Persistence Exponents via Perturbation Theory: AR(1)-Processes

Abstract

For AR(1)-processes $X_n=\rho X_{n-1}+\xi _n$, $n\in \mathbb {N}$, where $\rho \in \mathbb {R}$ and $(\xi _i)_{i\in \mathbb {N}}$ is an i.i.d. sequence of random variables, we study the so-called persistence probabilities $\mathbb {P}(X_0\ge 0,\ldots , X_N\ge 0)$ for $N\rightarrow \infty$. For a wide class of Markov processes a recent result (Aurzada et al. in Persistence exponents in Markov chains, arXiv preprint arXiv:1703.06447, 2017) shows that these probabilities decrease exponentially fast and that the rate of decay can be identified as an eigenvalue of some integral operator. We discuss a perturbation technique to determine a series expansion of the eigenvalue in the parameter $\rho$ for normally distributed AR(1)-processes.

Collection of Auxiliary Results from Perturbation Theory

1.1 Overview

The aim of this appendix is to give the reader a simple and almost self-contained introduction to the relevant results of perturbation theory for this paper. The appendix is based on the classical work [15].

Let X be a Banach space with norm $\Vert \cdot \Vert _{X}$. We write $\mathcal {L}(X)$ for the set of bounded linear operators on X. For $T\in \mathcal {L}(X)$ we will denote by $\Vert T\Vert$ the operator norm of T, i.e. $\Vert T\Vert :=\inf \{c\ge 0:\Vert Tx\Vert _{X}\le c\Vert x\Vert _{X} \text{ for } \text{ all } x\in X \}$. For a sequence $(T_n)_n$ of operators on X and an operator T on X, we write $\lim _{n\rightarrow \infty } T_n=T$ if $\lim _{n\rightarrow \infty }\Vert T_n-T\Vert = 0$. Moreover, let $D\subseteq \mathbb {C}$ be a domain.

Definition

The operator-valued function $\mathcal {T}:D\rightarrow \mathcal {L}(X)$ is called holomorphic if $\lim _{h\rightarrow 0}\frac{\mathcal {T}(t+h)-\mathcal {T}(t)}{h}$ exists for all $t\in D$.

Throughout the appendix, we will make use of properties of holomorphic operator-valued functions. Roughly speaking, the results of complex-valued functions can be generalized to operator-valued functions by considering scalar-valued functions defined via the dual space. For an overview of this topic we refer to [5, Sects. 3.3, 10.1] and [1, Appendix A].

A holomorphic operator-valued function on a disc can be expressed as a power series on this disc. Conversely, a convergent power series on a disc defines a holomorphic function. To simplify notation we continue to write $T_t$ for $\mathcal {T}(t)$. The key to prove Theorem 2 are the following results from perturbation theory.

Theorem 4

Assume that $\mathcal {T}:D\rightarrow \mathcal {L}(X)$ is holomorphic. Let $t_0\in D$ and $\lambda _0$ be an isolated eigenvalue of $T_{t_0}$ with algebraic multiplicity equal to one. Then there exists an open neighbourhood $U\subseteq D$ of $t_0$ and a holomorphic function $\lambda :U\rightarrow \mathbb {C}$ such that $\lambda _t$ is an eigenvalue of $T_t$ for $t\in U$.

In addition, there exists an open neighbourhood $U'\subseteq D$ of $t_0$ and a holomorphic function $g:U'\rightarrow X$ such that $g_t$ is an eigenvector of $T_t$ with eigenvalue $\lambda _t$ for $t\in U'\cap U$.

Theorem 5

Under the conditions of Theorem 4 with $T_t = \sum _{n\in \mathbb {N}}(t-t_0)^n T^{(n)}$ assume that $\Vert T^{(n)}\Vert \le ac^{n-1}$ for all $n\in \mathbb {N}$ with $a,c\ge 0$. The following lower bound for the radius of convergence r of the power series of $\lambda _t$ at $t_0$ holds:

$$\begin{aligned} r\ge \min _{\lambda \in \Gamma }\frac{1}{a\Vert R_0(\lambda )\Vert +c}, \end{aligned}$$

where $R_0(\cdot )$ is the resolvent operator of $T_{t_0}$ and $\Gamma$ is an arbitrary closed curve that lies outside $\Sigma (T_{t_0})$ with positive direction which encloses $\lambda _0$, where $\Sigma (T_{t_0})$ is the spectrum of $T_{t_0}$.

Corollary 6

Under the assumptions of Theorem 5, and if moreover X is a Hilbert space and $T_{t_0}$ is normal, we get

$$\begin{aligned} r\ge \frac{1}{\frac{2a}{d}+c}, \end{aligned}$$

where $d:={{\,\mathrm{dist}\,}}\left( \lambda _0,\Sigma (T_{t_0})\setminus \{\lambda _0\}\right)$.

1.2 Preliminaries

Lemma 7

Let $T\in \mathcal {L}(X)$ with $\Vert T\Vert <1$. Then the Neumann series $\sum _{n=0}^\infty T^n$ converges in the operator norm and we have

$$\begin{aligned} ({{\,\mathrm{Id}\,}}-T)^{-1} = \sum _{n=0}^\infty T^n. \end{aligned}$$

This is a well-known result in functional analysis and can be found for instance in [24, Propostion I.1.6].

Definition

For $T\in \mathcal {L}(X)$ the resolvent set $\sigma (T)$ is the set of all $\lambda \in \mathbb {C}$ such that $(T-\lambda {{\,\mathrm{Id}\,}})$ has a bounded inverse.

The operator-valued function $R:\sigma (T)\rightarrow \mathcal {L}(X),\ \lambda \mapsto (T-\lambda {{\,\mathrm{Id}\,}})^{-1}$ is called the resolvent operator of T.

For abbreviation, we write $T-\lambda$ instead of $T-\lambda {{\,\mathrm{Id}\,}}$ when no confusion can arise.

Lemma 8

We have

$$\begin{aligned} R(\lambda ')-R(\lambda ) = (\lambda '-\lambda )R(\lambda ')R(\lambda ) \end{aligned}$$

for all $\lambda ,\lambda '\in \sigma (T)$. In particular, $R(\lambda )$ and $R(\lambda ')$ commute.

Proof

The following easy computation shows the statement:

$$\begin{aligned} R(\lambda ')-R(\lambda )&= R(\lambda ')(T-\lambda )R(\lambda )-R(\lambda ')(T-\lambda ')R(\lambda )\\&= -R(\lambda ')\lambda R(\lambda ) + R(\lambda ')\lambda ' R(\lambda ) \\&= (\lambda '-\lambda )R(\lambda ')R(\lambda ). \end{aligned}$$

$\square$

Proposition

Let $\lambda ,\lambda _0\in \sigma (T)$ with $|\lambda -\lambda _0|<\Vert R(\lambda _0)\Vert ^{-1}$ then the so-called first Neumann series for the resolvent $\sum _{n=0}^\infty (\lambda -\lambda _0)^n R(\lambda _0)^{n+1}$ is convergent. In this case, we have

$$\begin{aligned} R(\lambda )= \sum _{n=0}^\infty (\lambda -\lambda _0)^n R(\lambda _0)^{n+1}. \end{aligned}$$

This shows that $R(\cdot )$ is holomorphic on $\sigma (T)$.

Proof

Let $\lambda ,\lambda _0\in \sigma (T)$. By Lemma 8 we obtain

$$\begin{aligned} R(\lambda ) = R(\lambda _0)\left( 1-(\lambda -\lambda _0)R(\lambda _0) \right) ^{-1}. \end{aligned}$$

If $\Vert (\lambda -\lambda _0)R(\lambda _0)\Vert <1$, Lemma 7 implies that

$$\begin{aligned} R(\lambda ) = R(\lambda _0)\sum _{n=0}^\infty (\lambda -\lambda _0)^n R(\lambda _0)^n, \end{aligned}$$

which proves the statement. $\square$

1.3 Perturbation of the Resolvent Operator

We define

$$\begin{aligned} R(t,\lambda )= (T_t-\lambda )^{-1} \end{aligned}$$

for any $(t,\lambda )$ with $\lambda \in \sigma (T_t)$. We already know from the last proposition that $R(t,\lambda )$ is holomorphic in $\lambda$ for each fixed t. Now we will show that $R(t,\lambda )$ is holomorphic in both variables.

Proposition

Let $\mathcal {T}:D\rightarrow \mathcal {L}(X)$ be holomorphic. Then $R(t,\lambda )$ is holomorphic in both variables t and $\lambda$.

Proof

Let $(t_0,\lambda _0)$ such that $\lambda _0\in \sigma (T_{t_0})$. Since $\mathcal {T}$ is holomorphic we can write $T_t = \sum _{n=0}^\infty T^{(n)}(t-t_0)^n$ for $|t-t_0|$ small. We have

$$\begin{aligned} T_t-\lambda&= T_{t_0}-\lambda _0 - (\lambda -\lambda _0)+\sum _{n=1}^\infty T^{(n)}(t-t_0)^n\\&= (T_{t_0}-\lambda _0)\Big (1-\Big (\lambda -\lambda _0-\sum _{n=1}^\infty T^{(n)}(t-t_0)^n\Big )R(t_0,\lambda _0) \Big ). \end{aligned}$$

Thus,

$$\begin{aligned} R(t,\lambda ) = (T_t-\lambda )^{-1} = R(t_0,\lambda _0)\Big (1-\Big (\lambda -\lambda _0-\sum _{n=1}^\infty T^{(n)}(t-t_0)^n\Big )R(t_0,\lambda _0) \Big )^{-1}. \end{aligned}$$

If

$$\begin{aligned} \Vert \big (\lambda -\lambda _0-\sum _{n=1}^\infty T^{(n)}(t-t_0)^n\big )R(t_0,\lambda _0)\Vert <1, \end{aligned}$$

(5)

the last expression can be written as a double series in $\lambda$ and t by Lemma  7. This condition is satisfied if

$$\begin{aligned} |\lambda -\lambda _0|+\sum _{n=1}^\infty |t-t_0|^n\Vert T^{(n)}\Vert < \Vert R(t_0,\lambda _0)\Vert ^{-1}, \end{aligned}$$

(6)

which is the case if $|\lambda -\lambda _0|$ and $|t-t_0|$ are small enough. $\square$

Remark

If we fix $\lambda$ and write $R(t,\lambda )$ as a power series in t at $t_0$ we get

$$\begin{aligned} R(t,\lambda ) = R(t_0,\lambda )\Big (\sum _{k=0}^\infty \big (-\sum _{n=1}^\infty T^{(n)}(t-t_0)^n R(t_0,\lambda )\big )^k\Big ). \end{aligned}$$

We can rewrite this as

$$\begin{aligned} R(t,\lambda ) = \sum _{k=0}^\infty (t-t_0)^k R^{(k)}(\lambda ), \end{aligned}$$

(7)

with

$$\begin{aligned} R^{(k)}(\lambda ) = \sum _{\begin{array}{c} k_1+\cdots +k_n=k\\ k_i\ge 1,\,n\ge 1 \end{array}} (-1)^n R(t_0,\lambda )T^{(k_1)}R(t_0,\lambda )T^{(k_2)}\cdots T^{(k_n)}R(t_0,\lambda ). \end{aligned}$$

The right-hand side of Eq. (7) is called the second Neumann series for the resolvent.

1.4 Perturbation of the Eigenprojection, Eigenvalue, and Eigenvector

Proposition

Let $T\in \mathcal {L}(X)$ and $\lambda _0$ be a simple isolated eigenvalue of $T_0$. Let $\Gamma$ be a closed curve in $\sigma (T)$ with positive direction which encloses $\lambda _0$. Then

$$\begin{aligned} P_0 := - \frac{1}{2\pi i}\int _\Gamma R(\lambda )\,\text{ d }\lambda \end{aligned}$$

is a projection on the eigenspace of $\lambda _0$.

Proof

We need to show that:

• (i) $P_0$ is a projection, i.e. $P_0^2=P_0$.

• (ii) $\mathbf {R}(P_0)= M_0$, where $\mathbf {R}(P_0)$ is the range of $P_0$ and $M_0$ is the eigenspace of $\lambda _0$.

Let $\Gamma '$ be a closed curve in $\sigma (T)$ with positive direction which encloses $\lambda _0$ and lies outside $\Gamma$. Then $\int _\Gamma R(\lambda )\,\text{ d }\lambda = \int _{\Gamma '} R(\lambda )\,\text{ d }\lambda$ and we have

$$\begin{aligned} (-2\pi i)^2 P_0^2&= \int _\Gamma R(\lambda )\,\text{ d }\lambda \cdot \int _{\Gamma '} R(\mu )\,\text{ d }\mu \\&= \int _{\Gamma '} \int _\Gamma R(\lambda ) R(\mu ) \,\text{ d }\lambda \,\text{ d }\mu \\&= \int _{\Gamma '} \int _\Gamma \frac{1}{\mu -\lambda } R(\mu )\,\text{ d }\lambda \,\text{ d }\mu - \int _\Gamma \int _{\Gamma '} \frac{1}{\mu -\lambda } R(\lambda )\,\text{ d }\mu \,\text{ d }\lambda \\&= \int _{\Gamma '} 0 \,\text{ d }\mu - \int _\Gamma 2\pi i R(\lambda )\,\text{ d }\lambda \\&= -2\pi i \int _\Gamma R(\lambda )\,\text{ d }\lambda \\&= (-2\pi i)^2 P_0, \end{aligned}$$

where the third equality follows by Lemma 8. This shows (i). To prove (ii), we begin by showing $M_0\subseteq \mathbf {R}(P_0)$. Let $x\in M_0$, i.e. $T(x)=\lambda _0 x$. Then

$$\begin{aligned} P_0(x)= -\frac{1}{2\pi i}\int _\Gamma (T-\lambda {{\,\mathrm{Id}\,}})^{-1}(x)\,\text{ d }\lambda = -\frac{1}{2\pi i}\int _\Gamma (\lambda _0-\lambda )^{-1}(x)\,\text{ d }\lambda = x. \end{aligned}$$

Now, we proceed to show that $\mathbf {R}(P_0)\subseteq M_0$. We compute

$$\begin{aligned} (-2\pi i)TP_0&= \int _\Gamma T(T-\lambda )^{-1}\,\text{ d }\lambda \\&= \int _\Gamma {{\,\mathrm{Id}\,}}+\lambda (T-\lambda )^{-1}\,\text{ d }\lambda \\&= \int _\Gamma \lambda (T-\lambda )^{-1}\,\text{ d }\lambda \\&= {{\,\mathrm{Res}\,}}_{\lambda _0}(\lambda (T-\lambda )^{-1})\\&= \lambda _0 {{\,\mathrm{Res}\,}}_{\lambda _0}((T-\lambda )^{-1})\\&= \lambda _0 \int _\Gamma R(\lambda )\,\text{ d }\lambda \\&= (-2\pi i)\lambda _0 P_0. \end{aligned}$$

Hence, for all $x\in X$ we get $P_0(x)\in M_0$, which completes the proof. $\square$

In what follows, we assume that $\mathcal {T}:D\rightarrow \mathcal {L}(X)$ is holomorphic and that $\lambda _0$ is an isolated eigenvalue of $T_{t_0}$, $t_0\in D$, with algebraic multiplicity equal to one. Let $\Gamma$ be a closed curve in $\sigma (T_{t_0})$ with positive direction which encloses $\lambda _0$.

Proposition

The operator-valued function $\mathcal {P}:D\rightarrow \mathcal {L}(X),\quad t\rightarrow P_t:=\mathcal {P}(t)=\int _\Gamma R(t,\lambda )\,\text{ d }\lambda$ is holomorphic at an open neighbourhood of $t_0.$ It holds that $\dim (P_0X)=\dim (P_tX)$.

Proof

Since $\inf _{\lambda \in \Gamma }\Vert R(t_0,\lambda )\Vert ^{-1}>0$, from (6) it follows that the second Neumann series for the resolvent is uniformly convergent for $\lambda$ if $|t-t_0|$ is sufficiently small. In particular, $R(\lambda ,t)$ is well-defined for such $(\lambda ,t)$ and thus, $\Gamma \subseteq \sigma (T_t)$. Hence, $P_t$ is well-defined for $|t-t_0|$ small and due to the proposition in A.3 we get that $P_t$ is holomorphic at an open neighbourhood of $t_0$. The equality of the dimensions follows by [15, Lemma I.4.10]. $\square$

Proof of Theorem 4

Combining the last two propositions we see that $P_t$ is the eigenprojection for $T_t$ on the eigenspace of a simple eigenvalue $\lambda _t$ and that $P_t$ is holomorphic in t. Accordingly, we deduce a perturbation series for the eigenvalue $\lambda _t$ via the formula $\lambda _t={{\,\mathrm{trace}\,}}(T_tP_t)$.

Let $g_{0}$ be an eigenfunction of $\lambda _{0}$. Then $g_t= P_tg_{0}\in P_tX$ and thus an element of the eigenspace of $\lambda _t$. If $g_t\ne 0$ it is an eigenfunction of $\lambda _t$. $P_t$ is holomorphic and hence $g_t$ is holomorphic. Since $g_{0}\ne 0$, we have that $g_t\ne 0$ at least for $|t-t_0|$ small. $\square$

1.5 Lower Bound for the Radius of Convergence

Let $T_t=\sum _{n=0}^\infty (t-t_0)^n T^{(n)}$ for $t\in D$ and $\lambda _0$ be an isolated eigenvalue of $T_{t_0}$ with algebraic multiplicity equal to one. It follows from Eq. (5) that for a fixed $\lambda$ the power series $\sum _{n=0}^\infty R^{(n)}(\lambda )(t-t_0)^n$ is convergent, if $\Vert R(t_0,\lambda )\sum _{n=1}^\infty T^{(n)}(t-t_0)^n\Vert <1$.

Proof of Theorem 5

Assume that $\Vert T^{(n)}\Vert \le ac^{n-1}$ with $a,c\ge 0$. Then the power series is convergent if

$$\begin{aligned} |t-t_0|< \frac{1}{\Vert R(t_0,\lambda )\Vert \cdot a+c}. \end{aligned}$$

Consequently, $P_t=\int _\Gamma R(t,\lambda ) \,\text{ d }\lambda$ can be expressed as a power series if $|t-t_0|< \inf _{\lambda \in \Gamma }\frac{1}{\Vert R(t_0,\lambda )\Vert \cdot a +c}$.

The so obtained lower bound of the radius of convergence of $P_t$, and therewith of $\lambda _t$, depends crucially on the chosen curve $\Gamma$. It is worthwhile to get this bound as large as possible.

Proof of Corollary 6

If $T_{t_0}$ is a normal operator on a Hilbert space we have that $R(t_0,\lambda )$ is normal and as a consequence the operator norm of $R(t_0,\lambda )$ is equal to the spectral radius of $R(t_0,\lambda )$. From this we conclude

$$\begin{aligned} \Vert R(t_0,\lambda )\Vert = {{\,\mathrm{dist}\,}}(\lambda ,\Sigma (T_{t_0}))^{-1}, \end{aligned}$$

where $\Sigma (T_{t_0})$ is the spectrum of $T_{t_0}$ and ${{\,\mathrm{dist}\,}}$ is the Hausdorff distance. Let $d:={{\,\mathrm{dist}\,}}(\lambda _0,\Sigma (T_{t_0})\setminus \{\lambda _0\})$ and let $\Gamma$ be a circle with radius $\frac{d}{2}$ and center $\lambda _0$. Then $\Vert R(t_0,\lambda )\Vert =\frac{2}{d}$ for every $\lambda \in \Gamma$ and we get

$$\begin{aligned} r\ge \frac{1}{\frac{2a}{d}+c}. \end{aligned}$$

$\square$