theta)}{c(\theta,a)\lambda^{n}}\frac{1}{a+1}\,\mathrm{d}y \\&=\int_{\{y>\theta x\}\cap[-a,1]}h(y)\,\frac{1}{a+1}\,\mathrm{d}y, \end{align}\] where we used Lemma 3 to find an integrable majorant for the exchange of limit and integral and we used the assumption in the last step. ◻

Proof. By Lemma 4 and the assumption, the function \(h\) satisfies equation ?? .
Let \(C_1\dots , C_m\in\mathcal{B}(\mathbb{R})\), \(\mathbf{x}\mathrel{\vcenter{:}}=(x_1,\dots,x_m)\in [-a,1]^m\) and set \(x_0\mathrel{\vcenter{:}}= y\). For fixed \(m\in\mathbb{N}\), we have, by the Markov property, \[\begin{align} &\lim_{n\to\infty}\mathbb{P}_{y}(M_1\in C_1\times\{y\},M_2\in C_2\times C_1,\dots,M_m\in C_m\times C_{m-1}~|\Omega_n) \\ &=\lim_{n\to\infty}\int_{[-a,1]^m}\mathbb{1}_{C_1\times\dots\times C_m}(\mathbf{x})\left[\prod_{i=1}^m p(x_{i-1},\mathrm{d}x_i)\mathbb{1}_S(x_i,x_{i-1})\right]\frac{\mathbb{P}_{x_m}(\Omega_{n-m})}{\mathbb{P}_{x_0}(\Omega_n)} \\&=\int_{[-a,1]^m}\mathbb{1}_{C_1\times\dots\times C_m}(\mathbf{x})\left[\prod_{i=1}^m p(x_{i-1},\mathrm{d}x_i)\mathbb{1}_S(x_i,x_{i-1})\right]\lim_{n\to\infty}\frac{\mathbb{P}_{x_m}(\Omega_{n-m})}{\mathbb{P}_{x_0}(\Omega_n)} \\&=\int_{[-a,1]^m}\mathbb{1}_{C_1\times\dots\times C_m}(\mathbf{x})\left[\prod_{i=1}^m \frac{1}{a+1}\,\mathrm{d}x_i\mathbb{1}_S(x_i,x_{i-1})\right]\frac{h(x_m)}{h(x_0)}\lambda^{-m} \\&=\int_{[-a,1]^m}\mathbb{1}_{C_1\times\dots\times C_m}(\mathbf{x})\left[\prod_{i=1}^m \mathbb{1}_S(x_i,x_{i-1})\frac{h(x_i)}{h(x_{i-1})}\frac{1}{\lambda}\frac{1}{a+1}\,\mathrm{d}x_i\right], \end{align}\] where we used Lemma 3 to find an integrable majorant for the exchange of limit and integral and we used the assumption on the asymptotics of \((p_n^{a,x}(\theta))\) in the third step. Thus, the process with density given by (?? ) is the limiting process \(M'\). One can check that the density is a Markov kernel using the eigenvalue equation ?? . ◻

Proof. From a recursion of the probabilities \((p_n^{a,y}(\theta))\), we will deduce a differential equation for the generating function which has the unique solution \(\frac{E\left(\theta,\frac{-\theta zy}{a+1}\right)}{E\left(\theta,\frac{-z}{a+1}\right)}\).
By the choice of the parameters in the blue region, we get that \(y\in[-a,1]\) implies \(\theta y\in[-a,1]\). Consider \(y\) such that \(\theta y\in[-a,1]\). For all \(n\geq 1\), it holds \[\begin{align} p_n^{a,y}(\theta)&=\frac{1}{(a+1)^n}\int_{\theta y}^1\int_{\theta x_2}^1\dots\int_{\theta x_n}^1 \mathrm{d}x_{n+1}\dots \mathrm{d}x_3\mathrm{d}x_2\\&=\frac{1}{a+1}\int_{\theta y}^1\left[\frac{1}{(a+1)^{n-1}}\int_{\theta x_2}^1\dots\int_{\theta x_n}^1 \mathrm{d}x_{n+1}\dots \mathrm{d}x_3\right]\mathrm{d}x_2=\frac{1}{a+1}\int_{\theta y}^1p_{n-1}^{a,x_2}(\theta)\mathrm{d}x_2. \end{align}\] Multiplying with \(z^n\), summing over \(n\) and taking the derivative with respect to \(y\) gives \[\begin{align} \frac{\partial}{\partial y}\hat{P}_{a,y,\theta}(z)\notag &=\frac{\partial}{\partial y}\sum_{n=1}^{\infty} p_n^{a,y}(\theta)\cdot z^n = \frac{1}{a+1}~ \frac{\partial}{\partial y}\int_{\theta y}^1\left(\sum_{n=1}^{\infty} p_{n-1}^{a,x_2}(\theta)\cdot z^n\right) \mathrm{d}x_2\notag \\ & =\frac{z}{a+1}~ \frac{\partial}{\partial y}\int_{\theta y}^1\hat{P}_{a,x_2,\theta}(z) \mathrm{d}x_2 \notag =\frac{-\theta z}{a+1}\hat{P}_{a,\theta y,\theta}(z), \end{align}\] where we used that \(p_0^{a,y}(\theta) z^0=1\) is constant in the first step and the fact that the integral is bounded and the series is absolutely convergent within the disk of convergence so that we can exchange the sum and the integral. In conclusion, we found a differential equation for the function \(\hat{P}_{a,y,\theta}(z)\) with the additional constraint \(\hat{P}_{a,\frac{1}{\theta},\theta}(z)=1\) (coming from the definition of the generating function and \(p_0^{a,y}(\theta)=1\) and \(p_n^{a,\frac{1}{\theta}}(\theta)=0\) for \(n\ge 1\)). The unique solution to this differential equation with the boundary condition is \(y\mapsto\frac{E\left(\theta,\frac{-\theta zy}{a+1}\right)}{E\left(\theta,\frac{-z}{a+1}\right)}\), as can be checked by taking derivatives (applying ?? ) and checking the additional constraint. Since \(\frac{1}{\lambda}\) is the unique zero with smallest modulus of the denominator and the numerator has no zero in \(\frac{1}{\lambda}\) and no singularities, the radius of convergence is given by \(\frac{1}{\lambda}\). ◻

Proof. We want to use Lemma 8, so we need to check that there is a function \(f(z)\) with \[\begin{align} \label{asymptotic95blue95proof} \sum_{n=0}^{\infty}p_n^{a,y}(\theta)z^n=\frac{f(z)}{z_0-z}~~~~\text{for all}~|z|<z_0=\frac{1}{\lambda} \end{align}\tag{5}\] which is analytic on \(|z|<z_1\) with some \(z_1>\frac{1}{\lambda}\) and \(f\left(z_0\right)\neq 0\).
Recall that \(\sum_{n=0}^{\infty}p_n^{a,y}(\theta)z^n=\frac{E\left(\theta,\frac{-\theta zy}{a+1}\right)}{E\left(\theta,\frac{-z}{a+1}\right)}\). In order to find \(f\) such that 5 holds, note that \(z_0\) is the zero with smallest modulus of the denominator \(g(z)\) of the generating function. We also know from Lemma 7 that \(z_0\) is a simple zero of \(g\) and that the zero with the next-largest modulus has modulus \(z_1>z_0\); so that \(g(z)/(z_0-z)\) is analytic on \(|z|<z_1\). Therefore \[u(z)\mathrel{\vcenter{:}}= E\left(\theta,\frac{-\theta zy}{a+1}\right)~~\text{ and }~~v(z)\mathrel{\vcenter{:}}= g(z)\frac{1}{z_0-z},\] are analytic on \(|z|<z_1\). We rewrite \(\sum_{n=0}^{\infty}p_n^{a,y}(\theta)z^n=\frac{u(z)}{v(z)}\frac{1}{z_0-z}\). In order to show that the function \(f(z)\mathrel{\vcenter{:}}=\frac{u(z)}{v(z)}\) satisfies the assumptions of Lemma 8 it remains to be seen that \(f(z_0)\neq 0\). Let us first compute \[\begin{align} v(z_0)=-g'(z_0)=-E\left(\theta,\frac{-\theta z_0}{a+1}\right)\frac{-1}{a+1} = \frac{E\left(\theta,\frac{-\theta z_0}{a+1}\right)}{a+1}, \end{align}\] which is non-zero: Indeed, if \(\theta>0\), we have \(0<\theta z_0<z_0\) and \(z_0\) is the smallest positive zero of \(z\mapsto E\left(\theta,\frac{-z}{a+1}\right)\). If \(\theta<0\), we have \(|\theta z_0|\le z_0\) and \(\theta z_0\neq z_0\), so \(\theta z_0\) is no zero of \(z\mapsto E\left(\theta,\frac{-z}{a+1}\right)\). In particular, this gives \[\begin{align} f(z_0)=\frac{a+1}{E\left(\theta,\frac{-\theta z_0}{a+1}\right)}E\left(\theta,\frac{-\theta yz_0}{a+1}\right)\neq 0, \end{align}\] where \(E\left(\theta,\frac{-\theta yz_0}{a+1}\right)\neq 0\): If \(\theta>0\), \(\theta y z_0\) is negative for \(y<0\) and thus cannot be a root. If \(y\in[0,1]\), we have \(0<\theta y z_0<z_0\). For \(\theta<0\), it holds \(|\theta y z_0|\le z_0\) with \(\theta y z_0\neq z_0\) by choice of \(a\). An application of Lemma 8 yields the claim. ◻

Proof of Theorem 2 (B). For \(y\in[-a,1]\), set \(c(\theta,a)\mathrel{\vcenter{:}}= \frac{a+1}{E\left(\theta,\frac{-\theta }{\lambda(a+1)}\right)}\). By Lemma 10, we have \(p_n^{a,y}(\theta)\sim c(\theta,a)h(y)\lambda^{n+1}\) as \(n\to\infty\). Lemma 5 then yields the claim and the formula for \(h(y)\) for \(y\in[-a,1]\). The claim about the formula for \(h(y)\) for the extended range \(\theta y \in [-a,1]\) follows from the eigenvalue equation. ◻

Proof. For \(y\le\frac{1}{\theta}\), we use \(X_2\sim U[-a,1]\) and get for \(n\geq 1\): \[\begin{align} p_n^{a,y}(\theta)\leq\mathbb{P}(X_2\ge1 ,X_3\geq\theta X_2,\dots,X_{n+1}\geq\theta X_n)=0. \end{align}\] Therefore, we have \(\hat{P}_{a,y,\theta}(z)=p_0^{a,y}(\theta)z^0=1\) for \(y\in\left[-a,\frac{1}{\theta}\right]\).
The case \(y>\frac{1}{\theta}\) is left. Here, we reduce the case to the shared borderline with the blue region, which is \(\theta\in[-1,0)\) and \(\tilde{a}=\frac{-1}{\theta}\). To do so, we condition the occurring random variables to be \(\geq\frac{1}{\theta}\). First, set \(A\mathrel{\vcenter{:}}= \{X_i\geq\frac{1}{\theta}~\forall~2\leq i\leq n+1\}\). The conditions \(\theta X_i\leq X_{i+1}\leq 1\) for all \(2\leq i\leq n\) and \(X_{n+1}\geq\theta X_n\geq \theta\geq\frac{1}{\theta}\) imply \(X_i\geq\frac{1}{\theta}\) for all \(2\leq i \leq n+1\), so we see that \[\begin{align} \mathbb{P}(\{X_2\geq\theta y,X_3\geq\theta X_2,\dots,X_{n+1}\geq\theta X_n\}\cap A^c)=0. \end{align}\] Now we use \(\tilde{X}_i \sim U\left[\frac{1}{\theta},1\right]\) and \(\nu\mathrel{\vcenter{:}}=\frac{1-\frac{1}{\theta}}{a+1}\) to write \[\begin{align} \label{reduction95green95blue95prob} p_n^{a,y}(\theta)&=\mathbb{P}(X_2\geq\theta y,X_3\geq\theta X_2,\dots,X_{n+1}\geq\theta X_n~|~A)\;\mathbb{P}(A)\notag \\&=\mathbb{P}(\tilde{X}_2\geq\theta y,\tilde{X}_3\geq\theta \tilde{X}_2,\dots,\tilde{X}_{n+1}\geq\theta \tilde{X}_n)\;\mathbb{P}\left(X_1\geq\frac{1}{\theta}\right)^n\notag \\&=\mathbb{P}(\tilde{X}_2\geq\theta y,\tilde{X}_3\geq\theta \tilde{X}_2,\dots,\tilde{X}_{n+1}\geq\theta \tilde{X}_n)\left(\frac{1-\frac{1}{\theta}}{a+1}\right)^n\notag \\&= p_n^{\frac{-1}{\theta},y}(\theta)\nu^n \end{align}\tag{6}\] and thus \[\begin{align} \hat{P}_{a,y,\theta}(z)&\notag = \sum_{n=0}^{\infty}p_n^{\frac{-1}{\theta},y}(\theta)\left(\nu z\right)^n =\hat{P}_{\frac{-1}{\theta},y,\theta}\left(\nu z\right). \end{align}\] Note that every starting point \(\frac{1}{\theta}< y\leq 1\) from the green region is also valid in the blue region for \(\theta\in[-1,0)\) and \(\tilde{a}=\frac{-1}{\theta}\), so we already computed \(\hat{P}_{\frac{-1}{\theta},y,\theta}(\cdot)\) in Lemma 9 for the blue region. Plugging this in, we obtain the claim. ◻

Proof. Analogously to the proof of Lemma 11, set \(A\mathrel{\vcenter{:}}= \{X_i\geq\frac{1}{\theta}~\forall~2\leq i\leq n+1\}\) and recall that the conditions \(\theta X_i\leq X_{i+1}\leq 1\) for all \(2\leq i\leq n\) and \(X_{n+1}\geq\theta X_n\geq \theta\geq\frac{1}{\theta}\) imply \(X_i\geq\frac{1}{\theta}\) for all \(2\leq i \leq n\). In particular, \(\mathbb{P}(\Omega_n\cap A^c)=0\). Moreover, define \(\tilde{M}_i\mathrel{\vcenter{:}}=(\tilde{X}_{i+1},\tilde{X}_i)\) for i.i.d. \(\tilde{X}_i \sim U\left[\frac{1}{\theta},1\right]\) for \(i\geq 2\), \(\tilde{X}_1:=y\), and \(\tilde{\Omega}_n\mathrel{\vcenter{:}}=\{\tilde{M}_1\in S,\dots, \tilde{M}_n\in S\}\), then we have \[\begin{align} &\mathbb{P}_{y}((M_k)_{1\leq k\leq m}\in\cdot~|\Omega_n) =\mathbb{P}_{y}((M_k)_{1\leq k\leq m}\in \cdot~|\Omega_n\cap A) =\mathbb{P}_{y}((\tilde{M}_k)_{1\leq k\leq m}\in\cdot~|\tilde{\Omega}_n). \end{align}\] Since \(\tilde{X}_i \sim U\left[\frac{1}{\theta},1\right]\) and \(\frac{1}{\theta}< y\leq 1\) is valid for the blue region, Theorem 2 (B) provides that the limiting process \(M'\) is a Markov chain with transition kernel \[\begin{align} p'(x_1,\mathrm{d}x_2)&\mathrel{\vcenter{:}}= \mathbb{1}_S(x_2,x_1)\frac{h(x_2)}{h(x_{1})}\frac{1}{\tilde{\lambda}}\frac{1}{1-\frac{1}{\theta}}\,\mathrm{d}x_2,~~~~~~~~x_1,x_2\in \left[\frac{1}{\theta},1\right], \end{align}\] where \(\tilde{\lambda}\) and \(h\) are the eigenvalue and eigenfunction from the blue region in the special case \(\tilde{a}=\frac{-1}{\theta}\), respectively given in 4 and by \[\begin{align} h(x)=\begin{cases} 0,~~&~\theta x \ge1, \\E\left(\theta,\frac{\theta^2 x}{\tilde{\lambda} (1-\theta)}\right),&\frac{1}{\theta}\leq \theta x< 1, \\E\left(\theta,\frac{1}{\tilde{\lambda} (1-\theta)}\right),&~ \theta x< \frac{1}{\theta}. \end{cases} \end{align}\] Finally, Corollary 12 completes the claim. ◻

Proof. The first statement is shown in Corollary 1 of [3]. The second statement is obtained by the same method. ◻

Proof. The idea of the proof follows the idea of Lemma 2 and Lemma 3 of [3]. We use the notation from Definition 14 and set \(A_1\mathrel{\vcenter{:}}=\{\theta y\leq X_2\}\), \(A_i\mathrel{\vcenter{:}}=\{\theta X_i\leq X_{i+1}\}\) for \(2\le i\le n\) and \(B_1\mathrel{\vcenter{:}}=\{\theta y\geq X_2\}\), \(B_i\mathrel{\vcenter{:}}=\{\theta X_i\geq X_{i+1}\}\) for \(2\leq i \leq n\). Note that due to the continuous distribution of the \(X_i\), the set \(B_i\) is the complement of the set \(A_i\) up to a null set. Applying this to the set \(A_n\), we get for \(n\geq 1\) \[\begin{align} p_n^{a,y}(\theta) &=\mathbb{P}(A_1\cap\dots\cap A_n) =\mathbb{P}(A_1\cap\dots\cap A_{n-1})-\mathbb{P}(A_1\cap\dots\cap A_{n-1}\cap B_n) \\&=p_{n-1}^{a,y}(\theta)-\mathbb{P}(A_1\cap\dots\cap A_{n-1}\cap B_n). \end{align}\] Using the fact that \(B_{n-1}\) is the complement of \(A_{n-1}\) up to a null set, we split the probability \(\mathbb{P}(A_1\cap\dots\cap A_{n-1}\cap B_n)\) and then use that \(B_n\) is independent of \(A_1\cap\dots \cap A_{n-2}\): \[\begin{align} \mathbb{P}(A_1\cap\dots\cap A_{n-1}\cap B_n) &=\mathbb{P}(A_1\cap\dots\cap A_{n-2}\cap B_n)-\mathbb{P}(A_1\cap\dots\cap A_{n-2}\cap B_{n-1}\cap B_n) \\&=\mathbb{P}(A_1\cap\dots\cap A_{n-2})\mathbb{P}( B_n)-\mathbb{P}(A_1\cap\dots\cap A_{n-2}\cap B_{n-1}\cap B_n) \\&=p_{n-2}^{a,y}(\theta)q_1^a(\theta)-\mathbb{P}(A_1\cap\dots\cap A_{n-2}\cap B_{n-1}\cap B_n). \end{align}\] We therefore have \[\begin{align} p_n^{a,y}(\theta)=p_{n-1}^{a,y}(\theta)-p_{n-2}^{a,y}(\theta)q_1^a(\theta)+\mathbb{P}(A_1\cap\dots\cap A_{n-2}\cap B_{n-1}\cap B_n). \end{align}\] We proceed this way of transforming the terms of the form \(\mathbb{P}(A_1\cap\dots\cap A_i\cap B_{i+1}\cap\dots\cap B_n)\). In particular, for \(n\geq 1\), we get the recursion \[\begin{align} p_n^{a,y}(\theta)=\left(\sum_{k=1}^n(-1)^{k-1}p_{n-k}^{a,y}(\theta)q_{k-1}^a(\theta)\right)+(-1)^nq_n^{a,y}(\theta). \end{align}\] Multiplying by \(z^n\) and taking the sum over \(n\), this becomes \[\begin{align} \hat{P}_{a,y,\theta}(z)-1 &=\sum_{n=1}^{\infty}p_n^{a,y}(\theta)z^n =\sum_{n=1}^{\infty}\sum_{k=1}^n(-1)^{k-1}p_{n-k}^{a,y}(\theta)q_{k-1}^a(\theta)z^n+\sum_{n=1}^{\infty}(-1)^nq_n^{a,y}(\theta)z^n \\&=\sum_{n=1}^{\infty}\sum_{k=1}^np_{n-k}^{a,y}(\theta)z^{n-k}zq_{k-1}^a(\theta)(-1)^{k-1}z^{k-1}+\sum_{n=1}^{\infty}(-1)^nq_n^{a,y}(\theta)z^n \\&=\sum_{k=1}^{\infty}\sum_{n=k}^{\infty}p_{n-k}^{a,y}(\theta)z^{n-k}zq_{k-1}^a(\theta)(-z)^{k-1}+\hat{Q}_{a,y,\theta}(-z)-1 \\&=\sum_{k=1}^{\infty}zq_{k-1}^a(\theta)(-z)^{k-1}\sum_{n=k}^{\infty}p_{n-k}^{a,y}(\theta)z^{n-k}+\hat{Q}_{a,y,\theta}(-z)-1 \\&=\hat{P}_{a,y,\theta}(z)\sum_{k=1}^{\infty}zq_{k-1}^a(\theta)(-z)^{k-1}+\hat{Q}_{a,y,\theta}(-z)-1 \\&=\hat{P}_{a,y,\theta}(z)z\hat{Q}_{a,\theta}(-z)+\hat{Q}_{a,y,\theta}(-z)-1. \end{align}\] Rearranging and dividing by \(1-z\hat{Q}_{a,\theta}(-z)\) gives the claim. ◻

Proof. The statement follows directly by using ?? to rewrite the terms on the right-hand side of ?? . ◻

Proof. Note that for \(\theta\in[-1,0)\) and \(a\in(0,-\theta)\) from the yellow region, we have that the pair \(\theta\) and \(\tilde{a}\mathrel{\vcenter{:}}=\frac{1}{a}>\frac{-1}{\theta}\) is part of the green region. Moreover, if \(y\in[-a,1]\), it holds that \(\tilde{y}\mathrel{\vcenter{:}}=\frac{-y}{a}\in[-\tilde{a},1]\). So we already calculated the value of \(\hat{P}_{\frac{1}{a},\tilde{y},\theta}(-z)\) in Lemma 11 in the green region, which we recall to be \[\begin{align} \hat{P}_{\frac{1}{a},\tilde{y},\theta}(-z) = \begin{cases} 1,~~&~-\frac{1}{a}\leq \tilde{y}<\frac{1}{\theta}, \\\frac{E\left(\theta,\frac{\theta z\tilde{y}a}{a+1}\right)}{E\left(\theta,\frac{za}{a+1}\right)},~~&~\frac{1}{\theta}\leq \tilde{y} \leq 1. \end{cases} \end{align}\] Furthermore, the value of \(\hat{P}_{\frac{1}{a},\theta}(-z)\) can be taken from Lemma 7 of the green region in [3]: \[\begin{align} \hat{P}_{\frac{1}{a},\theta}(-z) &=\frac{\theta +a}{\theta(a+1)}+\frac{E\left(\theta,\frac{za}{\theta(a+1)}\right)-E\left(\theta,\frac{za}{a+1}\right)}{-zE\left(\theta,\frac{za}{a+1}\right)}. \end{align}\] Plugging the last two displays (using \(\tilde{y}=\frac{-y}{a}\)) into ?? gives us the two claimed formulas. ◻

Proof. Let us find a candidate for \(z_0\) and then show that it is the solution of (7 ) with smallest modulus. For this purpose, we first consider the real solutions of (7 ). Define \(m_0>0\) to be the solution with smallest modulus of the equation \(E(\theta,\frac{-za}{a+1})=0\). This is a well-defined positive real number due to Lemma 7. Consider the function \(z\mapsto \frac{E\left(\theta,\frac{za}{\theta(a+1)}\right)}{E\left(\theta,\frac{za}{a+1}\right)}\) for real \(z\), which is the left-hand side of (7 ). This function has a pole at \(-m_0<0\), it has a zero at \(-\theta m_0\in(0,m_0]\), its value at \(0\) is \(1\), and on \((-m_0,m_0)\) it has no other zeros; all of this follows by Lemma 7. By continuity, this means that it attains only positive values on \((-m_0,-\theta m_0)\). On the other hand, the right hand side of (7 ) is a linear function with positive slope \(\frac{\theta+a}{\theta(a+1)}\), which must have an intersection with the left-hand side for positive arguments. Therefore, there exists a smallest solution \(z_0\in(0,-\theta m_0)\). Let us turn to solutions \(z\in\mathbb{C}\) of (7 ). We know, by Lemma 7, that \(E(\theta,\frac{za}{\theta(a+1)})\neq 0\) for \(z\in\mathbb{C}\) with \(|z|<-\theta m_0\) and likewise \(E(\theta,\frac{za}{a+1})\neq 0\) even for \(z\in\mathbb{C}\) with \(|z|<m_0\). This will imply the last part of the lemma’s claim. Further, it shows that the (complex) solutions of (7 ) for \(|z|<m_0\) are the same as the (complex) zeros of the function \[g(z) := 1 - z \,\frac{\theta+a}{\theta(a+1)} \cdot \frac{E\left(\theta,\frac{za}{a+1}\right)}{E\left(\theta,\frac{za}{\theta(a+1)}\right)}.\] It turns out that \(g\) can be written with the help of a series with positive coefficients: Indeed, for \(|-z/\theta|<m_0\), we have, by Lemma 11, \[g(z) = 1 - z \,\frac{\theta+a}{\theta(a+1)} \cdot \hat{P}_{\frac{1}{a},1,\theta}\left(\frac{-z}{\theta}\right) = 1 + \frac{\theta+a}{ a+1} \cdot \sum_{n=0}^\infty p_n^{\frac{1}{a},1}(\theta) (-1/\theta)^{n+1} z^{n+1} =: 1 - \tilde{g}(z),\] and the coefficients of the series of \(\tilde{g}\), namely the numbers \(\alpha_n:=-\frac{\theta+a}{a+1} p_n^{\frac{1}{a},1}(\theta) (-1/\theta)^{n+1}\), are all strictly positive. Note that the solutions to (7 ) with \(|z|<m_0\) correspond to the solutions of \(g(z)=0\), i.e.\(\tilde{g}(z)=1\).

We know from the construction above that \(g(z_0)=0\), i.e.\(\tilde{g}(z_0)=1\) and \(0<z_0<m_0\). Now, let \(z\in\mathbb{C}\) with \(|z|<z_0\). Then, using positive coefficients, \[|\tilde{g}(z)| \leq \tilde{g}(|z|) < \tilde{g}(z_0) = 1.\] This shows that there cannot be any solution of \(\tilde{g}(z_1)=1\) with \(z_1\in\mathbb{C}\) and \(|z_1|<z_0\). It remains to be seen that there are no solutions \(z_1\in\mathbb{C}\) with \(|z_1|=z_0\) either. Assume that \(z_1=e^{i\phi } z_0\) is such a solution. Then \[1=|\tilde{g}(z_1)| \leq \tilde{g}(|z_1|) = \tilde{g}(z_0) =1,\] and the triangle inequality step can only be an equality if all the terms in the sum have the same complex argument, i.e.\(\arg(\alpha_n z_1^{n+1})=\arg(\alpha_m z_1^{m+1})\) for all \(n,m\). Since the coefficients \(\alpha_n\) are positive, this is true if and only if the \(z_1^n\) have the same complex argument for all \(n\geq 1\), i.e.\(e^{i\phi n}\) is the same for all \(n\), which can only happen for \(\phi=0\), i.e.for \(z_1=z_0\). This shows that there are no solutions \(z_1\in \mathbb{C}\) of (7 ) with \(|z_1|\leq z_0\), \(z_1\neq z_0\). ◻

Proof. Similarly to the proof of Lemma 10, we construct a function \(f(z)\) with \[\begin{align} \label{gelb95f} \sum_{n=0}^{\infty}p_n^{a,y}(\theta)z^n=\frac{f(z)}{z_0-z},~~~~\text{for all}~|z|<z_0, \end{align}\tag{8}\] which is analytic on \(|z|<z_1\) with some \(z_0<z_1\) and \(f\! \left(z_0\right)\neq 0\). Applying Lemma 8 will then yield the claim.
Recall that the generating function in Lemma 18 has the structure \[\begin{align} \hat{P}_{a,y,\theta}(z)=\frac{u(z)}{g(z)} \end{align}\] with \(g(z)\mathrel{\vcenter{:}}= E\left(\theta,\frac{za}{\theta(a+1)}\right)-z\frac{\theta +a}{\theta(a+1)}E\left(\theta,\frac{za}{a+1}\right)\) and \(u(z)\) depending on the choice of \(y\): \[\begin{align} u(z)\mathrel{\vcenter{:}}=\begin{cases} E\left(\theta,\frac{-\theta zy}{a+1}\right),~~~&-a\leq y \leq \frac{-a}{\theta}, \\E\left(\theta,\frac{za}{a+1}\right),~~&\frac{-a}{\theta}<y\leq1. \end{cases} \end{align}\] Moreover, \(z_0>0\) is the root with the smallest absolute value of \(g\). Set \(v(z)\mathrel{\vcenter{:}}= g(z)\frac{1}{z_0-z}\). We show that the function \(f\) defined by \(f(z)\mathrel{\vcenter{:}}=\frac{u(z)}{v(z)}\) satisfies the assumptions stated above.

By Lemma 19, \(z_0\) is the unique zero of \(g\) of smallest modulus with all other zeros lying outside \(|z|<z_1\) for some \(z_1>z_0\). We will see below that \(g'(z_0)<0\), implying that \(z_0\) is a simple zero. Therefore, \(v\) is analytic and does not vanish on a disk \(|z|<z_1\) with \(z_1>z_0\). Further, \(u\) is entire so that \(f\) is analytic on \(|z|<z_1\). It remains to be seen that \(f(z_0)\neq 0\) in order to apply Lemma 8 and to compute \(f(z_0)\).

For this purpose, note that \(v(z_0)=-g'(z_0)\), which can be computed readily: \[\begin{align} g'(z_0) & = E\left(\theta,\frac{z_0 a}{a+1}\right) \frac{a}{\theta(a+1)} - \frac{\theta +a}{\theta(a+1)}E\left(\theta,\frac{z_0 a}{a+1}\right) - z_0 \frac{\theta +a}{\theta(a+1)}E\left(\theta,\frac{\theta z_0a}{a+1}\right) \frac{a}{a+1} \\ & = \frac{-1}{a+1} E\left(\theta,\frac{z_0 a}{a+1}\right) - z_0 \frac{a(\theta +a)}{\theta(a+1)^2}E\left(\theta,\frac{\theta z_0a}{a+1}\right) < 0, \end{align}\] because the \(E\)-terms are positive (due to Lemma 19) while the prefactors are negative since \(\theta\in[-1,0)\) and \(a<-\theta\).

It remains to show \(u(z_0)\neq 0\) with the help of Lemma 19. For \(-a\leq y \leq \frac{-a}{\theta}\), we have \(u(z)= E\left(\theta,\frac{-\theta zy}{a+1}\right)\) and \(|\theta yz_0|\le z_0 a\) and thus, \(u(z_0)= E\left(\theta,\frac{-\theta z_0y}{a+1}\right)\neq 0\). For \(\frac{-a}{\theta}<y\leq1\), \(u(z)= E\left(\theta,\frac{za}{a+1}\right)\) and \(u(z_0)= E\left(\theta,\frac{z_0a}{a+1}\right) \neq 0\). This gives us \(f(z_0)\neq 0\) as well as the value of \(f(z_0)=f(\frac{1}{\lambda})\) in both of the cases and hence Lemma 8 yields the claims. ◻

Proof of Theorem 2 (Y). There are three sets to be considered: For \(-a\leq \theta x\leq -a \theta\) (i.e.\(-a\leq x\leq -a/\theta\)), we know from Lemma 20 that \(p_n^{a,x}(\theta)\sim c(\theta,a)h(x)\lambda^{n+1}\) with
\(h(x)= E\left(\theta,\frac{-\theta x}{\lambda(a+1)}\right)\). Further, for \(\theta\leq \theta x \leq -a\) (i.e.\(-a/\theta\leq x\leq 1\)), we also know from Lemma 20 that \(p_n^{a,x}(\theta)\sim c(\theta,a)h(x)\lambda^{n+1}\) with \(h(x)= E\left(\theta,\frac{a}{\lambda(a+1)}\right)\). These two asymptotics are the input of Lemma 5 which shows the convergence claim to the Doob transformed process in Theorem 2 (Y).

It is left to prove the eigenvalue equation ?? for \(-a\theta<\theta x\leq 1\) (i.e.\(1/\theta\leq x < -a\), which is not needed for the Doob limit claim). Here, we directly verify the eigenvalue equation. We use \(-a\leq -a\theta\leq\theta x\leq 1\) as well as \(x<0\) to see that for \(-a\theta\leq\theta x\leq\frac{-a}{\theta}\), we have \[\begin{align} &\int_{\{y>\theta x\}\cap[-a,1]} h(y) \frac{1}{a+1}\,\mathrm{d}y \\&=\frac{1}{a+1}\int_{\theta x}^{\frac{-a}{\theta}}h(y)\mathrm{d}y+\frac{1}{a+1}\int_{\frac{-a}{\theta}}^1h(y)\mathrm{d}y \\&=\frac{1}{a+1}\int_{\theta x}^{\frac{-a}{\theta}}E\left(\theta,\frac{-\theta y}{\lambda(a+1)}\right)\mathrm{d}y+\frac{1}{a+1}\int_{\frac{-a}{\theta}}^1E\left(\theta,\frac{a}{\lambda(a+1)}\right)\mathrm{d}y \\&=\lambda E\left(\theta,\frac{-\theta x}{\lambda(a+1)}\right)-\lambda E\left(\theta,\frac{a}{\theta\lambda(a+1)}\right)+\frac{\theta+a}{\theta(a+1)}E\left(\theta,\frac{a}{\lambda(a+1)}\right) \\&=\lambda h(x), \end{align}\] where we used 7 for \(z_0=1/\lambda\) in the last step, and for \(\frac{-a}{\theta}<\theta x\leq 1\), \[\begin{align} &\int_{\{y>\theta x\}\cap[-a,1]} h(y) \frac{1}{a+1}\,\mathrm{d}y =\frac{1}{a+1}\int_{\theta x}^{1}h(y)\mathrm{d}y=\frac{1-\theta x}{a+1} E\left(\theta,\frac{a}{\lambda(a+1)}\right)=\lambda h(x). \qedhere \end{align}\] ◻

Proof of Lemma 21. The proof follows by induction. We start with the case \(p=0\), i.e.when \(-1<a<-\theta\). Then \(F(z)=1\) and \(u(z)=1-z\) is strictly decreasing on \(\mathbb{R}_{\ge0}\), in particular at its first zero \(z_0=1\).

The induction hypothesis (the case \(p-1\)) is that \(u_{\theta,a'}'(z')<0\) for all \(0\leq z'\leq z_0(a')\) for all \(a'\) with \(\theta^{p-1}\geq -a'>\theta^p\). We now show the same for \(p\), i.e.for \(\theta^p \geq -a>\theta^{p+1}\) we claim that \(u_{\theta,a}'(z)<0\) for all \(0\leq z\leq z_0(a)\).

For ease of notation, set \(a'\mathrel{\vcenter{:}}= a/\theta<a\) and \[\beta \mathrel{\vcenter{:}}= \frac{a+1}{a'+1} = \frac{\theta (a+1)}{\theta+a}>1.\] Let \(0\leq z\leq z_0(a)\). Set \(z'\mathrel{\vcenter{:}}= z/\beta\). Then, by Lemma 23, we get \(0\leq z'\leq z_0(a)/\beta\leq z_0(a')\). Rewriting the identity (?? ) in the new notation, one obtains \[\label{eqn:orangecomparisoncrucial} (a+1)(-u_{\theta,a}'(z)) = u_{\theta,a/\theta}\left( \theta \frac{z}{\beta} \right)+ a \cdot u_{\theta,a/\theta}\left( \frac{z}{\beta} \right) = u_{\theta,a'}\left( \theta z' \right)- (-a) u_{\theta,a'}\left( z' \right).\tag{12}\] By the induction hypothesis, we know that since \(0\leq z'\leq z_0(a')\), the function \(u_{\theta,a'}\) is strictly decreasing. In particular, \(u_{\theta,a'}\left( \theta z' \right)>u_{\theta,a'}\left( z' \right)\geq 0\) (exclude \(z'=0\) here). Since \((-a)\in[0,1)\), this implies (including again \(z'=0\)) \[u_{\theta,a'}\left( \theta z' \right)>(-a)u_{\theta,a'}\left( z' \right).\] Inserting this into (12 ) gives \(u_{\theta,a}'(z)<0\) for all \(0\leq z\leq z_0(a)\), in particular for \(z=z_0(a)\). ◻

Proof. Let \(y\in A_m\). For \(1\le m\le p\), we have \(1\ge\theta y\ge\frac{-a}{\theta^{m-1}}\ge-a\). Therefore, for \(n\ge 1\), \[\begin{align} p_{n}^{a,y}(\theta) &=\mathbb{P}(\theta y\leq X_2,\dots,\theta X_n\leq X_{n+1}) \notag \\&=\frac{1}{(a+1)^n}\int_{\theta y}^1\bigg[\int_{-a}^1\mathbb{1}_{\{\theta x_{2}\leq x_{3}\}}\dots \int_{-a}^1\mathbb{1}_{\{\theta x_{n}\leq x_{n+1}\}}\mathrm{d}x_{n+1}\dots \mathrm{d}x_3\bigg]\mathrm{d}x_2\notag \\&=\frac{1}{a+1}\int_{\theta y}^1 p_{n-1}^{a,x_2}(\theta)\mathrm{d}x_2.\notag \end{align}\] Multiplying with \(z^n\) and summing over \(n\), we get \[\begin{align} \hat{P}_{a,y,\theta}(z) &=1+\frac{z}{a+1}\int_{\theta y}^1\hat{P}_{a,x_2,\theta}(z)\mathrm{d}x_2\label{eqn:orange:integralequationhatp}. \end{align}\tag{13}\] Taking the derivative with respect to \(y\) gives us the equation (?? ). Note that for \(y\in A_m\), the function \(\hat{P}_{a,\theta y,\theta}(z)\) is the generating function in the region \(A_{m-1}\) since \(\theta y\in A_{m-1}\). Moreover, the function \(y\mapsto\hat{P}_{a,y,\theta}(z)\) is continuous on \(y\in[-a,1]\), so in particular at \(y=\frac{-a}{\theta^{m}}\), which implies ?? . ◻

Proof. First, consider \(y\in A_0\), which means that \(\theta y<-a\). So for \(n\geq 1\), we have \[\begin{align} p_{n}^{a,y}(\theta) &=\mathbb{P}(\theta y\leq X_2,\theta X_2\leq X_3,\dots,\theta X_n\leq X_{n+1}) = p_{n-1}^{a}(\theta), \end{align}\] so \(\hat{P}_{a,y,\theta}(z) =1+\sum_{n=1}^{\infty}p_{n}^{a,y}(\theta)z^n =1+z\hat{P}_{a,\theta}(z)\). The generating function \(\hat{P}_{a,\theta}(z)\) can be taken from Lemma 10 of [3] (cf.@eq:eqn:orange:gfintegrated ) so that \[\begin{align} \hat{P}_{a,y,\theta}(z)=1+\frac{zF(z)}{1-zF(z)}=\frac{1}{1-zF(z)}=c_{0,0}. \end{align}\] Using Lemma 24, we now get a closed form in case \(y\in A_1\): For \(y\in A_1\), we have \(\theta y\in A_0\) and thus, \[\begin{align} \hat{P}_{a,y,\theta}(z)=\frac{- z\theta}{a+1}c_{0,0} y+c_{0,0}+\frac{ -az}{a+1}c_{0,0}, \end{align}\] and the claim follows. ◻

Proof. We prove the corollary via induction. For \(y\in A_0\) and \(y\in A_1\), we have already verified the claim in Lemma 25. Assume that for \(\tilde{y}\in A_{m-1}\), we have \(\hat{P}_{a,\tilde{y},\theta}(z)=\sum_{i=0}^{m-1}c_{m-1,i}H_i(\tilde{y})\). In particular, this holds for \(\tilde{y}=\theta y\) when \(y\in A_m\). We show that the function \(y\mapsto\sum_{i=0}^mc_{m,i}H_i(y)\) on \(A_m\) satisfies the claims of Lemma 24. First, we check the differential equation: \[\begin{align} \frac{\partial}{\partial y}\sum_{i=0}^mc_{m,i}H_i(y) &=\sum_{i=1}^mc_{m,i}H_{i-1}(y) =\sum_{i=1}^m\frac{-z\theta^ic_{m-1,i-1}}{a+1}H_{i-1}(y) \\&=\frac{-z\theta}{a+1}\sum_{i=1}^mc_{m-1,i-1}H_{i-1}(\theta y) =\frac{-z\theta}{a+1}\sum_{i=0}^{m-1}c_{m-1,i}H_{i}(\theta y) \\&=\frac{-z\theta}{a+1}\hat{P}_{a,\theta y,\theta}(z). \end{align}\] To check continuity in \(\frac{-a}{\theta^m}\), we compute \(\lim_{\tilde{y}\nearrow\frac{-a}{\theta^{m}}}\hat{P}_{a,\tilde{y},\theta}(z)\) with the obtained formula for \(\hat{P}_{a,\tilde{y},\theta}(z)\) and \(\tilde{y}\in A_{m-1}\). A direct computation yields \[\begin{align} \lim_{\tilde{y}\nearrow\frac{-a}{\theta^{m}}}\hat{P}_{a,\tilde{y},\theta}(z) &=\sum_{i=0}^{m-1}c_{m-1,i}H_i\left(\frac{-a}{\theta^m}\right) \\&=\sum_{i=1}^mc_{m,i}H_i\left(\frac{-a}{\theta^{m}}\right)+\sum_{i=0}^{m-1}\left(c_{m-1,i}H_i\left(\frac{-a}{\theta^m}\right)-c_{m,i+1}H_{i+1}\left(\frac{-a}{\theta^{m}}\right)\right) \\&=\sum_{i=1}^mc_{m,i}H_i\left(\frac{-a}{\theta^{m}}\right)+c_{m,0}H_0\left(\frac{-a}{\theta^{m}}\right) \\&=\sum_{i=0}^mc_{m,i}H_i\left(\frac{-a}{\theta^{m}}\right), \end{align}\] which is equal to \(\hat{P}_{a,\frac{-a}{\theta^m},\theta}(z)\). This completes the proof. ◻

Proof. By induction, we have for \(m\ge 0\) and \(0\le i \le m\) \[\begin{align} \label{orange95ind95cmi} c_{m,i} =\left(\frac{-z}{a+1}\right)^i\theta^\frac{i(i+1)}{2}c_{m-i,0}. \end{align}\tag{14}\] Corollary 26 therefore yields (?? ). Plugging 14 into the formula for \(c_{k,0}\) for \(1\le k\le m\) yields the second part: \[\begin{align} c_{k,0} &=\sum_{i=0}^{k-1}\left(\frac{-z}{a+1}\right)^i\theta^\frac{i(i+1)}{2}c_{k-1-i,0}H_i\left(\frac{-a}{\theta^k}\right)-\left(\frac{-z}{a+1}\right)^{i+1}\theta^\frac{(i+2)(i+1)}{2}c_{k-1-i,0}H_{i+1}\left(\frac{-a}{\theta^k}\right) \\&=\sum_{i=0}^{k-1}\left(\frac{-z}{a+1}\right)^i\theta^\frac{i(i+1)}{2}c_{k-1-i,0}H_i\left(\frac{-a}{\theta^k}\right)\left(1-\frac{za\theta^{i+1-k}}{(a+1)(i+1)}\right) \\&=\sum_{i=0}^{k-1}\theta^{\frac{i(i+1)}{2}-ki}\left(\frac{za}{a+1}\right)^{i+1}\frac{1}{(i+1)!}\left(\frac{(a+1)(i+1)}{za}-\theta^{i+1-k}\right)c_{k-1-i,0}.\qedhere \end{align}\] ◻

Proof. Iterating ?? , we have for \(k\ge 1\): \[\begin{align} c_{k,0}&=\sum_{i=0}^{k-1}s_z(k,i)c_{k-1-i,0}\notag \\&=\sum_{i_1=0}^{k-1}s_z(k,i_1)\sum_{i_2=0}^{k-2-i_1}s_z(k-1-i_1,i_2)c_{k-2-i_1-i_2,0}\notag \\&=\frac{1}{1-zF(z)}\sum_{{\ell}=1}^k\sum_{\substack{i_1+\dots +i_{\ell}= k-{\ell}\\i_r\ge 0}}\prod_{r=1}^{\ell}s_z(k-(r-1)-\sum_{j=1}^{r-1}i_j,i_r),\notag \end{align}\] where we summed over all possible decompositions \(k=\ell+i_1+\dots+i_\ell\). Plugging this into Corollary 27 gives \[\begin{align} \hat{P}_{a,y,\theta}(z) &= \sum_{i=0}^m\frac{\theta^\frac{i(i+1)}{2}}{i!}\left(\frac{-zy}{a+1}\right)^ic_{m-i,0} \\&=\frac{1}{1\!-\!zF(z)}\!\sum_{i=0}^{m}\frac{\theta^\frac{i(i+1)}{2}}{i!}\left(\frac{-zy}{1\!+\!a}\right)^i\sum_{{\ell}=0}^{m-i}\sum_{\substack{i_1+\dots +i_{\ell}= m-i-{\ell}\\i_r\ge 0}}\prod_{r=1}^{\ell}s_z(m\!-\!i\!-\!(r\!-\!1)\!-\!\sum_{j=1}^{r-1}i_j,i_r). \qedhere \end{align}\] ◻

Proof. Let us start with the case \(m=0\). Then \(\theta y\leq -a\) and we have \[p_n^{a,y}(\theta) = p_{n-1}^a(\theta),\] so that the claim holds for \(m=0\) with \(\kappa(y)\mathrel{\vcenter{:}}= 1\). For \(m\geq 1\), we proceed by induction. Let \(y\in A_m\). Assume there exist \(\alpha=\alpha(y)\) and \(\beta=\beta(y)\) with \[\label{eqn:choicealphabeta} -a\leq \alpha < \beta \leq 1, \qquad \alpha\geq \theta y, \qquad \beta \in A_{m-1}.\tag{15}\] Then, by the Markov property and the monotonicity of \(x\mapsto p_{n-1}^{a,x}(\theta)\), \[\begin{align} p_n^{a,y}(\theta) = & \, \int_{-a}^1 \mathbb{1}_{x_2\geq \theta y} \, p_{n-1}^{a,x_2}(\theta) \, \frac{1}{a+1}\mathrm{d}x_2 \geq \, \int_{\alpha}^\beta \, p_{n-1}^{a,x_2}(\theta) \, \frac{1}{a+1}\mathrm{d}x_2 \\\geq & \, \int_{\alpha}^\beta \, p_{n-1}^{a,\beta}(\theta) \, \frac{1}{a+1}\mathrm{d}x_2 = \, \frac{\beta-\alpha}{a+1}\, p_{n-1}^{a,\beta}(\theta). \end{align}\] Since \(\beta\in A_{m-1}\), we can apply the induction hypothesis and estimate \[p_{n-1}^{a,\beta}(\theta)\geq \kappa(\beta) p_{(n-1)-(m-1)-1}^a(\theta)=\kappa(\beta) p_{n-m-1}^a(\theta),\] so that for \(y\in A_m\) we can set \(\kappa(y)\mathrel{\vcenter{:}}= \kappa(\beta(y)) \frac{\beta(y)-\alpha(y)}{a+1}\) and obtain the claim by combining the last two displays.

We still have to show that there are valid choices for \(\alpha=\alpha(y)\) and \(\beta=\beta(y)\) satisfying (15 ): set \[\alpha\mathrel{\vcenter{:}}= \theta y,\qquad \beta\mathrel{\vcenter{:}}= \frac{1}{2}\left( \theta y+\frac{-a}{\theta^m}\right).\] Then \(\alpha=\theta y \geq -a\), since \(m\geq 1\). Further, \[\beta - \alpha= \frac{1}{2}\left( \theta y+\frac{-a}{\theta^m}\right) - \theta y =\frac{1}{2}\left( -\theta y+\frac{-a}{\theta^m}\right) > \frac{1}{2}\left(-\theta \frac{-a}{\theta^{m+1}}+\frac{-a}{\theta^m}\right) = 0,\] which in particular means \(\beta > \alpha = \theta y \geq \frac{-a}{\theta^{m-1}}\). Finally, \[\beta = \frac{1}{2}\left( \theta y+\frac{-a}{\theta^m}\right) < \frac{1}{2}\left( \theta \frac{-a}{\theta^{m+1}}+\frac{-a}{\theta^m}\right) = \frac{-a}{\theta^m} \leq 1,\] so that indeed \(\beta\in A_{m-1}\). Note that in case \(m=p\), we additionally used the fact that \(\theta<\frac{-a}{\theta^p}\) in the inequalities above. ◻

Proof. Using Corollary 28, Lemma 29 and (10 ), we have for real \(0\leq z<z_0\) and with \(\kappa(y)>0\) that \[\begin{align} \frac{G_{a,y,\theta}(z)}{u(z)} & = \hat{P}_{a,y,\theta}(z) \geq \kappa(y) z^{m+1} \hat{P}_{a,\theta}(z) =\kappa(y) z^{m+1} \, \frac{F(z)}{u(z)}, \end{align}\] where \(u\) is as in (11 ). Since \(u(z)>0\) for all \(0\leq z<z_0\) (recall that \(z_0\) is the smallest positive real zero of \(u\) and \(u(0)=1\)), we know that for all \(0\leq z <z_0\) \[\begin{align} G_{a,y,\theta}(z) & \geq \kappa(y) z^{m+1}F(z), \end{align}\] and letting \(z\uparrow z_0\), we see that, using the definition of \(z_0\), \[\begin{align} G_{a,y,\theta}(z_0) & \geq \kappa(y) z_0^{m+1}F(z_0) = \kappa(y) z_0^{m+1}\, \frac{1}{z_0} > 0. \qedhere \end{align}\] ◻

Proof. Let \(z_1\in\mathbb{C}\) with \(|z_1|\leq z_0\) and \(u(z_1)=0\). Assume \(z_1\neq z_0\).

We start with some initial definitions. Recall that \(u\) is a polynomial of degree \(\leq p+1\) which has \(z_0\) and \(z_1\) as zeros. Denoting the multiplicity of the zero \(z_1\) by \(r\), we note that \(r\in\{1,\ldots,p\}\). Further, we define \[\begin{align} \label{eqn:defnphi} \phi_1(y) & :=\lim_{z\to z_1} (z_1-z)^r \hat{P}_{a,y,\theta}(z) = \frac{G_{a,y,\theta}(z_1)}{\lim_{z\to z_1} \frac{u(z)}{(z_1-z)^r} },\qquad y\in[-a,1]. \end{align}\tag{16}\] We note that \(G_{a,y,\theta}(z_1)\) is some finite number (which may be zero), because \(G_{a,y,\theta}(z)\) is a polynomial in \(z\), while the limit in the denominator exists and is non-zero because \(r\) is the exact multiplicity of \(z_1\) as a zero of \(u\). Note that \(\phi_1\) is not identically zero on \([-a,1]\). Indeed, at least for \(y\in A_0\) we have \[\phi_1(y)=\lim_{z\to z_1} (z_1-z)^r \hat{P}_{a,y,\theta}(z) = \lim_{z\to z_1} (z_1-z)^r \frac{1}{u(z)},\] because \(\hat{P}_{a,y,\theta}(z)=\frac{1}{u(z)}\); and the limit does not vanish, as argued above.

Define \[\begin{align} \phi_0(y):=\lim_{z\to z_0} (z_0-z) \hat{P}_{a,y,\theta}(z) =\frac{G_{a,y,\theta}(z_0)}{\lim_{z\to z_0} \frac{u(z)}{z_0-z} } = \frac{G_{a,y,\theta}(z_0)}{- \lim_{z\to z_0} \frac{u(z)-u(z_0)}{z-z_0} } =\frac{G_{a,y,\theta}(z_0)}{-u'(z_0)}, \end{align}\] and \(G_{a,y,\theta}(z_0)\) is some finite number, as argued above. Note that the last chain of equalities also shows that \(\phi_0\) is strictly positive and finite on \([-a,1]\), by Lemma 30 and Lemma 21.

We will now obtain eigenvalue equations for \(\phi_1\) and \(\phi_0\). It is known from (13 ) that, for \(y\in[-a,1]\), \[\label{eqn:orange:recursion} \hat{P}_{a,y,\theta}(z) =1+\frac{z}{a+1}\int_{\{x\geq \theta y\}\cap[-a,1]} \hat{P}_{a,x,\theta}(z)\mathrm{d}x.\tag{17}\] In fact, we showed this for \(y\in A_m\), \(m\geq 1\), only, but for \(y\in A_0\) it is easy to verify. We multiply (17 ) by \((z_1-z)^r\) and take the limit when \(z\to z_1\). Then, \[\begin{align} \label{eqn:orange:recursionphi} \phi_1(y) =\frac{z_1}{a+1}\int_{\{x\geq \theta y\}\cap[-a,1]} \phi_1(x) \mathrm{d}x. \end{align}\tag{18}\] Here, we exchanged the limit and the integral. This is justified by the dominated convergence theorem, as visible from (16 ): the fraction \(\frac{(z_1-z)^r}{u(z)}\) can be uniformly bounded when \(z\to z_1\), because \(z_1\) is a zero of \(u\) of exact multiplicity \(r\). Moreover, \(G_{a,x,\theta}(z)\) is a polynomial in \(z\) whose coefficients are uniformly bounded in \(x\in[-a,1]\). Hence the integrand is dominated by an integrable constant, and dominated convergence applies. Equation (18 ) implies that \(\phi_1\) is continuous.

Completely analogously (but multiplying this time (17 ) by \((z_0-z)\)), we obtain \[\begin{align} \label{eqn:orange:recursionh} \phi_0(y) =\frac{z_0}{a+1}\int_{\{x\geq \theta y\}\cap[-a,1]} \phi_0(x) \mathrm{d}x. \end{align}\tag{19}\] The last equation shows that \(\phi_0\) is in particular continuous. We can now iterate (18 ) to get \[\begin{align} \phi_1(y) & = \frac{z_1^n}{(a+1)^n} \int_{-a}^1 \int_{-a}^1 \cdots \int_{-a}^1 \mathbb{1}_{x_2\geq \theta y, x_3\geq \theta x_2, \ldots , x_{n+1} \geq \theta x_n } \mathrm{d}x_2 \ldots \mathrm{d}x_n \phi_1(x_{n+1}) \mathrm{d}x_{n+1} \notag \\ & = z_1^n \int_{-a}^1 \mathbb{P}( X_2 \geq \theta y, X_3\geq \theta X_2, \ldots, X_n\geq \theta X_{n-1}, x_{n+1} \geq \theta X_n) \phi_1(x_{n+1}) \,\frac{\mathrm{d}x_{n+1}}{a+1} \notag \\ & =: z_1^n \int_{-a}^1 K_n(y,x_{n+1}) \phi_1(x_{n+1}) \,\frac{\mathrm{d}x_{n+1}}{a+1}.\label{eqn:orange:iteratedphi} \end{align}\tag{20}\] Clearly, \(K_n(y,x_{n+1})>0\) for any \(y,x_{n+1}\in[-a,1]\) if \(n\) is large enough (in fact, \(n\geq p+1\) already works). By (19 ), the analogous fact is true for \(\phi_0\): \[\begin{align} \label{eqn:orange:iteratedh} \phi_0(y) & = z_0^n \int_{-a}^1 K_n(y,x_{n+1}) \phi_0(x_{n+1}) \,\frac{\mathrm{d}x_{n+1}}{a+1}. \end{align}\tag{21}\]

We finally show some decisive estimates between \(\phi_1\) and \(\phi_0\). Since \(\phi_0(y)>0\) for all \(y\in[-a,1]\), we can define \[M:=\max_{y\in[-a,1]} \frac{|\phi_1(y)|}{\phi_0(y)}.\] Since \(\phi_1\) and \(\phi_0\) are continuous, the definition of \(M\) allows to choose a \(y_\ast\in[-a,1]\) such that \[\begin{align} M = \frac{|\phi_1(y_\ast)|}{\phi_0(y_\ast)}. \end{align}\] Using (20 ) and (21 ) and the triangle inequality, for \(n\geq p+1\), we obtain \[\begin{align} M \phi_0(y_\ast) & = |\phi_1(y_\ast)| \notag \\ & = \left|z_1 ^n \int_{-a}^1 K_n(y_\ast,x_{n+1}) \phi_1(x_{n+1}) \,\frac{\mathrm{d}x_{n+1}}{a+1} \right| \notag \\ & \leq |z_1|^n \int_{-a}^1 K_n(y_\ast,x_{n+1}) |\phi_1(x_{n+1})| \, \frac{\mathrm{d}x_{n+1}}{a+1} \label{eqn:orange:triangleintegral} \\ & \leq z_0^n \int_{-a}^1 K_n(y_\ast,x_{n+1}) M \phi_0(x_{n+1}) \, \frac{\mathrm{d}x_{n+1}}{a+1} \notag \\ & = M \phi_0(y_\ast). \notag \end{align}\tag{22}\] Since \(M\) and \(\phi_0(y_\ast)\) are positive, the inequalities must in fact be equalities, so that in particular \(|z_1|=z_0\). Further, since \(K_n(y_\ast,x_{n+1})>0\) and the integrands \(\phi_1(x_{n+1})\) and \(M \phi_0(x_{n+1})\) are continuous, the definition of \(M\) and fact that \[\int_{-a}^1 K_n(y_\ast,x_{n+1}) (M \phi_0(x_{n+1})- |\phi_1(x_{n+1})|) \mathrm{d}x_{n+1} = 0\] imply that we must have \(|\phi_1(x)|=M \phi_0(x)\) for all \(x\in[-a,1]\). Additionally, the triangle inequality step (22 ) can only be true if the complex argument of \(\phi_1(x)\) is identical for Lebesgue almost all \(x\); and by continuity of \(\phi_1\), we must have \(\phi_1(x)=e^{i\psi} M \phi_0(x)\) for some \(\psi\in[0,2\pi)\). Using this complex representation of \(\phi_1(y)\) and (18 ) in the first and (19 ) in the last step, we obtain \[\begin{align} e^{i\psi} M \phi_0(y) & = \frac{z_1}{a+1} \int_{\{x\geq \theta y\}\cap[-a,1]} e^{i\psi} M \phi_0(x) \mathrm{d}x \\ & = \frac{z_1}{z_0} \, e^{i\psi} M\, \frac{z_0}{a+1} \int_{\{x\geq \theta y\}\cap[-a,1]} \phi_0(x) \mathrm{d}x = \frac{z_1}{z_0} e^{i\psi} M\, \phi_0(y), \end{align}\] which implies \(z_1=z_0\). The second part of the assertion follows simply by the fact that \(u\) is a polynomial, so that the different zeros are necessarily separated. ◻

Proof. Our goal is to use Lemma 8. By Corollary 28, we have with \(z_0=\frac{1}{\lambda}\) \[\begin{align} \hat{P}_{a,y,\theta}(z)&= \frac{G_{a,y,\theta}(z)}{1-zF(z)}=\frac{G_{a,y,\theta}(z)\frac{z_0-z}{1-zF(z)}}{z_0-z}, \end{align}\] and we have to show that for some \(z_1>z_0\), the function \(f(z)=G_{a,y,\theta}(z)\frac{z_0-z}{1-zF(z)}=\frac{G_{a,y,\theta}(z)(z_0-z)}{u(z)}\) is analytic on \(|z|<z_1\) and \(f(z_0)\neq 0\).

The fact that \(G_{a,y,\theta}(z)\) is entire follows directly from its definition. The fact that \(z\mapsto \frac{z_0-z}{1-zF(z)}\) is analytic on \(|z|<z_1\) for some \(z_1>z_0\) is a conclusion from Lemma 21 and Lemma 31. Further, \(G_{a,y,\theta}(z_0)\neq 0\) follows from Lemma 30. Now, Lemma 8 gives \[\begin{align} p_n^{a,y}(\theta)&\sim f(z_0)\lambda^{n+1}\\&=\lambda^{n+1}\frac{-1}{u'(z_0)}\sum_{i=0}^m\left(\frac{-z_0y}{a+1}\right)^i\frac{\theta^\frac{i(i+1)}{2}}{i!}\sum_{{\ell}=0}^{m-i}\sum_{\substack{i_1+\dots +i_{\ell}= m-i-{\ell}\\i_r\ge 0}}\prod_{r=1}^{\ell}s(m-i-(r-1)-\sum_{j=1}^{r-1}i_j,i_r) \\&=\lambda^{n+1}\sum_{i=0}^my^ic_i(\theta,\lambda,a,m) \end{align}\] with coefficients given in (?? ). ◻

Proof of Theorem 2 (O). Lemma 32 gives \(p_n^{a,y}(\theta)\sim \lambda^{n+1}\sum_{i=0}^my^ic_i(\theta,\lambda,a, m)=\lambda^{n+1} h(y)\) for \(0\le m\le p\) and \(y\in A_m\) and \(h\) as defined in the claim of Theorem 2 (O). Like in the other regions, we use Lemma 5 to conclude to the Doob limit. The theorem also states the value of the eigenfunction for the case \(\theta x\in A_p=[\frac{-a}{\theta^p},1]\). That part does not follow from the asymptotics and is not required for the application of Lemma 5. The value of \(h\) for this case can be verified directly from the eigenvalue equation. ◻