Fredholm Determinant of an Integral Operator Driven by a Diffusion Process

Abstract

This article aims to give a formula for differentiating, with respect to 𝑇, an expression of the form 𝜆(𝑇,𝑥)∶=𝔼_𝑥[𝑓(𝑋_𝑇)𝑒^{−∫𝑇0𝑉(𝑋𝑠)𝑑𝑠}(det(𝐼+𝐾_𝑋,𝑇))^𝑃], where 𝑝≥0 and 𝑋 is a diffusion process starting from 𝑥, taking values in a manifold, and the expectation is taken with respect to the law of this process. 𝐾_𝑋,𝑇∶𝐿²([0,𝑇)→ℝ^𝑁)→𝐿²([0,𝑇)→ℝ^𝑁) is a trace class operator defined by 𝐾_𝑋,𝑇𝑓(𝑠)=∫^𝑇0𝐻(𝑠⋀𝑡)Γ(𝑋(𝑡))𝑓(𝑡)𝑑𝑡, where 𝐻, Γ are locally Lipschitz, positive 𝑁×𝑁 matrices.

1. Introduction

Suppose we have a differentiable manifold 𝑀 of dimension 𝑑. By the Whitney's embedding theorem, there exists an embedding 𝑖∶𝑀→ℝ^𝑁′ such that 𝑖(𝑀) is a closed subset of ℝ^𝑁′. It turns out that 𝑁^=2𝑑+1 will do. We will identify 𝑀 with the image 𝑖(𝑀) and assume that 𝑀 itself is a closed submanifold of ℝ^𝑁′. We will also assume that 𝑀 does not have a boundary. Let 𝑀∪{𝜕_𝑀} be a one-point compactification of 𝑀.

Definition 1.1. An 𝑀-valued path 𝜔 with explosion time 𝑒=𝑒(𝜔)>0 is a continuous map 𝜔∶[0,∞)→𝑀∪{𝜕_𝑀} such that 𝜔_𝑡∈𝑀 for 0≤𝑡<𝑒 and 𝜔_𝑡=𝜕_𝑀 for all 𝑡≥𝑒 if 𝑒<∞. The space of 𝑀-valued paths with explosion time is called the path space of 𝑀 and is denoted by 𝑊(𝑀).

Let (Ω,ℱ_∗,ℙ) be a filtered probability space and let  𝐿 be a smooth second-order elliptic operator on 𝑀. Using the coordinates of the ambient space {𝑥₁,…,𝑥_𝑁^′}, and extending 𝐿 smoothly to 𝐿 in the ambient space, we may write𝐿=12^𝑁′_𝑖,𝑗=1𝐴_𝑖𝑗𝜕²𝜕𝑥_𝑖𝜕𝑥_𝑗+^𝑁′_𝑖=1𝑏_𝑖𝜕𝜕𝑥_𝑖,(1.1)with 𝜎∶=√𝐴, 𝐴 a positive matrix. Since 𝐴 is smooth, its square root is locally Lipschitz. Construct a time homogeneous Itô diffusion process 𝑋∶Ω→𝑊(𝑀) which solves the following stochastic differential equation:𝑑𝑋_𝑠=𝑏𝑋_𝑠𝑑𝑠+𝜎𝑋^𝑠𝑑𝐵^𝑠,𝑠≥𝑡;𝑋_𝑡=𝑥(1.2)in the ambient space ℝ^𝑁′, where 𝐵_𝑠 is 𝑁^-dimensional Euclidean-Brownian motion and 𝑏∶ℝ^𝑁′→ℝ^𝑁′,𝜎∶ℝ^𝑁′→ℝ^{𝑁′×𝑁′} such that‖‖𝑏(𝑥)−𝑏(𝑦)‖‖+‖‖𝜎(𝑥)−𝜎(𝑦)‖‖≤𝐷(𝑅)||𝑥−𝑦||(1.3)for some constant 𝐷(𝑅) dependent on an open ball centered at 𝑥 with radius 𝑅. Till the explosion time 𝑒(𝑋), 𝑋_𝑠∈𝑀 for 0≤𝑠<𝑒(𝑋). On 𝑀, 𝐿=𝐿. Furthermore, 𝜇^𝑋∶=ℙ∘𝑋⁻¹ is an 𝐿-diffusion measure on 𝑊(𝑀). As a result, we use 𝜇^𝑋 to be the probability measure on 𝑊(𝑀). Refer to [1] for a more detailed description.

Fix some positive integer 𝑁. Let 𝑀_𝑁(ℝ) and 𝒮_𝑁(ℝ) be the spaces of 𝑁×𝑁 real-valued matrices and symmetric 𝑁×𝑁 real-valued matrices, respectively. Also let 𝑀^+𝑁(ℝ)⊆𝒮_𝑁(ℝ) be the space of nonnegative matrices. Suppose that 𝐻∶[0,∞)→𝑀^+𝑁(ℝ) is locally Lipschitz with 0≤𝐻(𝑠)<𝐻(𝑡) if 𝑠<𝑡, 𝐻(0)=0, and Γ∶𝑀→𝑀^+𝑁(ℝ) is locally Lipschitz. Also assume that sup_𝑥∈𝑀‖Γ(𝑥)‖ is bounded, where ‖⋅‖ is the operator norm. Define 𝑀_Γ(𝑋) as the multiplication operator with Γ(𝑋) and Υ_𝑇 as the integral operator with kernel 𝐻(𝑠⋀𝑡), that is, for any 𝑓∈𝐿²([0,𝑇]→ℝ^𝑁),Υ^𝑇𝑓(𝑠)=^𝑇0𝐻(𝑠𝑡)𝑓(𝑡)𝑑𝑡,(1.4)where 𝑠⋀𝑡 is the minimum of 𝑠 and 𝑡. Note that under the assumptions on 𝐻, Υ_𝑇 is a positive operator and is trace class (see Proposition 3.1.).

Consider the following integral operator: 𝐾_𝑋,𝑇∶𝐿²([0,𝑇]→ℝ^𝑁)→𝐿²([0,𝑇]→ℝ^𝑁),𝐾^𝑋,𝑇𝑓(𝑠)=Υ𝑀_Γ(𝑋)𝑓(𝑠)=^𝑇0𝐻(𝑠𝑡)Γ𝑋(𝑡)𝑓(𝑡)𝑑𝑡.(1.5)It is a fact that for any trace class operator 𝐾, if we let ‖𝐾‖₁ denote the trace of |𝐾|, then‖‖𝐾_𝑋,𝑇‖‖₁≤sup_𝑥∈𝑀‖‖Γ(𝑥)‖‖‖‖Υ‖‖₁≤𝐶^𝑇0tr𝐻(𝑠)𝑑𝑠,(1.6)for some constant 𝐶. Here, tr means taking the trace of a matrix. Thus 𝐾_𝑋,𝑇 is trace class. Therefore,|||det𝐼+𝐾_𝑋,𝑇|||≤exp‖‖𝐾^𝑋,𝑇‖‖₁≤exp𝐶^𝑇0tr𝐻(𝑠)𝑑𝑠.(1.7)Hence the Fredholm determinant is bounded for each 𝑇.

Let 𝑓,𝑉 be continuous bounded functions taking 𝑥∈𝑀∪{𝜕𝑀}↦ℝ. Fix some number 𝑝≥0. Define a function 𝜆∶[0,∞)×𝑀→ℝ by𝜆(𝑇,𝑥)∶=𝔼_𝑥[𝑓𝑋_𝑇𝑒^{−∫𝑇0𝑉(𝑋𝑠)𝑑𝑠}det𝐼+𝐾^𝑋,𝑇^𝑝],(1.8)where the expectation is taken with respect to 𝜇^𝑋 and the paths start from 𝑥. Note that 𝜆 is finite for any 𝑥,𝑇 from the above discussion. Let ℎ∶[0,∞)×𝑀×𝑀_𝑁(ℝ)→𝑀_𝑁(ℝ) such thatℎ(𝑡,𝑥,𝑣)=𝑣−𝐻(𝑡)Γ(𝑥)𝑣−𝐻(𝑡).(1.9)The main result is as follows.

Theorem 1.2. Let (𝑡,𝑥,𝑣)∈[0,∞)×𝑀×𝒮_𝑁(ℝ). Then 𝜆(𝑇,𝑥)=𝑒^𝑇𝐻𝑓(0,𝑥,0),(1.10)where 𝐻=𝐿+𝜕𝜕𝑡+^𝑁_𝑖,𝑗=1ℎ(𝑡,𝑥,𝑣)_𝑖𝑗𝜕𝜕𝑣^𝑖𝑗−𝑉(𝑥)+𝑝𝑡𝑟𝐻(𝑡)Γ(𝑥)−𝑣Γ(𝑥).(1.11)Here, ℎ(𝑡,𝑥,𝑣)_𝑖𝑗 is the 𝑖,𝑗 component of the matrix ℎ.

Clearly, 𝜆 is not in the Feynman-Kac formula form using the process 𝑋. The idea is to construct a diffusion process 𝑊_𝑠=(𝑠,𝑋_𝑠,𝑍_𝑠)∈[0,∞)×𝑀×𝒮_𝑁(ℝ) such thatdet𝐼+𝐾_𝑋,𝑇=exp^𝑇0𝐺𝑊_𝑠𝑑𝑠,(1.12)and 𝐺 is given by𝐺(𝑡,𝑥,𝑣)=tr𝐻(𝑡)Γ(𝑥)−𝑣Γ(𝑥).(1.13)If we can achieve this, then our result follows from a simple application of the Feynman-Kac formula. Proving (1.12) requires the following 2 steps.

First, we have to prove an essential formula for the derivative of logdet(𝐼+𝑧𝐾_𝑋,𝑇) with respect to 𝑇, given by𝑑𝑑𝑇logdet𝐼+𝑧𝐾^𝑋,𝑇=tr𝑧𝐻(𝑇)Γ𝑋^𝑇−𝑍^𝑧,𝑇Γ𝑋_𝑇,(1.14)where 𝑧 is a complex number and 𝑍_𝑧,𝑇 is some adapted process. For a precise definition of 𝑍_𝑧,𝑇, see (4.1), with 𝐾_𝑋,𝑠 replaced by 𝑧𝐾_𝑋,𝑠. When 𝑧=1, 𝑍_1,𝑇=𝑍_𝑇. The goal is to show that the formula holds for 𝑧=1 by analytic continuation.

Fix a time 𝑇. By making |𝑧| small such that ‖𝑧𝐾_𝑋,𝑇‖<1, we can use the perturbation formula and apply it to the determinant; see (2.3). Differentiating this equation with respect to 𝑇 will give us (1.14). By analytic continuation, we can extend the formula in some domain 𝑂⊆ℂ containing the origin, provided we avoid the poles of the resolvent of 𝑧𝐾_𝑋,𝑠 for all 𝑠≤𝑇. If 1∈𝑂, then (1.14) holds with 𝑧=1. By integrating both sides and raising to the exponent, we will get (1.12). Note that if ‖𝐾_𝑋,𝑇‖<1 for some time 𝑇, (1.14) and hence (1.12) hold. The details are given in Sections 2 and 3.

Now assume that (1.14) holds with 𝑧=1. The second step consists of constructing a diffusion process 𝑊 from 𝑋 by using a stochastic differential equation. To do this, we differentiate 𝑍_𝑇 with respect to 𝑇 and show that it satisfies the differential equation𝑑𝑍_𝑇=ℎ𝑇,𝑋_𝑇,𝑍_𝑇=𝑍^𝑇−𝐻(𝑇)Γ𝑋_𝑇𝑍^𝑇−𝐻(𝑇)𝑑𝑇,(1.15)and hence 𝑊_𝑇=(𝑇,𝑋_𝑇,𝑍_𝑇) satisfies the following stochastic differential equation:𝑑𝑊_𝑇=1,𝑏𝑋_𝑇,ℎ𝑊^𝑇𝑑𝑇+0,𝜎𝑋^𝑇,0𝑑𝐵^𝑇,(1.16)with explosion time 𝑒(𝑊). From this stochastic differential equation, it is clear that 𝑊 is a diffusion process and by replacing the Fredholm determinant by the formula in (1.12), 𝜆(𝑇,𝑥) can be written as a Feynman-Kac form using this process 𝑊. However, if 𝑒(𝑊)<𝑇<𝑒(𝑋), then (1.12) may fail to hold.

The positivity of 𝐻 and Γ are used to show that 𝑍_1,𝑇=𝑍_𝑇 exists for all time 𝑇 and hence (1.14) holds at 𝑧=1. This will also imply that 𝑒(𝑊)=𝑒(𝑋). In particular, only the positivity of 𝐻 is required to show that 𝐾_𝑋,𝑇 is a trace class operator. To avoid 𝑒(𝑊)<𝑒(𝑋), we can restrict ourselves to small time 𝑇 such that (2.24) holds true.

We can weaken our assumptions on 𝐻 and Γ by not insisting that they are symmetric matrices. If we only assume that 𝐾_𝑋,𝑇 is trace class, then we can replace 𝒮_𝑁(ℝ) with 𝑀_𝑁(ℝ). Under these weaker assumptions, we have the following result.

Theorem 1.3. Suppose that, for a given locally Lipschitz 𝐻 and Γ, 𝐾_𝑋,𝑇 is trace class. Assume that there exists some constant 𝐶 such that ‖‖su𝐻p_𝑥∈𝑀(𝑠‖‖)Γ‖‖(𝑥≤)‖‖𝐶≤𝑠,𝐶𝑠<≥∞0,.(1.17) Let (𝑡,𝑥,𝑣)∈[0,∞)×𝑀×𝑀_𝑁(ℝ). Then for all 𝑇<1/𝐶², 𝜆(𝑇,𝑥)=𝑒^𝑇𝐻𝑓(0,𝑥,0),(1.18)where 𝐻=𝐿+𝜕𝜕𝑡+^𝑁_𝑖,𝑗=1ℎ(𝑡,𝑥,𝑣)_𝑖𝑗𝜕𝜕𝑣^𝑖𝑗−𝑉(𝑥)+𝑝𝑡𝑟𝐻(𝑡)Γ(𝑥)−𝑣Γ(𝑥).(1.19)

From Lemma 2.4, using the assumptions on 𝐻 and Γ, ‖𝐾_𝑋,𝑇‖≤𝐶²𝑇. If 𝑇<1/𝐶², then the norm is less than 1. Hence (1.14) holds and thus (1.12) holds true.

2. Functional Analytic Tools

Notation 2.1. Suppose that 𝐾 is an integral operator, acting on 𝐿²([0,𝑇]→ℝ^𝑁)↦𝐿²([0,𝑇]→ℝ^𝑁). We will write 𝐾𝑓 to mean (𝐾𝑓)(𝑠)=^𝑇0𝐾(𝑠,𝑡)𝑓(𝑡)𝑑𝑡,(2.1)where 𝑇<∞. To distinguish the operator 𝐾 from its kernel, we will write 𝐾(𝑠,𝑡) to refer to its kernel. This may be confusing, but it is used to avoid too many symbols. In this article, our integral operator is always trace class and the kernel is a continuous 𝑁×𝑁 matrix-valued function. By abuse of notation, 𝐾^𝑛(𝑠,𝑡) refers to the kernel of the integral operator 𝐾^𝑛.

Notation 2.2. We will use tr to denote the trace of a matrix and Tr to denote taking the trace of a trace class operator. ‖⋅‖ will denote the operator norm.

It is well known that for a trace class operator 𝐴 and 𝑧∈ℂ, Trlog(𝐼+𝑧𝐴) is a meromorphic function and has singularities at points 𝑧 such that −𝑧⁻¹∈𝜎(𝐴). Define the determinant det(𝐼+𝑧𝐴), given bydet(𝐼+𝑧𝐴)=𝑒^{Trlog(𝐼+𝑧𝐴)}.(2.2)However, this determinant, also known as the Fredholm determinant of 𝐴, is analytic in 𝑧 because the singularities 𝑧 such that −𝑧⁻¹∈𝜎(𝐴) are removable; see [2, Lemma 16].

We want to differentiate the function logdet(𝐼+𝐾_𝑇) with respect to 𝑇, where we write 𝐾_𝑇 to denote the dependence on the domain [0,𝑇]. If the kernel of 𝐾_𝑇 is given by 𝐾(𝑠,𝑡), then for small 𝑧 such that ‖𝑧𝐾_𝑇‖<1, using the perturbation formula,det𝐼+𝑧𝐾_𝑇=exp(^∞_𝑛=1(−1)^𝑛+1𝑛Tr𝑧𝐾^𝑇^𝑛).(2.3)If we let 𝑟=‖𝐾_𝑇‖<1, then^∞_𝑛=1|||(−1)^𝑛+1𝑛Tr𝐾^𝑛𝑇|||≤^∞_𝑛=1𝑟^𝑛−1𝑛‖‖𝐾^𝑇‖‖₁≤‖‖𝐾^𝑇‖‖₁1−𝑟.(2.4)Thus the series in the exponent converges absolutely.

We will define the resolvent 𝑅_𝑇 by 𝐾_𝑇(𝐼+𝐾_𝑇)⁻¹. Since we can write 𝐾_𝑇(𝐼+𝐾_𝑇)⁻¹=𝐾_𝑇−𝐾_𝑇(𝐼+𝐾_𝑇)⁻¹𝐾_𝑇, the kernel of 𝑅_𝑇 can be written as𝑅_𝑇(𝑠,𝑡)=𝐾(𝑠,𝑡)−𝐾_𝑇𝐼+𝐾^𝑇⁻¹𝐾_𝑇(𝑠,𝑡).(2.5)When we write 𝐾_𝑇(𝐼+𝐾_𝑇)⁻¹𝐾_𝑇(𝑠,𝑡), we mean𝐾_𝑇𝐼+𝐾^𝑇⁻¹𝐾_𝑇(𝑠,𝑡)=^𝑇0𝐾(𝑠,𝑢)𝐼+𝐾_𝑇⁻¹𝐾(⋅,𝑡)(𝑢)𝑑𝑢.(2.6)We will also write the resolvent of the operator 𝑧𝐾_𝑇, 𝑧∈ℂ as 𝑅_𝑇,𝑧. One more point to note is that we assume that 𝐾(⋅,0)=𝐾(0,⋅)=0.

Now the operator 𝐾_𝑠 is an operator defined on different Hilbert spaces 𝐿²[0,𝑠]. Therefore, we will now think of our operator 𝐾_𝑠 as acting on 𝐿²[0,𝑇], defined as𝐾^𝑠𝑓(𝑢)=^𝑇0𝐾(𝑢,𝑣)𝜒_[0,𝑠](𝑣)𝑓(𝑣)𝑑𝑣,(2.7)where 𝜒 is the characteristic function. Hence now our operator 𝐾_𝑠 has a kernel 𝐾(𝑢,𝑣)𝜒_[0,𝑠](𝑣) dependent on the parameter 𝑠. Note that our 𝐾_𝑠 is continuous in the 𝑢 variable but is discontinuous at 𝑣=𝑠. Thus when we write 𝐾_𝑠(𝑢,𝑠), we mean𝐾_𝑠(𝑢,𝑠)∶=lim_𝑣↑𝑠𝐾_𝑠(𝑢,𝑣).(2.8)

Definition 2.3. Let 𝐾(⋅,⋅) be a continuous matrix-valued function and let  ‖𝐾(⋅,⋅)‖ be the matrix norm of 𝐾(⋅,⋅). Define 𝐶_𝑇 to be the maximum value of ‖𝐾(⋅,⋅)‖ on [0,𝑇]×[0,𝑇].

The next lemma allows us to control the operator norm of the operator by controlling the sup norm of the kernel.

Lemma 2.4. For 0≤𝑠<𝑠^≤𝑇, ‖‖𝐾_𝑠′−𝐾_𝑠‖‖≤𝐶_𝑇𝑇𝑠^−𝑠.(2.9)

Proof. For 𝑓,𝑔∈𝐿² and any 𝑠∈[0,𝑇],|||𝐾_𝑠−𝐾_𝑠′𝑓,𝑔|||≤≤^𝑇0‖‖𝑔(𝑢)^𝑇𝜒_{[𝑠,𝑠′]}(𝑣)𝐾(𝑢,𝑣)𝑓(𝑣)‖‖𝑑𝑣𝑑𝑢_𝑇0𝜒_{[𝑠,𝑠′]}(𝑣)𝐶_𝑇≤𝐶_𝑇|||𝑔(𝑢)||||||𝑓(𝑣)|||𝑑𝑣𝑑𝑢^𝑇0|||𝑔(𝑢)|||𝑑𝑢^𝑇0≤𝐶_𝑇𝜒_{[𝑠,𝑠′]}(𝑢)|||𝑓(𝑢)|||𝑑𝑢^𝑇0|||𝑔(𝑢)|||²𝑑𝑢⋅^𝑇01𝑑𝑢^1/2^𝑇0|||𝑓(𝑢)|||²𝑑𝑢⋅^𝑇0𝜒_{[𝑠,𝑠′]}(𝑢)𝑑𝑢^1/2≤𝐶_𝑇𝑇𝑠^−𝑠‖‖𝑓‖‖₂‖‖𝑔‖‖₂.(2.10)Hence,‖‖𝐾_𝑠′−𝐾_𝑠‖‖≤𝐶_𝑇𝑇𝑠^−𝑠(2.11)for all 0≤𝑠<𝑠^≤𝑇.

Lemma 2.5. Fix a 𝑧∈ℂ. For any 𝑇 such that |𝑧𝐶_𝑇𝑇|<1 and if 𝐾(𝑠,𝑡) is continuous, then 𝑑𝑑𝑇logdet𝐼+𝑧𝐾^𝑇=𝑡𝑟𝑅^𝑇,𝑧(𝑇,𝑇).(2.12)

Proof. Since 𝑧 is fixed, we will replace 𝑧𝐾_𝑇 by 𝐾_𝑇 and hence assume that |𝐶_𝑇𝑇|<1. Lemma 2.4 tells us that ‖𝐾_𝑇‖<1 and thus (2.3) holds true. Taking log on both sides of (2.3), we havelogdet𝐼+𝐾_𝑇=^∞_𝑛=1(−1)^𝑛+1𝑛Tr𝐾^𝑛𝑇.(2.13)NowTr𝐾^𝑛𝑇=^𝑇0⋯^𝑇0tr^𝑛_𝑖=1𝐾𝑠_𝑖,𝑠_𝑖+1𝑑𝑠¹⋯𝑑𝑠_𝑛,(2.14)where 𝑠_𝑛+1=𝑠₁. Differentiate with respect to 𝑇 and using the fundamental theorem of calculus, we get𝑑𝑑𝑇^𝑇0⋯^𝑇0tr^𝑛_𝑖=1𝐾𝑠_𝑖,𝑠_𝑖+1𝑑𝑠¹⋯𝑑𝑠_𝑛=𝑛tr^𝑇0⋯^𝑇0𝐾𝑇,𝑠₂(^𝑛−1_𝑖=2𝐾𝑠_𝑖,𝑠_𝑖+1)𝐾𝑠^𝑛,𝑇𝑑𝑠₂⋯𝑑𝑠_𝑛=𝑛tr𝐾^𝑛𝑇(𝑇,𝑇).(2.15)Let 𝐶_𝑇𝑇=𝛼. Thus,^∞_𝑛=1|||tr𝐾_𝑛𝑇(𝑇,𝑇)|||≤𝑁^∞_𝑛=1‖‖𝐾^𝑛𝑇(𝑇,𝑇)‖‖≤𝑁^∞_𝑛=1𝐶^𝑛𝑇𝑇^𝑛−1≤𝑁𝐶_𝑇∞_𝑛=0𝛼^𝑛=𝑁𝐶^𝑇1−𝛼<∞.(2.16)Thus𝑑𝑑𝑇logdet𝐼+𝐾^𝑇=^∞_𝑛=1𝑑𝑑𝑇((−1𝑛)^𝑛+1tr^𝑇0⋯^𝑇0𝑛_𝑖=1=tr_∞𝐾𝑠_𝑖,𝑠_𝑖+1𝑑𝑠¹⋯𝑑𝑠_𝑛)_𝑛=1=tr𝐾_𝑇(−1)^𝑛+1𝐾^𝑛𝑇(𝑇,𝑇)=tr𝑅_𝑇(𝐼+𝐾_𝑇)⁻¹(𝑇,𝑇)(𝑇,𝑇).(2.17)

Lemma 2.6. Let 𝐾(⋅,⋅) be continuous. For all 𝑧∈ℂ such that −𝑧⁻¹∉𝜎(𝐾_𝑠) for all 𝑠∈[0,𝑇], then 𝑡𝑟𝑅_⋅,𝑧(⋅,⋅) is continuous.

Proof. Fix a 𝑧 and write 𝑧𝐾_𝑠 as 𝐾_𝑠. By assumption, 𝐼+𝐾_𝑠 is invertible for all 𝑠. By Lemma 2.4, 𝐾_𝑠→𝐾_𝑠0 as 𝑠→𝑠₀. Note that𝐼+𝐾_𝑠=𝐼+𝐾_𝑠0+𝐾_𝑠−𝐾_𝑠0=𝐼+𝐾_𝑠0𝐼+𝐼+𝐾^𝑠₀⁻¹𝐾^𝑠−𝐾_𝑠0,(2.18)and if we let 𝐺_𝑠=𝐾_𝑠−𝐾_𝑠0, then𝐼+𝐾^𝑠⁻¹=𝐼+𝐼+𝐾_𝑠𝑜⁻¹𝐺_𝑠⁻¹𝐼+𝐾^𝑠_𝑜⁻¹.(2.19)By the open mapping theorem, because 𝐼+𝐾_𝑠𝑜 is a surjective continuous map, it is an open map. Therefore, its inverse is a bounded operator. Since 𝐺_𝑠=𝐾_𝑠−𝐾_𝑠0→0, thus (𝐼+(𝐼+𝐾_𝑠𝑜)⁻¹𝐺_𝑠)⁻¹→𝐼 and hence (𝐼+𝐾_𝑠)⁻¹ converges to (𝐼+𝐾_𝑠0)⁻¹ as 𝑠→𝑠₀. This shows that (𝐼+𝐾_𝑠)⁻¹ is continuous in 𝑠. Note that𝑅_𝑠(𝑠,𝑠)=𝐾(𝑠,𝑠)−𝐾_𝑠𝐼+𝐾^𝑠⁻¹𝐾(𝑠,𝑠).(2.20)Since 𝐾(𝑠,𝑠), 𝐾_𝑠 and (𝐼+𝐾_𝑠)⁻¹ are continuous in 𝑠, hence 𝑅_𝑠(𝑠,𝑠) is continuous in 𝑠.

Lemma 2.7. Let 𝐾(⋅,⋅) be continuous. If there exists an open-connected set 𝑂 containing 0 such that for 𝑧∈𝑂, (𝐼+𝑧𝐾_𝑠)⁻¹ is analytic for all 𝑠∈[0,𝑇], then logdet𝐼+𝑧𝐾_𝑇=^𝑇0𝑡𝑟𝑅_𝑠,𝑧(𝑠,𝑠)𝑑𝑠(2.21)for all 𝑧∈𝑂.

Proof. For all 𝑧∈𝑂, 𝐼+𝑧𝐾_𝑠 is invertible for all 𝑠∈[0,𝑇] and hence tr𝑅_𝑠,𝑧(𝑠,𝑠) is analytic in 𝑂. Therefore, it follows that ∫^𝑇0tr𝑅_𝑠,𝑧(𝑠,𝑠)𝑑𝑠 is analytic in 𝑂, because𝑑𝑑𝑧^𝑇0tr𝑅_𝑠,𝑧(𝑠,𝑠)𝑑𝑠=^𝑇0𝑑𝑑𝑧tr𝑅^𝑠,𝑧(𝑠,𝑠)𝑑𝑠.(2.22)By Lemma 2.4, ‖𝐾_𝑠‖≤𝑇𝐶_𝑇 for all 𝑠∈[0,𝑇]. Thus if we choose 𝑈∶={𝑧||𝑧|<1/(𝑇𝐶_𝑇)}, then 𝑈 is an open set containing 0 and for 𝑧∈𝑈, ‖𝑧𝐾_𝑠‖<1 for all 𝑠∈[0,𝑇]. From Lemma 2.5, for 𝑧∈𝑈,logdet𝐼+𝑧𝐾_𝑇=^𝑇0tr𝑅_𝑠,𝑧(𝑠,𝑠)𝑑𝑠.(2.23)Since logdet(𝐼+𝑧𝐾_𝑇) is also analytic in 𝑂 and agrees with ∫^𝑇0tr𝑅_𝑠,𝑧(𝑠,𝑠)𝑑𝑠 in 𝑈, it follows that both functions are equal for all 𝑧∈𝑂.

The proof in the previous theorem gives us the existence of a small neighbourhood containing 0 such that (2.21) holds. Hence we have the following corollary.

Corollary 2.8. Fix 𝑇>0 and 𝐶_𝑇=sup_{𝑠,𝑡∈[0,𝑇]}‖𝐾(𝑠,𝑡)‖. There exists an open set 𝑈_𝑇∶={𝑧||𝑧|<1/(𝑇𝐶_𝑇)} such that (2.21) holds for all 𝑧∈𝑈_𝑇.

Corollary 2.9. Let 𝑂 be an open-connected set as in Lemma 2.7 such that 1∈𝑂. Then for 𝑠∈(0,𝑇), 𝑑𝑑𝑠logdet𝐼+𝐾^𝑠=𝑡𝑟𝑅^𝑠(𝑠,𝑠).(2.24)

Proof. The corollary follows from differentiating (2.21). By Lemma 2.6, tr𝑅_𝑠(𝑠,𝑠) is continuous and hence the fundamental theorem of calculus applies.

3. Fredholm Determinant

The kernel we are interested in is 𝐾_𝑋,𝑇=𝐻(𝑠⋀𝑡)Γ(𝑋(𝑡)), for some process 𝑋. More generally, the kernel we are interested in is of the form 𝐾_𝑇(𝑠,𝑡)=𝐻(𝑠⋀𝑡)Λ(𝑡) for some continuous matrix-valued Λ. The Hilbert space is 𝐿²([0,𝑇]→ℝ^𝑁) for some positive number 𝑇. Without any ambiguity, we will in future write this space as 𝐿². We will also use ‖⋅‖₂ to denote the 𝐿² norm.

Proposition 3.1. If 𝐻 is continuous, 0≤𝐻(𝑠)≤𝐻(𝑡) for any 𝑠≤𝑡 and Λ continuous, then Υ_𝑇 as defined in Section 1 is a trace class operator.

To prove this result, we need the following theorem, which is [3, Theorem 2.12].

Theorem 3.2. Let 𝜇 be a Baire measure on a locally compact space 𝑋. Let 𝐾 be a function on 𝑋×𝑋 which is continuous and Hermitian positive, that is, ^𝐽_𝑖,𝑗=1𝑧_𝑖𝑧_𝑗𝐾𝑥_𝑖,𝑥_𝑗≥0(3.1)for any 𝑥₁,…,𝑥_𝐽∈𝑋, 𝑧₁,…,𝑧_𝐽∈ℂ^𝐽 and for any 𝐽. Then 𝐾(𝑥,𝑥)≥0 for all 𝑥. Suppose that, in addition, 𝐾(𝑥,𝑥)𝑑𝜇(𝑥)<∞.(3.2)Then there exists a unique trace class integral operator 𝐴 such that (𝐴‖‖𝑓𝐴)(‖‖𝑥)=𝐾(𝑥,𝑦)𝑓(𝑦)𝑑𝜇(𝑦),₁=𝐾(𝑥,𝑥)𝑑𝜇.(3.3)

Proof of Proposition 3.1. Let 𝑋=[0,𝑇] and 𝜇 be Lebesgue measure. Using Theorem 3.2, it suffices to show that 𝐻(𝑠⋀𝑡) is Hermitian positive. Let 𝑧₁,𝑧₂,…,𝑧_𝐽 be any complex column vectors. Note that there are 𝑁 entries in each column and the entries are complex valued. Let 𝑠₁,…,𝑠_𝐽∈[0,𝑇]. The proof is obtained using induction. Clearly, when 𝐽=1, it is trivial. Suppose it is true for all values from 𝑘=1,2,…,𝐽−1. By relabelling, we can assume that 𝑠₁≤𝑠_𝑘, 𝑘=2,…,𝐽. Hence 𝑠₁⋀𝑠_𝑘=𝑠₁ for any 𝑘. Let ⟨⋅,⋅⟩ be the usual dot product. Then^𝐽_𝑗=1𝐻𝑠¹𝑠^𝑗𝑧^𝑗,𝑧₁+^𝐽_𝑗=1𝐻𝑠^𝑗𝑠¹𝑧¹,𝑧_𝑗=^𝐽_𝑗=1𝐻𝑠¹𝑧^𝑗,𝑧₁+^𝐽_𝑗=1𝐻𝑠¹𝑧¹=⟨_𝐽,𝑧_𝑗_𝑗=1𝐻𝑠₁𝑧^𝑗,^𝐽_𝑗=1𝑧_𝑗⟩−^𝐽_𝑖_,𝑗=2𝐻𝑠¹𝑧^𝑗,𝑧_𝑖.(3.4)Therefore,^𝐽_𝑖,𝑗=1𝐻𝑠^𝑖𝑠^𝑗𝑧^𝑗,𝑧_𝑖=⟨^𝐽_𝑗=1𝐻𝑠₁𝑧^𝑗,^𝐽_𝑗=1𝑧_𝑗⟩−^𝐽_𝑖_,𝑗=2𝐻𝑠¹𝑧^𝑗,𝑧_𝑖+^𝐽_𝑖_,𝑗=2𝐻𝑠^𝑖𝑠^𝑗𝑧^𝑗=⟨_𝐽,𝑧_𝑖_𝑗=1𝐻𝑠₁𝑧^𝑗,^𝐽_𝑗=1𝑧_𝑗⟩+^𝐽_𝑖_,𝑗=2𝐻𝑠^𝑖𝑠^𝑗−𝐻𝑠¹𝑧^𝑗,𝑧_𝑖.(3.5)Since 𝑠_𝑖⋀𝑠_𝑗≥𝑠₁,𝑖=2,…,𝐽 and hence 𝐻(𝑠_𝑖⋀𝑠_𝑗)≥𝐻(𝑠₁) by assumption on 𝐻. Thus by induction hypothesis, (replace 𝐻 by 𝐻(⋅)−𝐻(𝑠₁)),^𝐽_𝑖,𝑗=2𝐻𝑠^𝑖𝑠^𝑗−𝐻𝑠¹𝑧^𝑗,𝑧_𝑖≥0(3.6)and hence^𝐽_𝑖,𝑗=1𝐻𝑠^𝑖𝑠^𝑗𝑧^𝑗,𝑧_𝑖≥0.(3.7)

Notation 3.3. By abuse of notation, ⟨⋅,⋅⟩_𝑇 will denote integration over [0,𝑇], ⟨𝑓,𝑔⟩_𝑇∶=^𝑇0𝑓(𝑢)⋅𝑔(𝑢)𝑑𝑢,(3.8) where ⋅ should be interpreted as matrix multiplication or inner product, depending on the context. To ease the notation, we will write in future 𝐾(𝑇)∶=𝐾(𝑇,𝑇).

Remark 3.4. If we assume that 𝐻(𝑠)−𝐻(𝑡) is strictly positive if 𝑠>𝑡, which is the case we are interested in this article, then the proof of Proposition 3.1 shows that the operator Υ_𝑇 with kernel 𝐻(𝑠⋀𝑡) is strictly positive, that is, ⟨Υ_𝑇𝑓,𝑓⟩_𝑇>0 if ⟨𝑓,𝑓⟩_𝑇>0. This follows using a Riemann lower sum approximation on a double integral and that for any complex vectors 𝑧₁,…,𝑧_𝐽,^𝐽_𝑖,𝑗=1𝐻𝑠^𝑖𝑠^𝑗𝑧^𝑗,𝑧_𝑖𝑇²𝐽²>0(3.9)under the strict positivity assumptions.

The next proposition is a crucial statement. For the time being, we will assume that (𝐼+𝐾_𝑇)⁻¹ exists for any time 𝑇 without any further justification. Later on, we will prove that for our operator 𝐾_𝑋,𝑇, this is true; (see Proposition 5.2.). Writing in our new notation, we obtain the next proposition from Corollary 2.9.

Proposition 3.5. Let 𝐾_𝑇 be an integral operator with kernel 𝐾_𝑇(𝑠,𝑡)=𝐻(𝑠⋀𝑡)Λ(𝑡) for some continuous matrix-valued Λ: 𝑑𝑑𝑇logdet𝐼+𝐾^𝑇=𝑡𝑟𝐾(𝑇)−𝐻Λ,𝐼+𝐾^𝑇⁻¹𝐻_𝑇Λ(𝑇).(3.10)

Proof. We will use (2.5). So𝑅_𝑇(𝑇,𝑇)=𝐾(𝑇,𝑇)−𝐾(𝑇,⋅),𝐼+𝐾_𝑇⁻¹𝐾(⋅,𝑇)_𝑇=𝐾(𝑇,𝑇)−𝐻(⋅)Λ(⋅),𝐼+𝐾_𝑇⁻¹𝐻(⋅)Λ(𝑇)_𝑇=𝐾(𝑇)−𝐻Λ,𝐼+𝐾_𝑇⁻¹𝐻_𝑇Λ(𝑇).(3.11)Taking trace completes the proof.

Definition 3.6. Let 𝑍_𝑇=𝐻Λ,𝐼+𝐾_𝑇⁻¹𝐻_𝑇.(3.12)

To ease the notation, we will now write 𝐿∶=(𝐼+𝐾_𝑇)⁻¹ and 𝐿(𝑠,𝑡) to be the kernel of 𝐿. Note that in future we will drop the subscript 𝑇 from the operator 𝐾 and it should be understood that 𝐾 is dependent on 𝑇. Operators with a prime will denote its derivative with respect to 𝑇. Our task now is to differentiate 𝑍_𝑇.

Define a distributional kernel𝜌(𝑠,𝑡)=𝑅(𝑠,𝑡)−𝛿(𝑠−𝑡),(3.13)where 𝛿 is the Dirac delta function and 𝑅 is the resolvent.

For any operator depending smoothly on some parameter 𝑇, we have the differentiation formula𝐿^𝑇=𝑑𝑑𝑇𝐼+𝐾^𝑇⁻¹=−𝐼+𝐾_𝑇⁻¹𝐾^𝑇𝐼+𝐾^𝑇⁻¹.(3.14)For the integral operator 𝐾^, its kernel is given, by the fundamental theorem of calculus, by𝐾^(𝑠,𝑡)=𝐾(𝑠,𝑇)𝛿(𝑡−𝑇);(3.15)and hence combining with (3.14), we have𝐿^(𝑠,𝑡)==−−𝐿𝐿𝐾𝐾^_𝐿(𝑠,𝑡)(𝑠,𝑡)−𝐿𝐾(𝑠,𝑇)𝛿(⋅−𝑇),−𝐾𝐿(⋅,𝑡)_𝑇=−𝐿𝐾(𝑠,𝑇)𝛿(𝑡−𝑇)+𝐿𝐾(𝑠,𝑇)𝛿(⋅−𝑇),𝐾𝐿(⋅,𝑡)_𝑇==−𝑅𝑅(𝑠(,𝑠𝑇,)𝑇𝜌)(𝛿𝑇(,𝑡𝑡−).𝑇)+𝑅(𝑠,𝑇)𝑅(𝑇,𝑡)(3.16)

Notation 3.7. Let 𝐾 be an integral operator with kernel 𝐾(𝑠,𝑡). We define the adjoint 𝐾^∗ by 𝐾^∗𝑓)(𝑡=^𝑇0𝑓(𝑠)𝐾(𝑠,𝑡)𝑑𝑡.(3.17)Here, 𝑓 is an 𝑁×𝑁 matrix-valued function. We will also write Λ_𝑇=Λ(𝑇) and 𝐻_𝑇=𝐻(𝑇).

The following lemma defines the relationship between 𝑅 and 𝐿.

Lemma 3.8. It holds that 𝑅𝑅((𝑠𝑇,,𝑇𝑡))==(𝐿𝐿𝐻_∗)(𝑠)Λ_𝑇,𝐻Λ(𝑡).(3.18)

Proof. We write the identity operator 𝐼 as 𝐼𝑓(𝑠)=𝑓(𝑠)=∫^𝑇0𝛿(𝑠−𝑡)𝑓(𝑡)𝑑𝑡, where 𝛿 is Dirac delta function. Then𝑅(𝑠,𝑇)=𝑅𝐿((𝑇𝑠,,𝑡⋅)),=𝐾(⋅𝐾,𝑇()𝑇,_𝑇⋅),=𝐿(⋅𝐿,𝑡()𝑠,_𝑇⋅),𝐻(⋅⋀𝑇)Λ_𝑇_𝑇=𝐿(𝑠,⋅),𝐻(⋅)Λ_𝑇_𝑇=(𝐿𝐻)(𝑠)Λ_𝑇,=𝐻(𝑇⋀⋅)Λ(⋅),𝐿(⋅,𝑡)_𝑇=𝐿^∗𝐻Λ(𝑡).(3.19)

Theorem 3.9. 𝑍_𝑇 satisfies the following differential equation: 𝑍^𝑇=𝑍_𝑇−𝐻_𝑇Λ^𝑇𝑍^𝑇−𝐻_𝑇.(3.20)

Proof. Now by definition of 𝑍_𝑇,𝑍_𝑇=𝐻Λ,𝐼+𝐾_𝑇⁻¹𝐻_𝑇=⟨𝐻Λ,𝐿𝐻⟩_𝑇=𝐿^∗𝐻Λ,𝐻_𝑇.(3.21)Using (3.16) and from Lemma 3.8, 𝐿^𝐻(𝑠)=𝑅(𝑠,𝑇)𝜌(𝑇,⋅),𝐻(⋅)_𝑇=𝑅(𝑠,𝑇)𝑅(𝑇,⋅),𝐻(⋅)_𝑇−𝑅(𝑠,𝑇)𝐻_𝑇=(𝐿𝐻)(𝑠)Λ_𝑇𝐿^∗𝐻Λ,𝐻−(𝐿𝐻)(𝑠)Λ_𝑇𝐻_𝑇=(𝐿𝐻)(𝑠)Λ_𝑇𝑍_𝑇−(𝐿𝐻)(𝑠)Λ_𝑇𝐻_𝑇.(3.22)Hence differentiating with respect to 𝑇, using the fundamental theorem and (3.22), gives𝑑𝑑𝑇𝑍^𝑇=(𝐻Λ𝐿𝐻)(𝑇)+𝐻Λ,𝐿^𝐻⟩_𝑇=(𝐻Λ𝐿𝐻)(𝑇)+(𝐻Λ)(⋅),(𝐿𝐻)(⋅)Λ_𝑇𝑍_𝑇−(𝐿𝐻)(⋅)Λ_𝑇𝐻_𝑇_𝑇=(𝐻Λ𝐿𝐻)(𝑇)−𝑍_𝑇Λ_𝑇𝐻_𝑇+𝑍_𝑇Λ_𝑇𝑍_𝑇.(3.23)But(𝐾𝐿𝐻)(𝑇)==𝐾𝐻((𝐼𝑇+𝐾⋅))Λ⁻¹(𝐻⋅),((𝐿𝑇𝐻))(⋅)_𝑇=⟨𝐻Λ,𝐿𝐻⟩_𝑇=𝑍_𝑇.(3.24)Hence(𝐻Λ𝐿𝐻)(𝑇)==((𝐻𝐻ΛΛ𝐻𝐻))((𝑇𝑇))−−((𝐻𝐻ΛΛ𝐾)(𝑇𝐿)𝐻𝑍_𝑇)(𝑇).(3.25)Therefore𝑑𝑑𝑇𝑍^𝑇=(𝐻Λ𝐻)(𝑇)−(𝐻Λ)(𝑇)𝑍_𝑇−𝑍_𝑇Λ_𝑇𝐻_𝑇+𝑍_𝑇Λ_𝑇𝑍_𝑇=𝑍_𝑇−𝐻_𝑇Λ^𝑇𝑍^𝑇−𝐻_𝑇.(3.26)This completes the proof.

4. Integral Operator Driven by a Diffusion Process

Now back to the integral operator 𝐾_𝑋,𝑇 defined in Section 1. Define a process 𝑍_𝑠∶Ω→𝑀_𝑁(ℝ),𝑍_𝑠=𝐻Γ_𝑋,𝐼+𝐾_𝑋,𝑠⁻¹𝐻_𝑠,(4.1)where Γ_𝑋∶=Γ(𝑋) (see Notation 3.3 for the definition of the angle brackets). From the definition, it is clear that 𝑍_𝑠 is ℱ_∗ adapted. In fact, 𝑍_𝑠 is a symmetric matrix under the usual assumptions on 𝐻 and Γ.

Proposition 4.1. If 𝐻 and Γ are symmetric matrices, then 𝑍 is symmetric as a matrix.

Proof. Since 𝑠 is fixed, we will drop the subscript 𝑠. Also fix an 𝜔∈Ω, so we will also drop the subscript 𝑋. Let 𝐾^∗ be the adjoint of 𝐾 with kernel Γ(𝑠)𝐻(𝑠⋀𝑡). By assumption of symmetry and by definition,𝑍^𝑇==⟨𝐻𝐻Γ,𝐻𝐼⟩+−𝐾^∗𝐻𝐾⁻¹_∗,Γ𝐻=⟨𝐻Γ,𝐻⟩−𝐻Γ𝐾(𝐼𝐼++𝐾𝐾^∗⁻¹₋₁,Γ𝐻=𝐻Γ𝐼+𝐾₋₁,𝐻=𝐻Γ,𝐼+𝐾₋₁,𝐻𝐻=𝑍.(4.2)

Theorem 4.2. Let 𝑋_𝑠 be an 𝐿-diffusion process satisfying (1.2) and 𝐻∶[0,∞)→𝑀^+𝑁(ℝ) and let Γ∶𝑀→𝑀^+𝑁(ℝ) be continuous. Further assume that 𝐻(𝑠)≥𝐻(𝑡)≥0 for 𝑠≥𝑡. Let 𝐾_𝑋,𝑠 be an integral operator defined by (1.5) and 𝑍_𝑠=𝐻Γ_𝑋,𝐼+𝐾_𝑋,𝑠⁻¹𝐻_𝑠.(4.3)Let 𝑒(𝑍) be the explosion time of 𝑍. Then for 𝑠<𝑒(𝑍), 𝑊_𝑠=(𝑠,𝑋_𝑠,𝑍_𝑠)∶Ω→[0,∞)×𝑀×𝒮_𝑁(ℝ) satisfies the following stochastic differential equation: 𝑑𝑊_𝑠=1,𝑏𝑋_𝑠,ℎ𝑊^𝑠𝑑𝑠+0,𝜎𝑋^𝑠,0𝑑𝐵^𝑠,𝑠≥𝑡;𝑊_𝑡=(𝑡,𝑥,𝑣),(4.4)where ℎ(𝑡,𝑥,𝑣)=𝑣−𝐻(𝑡)Γ(𝑥)𝑣−𝐻(𝑡).(4.5)

Proof. In the ambient space ℝ^𝑁′, (𝑠,𝑋_𝑠) is a diffusion satisfying the stochastic differential equation of the form𝑑𝑠,𝑋_𝑠=1,𝑏𝑋^𝑠𝑑𝑠+0,𝜎𝑋^𝑠𝑑𝐵^𝑠.(4.6)Now by Theorem 3.9, 𝑍_𝑠 satisfies the following differential equation:𝑑𝑍_𝑠=ℎ𝑠,𝑋_𝑠,𝑍_𝑠𝑑𝑠,(4.7)where ℎ is defined by (4.5) and by Proposition 4.1, 𝑍_𝑠∈𝒮_𝑁(ℝ). Thus we can write𝑑𝑠,𝑋_𝑠,𝑍_𝑠=1,𝑏𝑋^𝑠,ℎ𝑠,𝑋^𝑠,𝑍_𝑠𝑑𝑠+0,𝜎𝑋^𝑠,0𝑑𝐵^𝑠,(4.8)which is (4.4). The existence of 𝑍_𝑠 for small time is guaranteed by Lemma 2.4.

Lemma 4.3. If Γ is locally Lipschitz, then 𝑊_𝑠 is the unique solution (path-wise) to (4.4).

Proof. Now by the definition of 𝑋_𝑠, 𝑏 and 𝜎 are locally Lipschitz. However, since Γ is locally Lipschitz on the manifold 𝑀 with bounded operator norm, it follows that ℎ is locally Lipschitz. Therefore (4.4) has a unique solution and is given by 𝑊_𝑠.

5. Long-Time Existence of 𝑊_𝑠

We had addressed the existence and uniqueness of the solution to (4.5), given by 𝑊_𝑠=(𝑠,𝑋_𝑠,𝑍_𝑠), with 𝑒(𝑊)≤𝑒(𝑋). We will now give sufficient conditions for 𝑒(𝑊)=𝑒(𝑋).

Proposition 5.1. Suppose that the integral operator Υ_𝑇 with kernel 𝐻(𝑠⋀𝑡) is a strictly positive operator and Γ is a symmetric nonnegative matrix. Then for 𝑧∈ℂ such that 𝑅𝑒(𝑧)≥0, (𝐼+𝑧𝐾_𝑋,𝑇)⁻¹ exists for all 𝑇<𝑒(𝑋).

Proof. When 𝑧=0 is trivial. So assume 𝑧≠0. Fix an 𝜔∈Ω and any 𝑇<𝑒(𝑋(𝜔)). Since 𝐾_𝑋=𝐾_𝑋,𝑇 is a compact operator, it suffices to show that the kernel of 𝐼+𝑧𝐾_𝑋 is 0. Write Γ(𝑋(𝜔))=Γ and 𝐾_𝑋(𝜔)=𝐾. Let 0≠𝑣∈𝐿² such that ⟨𝑣,𝑣⟩>0 and Γ𝑣=0. Then (𝐼+𝑧𝐾_𝑋)𝑣=𝑣 is nonzero. Hence we can assume that ⟨Γ𝑣,Γ𝑣⟩>0. Note that 𝐾=Υ𝑀_Γ and (𝑀_Γ+𝑀_ΓΥ𝑀_Γ) is a symmetric operator (see Section 1 for definitions of Υ and 𝑀_Γ.) Therefore,𝑀^Γ+𝑧𝑀_ΓΥ𝑀_Γ𝑣,𝑣==⟨Γ𝑀𝑣,_Γ𝑣𝑣⟩,𝑣+𝑧+⟨Υ𝑧Γ𝑣𝑀,Γ_Γ𝑣Υ⟩𝑀._Γ𝑣,𝑣(5.1)Since Γ is a nonnegative matrix, ⟨Γ𝑣,𝑣⟩≥0, and because Υ is a strictly positive operator, ⟨ΥΓ𝑣,Γ𝑣⟩>0. If Re(𝑧)>0,Re𝑀_Γ𝑣,𝑣+𝑧Υ𝑀_Γ𝑣,𝑀_Γ𝑣>0.(5.2)Otherwise, Im(𝑧)≠0 and hence we haveIm(𝑧)Υ𝑀_Γ𝑣,𝑀_Γ𝑣≠0.(5.3)Either way, if Re(𝑧)≥0,𝑀^Γ+𝑧𝑀_ΓΥ𝑀_Γ𝑣=𝑀^Γ𝐼+𝑧Υ𝑀^Γ𝑣(5.4)is nonzero, and therefore (𝐼+𝑧Υ𝑀_Γ)𝑣 is nonzero. Thus for any nonzero 𝐿²[0,𝑇] function 𝑣, (𝐼+𝑧Υ𝑀_Γ)𝑣 is never zero and since 𝜔 is arbitrary, hence 𝐼+𝑧𝐾_𝑋 is invertible for any 𝜔∈Ω.

Proposition 5.2. Suppose that the usual assumptions on Γ and 𝐻 hold. Then (𝐼+𝐾_𝑋,𝑠)⁻¹ exists for all 0≤𝑠<𝑒(𝑋). Furthermore, (3.10) holds for all 0≤𝑠<𝑒(𝑋).

Proof. Under the assumptions on Γ and 𝐻, Proposition 3.1 and Remark 3.4 will imply Proposition 5.1. Fix an 𝜔∈Ω, a 𝑇<𝑒(𝑋(𝜔)) and let 𝐶_𝑇 be as defined in Lemma 2.4. Then on 𝑈={𝑧∣|𝑧|<1/𝐶_𝑇}, (𝐼+𝑧𝐾_𝑋,𝑠) is invertible for all 𝑠∈[0,𝑇]. Hence 𝑂=𝑈∪{𝑧∣Re(𝑧)>0} is an open-connected set containing 0, and (𝐼+𝑧𝐾_𝑋,𝑠)⁻¹ exists for all 𝑠∈[0,𝑇]. In particular, at 𝑧=1. Then the assumptions in Corollary 2.9 are met, and hence (3.10) holds.

6. Proof of Main Result

The proof of Theorem 1.2 now follows from Theorem 4.2 and Proposition 5.2. Integrating (3.10), we have for 𝑇<𝑒(𝑋),logdet𝐼+𝐾_𝑋,𝑇=^𝑇0tr𝐻(𝑠)Γ𝑋_𝑠𝑑𝑠−𝑍^𝑠Γ𝑋_𝑠𝑑𝑠.(6.1)For (𝑡,𝑥,𝑣)∈[0,∞)×𝑀×𝒮_𝑁(ℝ), defineΨ(𝑇,𝑡,𝑥,𝑣)=𝔼_{(𝑡,𝑥,𝑣)}𝑓𝑋^𝑇exp−^𝑇0𝑉𝑋_𝑠−𝑝tr𝐻(𝑠)Γ𝑋^𝑠)−𝑍_𝑠Γ𝑋_𝑠𝑑𝑠(6.2)and observe thatΨ(𝑇,0,𝑥,0)=𝔼_𝑥[𝑓𝑋_𝑇𝑒^{−∫𝑇0𝑉(𝑋𝑠)𝑑𝑠}det𝐼+𝐾^𝑋,𝑇^𝑝]=𝜆(𝑇,𝑥).(6.3)By the Feynman-Kac formula, Ψ satisfies the following partial differential equation:𝜕𝜕𝑇Ψ(𝑇,𝑡,𝑥,𝑣)=𝐻Ψ(𝑇,𝑠,𝑥,𝑣).(6.4)ThusΨ(𝑇,𝑡,𝑥,𝑣)=𝑒^𝑇𝐻𝑓(𝑡,𝑥,𝑣),(6.5)and therefore from (6.3),𝜆(𝑇,𝑥)=𝑒^𝑇𝐻𝑓(0,𝑥,0).(6.6)