Sensitivity analysis of long-term cash flows

Abstract

This paper conducts a sensitivity analysis of long-term cash flows. The price of the cash flow at time zero is given by the pricing operator of a Markov diffusion acting on the cash flow function. We study the extent to which the price of the cash flow is affected by small perturbations of the underlying Markov diffusion. The main tool is the Hansen–Scheinkman decomposition, which is a method to express the cash flow in terms of eigenvalues and eigenfunctions of the pricing operator. By incorporating techniques developed by Fournié et al. (Finance Stoch. 3:391–412, 1999), the sensitivities of long-term cash flows can be represented via simple expressions in terms of eigenvalues and eigenfunctions.

Appendices

Appendix A: Proof of Theorem 4.8

Propositions A.1 and A.2 are essential steps for proving Theorem 4.8. Proposition A.1 is a generalization of Fournié et al. [12, Proposition 3.1]. We modify their proof. Recall the functions $\overline{k}_{\epsilon}$ and $\overline{k}$ defined in (4.6).

Proposition A.1

Let $(b_{\epsilon},\sigma,r_{\epsilon},f_{\epsilon})$ and $\xi$ be a quadruple of functions and an initial value, respectively, satisfying Assumptions 4.1 and 4.2. Assume that both $\phi_{\epsilon}(x)$ and $\nabla\phi_{\epsilon}(x)$ (thus, $k_{\epsilon}(x)$) are continuously differentiable in $\epsilon$ on $I$ for each $x$ and that there exist functions $g,\psi:\mathbb{R}^{d}\rightarrow\mathbb{R}$ satisfying (4.7) and (4.8) for $(\epsilon,x)$ in $I\times \mathbb{R}^{d}$. Suppose that for each $T>0$, there exist positive constants $\epsilon _{0}$, $\epsilon_{1}$, $p$, $q$ with $p\geq2$ and $1/p+1/q=1$ such that

$$\begin{aligned} \mathbb{E}_{\xi}^{\mathbb{Q}}\big[e^{\epsilon_{0}\int_{0}^{T}g^{2}(X_{s})\,ds}\big]&< \infty , \end{aligned}$$

(A.1)

$$\begin{aligned} \mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[\int_{0}^{T}g^{p+\epsilon_{1}}(X_{s})\,ds\bigg]&< \infty , \end{aligned}$$

(A.2)

$$\begin{aligned} \mathbb{E}_{\xi}^{\mathbb{Q}}[\psi^{q}(X_{T})]&< \infty . \end{aligned}$$

(A.3)

Then the partial derivative $\frac{\partial}{\partial\epsilon}\mathbb {E}_{\xi}^{\mathbb{Q}^{\epsilon}}[(f_{\eta}/\phi_{\eta})(X_{T}^{\epsilon})]$ exists and

$$ \begin{aligned} \frac{\partial}{\partial\epsilon}\mathbb{E}_{\xi}^{\mathbb{Q}^{\epsilon}}[(f_{\eta}/\phi_{\eta})(X_{T}^{\epsilon})]=\mathbb{E}_{\xi}^{\mathbb{Q}^{\epsilon}} \bigg[(f_{\eta}/\phi_{\eta})(X_{T}^{\epsilon})\int_{0}^{T}\overline {k}_{\epsilon}(X_{s}^{\epsilon})\, dW_{s}^{\epsilon}\bigg]. \end{aligned} $$

(A.4)

Moreover, this partial derivative is continuous in $(\eta,\epsilon)$ on $I^{2}$.

The proof is organized as follows.

(I) First, prove (A.4) for $\epsilon=0$, that is,

$$ \frac{\partial}{\partial\epsilon} \bigg| _{\epsilon=0}\mathbb{E}_{\xi}^{\mathbb{Q_{\epsilon}}} [(f_{\eta}/\phi_{\eta})(X_{T}^{\epsilon})] =\mathbb{E}_{\xi}^{\mathbb{Q}} \bigg[(f_{\eta}/\phi_{\eta})(X_{T})\int_{0}^{T}\overline{k}(X_{s})\, dW_{s}\bigg]. $$

(A.5)

We conduct the following sub-steps to show the above equality.

(i) Show that

$$ \frac{\partial}{\partial\epsilon} \bigg| _{\epsilon=0}\mathbb{E}_{\xi}^{\mathbb{Q}^{\epsilon}} [(f_{\eta}/\phi_{\eta})(X_{T}^{\epsilon})] =\lim_{\epsilon\to0}\,\mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[(f_{\eta}/\phi _{\eta})(X_{T})\int_{0}^{T} \!Z^{\epsilon}_{s}\,\ell_{\epsilon}(X_{s})\,dW_{s}\bigg] $$

for the first order approximation $\ell_{\epsilon}$ of $k_{\epsilon}$ in the $\epsilon$-perturbation, and for an exponential martingale $Z^{\epsilon}$. Both $\ell_{\epsilon}$ and $Z^{\epsilon}$ are defined later. Then it is enough to show that

$$ \lim_{\epsilon\to0}\,\mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[(f_{\eta}/\phi_{\eta})(X_{T})\int_{0}^{T}\!\big(Z^{\epsilon}_{s}\ell_{\epsilon}(X_{s}) -\overline {k}(X_{s})\big)\,dW_{s}\bigg]=0, $$

(A.6)

which gives the result stated in (A.5).

(ii) To show (A.6), it suffices to show that $\int_{0}^{T}(Z^{\epsilon}_{s}\ell_{\epsilon}(X_{s}) -\overline{k}(X_{s}))\,dW_{s}\to 0$ in $L^{p}$ as $\epsilon\to0$. Observing the equality

$$\begin{aligned} \int_{0}^{T}\!\big(Z^{\epsilon}_{s}\ell_{\epsilon}(X_{s}) -\overline{k}(X_{s})\big)\,dW_{s} =\int_{0}^{T}\!(Z^{\epsilon}_{s}-1)\ell_{\epsilon}(X_{s})\,dW_{s}+\int_{0}^{T}\!(\ell _{\epsilon}-\overline{k})(X_{s})\,dW_{s}, \end{aligned}$$

we prove that the two terms on the right-hand side converge to zero in $L^{p}$ as $\epsilon\to0$.

(II) Using (A.5), verify (A.4) for arbitrary $\epsilon\in I$.

(III) Prove that the partial derivative in (A.4) is continuous in $(\eta,\epsilon)$ on $I^{2}$.

Proof

The proof of Proposition A.1 is given in several steps.

Step (I)-(i). From (2.4) and (2.5), the process $(W_{t}^{\epsilon})_{t\ge0}:=(B_{t}-\int_{0}^{t}\varphi_{\epsilon}(X_{s}^{\epsilon})\,ds)_{t\ge0}$ is a $\mathbb{Q}^{\epsilon}$-Brownian motion, and the $\mathbb{Q}^{\epsilon}$-dynamics of $X^{\epsilon}$ is

$$ \begin{aligned} dX_{t}^{\epsilon}&=(b_{\epsilon}+\sigma\varphi_{\epsilon})(X_{t}^{\epsilon})\, dt+\sigma (X_{t})\, dW_{t}^{\epsilon}=(\sigma k_{\epsilon})(X_{t}^{\epsilon})\,dt+\sigma(X_{t}^{\epsilon})\, dW_{t}^{\epsilon}\;. \end{aligned} $$

Because $k_{\epsilon}$ is continuously differentiable in $\epsilon$ on $I$, by the Taylor expansion, we write $k_{\epsilon}=k+\epsilon\ell _{\epsilon}$ for some $d\times1$ vector function $\ell_{\epsilon}$. The $\mathbb{Q}^{\epsilon}$-dynamics of $X^{\epsilon}$ is given by

$$ dX_{t}^{\epsilon}=(\sigma k+\epsilon\sigma\ell_{\epsilon})(X_{t}^{\epsilon})\, dt+\sigma(X_{t}^{\epsilon})\,dW_{t}^{\epsilon}. $$

By Assumption (A.1), the process

$$ (Z^{\epsilon}_{t})_{0\leq t\leq T} :=\big(e^{\epsilon\int_{0}^{t}\ell_{\epsilon}(X_{s})\,dW_{s} -\frac{\epsilon^{2}}{2}\int_{0}^{t}|\ell_{\epsilon}|^{2}(X_{s})\,ds }\big)_{0\leq t\leq T} $$

is a martingale for small $\epsilon$ because the Novikov condition is satisfied. Here we use the mean-value theorem to estimate for some $\epsilon^{*}\in I$ that

$$ |\ell_{\epsilon}(x)|=\left|\frac{k_{\epsilon}(x)-k(x)}{\epsilon}\right |=\left| \frac{\partial}{\partial\epsilon} \Big| _{\epsilon=\epsilon ^{*}}k_{\epsilon}(x)\, \right|\leq g(x). $$

By Girsanov’s theorem, we have $\mathbb{E}_{\xi}^{\mathbb{Q}^{\epsilon}} [(f_{\eta}/\phi_{\eta})({X}_{T}^{\epsilon})]=\mathbb{E}_{\xi}^{\mathbb {Q}}[(f_{\eta}/\phi_{\eta})(X_{T})\,Z^{\epsilon}_{T}]$ and so

$$\begin{aligned} \frac{\partial}{\partial\epsilon} \bigg| _{\epsilon=0}\mathbb{E}_{\xi}^{\mathbb{Q}^{\epsilon}} [(f_{\eta}/\phi_{\eta})(X_{T}^{\epsilon})] &= \frac{\partial}{\partial\epsilon} \bigg| _{\epsilon=0}\mathbb {E}_{\xi}^{\mathbb{Q}}[(f_{\eta}/\phi_{\eta})(X_{T})\,Z^{\epsilon}_{T}] \\ &=\lim_{\epsilon\to0}\,\mathbb{E}_{\xi}^{\mathbb{Q}} \bigg[(f_{\eta}/\phi _{\eta})(X_{T})\,\frac{Z^{\epsilon}_{T}-1}{\epsilon}\bigg] \\ &=\lim_{\epsilon\to0}\,\mathbb{E}_{\xi}^{\mathbb{Q}} \bigg[(f_{\eta}/\phi _{\eta})(X_{T})\int_{0}^{T}Z^{\epsilon}_{s}\,\ell_{\epsilon}(X_{s})\,dW_{s}\bigg]. \end{aligned}$$

(A.7)

For the last equality, we used $\frac{Z^{\epsilon}_{T}-1}{\epsilon}=\int _{0}^{T}Z^{\epsilon}_{s}\,\ell_{\epsilon}(X_{s})\,dW_{s}$, which is easily obtained by the Itô formula. From (A.7), it suffices to prove that

$$ \lim_{\epsilon\to0}\,\mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[(f_{\eta}/\phi_{\eta})(X_{T})\int_{0}^{T}\big(Z^{\epsilon}_{s}\ell_{\epsilon}(X_{s}) -\overline {k}(X_{s})\big)\,dW_{s}\bigg]=0, $$

which gives (A.5).

Step (I)-(ii). From the condition $\mathbb{E}_{\xi}^{\mathbb{Q}}[(f_{\eta}/\phi_{\eta})^{q}(X_{T})]\le\mathbb{E}_{\xi}^{\mathbb{Q}}[\psi^{q}(X_{T})]<\infty$, by the Hölder inequality, it is enough to show that $\int_{0}^{T}(Z^{\epsilon}_{s}\ell_{\epsilon}(X_{s}) -\overline{k}(X_{s}))\,dW_{s}$ converges to 0 in $L^{p}$ as $\epsilon\to0$. Using the equality

$$\begin{aligned} \int_{0}^{T}\big(Z^{\epsilon}_{s}\ell_{\epsilon}(X_{s}) -\overline{k}(X_{s})\big)\,dW_{s} &=\int_{0}^{T}(Z^{\epsilon}_{s}-1)\ell_{\epsilon}(X_{s})\,dW_{s} \\ &\phantom{=:}+\int_{0}^{T}(\ell_{\epsilon}-\overline{k})(X_{s})\,dW_{s}, \end{aligned}$$

(A.8)

we show that each term on the right-hand side goes to zero in $L^{p}$. For the second, we use dominated convergence. Because $|\ell_{\epsilon}-\overline{k}|^{p}\leq c\, (|\ell_{\epsilon}|^{p}+|\overline{k}|^{p})\leq2cg^{p}$ for some positive constant $c$ and $\mathbb{E}_{\xi}^{\mathbb{Q}}[\int_{0}^{T}g^{p}(X_{s})\,ds]$ is finite, we have

$$ \begin{aligned} \mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[\bigg|\int_{0}^{T}(\ell_{\epsilon}-\overline {k})(X_{s})\,dW_{s}\bigg|^{p}\bigg] &\leq c_{q}\,\mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[\bigg(\int_{0}^{T}|\ell _{\epsilon}-\overline{k}|^{2}(X_{s})\,ds\bigg)^{\frac{p}{2}}\bigg]\\&\leq c_{q}\, T^{\frac{p}{2}-1}\, \mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[\int_{0}^{T}|\ell _{\epsilon}-\overline{k}|^{p}(X_{t})\,dt\bigg]\longrightarrow0 \end{aligned} $$

as $\epsilon\rightarrow0$ for some positive constant $c_{q}$ which is independent of $\epsilon$. The Burkholder–Davis–Gundy and Jensen inequalities were used.

For the first term on the right-hand side of (A.8), we prove that

$$ \lim_{\epsilon\rightarrow0}\,\mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[\bigg|\int _{0}^{T}(Z^{\epsilon}_{s}-1)\,\ell_{\epsilon}(X_{s})\,dW_{s}\bigg|^{p}\bigg]=0. $$

Let $r>0$ be such that $1/r+1/(1+\epsilon_{1}/p)=1$. Applying the Burkholder–Davis–Gundy, Jensen and Hölder inequalities, we have

$$ \begin{aligned} &\mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[\bigg|\int_{0}^{T}(Z^{\epsilon}_{s}-1)\,\ell _{\epsilon}(X_{s})\,dW_{s}\bigg|^{p}\bigg]\\ &\leq c_{q}\,\mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[\bigg(\int_{0}^{T}(Z^{\epsilon}_{s}-1)^{2}\,|\ell_{\epsilon}|^{2}(X_{s})\,ds\bigg)^{\frac{p}{2}}\bigg]\\ &\leq c_{q}\,T^{\frac{p}{2}-1}\,\mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[\int _{0}^{T}|Z^{\epsilon}_{s}-1|^{p}\,|\ell_{\epsilon}|^{p}(X_{s})\,ds \bigg] \\ &\leq c_{q}\,T^{\frac{q}{2}-1}\bigg(\mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[\int _{0}^{T}|Z^{\epsilon}_{s}-1|^{pr}\,ds\bigg]\bigg)^{\frac{1}{r}}\,\bigg(\mathbb {E}_{\xi}^{\mathbb{Q}}\bigg[\int_{0}^{T}|\ell_{\epsilon}|^{p+\epsilon_{1}}(X_{s})\, ds\bigg]\bigg)^{\frac{p}{p+\epsilon_{1}}}\\ &\leq c_{q}\,T^{\frac{q}{2}-1}\bigg(\mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[\int _{0}^{T}|Z^{\epsilon}_{s}-1|^{pr}\,ds\bigg]\bigg)^{\frac{1}{r}}\,\bigg(\mathbb {E}_{\xi}^{\mathbb{Q}}\bigg[\int_{0}^{T}g^{p+\epsilon_{1}}(X_{s})\,ds\bigg]\bigg)^{\frac{p}{p+\epsilon_{1}}}. \end{aligned} $$

In the last inequality, since $\mathbb{E}_{\xi}^{\mathbb{Q}}[\int _{0}^{T}g^{p+\epsilon_{1}}(X_{s})\,ds]$ is finite from Assumption (A.2), it suffices to prove that $\mathbb{E}_{\xi}^{\mathbb {Q}}[\int_{0}^{T}|Z^{\epsilon}_{s}-1|^{pr}\,ds]\to0$ as $\epsilon\to0$. Choose a positive even integer $m$ such that $m>pr$. We show that $\mathbb{E}_{\xi}^{\mathbb{Q}}[\int_{0}^{T}(Z^{\epsilon}_{s}-1)^{m}\,ds]$ converges to zero as $\epsilon\to0$. Observe that

$$ \begin{aligned} (Z^{\epsilon}_{t}-1)^{m}=\sum_{j=0}^{m}{m \choose j} (-1)^{j}(Z^{\epsilon}_{t})^{j}. \end{aligned} $$

(A.9)

It is enough to show that $\int_{0}^{T}\mathbb{E}_{\xi}^{\mathbb {Q}}[(Z^{\epsilon}_{t})^{j}]\,dt \to T$ as $\epsilon\rightarrow0$ for $j=0,1,\dots,m$ because then

$$\begin{aligned} \mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[\int_{0}^{T}(Z^{\epsilon}_{s}-1)^{m}\,ds\bigg] &=\sum_{j=0}^{m}{m \choose j} (-1)^{j}\int_{0}^{T} \mathbb{E}_{\xi}^{\mathbb {Q}}[(Z^{\epsilon}_{t})^{j}]\,dt \\ &\longrightarrow T\sum_{j=0}^{m}{m\choose j} (-1)^{j}=0. \end{aligned}$$

To achieve this, the dominated convergence theorem is used. We prove that the expectation $\mathbb{E}_{\xi}^{\mathbb{Q}}[(Z^{\epsilon}_{t})^{j}]$ is uniformly bounded by a constant for small $\epsilon$ and $0\leq t\leq T$, and that $\mathbb{E}_{\xi}^{\mathbb{Q}}[(Z^{\epsilon}_{t})^{j}]$ converges to 1 as $\epsilon\to0$ for each fixed $t$. Observe that

$$ \begin{aligned} &\mathbb{E}_{\xi}^{\mathbb{Q}}[(Z^{\epsilon}_{t})^{j}]\\ &=\mathbb{E}_{\xi}^{\mathbb{Q}}\big[e^{j\epsilon\int_{0}^{t}\ell_{\epsilon}(X_{s})\,dW_{s} -\frac{1}{2}j\epsilon^{2}\int_{0}^{t}|\ell_{\epsilon}|^{2}(X_{s})\,ds }\big]\\ &=\mathbb{E}_{\xi}^{\mathbb{Q}} \big[e^{j\epsilon\int_{0}^{t}\ell_{\epsilon}(X_{s})\,dW_{s} -j^{2}\epsilon^{2}\int_{0}^{t}|\ell_{\epsilon}|^{2}(X_{s})\,ds} e^{ j(j-1/2)\epsilon ^{2}\int_{0}^{t}|\ell_{\epsilon}|^{2}(X_{s})\,ds }\big] \\ &\leq\big(\mathbb{E}_{\xi}^{\mathbb{Q}}\big[e^{2j\epsilon\int_{0}^{t}\ell _{\epsilon}(X_{s})\,dW_{s} -2j^{2}\epsilon^{2}\int_{0}^{t}|\ell_{\epsilon}|^{2}(X_{s})\,ds}\big] \big)^{\frac {1}{2}} \, \big(\mathbb{E}_{\xi}^{\mathbb{Q}}\big[e^{j(2j-1)\epsilon^{2}\int _{0}^{t}|\ell_{\epsilon}|^{2}(X_{s})\,ds}\big]\big)^{\frac{1}{2}}\\ &= \big(\mathbb{E}_{\xi}^{\mathbb{Q}}\big[e^{j(2j-1)\epsilon^{2}\int _{0}^{t}|\ell_{\epsilon}|^{2}(X_{s})\,ds}\big] \big)^{\frac{1}{2}} \\ &\leq\big(\mathbb{E}_{\xi}^{\mathbb{Q}}\big[e^{j(2j-1)\epsilon^{2}\int _{0}^{t}g^{2}(X_{s})\,ds}\big] \big)^{\frac{1}{2}}; \end{aligned} $$

the last equality uses that the first term is a martingale for small $\epsilon$. By choosing $I$ smaller if necessary, we may assume that $j(2j-1)\epsilon^{2}\le\epsilon_{0}$ for all $\epsilon\in I$ and $j=0,1,\dots,m$. For $0\le t\le T$ and $\epsilon\in I$, we have

$$ \mathbb{E}_{\xi}^{\mathbb{Q}}[(Z^{\epsilon}_{t})^{j}] \leq\big(\mathbb{E}_{\xi}^{\mathbb{Q}}\big[e^{\epsilon_{0}\int _{0}^{T}g^{2}(X_{s})\,ds}\big]\big)^{\frac{1}{2}}. $$

(A.10)

Thus, for $\epsilon\in I$ and $0\leq t\leq T$, the expectation $\mathbb{E}_{\xi}^{\mathbb{Q}}[(Z^{\epsilon}_{t})^{j}]$ is uniformly bounded by the constant $(\mathbb{E}_{\xi}^{\mathbb{Q}}[e^{\epsilon_{0}\int _{0}^{T}g^{2}(X_{s})\,ds}])^{\frac{1}{2}}$ which is a finite number by Assumption (A.1).

We now show that $\mathbb{E}_{\xi}^{\mathbb{Q}}[(Z^{\epsilon}_{t})^{j}]$ converges to 1 as $\epsilon\to0$ for fixed $t\in[0,T]$. Apply the dominated convergence theorem to $e^{j(2j-1)\epsilon^{2}\int_{0}^{t}g^{2}(X_{s})\,ds}$ as $\epsilon\to0$. Because this is dominated by $e^{\epsilon_{0}\int_{0}^{T}g^{2}(X_{s})\,ds}$ whose expectation is finite, we know that $\mathbb{E}_{\xi}^{\mathbb{Q}}[e^{j(2j-1)\epsilon^{2}\int _{0}^{t}g^{2}(X_{s})\,ds}]$ converges to 1 as $\epsilon\to0$. It follows that

$$\begin{aligned} 1=\mathbb{E}_{\xi}^{\mathbb{Q}}\Big[\liminf_{\epsilon\rightarrow 0}(Z^{\epsilon}_{t})^{j}\Big] &\leq\liminf_{\epsilon\rightarrow0}\mathbb{E}_{\xi}^{\mathbb {Q}}[(Z^{\epsilon}_{t})^{j}] \\ &\leq \limsup_{\epsilon\rightarrow0}\mathbb{E}_{\xi}^{\mathbb{Q}}[(Z^{\epsilon}_{t})^{j}] \\ &\leq\lim_{\epsilon\rightarrow0}\mathbb{E}_{\xi}^{\mathbb{Q}}\big[e^{j(2j-1)\epsilon^{2}\int_{0}^{t}g^{2}(X_{s})\,ds}\big]=1. \end{aligned}$$

(A.11)

This completes the proof.

Step (II). We show (A.4) for arbitrary $\epsilon \in I$. Fix $\epsilon\in I$ and choose an open interval $J$ at 0 small enough so that $\epsilon+J$ is still in $I$. To utilize (A.5), we introduce another parameter $h$ in the interval $J$. Consider the quadruple of functions $(b_{\epsilon+h},\sigma,r_{\epsilon +h},f_{\epsilon+h})$ and $\xi$ with perturbation parameter $h$. This quadruple and initial value satisfy the hypothesis of Proposition A.1 because $\epsilon+J$ is a subset of $I$. Thus from the result of Step (I), we know that $\mathbb{E}_{\xi}^{\mathbb{Q}^{\epsilon+h}} [(f_{\eta}/\phi_{\eta})(X_{T}^{\epsilon+h})]$ is differentiable at $h=0$ and

$$ \frac{\partial}{\partial\epsilon} \bigg| _{h =0}\mathbb{E}_{\xi}^{\mathbb {Q}^{\epsilon+h}} [(f_{\eta}/\phi_{\eta})(X_{T}^{\epsilon+h})] =\mathbb{E}_{\xi}^{\mathbb {Q}^{\epsilon}} \bigg[(f_{\eta}/\phi_{\eta})(X_{T}^{\epsilon})\int_{0}^{T}\overline {k}_{\epsilon}(X_{s}^{\epsilon})\, dW_{s}^{\epsilon}\bigg] $$

for $\overline{k}_{\epsilon}(x)= \frac{\partial}{\partial\epsilon} k_{\epsilon}(x) =\frac{\partial}{\partial h}|_{h=0} \, k_{\epsilon+h}(x)$. This gives (A.4).

Step (III). We claim that the partial derivative

$$ \frac{\partial}{\partial\epsilon}\mathbb{E}_{\xi}^{\mathbb{Q}^{\epsilon}} [(f_{\eta}/\phi_{\eta})(X_{T}^{\epsilon})]=\mathbb{E}_{\xi}^{\mathbb{Q}^{\epsilon}} \bigg[(f_{\eta}/\phi_{\eta})(X_{T}^{\epsilon})\int_{0}^{T}\overline {k}_{\epsilon}(X_{s}^{\epsilon})\, dW_{s}^{\epsilon}\bigg] $$

is continuous in $(\eta,\epsilon)$ on $I^{2}$. Without loss of generality, we prove the continuity at the origin $(\eta,\epsilon)=(0,0)$. Observe that

$$ \begin{aligned} &\mathbb{E}_{\xi}^{\mathbb{Q}^{\epsilon}} \bigg[(f_{\eta}/\phi_{\eta})(X_{T}^{\epsilon})\int_{0}^{T}\overline {k}_{\epsilon}(X_{s}^{\epsilon})\, dW_{s}^{\epsilon}\bigg]\\ &=\mathbb{E}_{\xi}^{\mathbb{Q}} \bigg[(f_{\eta}/\phi_{\eta})(X_{T})\bigg(\int_{0}^{T}\overline{k}_{\epsilon}(X_{s})\, dW_{s}+\epsilon\int_{0}^{T}(\overline{k}_{\epsilon}\ell_{\epsilon})(X_{s})\,ds\bigg) Z_{T}^{\epsilon}\bigg]. \end{aligned} $$

For convenience, define

$$ \begin{aligned} &H_{T}^{\epsilon}:=\bigg(\int_{0}^{T}\overline{k}_{\epsilon}(X_{s})\, dW_{s}+\epsilon\int_{0}^{T}(\overline{k}_{\epsilon}\ell_{\epsilon})(X_{s})\, ds\bigg) Z_{T}^{\epsilon},\\ &H_{T}:=H_{T}^{0}. \end{aligned} $$

We want to show that as $(\eta,\epsilon)\to(0,0)$,

$$ \mathbb{E}_{\xi}^{\mathbb{Q}} [(f_{\eta}/\phi_{\eta})(X_{T}) H_{T}^{\epsilon}] \to\mathbb{E}_{\xi}^{\mathbb{Q}} [(f/\phi)(X_{T}) H_{T}]. $$

Since $\mathbb{E}_{\xi}^{\mathbb{Q}}[\psi^{q}(X_{T})]$ is finite, from the dominated convergence theorem, we know that $(f_{\eta}/\phi_{\eta})(X_{T})$ converges to $(f/\phi)(X_{T})$ in $L^{q}$ as $\eta\to0$. Thus it suffices to show that $H_{T}^{\epsilon}$ converges to $H_{T}$ in $L^{p}$ as $\epsilon\to0$. This can be obtained by showing that

$$\begin{aligned} Z_{T}^{\epsilon}\int_{0}^{T}\overline{k}_{\epsilon}(X_{s})\, dW_{s} &\longrightarrow \int_{0}^{T}\overline{k}(X_{s})\, dW_{s} , \end{aligned}$$

(A.12)

$$\begin{aligned} \epsilon Z_{T}^{\epsilon}\int_{0}^{T}(\overline{k}_{\epsilon}\ell_{\epsilon})(X_{s})\,ds &\longrightarrow0 \end{aligned}$$

(A.13)

in $L^{p}$ as $\epsilon\to0$.

To prove (A.12), choose a sufficiently large positive even integer $m$ such that $\frac{1}{p+\epsilon_{1}}+\frac{1}{m}<\frac{1}{p}$. From the generalized Hölder inequality, it suffices to show that as $\epsilon\to0$,

$$\begin{aligned} \int_{0}^{T}\overline{k}_{\epsilon}(X_{s})\, dW_{s} &\longrightarrow\int_{0}^{T}\overline{k}(X_{s})\, dW_{s} \qquad\mbox{in } L^{p+\epsilon_{1}},\\ Z_{T}^{\epsilon}&\longrightarrow1 \qquad\mbox{in } L^{m} . \end{aligned}$$

The $L^{m}$-convergence is obtained from (A.9) and the fact that $\lim_{\epsilon\to0}\mathbb{E}_{\xi}^{\mathbb{Q}}[(Z^{\epsilon}_{T})^{j}]=1$ as shown in (A.11) for $j=0,1,\dots,m$. For the $L^{p+\epsilon_{1}}$-convergence, observe that

$$ \mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[\bigg|\int_{0}^{T}(\overline{k}_{\epsilon}-\overline{k})(X_{s})\, dW_{s}\bigg|^{p+\epsilon_{1}}\bigg]\leq c_{p+\epsilon_{1}}\,\mathbb{E}_{\xi}^{\mathbb{Q}} \bigg[\int_{0}^{T}|\overline {k}_{\epsilon}-\overline{k}|^{p+\epsilon_{1}}(X_{s})\,ds\bigg], $$

where $c_{p+\epsilon_{1}}$ is the constant from the Burkholder–Davis–Gundy inequality. Due to Assumption (A.2), the dominated convergence theorem says that

$$ \mathbb{E}_{\xi}^{\mathbb{Q}} \bigg[\int_{0}^{T}|\overline{k}_{\epsilon}-\overline {k}|^{p+\epsilon_{1}}(X_{s})\,ds\bigg] \longrightarrow0 \qquad\mbox{as } \epsilon\to0. $$

To prove (A.13), it is enough to show that $Z_{T}^{\epsilon}\int_{0}^{T}(\overline{k}_{\epsilon}\ell_{\epsilon})(X_{s})\,ds$ is uniformly bounded in $\epsilon$ on $I$ in $L^{p}$. This is achieved from

$$ \begin{aligned} \mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[\bigg|Z_{T}^{\epsilon}\int_{0}^{T}(\overline {k}_{\epsilon}\ell_{\epsilon})(X_{s})\,ds \bigg|^{p}\bigg] &\leq\mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[(Z_{T}^{\epsilon})^{p}\bigg(\int _{0}^{T}g^{2}(X_{s})\,ds\bigg)^{p} \bigg]\\ &\leq(\mathbb{E}_{\xi}^{\mathbb{Q}}[(Z_{T}^{\epsilon})^{2p}])^{1/2} \bigg(\mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[\Big(\int_{0}^{T}g^{2}(X_{s})\,ds\Big)^{2p}\bigg]\bigg)^{1/2} . \end{aligned} $$

By using the same method as in (A.10), the first expectation is bounded uniformly in $\epsilon$ on $I$ by a constant times $(\mathbb {E}_{\xi}^{\mathbb{Q}}[e^{\epsilon_{0}\int_{0}^{T}g^{2}(X_{s})\,ds}])^{\frac{1}{2}}$. The second expectation is finite since $\mathbb{E}_{\xi}^{\mathbb{Q}}[e^{\epsilon_{0}\int _{0}^{T}g^{2}(X_{s})\,ds}]$ is finite by Assumption (A.1). □

The following proposition says that an exponential bound on the expectation of $e^{Y_{T}}$ in time $T$ guarantees a $T^{p}$-order bound on the expectation of $Y_{T}^{p}$ in time $T$ for any positive constant $p$.

Proposition A.2

Let $(Y_{t})_{t\geq0}$ be a nonnegative stochastic process and $p$ a positive constant. Suppose that there are positive constants $a$ and $c$ such that for all $T>0$,

$$ \mathbb{E}[e^{Y_{T}}]\leq c e^{aT}. $$

Then there exists a positive constant $d$ such that for all sufficiently large $T>0$,

$$ \mathbb{E}[Y_{T}^{p}]\leq d T^{p}. $$

Proof

It suffices to show that $\limsup_{T\rightarrow\infty}\frac{\mathbb{E}[Y_{T}^{p}]}{T^{p}}$ is finite. If this fails, there exists a sequence $(T_{n})_{n\in\mathbb{N}}$ such that $T_{n}\to\infty$ and $b_{n}:= \frac{\mathbb{E}[Y_{T_{n}}^{p}]}{T_{n}^{p}}\to\infty$ as $n\rightarrow\infty$. Let $\hat{p}$ be a nonnegative integer such that $\hat{p}< p\leq \hat{p}+1$. For any nonnegative random variable $Y$, we know that

$$ \begin{aligned} \mathbb{E}[e^{Y}] &=\sum_{j=0}^{\infty}\frac{\mathbb{E}[Y^{j}]}{j!} \geq\sum_{j=\hat{p}+1}^{\infty}\frac{\mathbb{E}[Y^{j}]}{j!} \geq\sum_{j=\hat{p}+1}^{\infty}\frac{(\mathbb{E}[Y^{p}])^{\frac {j}{p}}}{j!}\\ &=\sum_{j=0}^{\infty}\frac{(\mathbb{E}[Y^{p}])^{\frac{j}{p}}}{j!}-\sum _{j=0}^{\hat{p}}\frac{(\mathbb{E}[Y^{p}])^{\frac{j}{p}}}{j!}=e^{(\mathbb {E}[Y^{p}])^{\frac{1}{p}}}-\sum_{j=0}^{\hat{p}}\frac{(\mathbb {E}[Y^{p}])^{\frac{j}{p}}}{j!} \;. \end{aligned} $$

Here, we used the Taylor expansion and Jensen inequality. Substituting $Y=Y_{T_{n}}$, because $\mathbb{E}[Y_{T_{n}}^{p}]\to\infty$ as $n\to\infty$ and an exponential growth is faster than a polynomial growth, it follows that

$$ \mathbb{E}[e^{Y_{T_{n}}}]\geq e^{(\mathbb{E}[Y_{T_{n}}^{p}])^{\frac {1}{p}}}-\sum_{j=0}^{\hat{p}}\frac{(\mathbb{E}[Y_{T_{n}}^{p}])^{\frac {j}{p}}}{j!}\geq\frac{1}{\,2\,}e^{(\mathbb{E}[Y_{T_{n}}^{p}])^{\frac {1}{p}}}=\frac{1}{\,2\,}e^{b_{n}^{\,\frac{1}{p}} T_{n}} $$

for sufficiently large $n$. From the assumption, we obtain

$$ c e^{aT_{n}}\geq\mathbb{E}[e^{Y_{T_{n}}}]\geq\frac{1}{\,2\,}e^{b_{n}^{\,\frac {1}{p}} T_{n}}, $$

which is a contradiction because $\lim_{n\rightarrow\infty}b_{n}=\infty$. □

Proof of Theorem 4.8

By Proposition A.1, the existence and continuity of the partial derivative is directly obtained. Due to (A.5), it suffices to show that

$$ \lim_{T\to\infty}\frac{1}{T} \,\mathbb{E}_{\xi}^{\mathbb{Q}} \bigg[(f/\phi)(X_{T})\int_{0}^{T}\overline{k}(X_{s})\, dW_{s}\bigg]=0. $$

Since the growth of $\mathbb{E}_{\xi}^{\mathbb{Q}}[e^{\epsilon_{0}\int _{0}^{T}g^{2}(X_{s})\,ds}]$ is at most exponential, Proposition A.2 says that the growth of $\mathbb{E}_{\xi}^{\mathbb{Q}}[(\int_{0}^{T}g^{2}(X_{s})\, ds)^{\frac {p}{2}}]$ is at most of order $T^{\frac{p}{2}}$. Thus there is a constant $d_{p}$, depending on $p$ but not on $T$, such that for sufficiently large $T$,

$$ \begin{aligned} \bigg(\mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[\Big(\int_{0}^{T}|\overline {k}|^{2}(X_{s})\,ds\Big)^{\frac{p}{2}}\bigg]\bigg)^{\frac{2}{p}} &\leq\bigg(\mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[\Big(\int _{0}^{T}g^{2}(X_{s})\, ds\Big)^{\frac{p}{2}}\bigg]\bigg)^{\frac{2}{p}}\leq d_{p}T . \end{aligned} $$

Using Hölder and Burkholder–Davis–Gundy inequalities, it follows that

$$ \begin{aligned} &\frac{1}{T}\mathbb{E}_{\xi}^{\mathbb{Q}} \bigg[\bigg|(f/\phi)(X_{T})\int_{0}^{T}\overline{k}(X_{s})\, dW_{s}\bigg|\bigg]\\ &\leq\frac{1}{T}\big(\mathbb{E}_{\xi}^{\mathbb{Q}} [(f/\phi)^{q}(X_{T})] \big)^{\frac{1}{q}}\,\bigg(\mathbb{E}_{\xi}^{\mathbb{Q}} \bigg[\bigg|\int_{0}^{T}\overline{k}(X_{s})\, dW_{s}\bigg|^{p}\bigg]\bigg)^{\frac{1}{p}}\\ &\leq\frac{c_{q}}{T}\big(\mathbb{E}_{\xi}^{\mathbb{Q}} [(f/\phi)^{q}(X_{T})] \big)^{\frac{1}{q}}\,\bigg(\mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[\Big(\int_{0}^{T}|\overline{k}|^{2}(X_{s})\, ds\Big)^{\frac{p}{2}}\bigg]\bigg)^{\frac{1}{p}}\\ &\leq\frac{c_{q}}{T}\big(\mathbb{E}_{\xi}^{\mathbb{Q}} [(f/\phi)^{q}(X_{T})] \big)^{\frac{1}{q}}\,(d_{p}T)^{\frac{1}{2}}\\ &= \frac{c_{q}d_{p}^{\frac{1}{2}}}{T^{\frac{1}{2}}}\big(\mathbb{E}_{\xi}^{\mathbb{Q}} [(f/\phi)^{q}(X_{T})] \big)^{\frac{1}{q}} \longrightarrow0 \end{aligned} $$

(A.14)

as $T\rightarrow\infty$ because $\mathbb{E}_{\xi}^{\mathbb{Q}}[(f/\phi )^{q}(X_{T})]$ is bounded uniformly in $T$ on $[0,\infty)$. This completes the proof. □

Appendix B: Proof of Theorem 4.10

Proof

Let $c_{1}$ be a positive constant such that $\mathbb{E}_{\xi}^{\mathbb{Q}}[e^{\epsilon_{0}\,g^{2}(X_{T})}]\leq c_{1}$ for all $T\geq0$. Replacing $q$ by a sufficiently small positive number, we may assume that $1< q\leq2$, and that the constant $p$ defined by $1/p+1/q=1$ (thus, $p\ge2$) is an even integer.

For a fixed $T>0$, we first show that the conditions of Proposition A.1 are satisfied with these constants $p$ and $q$. Since (A.3) is already assumed to hold, it remains to prove (A.1) and (A.2). The expectation $\mathbb{E}_{\xi}^{\mathbb{Q}}[e^{\frac{1}{T}\int_{0}^{T}\epsilon_{0}\, g^{2}(X_{s})\,ds}]$ is bounded uniformly in $T$ on $[0,\infty)$ because

$$\begin{aligned} \mathbb{E}_{\xi}^{\mathbb{Q}}\big[e^{\frac{1}{T}\int_{0}^{T}\epsilon_{0}\, g^{2}(X_{s})\,ds}\big] &\leq \mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[\frac{1}{T}\int_{0}^{T} e^{\epsilon _{0}\,g^{2}(X_{s})}ds\bigg] \\ &=\frac{1}{T}\int_{0}^{T}\mathbb{E}_{\xi}^{\mathbb{Q}}[e^{\epsilon_{0}\, g^{2}(X_{s})}]\,ds\leq c_{1}. \end{aligned}$$

(B.1)

Replacing $\epsilon_{0}/T$ by $\epsilon_{0}$, (A.1) is satisfied. For (A.2), observe that for any $n\in\mathbb{N}$ such that $2n>p+1$, the expectation $\mathbb{E}_{\xi}^{\mathbb{Q}}[\int _{0}^{T}g^{2n}(X_{s})\,ds]$ is finite since

$$ \mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[\int_{0}^{T}(\epsilon_{0}g^{2})^{n}(X_{s})\, ds\bigg]\leq\mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[\int_{0}^{T}n! e^{\epsilon_{0}\, g^{2}(X_{s})}\,ds\bigg]\leq Tn!\,c_{1}. $$

Thus the expectation $\mathbb{E}_{\xi}^{\mathbb{Q}}[\int_{0}^{T}g^{p+1}(X_{s})\,ds]$ is also finite, and all conditions of Proposition A.1 are satisfied. Therefore the partial derivative $\frac{\partial}{\partial\epsilon }\mathbb{E}_{\xi}^{\mathbb{Q_{\epsilon}}}[(f_{\eta}/\phi_{\eta})(X_{T}^{\epsilon})]$ exists and is continuous in $(\eta,\epsilon)$ on $I^{2}$. Moreover,

$$ \frac{\partial}{\partial\epsilon} \bigg| _{\epsilon=0}\mathbb{E}_{\xi}^{\mathbb{Q_{\epsilon}}} [(f/\phi)(X_{T}^{\epsilon})] =\mathbb{E}_{\xi}^{\mathbb{Q}} \bigg[(f/\phi)(X_{T})\int_{0}^{T}\overline{k}(X_{s})\, dW_{s}\bigg] . $$

Now it remains to show that as $T\rightarrow\infty$,

$$ \frac{1}{T} \frac{\partial}{\partial\epsilon} \bigg| _{\epsilon =0}\mathbb{E}_{\xi}^{\mathbb{Q_{\epsilon}}} [(f/\phi)(X_{T}^{\epsilon})] =\frac{1}{T}\mathbb{E}_{\xi}^{\mathbb{Q}} \bigg[(f/\phi)(X_{T})\int_{0}^{T}\overline{k}(X_{s})\, dW_{s}\bigg] \longrightarrow0. $$

From (A.14), it is enough to prove that the term $\mathbb{E}_{\xi}^{\mathbb{Q}}[(\int_{0}^{T}g^{2}(X_{s})\, ds)^{\frac {p}{2}}]$ grows at most like $T^{\frac{p}{2}}$ as $T\to\infty$. Recall that $p$ is a positive even integer. Using (B.1) and the inequality $x^{\frac{p}{2}}\leq (p/2)!\,e^{x}$ for $x>0$, it follows that

$$ \mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[\bigg(\frac{1}{T}\int_{0}^{T}\epsilon _{0}g^{2}(X_{t})\,dt\bigg)^{\frac{p}{2}}\bigg]\leq (p/2)!\, \mathbb{E}_{\xi}^{\mathbb{Q}}\big[e^{\frac{1}{T}\int_{0}^{T}\epsilon_{0}\,g^{2}(X_{s})\,ds}\big] \leq (p/2)!\,c_{1}, $$

which gives the desired result. This completes the proof. □

Appendix C: Proof of Proposition 4.12

The pricing operator $\mathcal{P}^{\epsilon}$ in (4.10) is

$$ \mathcal{P}_{T}^{\epsilon}f(x)=\mathbb{E}_{X_{0}^{\epsilon}=x}^{\mathbb{P}}\big[e^{-\int_{0}^{T} r_{\epsilon}(X_{s}^{\epsilon})\,ds} f(X_{T}^{\epsilon})\big]. $$

Define an operator $\mathcal{P}_{T}^{\mbox{Lamp}\,\epsilon}$ corresponding to the Lamperti transformation by

$$ \mathcal{P}_{T}^{\mbox{Lamp}\,\epsilon}F(\zeta)= \mathbb{E}_{U_{0}^{\epsilon}=\zeta}^{\mathbb{P}}\big[e^{-\int_{0}^{T}R_{\epsilon}(U_{s}^{\epsilon})\,ds}F(U_{T}^{\epsilon})\big]. $$

Lemma C.1

Let $\beta$ be a real number and $h$ a positive function on ℝ. The following statements are equivalent:

• The pair $(e^{-\beta T},h)$ is an eigenpair of $\mathcal {P}_{T}^{\epsilon}$, that is,

  $$ \mathcal{P}_{T}^{\epsilon}h(x)=e^{-\beta T}h(x), \qquad x\in\mathbb{R}. $$

• The pair $(e^{-\beta T},h\circ v_{\epsilon})$ is an eigenpair of $\mathcal{P}_{T}^{\mbox{Lamp}\,\epsilon}$, that is,

  $$ \mathcal{P}_{T}^{\mbox{Lamp}\,\epsilon} (h\circ v_{\epsilon})(\zeta )=e^{-\beta T}(h\circ v_{\epsilon})(\zeta), \qquad\zeta\in\mathbb{R}. $$

Proof

The proof is straightforward from

$$ \begin{aligned} \mathcal{P}_{T}^{\epsilon}h(x) &=\mathbb{E}_{X_{0}^{\epsilon}=x}^{\mathbb{P}}\big[e^{-\int_{0}^{T} r_{\epsilon}(X_{s}^{\epsilon})\,ds} h(X_{T}^{\epsilon})\big]\\ &=\mathbb{E}_{U_{0}^{\epsilon}=u_{\epsilon}(x)}^{\mathbb{P}}\big[e^{-\int _{0}^{T}R_{\epsilon}(U_{s}^{\epsilon})ds}\,(h\circ v_{\epsilon})(U_{T}^{\epsilon})\big]=\mathcal{P}_{T}^{\mbox{Lamp}\,\epsilon }(h\circ v_{\epsilon})(\zeta) \end{aligned} $$

and $\zeta=u_{\epsilon}(x)$. □

Proof of Proposition 4.12

We show the “only if” direction. The “if” direction can be shown in a similar way, so we omit it. Assume that the quadruple $(b_{\epsilon},\sigma_{\epsilon},r_{\epsilon},f_{\epsilon})$ and $\xi$ satisfy Assumptions 4.1 and 4.2. Define $X^{\epsilon},\mathcal{P}^{\epsilon}, M^{\epsilon},\mathbb{Q}^{\epsilon},(\lambda_{\epsilon},\phi_{\epsilon}),\varphi_{\epsilon}$ accordingly. It is easy to check that the quadruple $(\delta_{\epsilon},1,R_{\epsilon},F_{\epsilon})$ satisfies Assumption 4.1. We show that the quadruple $(\delta_{\epsilon},1,R_{\epsilon},F_{\epsilon})$ and $\zeta _{\epsilon}$ satisfy Assumption 4.2 (that is, Assumptions 1.1–1.3 and 2.1–2.4). Assumption 1.1 is satisfied because $U^{\epsilon}$ defined in (4.12) is a strong solution of the SDE (4.13), and because the strong solution $X^{\epsilon}$ of the SDE (4.11) (thus, the strong solution $U^{\epsilon}$ of the SDE (4.13)) is unique and non-explosive. Assumptions 1.2 and 1.3 are trivial. For Assumption 2.1, we observe that the pair $(e^{-\lambda_{\epsilon}T},\phi_{\epsilon}\circ v_{\epsilon})$ is an eigenpair of $\mathcal{P}_{T}^{\mbox{Lamp}\,\epsilon}$ by Lemma C.1. From (2.2), it follows that the corresponding martingale is

$$\begin{aligned} M_{t}^{\mbox{Lamp}\,\epsilon} &:=e^{\lambda_{\epsilon}t-\int_{0}^{t} R_{\epsilon}(U_{s}^{\epsilon})\,ds}\, \frac{(\phi_{\epsilon}\circ v_{\epsilon})(U_{t}^{\epsilon})}{(\phi_{\epsilon}\circ v_{\epsilon})(\zeta_{\epsilon})} \\ &\phantom{:} = e^{\lambda_{\epsilon}t-\int_{0}^{t} r_{\epsilon}(X_{s}^{\epsilon})\,ds}\,\frac{\phi_{\epsilon}(X_{t}^{\epsilon})}{\phi_{\epsilon}(\xi)} =M_{t}^{\epsilon}, \qquad0\leq t\leq T. \end{aligned}$$

The recurrent eigenmeasure $\mathbb{Q}^{\mbox{Lamp}\,\epsilon}$ defined by this martingale $M^{\mbox{Lamp}\,\epsilon}$ satisfies

$$ \frac{d\mathbb{Q}^{\mbox{Lamp}}_{\epsilon}}{d\mathbb{P}^{\quad\; }} \bigg| _{\mathcal{F}_{T}}=M_{T}^{\mbox{Lamp}\,\epsilon }=M_{T}^{\epsilon}=\frac{d\mathbb{Q}^{\epsilon}}{d\mathbb{P}\,} \bigg| _{\mathcal{F}_{T}}. $$

Thus, $\mathbb{Q}^{\mbox{Lamp}\,\epsilon}=\mathbb{Q}^{\epsilon}$ on $\mathcal{F}_{T}$ for each $T\geq0$. It follows that Assumption 2.1 is satisfied because the recurrence of $X^{\epsilon}$ implies the recurrence of $U^{\epsilon}$. Assumption 2.2 is clearly satisfied since the recurrent eigenfunction $\phi_{\epsilon}\circ v_{\epsilon}$ is twice continuously differentiable. Assumptions 2.3 and 2.4 are directly obtained from $\mathbb{Q}^{\mbox{Lamp}\,\epsilon}=\mathbb{Q}^{\epsilon}$ on $\mathcal{F}_{T}$ for all $T$. By Lemma C.1 and the above argument, the two corresponding recurrent eigenvalues coincide. □

Appendix D: The CIR model

4.1 D.1 Hansen–Scheinkman decomposition

First, we show that $(b,\sigma,r,f)$ and $\xi$ satisfy Assumptions 4.1 and 4.2 (that is, Assumptions 1.1–1.3 and 2.1–2.4). We only prove Assumptions 2.1–2.4; the others are trivial. It can be shown that the pair $(\lambda,\phi(x)):=(\theta\kappa,e^{-\kappa x})$ is the recurrent eigenpair, where $\kappa:=\frac{\sqrt{a^{2}+2\sigma^{2}}-a}{\sigma^{2}}$ (see [23, Sect. 6.1.1]). This proves Assumptions 2.1 and 2.2. Consider the recurrent eigenmeasure ℚ. The corresponding Girsanov kernel is $\varphi(X_{t})=-\sigma\kappa\sqrt{X_{t}}$, and the ℚ-dynamics of $X$ is

$$ dX_{t}= ( \theta-\sqrt{a^{2}+2\sigma^{2}}\,X_{t})dt + \sigma\sqrt{X_{t}}\, dW_{t}, \qquad X_{0}=\xi. $$

(D.1)

Here, $W$ is a ℚ-Brownian motion. This process is a re-parametrized CIR model. It is well known that the CIR model has an invariant distribution $\nu$, which implies Assumption 2.3. For convenience, we define $b:=\sqrt{a^{2}+2\sigma^{2}}$.

To show Assumption 2.4, consider the ℚ-density function $\ell(x;t)$ of $X_{t}$ which is

$$ \ell(x;t):=h_{t}\,e^{-u-v}\Big(\frac{v}{u}\Big)^{q/2}I_{q}(2\sqrt{uv}), $$

where $I_{q}$ is the modified Bessel function of the first type of order $q$ and

$$ h_{t}=\frac{2b}{\sigma^{2}(1-e^{-bt})}, \quad q=\frac{2\theta}{\sigma^{2}}-1, \quad u=h_{t}\xi e^{-bt}, \quad v=h_{t}x. $$

After slightly rewriting, we find

$$ \ell(x;t)=k_{t}\,h_{t}\,e^{-h_{t}x}x^{q/2}I_{q}(2h_{t}e^{-bt/2}\sqrt{\xi x}). $$

(D.2)

Here, $k_{t}=e^{-h_{t}\xi e^{-bt}}(\xi e^{-bt})^{-q/2}$ and

$$ I_{q}(z)=\frac{(z/2)^{q}}{\pi^{1/2}\,\Gamma(q+1/2)}\int_{0}^{\pi}(e^{z\cos u}\sin^{2q}u)\,du\leq \frac{\pi^{1/2}(z/2)^{q}e^{z}}{\Gamma(q+1/2)} \;. $$

(D.3)

For large $t>0$, we have $\ell(x;t)\leq B\,e^{-h_{t}x} x^{q}e^{2h_{t}\sqrt{\xi r}}$ for a positive constant $B$. To show Assumption 2.4 for the function $f$, whose growth is at most polynomial, it suffices to prove (D.4) below for a constant $c>\kappa$ because the growth rate of $(f/\phi)(x)=f(x)e^{\kappa x}$ is less than the growth rate of $e^{cx}$ as $x\to\infty$. Choose a constant $c$ such that $\kappa=\frac{b-a}{\sigma^{2}}< c<\frac {2b}{\sigma^{2}}$. Because $\frac{2b}{\sigma^{2}}< h_{t}$, we know that $e^{cx}\ell(x;t)$ is dominated by $Be^{(c-\frac{2b}{\sigma^{2}})x} x^{q}e^{2h_{1}\sqrt{\xi x}}$, whose integral over $(0,\infty)$ is finite. By the dominated convergence theorem, it follows that

$$ \mathbb{E}_{\xi}^{\mathbb{Q}} [e^{cX_{t}}]=\int_{0}^{\infty}e^{cx}\ell(x;t)\,ds\longrightarrow\int_{0}^{\infty}e^{cx}\ell (x;\infty)\,dx=\int e^{cx}\,d\nu(x), $$

(D.4)

where $\ell(x;\infty)=\lim_{t\rightarrow\infty} \ell(x;t)$, which is equal to the invariant density function of $X$ under ℚ. For more details regarding the density of the CIR model, see [17, Proposition 6.3.2.1].

In summary, we have shown that the quadruple of functions $(b,\sigma,r,f)$ and the initial value $\xi$, which were defined in Sect. 5.1, satisfy Assumptions 4.1 and 4.2. From now on, in the context of the CIR model, the notations $X$, $\mathcal{P}$, ℒ, $M$, ℚ, $(\lambda,\phi)$, $\varphi$, $\nu$ are self-explanatory.

4.2 D.2 Sensitivity of $\xi$

In this section, we show the result on the long-term sensitivity of $\xi$ in (5.3). By Theorem 3.1, it suffices to show that $\mathbb{E}_{\xi}^{\mathbb{Q}}[(f/\phi)(X_{T})]$ is continuously differentiable in $\xi$ and that $\frac{\partial}{\partial\xi }\mathbb{E}_{\xi}^{\mathbb{Q}}[(f/\phi)(X_{T})]\rightarrow0$ as $T\rightarrow\infty$. Both can be easily justified by looking at the ℚ-density function $\ell(x;T)$ of $X_{T}$. Indeed,

$$\begin{aligned} \lim_{T\rightarrow\infty}\frac{\partial}{\partial\xi}\mathbb{E}_{\xi }^{\mathbb{Q}}[(f/\phi)(X_{T})] &=\lim_{T\rightarrow\infty}\frac{\partial}{\partial\xi}\int_{0}^{\infty}(f/\phi)(x) \,\ell(x;T)\,dx \\ &=\lim_{T\rightarrow\infty}\int_{0}^{\infty}(f/\phi)(x)\,\frac{\partial \ell(x;T)}{\partial\xi}\,dx \\ &=\int_{0}^{\infty}(f/\phi)(x)\,\lim_{T\rightarrow\infty}\frac{\partial \ell(x;T)}{\partial\xi}\,dx=0. \end{aligned}$$

(D.5)

The interchange of differentiation and integration in the second equality can be checked by a standard argument. For the last equality, we used the following lemma.

Lemma D.1

Let $\ell(x;t)$ be the ℚ-density function of $X_{t}$ given in (D.2). Then

$$ \lim_{t\rightarrow\infty}\frac{\partial\ell(x;t)}{\partial\xi}=0. $$

Proof

From the ℚ-density function

$$ \ell(x;t)=e^{-h_{t}\xi e^{-bt}}(\xi e^{-bt})^{-q/2}\,h_{t}\, e^{-h_{t}x}x^{q/2}I_{q}(2h_{t}e^{-bt/2}\sqrt{\xi x}), $$

we have

$$ \begin{aligned} \frac{\partial\ell(x;t)}{\partial\xi} =\bigg(-h_{t}e^{-bt}-\frac{q}{2\xi}+\frac{1}{2}\sqrt{\frac{x}{\xi}}\, h_{t}e^{-bt/2}\frac{I_{q-1}(z)+I_{q+1}(z)}{I_{q}(z)}\bigg)\ell(x;t), \end{aligned} $$

(D.6)

where $z=2h_{t}e^{-bt/2}\sqrt{\xi x}$. Here, we used the equality $I_{q}'(\cdot)=\frac{1}{2}(I_{q-1}(\cdot )+I_{q+1}(\cdot))$. Observe that $z\rightarrow0$ as $t\rightarrow\infty$. The modified Bessel function $I_{q}$ of order $q$ satisfies $\lim_{z\rightarrow0}\frac{I_{q}(z)}{\;\frac{(z/2)^{q}}{\Gamma(q+1)}\; }=1$; thus

$$\begin{aligned} \lim_{t\rightarrow\infty}h_{t}e^{-bt/2}\frac{I_{q-1}(z)+I_{q+1}(z)}{I_{q}(z)} &=\lim_{t\rightarrow\infty}h_{t}e^{-bt/2}\frac{\frac{(h_{t}e^{-bt/2}\sqrt {\xi x})^{q-1}}{\Gamma(q)}+\frac{(h_{t}e^{-bt/2}\sqrt{\xi x})^{q+1}}{\Gamma(q+2)}}{\frac{(h_{t}e^{-bt/2}\sqrt{\xi x})^{q}}{\Gamma (q+1)}}\\ &=\frac{q}{\sqrt{\xi x}}. \end{aligned}$$

In conclusion, we have $\lim_{t\rightarrow\infty}\frac{\partial\ell (x;t)}{\partial\xi}=0$ from (D.6). □

4.3 D.3 Sensitivity of $\theta$

In this section, we analyze the long-term sensitivity of $\theta$ in the CIR model. As discussed in (5.2) in Sect. 5.1, the parameter $\theta$ can be regarded as a perturbation parameter. The aim is to show the long-term sensitivity of $\theta$ in (5.3) using Corollary 4.11. Only the hypothesis of Theorem 4.8 will be checked because the other conditions are easy to prove. If we define the functions

$$ k(x)=\frac{\theta}{\sigma\sqrt{|x|}}-\frac{\sqrt{a^{2}+2\sigma ^{2}}}{\sigma}\,\sqrt{|x|}, \qquad g(x)=\frac{1}{\sigma\sqrt{|x|}}, $$

(D.7)

then $k$ and $g$ satisfy (4.6) and (4.7) in Sect. 4.1. Condition (i) in Theorem 4.8 can be proved via Proposition D.2 below. For (ii) and (iii), we set $p=q=2$. Let $\epsilon_{1}$ be a positive number such that $\epsilon_{1}<\frac {2\theta}{\sigma^{2}}-1$. By the same method as in (D.4), we get

$$ \begin{aligned} \mathbb{E}_{\xi}^{\mathbb{Q}}[g^{2+\epsilon_{1}}(X_{t})] &=\mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[\bigg(\frac{1}{\sigma\sqrt{X_{t}}}\bigg)^{2+\epsilon_{1}}\bigg] =\int_{0}^{\infty}\bigg(\frac{1}{\sigma\sqrt{x}}\bigg)^{2+\epsilon_{1}}\ell (x;t)\,dx \end{aligned} $$

(D.8)

is bounded uniformly in $t$ on $[0,\infty)$. Thus $\mathbb{E}_{\xi}^{\mathbb{Q}}[\int_{0}^{T}g^{2+\epsilon_{1}}(X_{t})\,dt]$ is finite for each $T$, which means that (ii) is satisfied. For (iii), we put $\psi=f/\phi$ since $f$ and $\phi$ are independent of $\theta$. Using the method in (D.4), we find that

$$ \mathbb{E}_{\xi}^{\mathbb{Q}}[(f/\phi)^{2}(X_{T})]\longrightarrow\int(f/\phi)^{2}(x)\,d\nu(x) $$

as $T\to\infty$ since the exponential growth rate of $(f/\phi)^{2}(x)$ is ${2(b-a)x}/{\sigma^{2}}$ as $x\to\infty$.

Proposition D.2

For any $\epsilon_{0}$ with $0<\epsilon_{0}\leq\frac{1}{2}(\frac{\sigma }{2}-\frac{\theta}{\sigma})^{2}$, there exist constants $a$ and $c$ such that for all $T>0$,

$$ \mathbb{E}_{\xi}^{\mathbb{Q}}\big[e^{\epsilon_{0}\int_{0}^{T}(1/X_{s})\,ds}\big] \leq c \,e^{aT}. $$

Proof

We modify the proof found in [1, Appendix C]. Their proof evaluates the above expectation under the condition $\epsilon_{0}<0$, whereas our proof evaluates the expectation under the condition $\epsilon_{0}>0$. Our proof is given in several steps.

(i) Let $Y:=1/X$. We find a positive function $V(y,t) = V (y,t; T)$ on $\mathbb{R}_{+}\times[0,T]$ such that

$$ V(Y_{t},t)e^{\epsilon_{0}\int_{0}^{t}Y_{s}\,ds}, \qquad0\leq t\leq T, $$

is a local martingale and $V(y,T)$ is a constant independent of $y$ and $T$.

(ii) Show that the function $V$ satisfies

$$ V(y,0) = V (y, 0 ; T) \leq c_{1} e^{-\gamma b T} $$

for positive constants $c_{1}$, $\gamma$, $b$. In other words, the decay rate of the function $V(y,0)$ is less than or equal to an exponential rate in $T$.

(iii) Because $V(Y_{t},t)e^{\epsilon_{0}\int_{0}^{t}Y_{s}\,ds}$, $0\leq t\leq T$, is a positive local martingale, it is a supermartingale. Thus we have

$$ \begin{aligned} c_{1} e^{-a T}\geq V(Y_{0},0) &\geq\mathbb{E}_{\xi}^{\mathbb{Q}}\big[V(Y_{T},T)e^{\epsilon_{0}\int_{0}^{T}Y_{s}\, ds}\big]\geq\mbox{const. }\mathbb{E}_{\xi}^{\mathbb{Q}}\big[e^{\epsilon_{0}\int_{0}^{T}Y_{s}\,ds}\big], \end{aligned} $$

which is the desired result.

Step (i). From (D.1), the ℚ-dynamics of $X$ is $dX_{t}=( \theta-bX_{t})dt + \sigma\sqrt{X_{t}}\, dW_{t}$, where $b=\sqrt{a^{2}+2\sigma^{2}}$. Define $Y:=1/X$. The Itô formula yields

$$ dY_{t}=\big((\sigma^{2}-\theta)Y_{t}+b\big)Y_{t}\,dt-\sigma Y_{t}^{3/2}\,dW_{t}. $$

We want to find a positive function $V(y,t) = V (y,t;T)$ on $\mathbb {R}^{+}\times[0,T]$ such that

$$ V(Y_{t},t)e^{\epsilon_{0}\int_{0}^{t}Y_{s}\,ds}, \qquad0\leq t\leq T, $$

is a local martingale and $V(y,T)$ is a constant independent of $y$ and $T$. Such a function $V(y,t)$ should satisfy

$$ V_{t}+\frac{1}{2}\sigma^{2}x^{3}V_{xx}+\big((\sigma^{2}-\theta)x+b\big)xV_{x}+\epsilon_{0}xV=0. $$

(D.9)

We make the ansatz $V(y,t)=f(x)x^{\gamma}$, where $x=a(t)/y$, and compute

$$ \begin{aligned} &V_{y}=-\frac{1}{a(t)}f'(x)x^{\gamma+2}-\frac{\gamma}{a(t)}f(x)x^{\gamma +1}, \\ &V_{yy}=\frac{1}{a^{2}(t)}f''(x)x^{\gamma+4}+\frac{2(\gamma +1)}{a^{2}(t)}f'(x)x^{\gamma+3}+\frac{\gamma(\gamma +1)}{a^{2}(t)}f(x)x^{\gamma+2}, \\ &V_{t}=\frac{a'(t)}{a(t)}f'(x)x^{\gamma+1}+\frac{a'(t)}{a(t)}\gamma f(x)x^{\gamma}. \end{aligned} $$

Then (D.9) gives

$$ \begin{aligned} 0&=\frac{1}{2}\sigma^{2}a(t)x^{\gamma+1}f''(x)\\ &\phantom{=:}+\bigg(\frac{a'(t)}{a(t)}x^{\gamma+1}-bx^{\gamma +1}-(\sigma^{2}-\theta)a(t)x^{\gamma} +\sigma^{2}(\gamma+1)a(t)x^{\gamma}\bigg)f'(x)\\ &\phantom{=:}+ \bigg(\frac{a'(t)}{a(t)}\gamma x^{\gamma}-b\gamma x^{\gamma} +\frac{1}{2}\sigma^{2}\gamma(\gamma+1)a(t)x^{\gamma-1} \\ &\phantom{=:+\bigg(}-(\sigma^{2}-\theta)\gamma a(t)x^{\gamma-1}+\epsilon _{0}a(t)x^{\gamma-1}\bigg)f(x). \end{aligned} $$

Assume that $a(t)$ and $\gamma$ satisfy

$$ \left\{\quad \begin{aligned} &\frac{a'(t)}{a(t)}-b=a(t), \\ &\frac{1}{2}\sigma^{2}\gamma(\gamma+1)-(\sigma^{2}-\theta)\gamma+\epsilon_{0}=0, \end{aligned} \right. $$

(D.10)

so that the above equation becomes $\frac{1}{2}\sigma^{2}xf''(x)+(x+\sigma^{2}\gamma+\theta)f'(x)+\gamma f(x)=0$. We define a new variable $z$ such that $x=-\frac{1}{2}\sigma^{2}z$ and define the function $g$ as $g(z):=f(x)$. Then $zg''(z)+(\kappa -z)g'(z)-\gamma g(z)=0$, where $\kappa:=2(\gamma+\frac{\theta}{\sigma^{2}})$. It is well known that the standard confluent hypergeometric function $f(x)=g(z)=M(\gamma,\kappa;z)$ is a solution of this equation.

We now want to find an explicit expression for the function $V(y,t)$. (D.10) yields $a(t)=\frac{b}{e^{b(T-t)}-1}$ for $0\leq t\leq T$ and

$$ \gamma=\frac{1}{2}-\frac{\theta}{\sigma^{2}}+\sqrt{\bigg(\frac{1}{2}-\frac {\theta}{\sigma^{2}}\bigg)^{2}-\frac{2\epsilon_{0}}{\sigma^{2}}}. $$

The number $\gamma$ is a real number because of the assumption that $0<\epsilon_{0}\leq\frac{1}{2}(\frac{\sigma}{2}-\frac{\theta}{\sigma})^{2}$. We know that $\kappa=2(\gamma+\frac{\theta}{\sigma^{2}})>0$ and, by using the Feller condition in the CIR process, $\gamma<0$. The solution $V(y,t)$ is

$$\begin{aligned} V(y,t)=f(x)x^{\gamma}&=g(z)\bigg(-\frac{1}{2}\sigma^{2}z\bigg)^{\gamma} \\ &=\bigg(\frac{1}{2}\sigma^{2}\bigg)^{\gamma}M(\gamma,\kappa;z)(-z)^{\gamma} \\ &=\bigg(\frac{1}{2}\sigma^{2}\bigg)^{\gamma}M(\kappa-\gamma,\kappa ;-z)(-z)^{\gamma}e^{z}, \end{aligned}$$

(D.11)

where $z=-\frac{2x}{\sigma^{2}}=-\frac{2a(t)}{\sigma^{2}y}=-\frac{2b}{\sigma ^{2}(e^{b(T-t)}-1)y}$. Here, we have used the equality $M(\gamma,\kappa;z)=M(\kappa -\gamma,\kappa;-z)e^{z}$.

We now show that $V(y,T)$ is a constant independent of $y$ and $T$. By direct calculation, where $\Gamma$ is the gamma function,

$$ \begin{aligned} \lim_{t\rightarrow T}V(y,t) &=\bigg(\frac{1}{2}\sigma^{2}\bigg)^{\gamma}\,\lim_{z\rightarrow-\infty} M(\kappa-\gamma,\kappa;-z)(-z)^{\gamma}e^{z}\\ &=\bigg(\frac{1}{2}\sigma^{2}\bigg)^{\gamma}\,\lim_{u\rightarrow\infty} M(\kappa-\gamma,\kappa;u)u^{\gamma}e^{-u} \\ &=\bigg(\frac{1}{2}\sigma^{2}\bigg)^{\gamma}\,\frac{\Gamma(\kappa)}{\Gamma (\kappa-\gamma)\Gamma(\gamma)}\,\lim_{u\rightarrow\infty}u^{\gamma}e^{-u} \int_{0}^{1} e^{us}s^{\kappa-\gamma-1}(1-s)^{\gamma-1}\,ds \\ &=\bigg(\frac{1}{2}\sigma^{2}\bigg)^{\gamma}\,\frac{\Gamma(\kappa)}{\Gamma (\kappa-\gamma)\Gamma(\gamma)}\,\lim_{u\rightarrow\infty}u^{\gamma}\int_{0}^{1} e^{-us}(1-s)^{\kappa-\gamma-1}s^{\gamma-1}\,ds \\ &=\bigg(\frac{1}{2}\sigma^{2}\bigg)^{\gamma}\,\frac{\Gamma(\kappa)}{\Gamma (\kappa-\gamma)\Gamma(\gamma)}\,\lim_{u\rightarrow\infty} \int_{0}^{u} e^{-t}(1-t/u)^{\kappa-\gamma-1}t^{\gamma-1}\,dt \\ &=\bigg(\frac{1}{2}\sigma^{2}\bigg)^{\gamma}\,\frac{\Gamma(\kappa)}{\Gamma (\kappa-\gamma)\Gamma(\gamma)}\,\int_{0}^{\infty}e^{-t}t^{\gamma-1}\,dt \\ &=\bigg(\frac{1}{2}\sigma^{2}\bigg)^{\gamma}\,\frac{\Gamma(\kappa)}{\Gamma (\kappa-\gamma)} . \end{aligned} $$

Step (ii). We now show that the function $V(y,0)$ satisfies $V(y,0)\leq c_{1} e^{-\gamma b T}$ for some positive constant $c_{1}$ independent of $T$. By substituting $t = 0$ in (D.11), we get

$$ V(y,0)=c_{2}(T;y)\bigg(\frac{\sigma^{2}(1-e^{-bT})y}{2b}\bigg)^{-\gamma} e^{-\gamma bT}, $$

where

$$ c_{2}(T;y):= \bigg(\frac{1}{2}\sigma^{2}\bigg)^{\gamma}M\bigg(\kappa-\gamma,\kappa ;\frac{2b}{\sigma^{2}(e^{bT}-1)y}\bigg)\,e^{-\frac{2b}{\sigma ^{2}(e^{bT}-1)y}}. $$

Now observe that the quantity $c_{2}(T;x)$ is bounded uniformly for large $T$ because we have $\lim_{u\rightarrow0}M(\kappa-\gamma,\kappa,u)=1$. This gives the desired result.

Step (iii). Because $V(Y_{t},t)e^{\epsilon_{0}\int_{0}^{t}Y_{s}\,ds}$, $0\leq t\leq T$, is a positive local martingale, it is a supermartingale. Thus we have

$$\begin{aligned} \bigg(\frac{1}{2}\sigma^{2}\bigg)^{\gamma}\,\frac{\Gamma(\kappa)}{\Gamma (\kappa-\gamma)} \mathbb{E}_{\xi}^{\mathbb{Q}}\big[e^{\epsilon_{0}\int_{0}^{T}Y_{s}\,ds}\big] &=\mathbb{E}_{\xi}^{\mathbb{Q}}\big[V(Y_{T},T)e^{\epsilon_{0}\int_{0}^{T}Y_{s}\, ds}\big]\\ &\leq V(Y_{0},0)\leq c_{1} e^{-\gamma b T}. \end{aligned}$$

This completes the proof. □

4.4 D.4 Sensitivity of $a$

This section analyzes the long-term sensitivity of $a$ in the drift coefficient of the CIR model. The parameter $a$ can be regarded as a perturbation parameter. The purpose is to show the long-term sensitivity of $a$ in (5.3) by applying Corollary 4.11. Only the hypothesis of Theorem 4.8 will be checked because the other conditions are easy to prove. From the function $k$ in (D.7), we define $g(x):=\frac{1}{\sigma}\sqrt{|x|}$ so that for all $a$,

$$ \left|\frac{\partial k(x)}{\partial a}\right| = \left|\frac{a\sqrt{|x|}}{\sigma\sqrt{a^{2}+2\sigma^{2}}}\right|\leq\frac {1}{\sigma}\sqrt{|x|} =g(x). $$

Then the functions $k$ and $g$ satisfy (4.6) and (4.7) in Sect. 4.1. For condition (i) in Theorem 4.8, it is sufficient to show that there exist constants $c$ and $d$ such that

$$ \begin{aligned} \mathbb{E}_{\xi}^{\mathbb{Q}}\big[e^{\int_{0}^{T}X_{s}\,ds}\big] \leq c \,e^{dT} \end{aligned} $$

(D.12)

for all $T$. This is proved in [26, Lemma 3.1]. For (ii) and (iii), we set $p=q=2$ and let $\epsilon_{1}=2$. Then it can easily be shown that the expectation

$$ \begin{aligned} \mathbb{E}_{\xi}^{\mathbb{Q}}[g^{2+\epsilon_{1}}(X_{t})]=\frac{1}{\sigma ^{4}}\mathbb{E}_{\xi}^{\mathbb{Q}}[X_{t}^{2}] \end{aligned} $$

(D.13)

is bounded uniformly in $t$ on $[0,\infty)$ because $X$ is a CIR process. Therefore the expectation $\mathbb{E}_{\xi}^{\mathbb{Q}}[\int _{0}^{T}g^{2+\epsilon_{1}}(X_{t})\,dt]$ is finite for each $T$, which implies (ii). For (iii), we define $\psi(x)=f(x)e^{cx}$ for a constant $c$ such that ${(b-a)}/{\sigma^{2}}< c<{b}/{\sigma^{2}}$, and then (4.8) follows. By using the method in (D.4), we obtain condition (iii) since the exponential growth rate of $\psi^{2}(x)$ is $2cx$ as $x\to\infty$.

4.5 D.5 Sensitivity of $\sigma$

This section conducts a sensitivity analysis with respect to the variable $\sigma$ in the CIR model. The parameter $\sigma$ in the diffusion term can be regarded as a perturbation parameter. Define the quadruple $(\delta,1,R,F)$ and $\zeta$ by the Lamperti transformation given in Sect. 4.2.1. Let $u(x):=\int _{0}^{x}\frac{1}{\sigma\sqrt{|y|}}\,dy=\frac{2}{\sigma}\sqrt{x}$ for $x>0$; then

$$ \delta(u)= \bigg(\frac{2\theta}{\sigma^{2}}-\frac{1}{2}\bigg)\frac {1}{u}-\frac{au}{2}, \quad R(u)=\sigma^{2}u^{2}/4, \quad F(u)=f(\sigma ^{2}u^{2}/4), \quad\zeta=\frac{2}{\sigma}\sqrt{\xi}. $$

Because Sect. D.1 shows that the quadruple $(b,\sigma,r,f)$ and the initial value $\xi$ satisfy Assumptions 4.1 and 4.2, the quadruple $(\delta,1,R,F)$ and $\zeta$ also satisfy Assumptions 4.1 and 4.2 by Proposition 4.12. The notations $U$, ℚ, $(\lambda,\Phi)$ are now self-explanatory. The recurrent eigenfunction and the payoff function are $\Phi(u)=\phi(\sigma^{2}u^{2}/4)$ and $F(u)=f(\sigma^{2}u^{2}/4)$, respectively.

The goal of this section is to show the long-term sensitivity of $\sigma$ in (5.3) by using Theorem 4.13. Conditions (i) and (ii) in Theorem 4.13 are discussed below, and condition (iii) is proved in Proposition D.4. Observe that $\lambda={\theta(\sqrt{a^{2}+2\sigma^{2}}-a)}/{\sigma^{2}}$ and $\Phi(\zeta)=\phi(\sigma^{2}\zeta^{2}/4)$ are continuously differentiable in $\sigma$, which with Proposition D.3 below gives (i). To check (ii), we apply Theorem 4.8. Recall $b=\sqrt{a^{2}+2\sigma^{2}}$, and define $k(u)=(\frac{2\theta}{\sigma^{2}}-\frac{1}{2})\frac{1}{u}-\frac{bu}{2}$ and $g(u)=C(u+\frac{1}{u})$ for sufficiently large $C>0$ such that $|\frac{\partial}{\partial\sigma}k(u)| \leq C(\frac{1}{u}+u)=g(u)$. Then the functions $k$ and $g$ satisfy (4.6) and (4.7) in Sect. 4.1. Observe that $g^{2}(U_{t})\leq C_{1}(\frac{1}{X_{t}}+{X_{t}})$ for sufficiently large $C_{1}>0$. To prove the exponential condition (i) in Theorem 4.8, it suffices to show that there exist positive constants $a$, $c$ and $\epsilon_{0}$ such that $\mathbb{E}_{\xi}^{\mathbb{Q}}[e^{\epsilon_{0}\int_{0}^{T}(X_{s}+\frac{1}{X_{s}})\,ds}] \leq c \,e^{aT}$ for all $T>0$. This can be shown by combining Proposition D.2 and (D.12). Condition (ii) of Theorem 4.8 can be confirmed with $p=2$ and $0<\epsilon_{1}<\min\{\frac{2\theta }{\sigma^{2}}-1,2\}$ by combining the methods in (D.8) and (D.13). To check (iii) of Theorem 4.8, choose a real number $c$ such that $(b-a)/4< c< b/4$ and define $\psi(u):=e^{cu^{2}}$. For sufficiently large $u$,

$$ F(u)/\Phi(u)=f(\sigma^{2}u^{2}/4)e^{\kappa\sigma^{2}u^{2}/4} =f(\sigma^{2}u^{2}/4)e^{(b-a)u^{2}/4}\leq \psi(u) $$

since $f$ is of polynomial growth. Using the method in (D.4), it is easy to show that the expectation $\mathbb{E}^{\mathbb{Q}}[\psi ^{2}(U_{T})]=\mathbb{E}^{\mathbb{Q}}[e^{2cU_{T}^{2}}]=\mathbb{E}^{\mathbb{Q}}[e^{\frac{8c}{\sigma^{2}}X_{T}}]$ is bounded uniformly in $T$ on $[0,\infty)$ because ${8c}/{\sigma^{2}}<{2b}/{\sigma^{2}}$. This proves (iii) of Theorem 4.8 with $q=2$.

Proposition D.3 below is useful for checking (i) in Theorem 4.13. The parameter $\sigma$ is a variable appearing in $F/\Phi$, $\zeta$ and the dynamics of $U$. We temporarily employ a new parameter $s$ to distinguish the parameter $\sigma$ in $F/\Phi$ and $\zeta$ from the parameter $\sigma$ in the dynamics of $U$. Define

$$\begin{aligned} &\eta(s)=\frac{\sqrt{a^{2}+2s^{2}}-a}{s^{2}}, \quad \pi_{s}(r)=e^{-\eta(s)r}, \quad \Pi_{s}(u)=\pi_{s}(s^{2}u^{2}/4), \\ &G_{s}(u)=f(s^{2}u^{2}/4), \quad q(s)=\frac{2}{s}\sqrt{\xi} , \end{aligned}$$

(D.14)

so that $\mathbb{E}_{\zeta}^{\mathbb{Q}} [(F/\Phi)(U_{T})]=\mathbb{E}_{q(\sigma)}^{\mathbb{Q}}[(G_{\sigma}/\Pi_{\sigma})(U_{T})]$.

Proposition D.3

Fix $\sigma_{0} > 0$. The partial derivative $\frac{\partial}{\partial s}\mathbb {E}_{q(s)}^{\mathbb{Q}}[(G_{s}/\Pi_{s})(U_{T})]$ exists and is continuous in $(s,\sigma)$ in a neighborhood of $(\sigma _{0},\sigma_{0})$. Moreover, we have

$$ \lim_{T\to\infty}\frac{1}{T} \frac{\partial}{\partial s} \bigg| _{s=\sigma} \mathbb{E}_{q(s)}^{\mathbb{Q}}[(G_{s}/\Pi_{s})(U_{T})]=0 $$

for any $\sigma>0$ in a neighborhood of $\sigma_{0}$.

Proof

The proof is given in several steps.

(i) Define a process $Z=(Z_{t})_{t\geq0}$ by $Z_{t}=Z_{t}(s)=s^{2}U_{t}^{2}/4$ so that

$$ \mathbb{E}_{q(s)}^{\mathbb{Q}}[(G_{s}/\Pi_{s})(U_{T})] =\mathbb{E}_{\xi}^{\mathbb{Q}}[(f/\pi_{s})(Z_{T})]. $$

The right-hand side is more manageable.

(ii) Show that the partial derivative $\frac{\partial}{\partial s}\mathbb{E}_{\xi}^{\mathbb{Q}}[(f/\pi_{s})(Z_{T})]$ exists and that

$$ \frac{\partial}{\partial s}\mathbb{E}_{\xi}^{\mathbb{Q}}[(f/\pi_{s})(Z_{T})] =\int_{0}^{\infty}f(z)\,\frac{\partial}{\partial s} \frac{\ell(z;T,s)}{\pi _{s}(z)}\,dz $$

for $(s,\sigma)$ near $(\sigma_{0},\sigma_{0})$, where $\ell(z;t,s)$ is the density function of $Z_{t}$. Then deduce that this partial derivative is continuous in $(s,\sigma)$ in a neighborhood of $(\sigma_{0},\sigma_{0})$.

(iii) Finally, we show that

$$ \int_{0}^{\infty}f(z)\,\frac{\partial}{\partial s} \bigg| _{s=\sigma} \frac {\ell(z;T,s)}{\pi_{s}(z)}\,dz $$

converges to a finite constant as $T\rightarrow\infty$, which gives the desired result.

Step (i). Define the process $Z=(Z_{t})_{t\geq0}$ by $Z_{t}=Z_{t}(s)=s^{2}U_{t}^{2}/4$ so that we obtain $(G_{s}/\Pi_{s})(U_{T})=(f/\pi_{s})(s^{2}U^{2}_{T}/4)=(f/\pi_{s})(Z_{T})$ and $Z_{0}=\xi$. Then we have $\mathbb{E}_{q(s)}^{\mathbb{Q}}[(G_{s}/\Pi_{s})(U_{T})] =\mathbb{E}_{\xi}^{\mathbb{Q}}[(f/\pi_{s})(Z_{T})]$. The Itô formula gives

$$ dZ_{t}=\bigg(\frac{\theta s^{2}}{\sigma^{2}}-bZ_{t}\bigg)\,dt+s\sqrt{Z_{t}}\, dW_{t}, \qquad Z_{0}=\xi. $$

It is noteworthy that both the parameters $\sigma$ and $s$ are components in the dynamics of $Z$, but we are only interested in the sensitivity of $s$. One of the notable properties of this process $Z$ is that the initial value is not perturbed.

Step (ii). The process $Z$ is a CIR process and the density function of $Z_{t}$ (from (D.2)) is

$$ \ell(z;t,s)=e^{-h_{t}\xi e^{-bt}}(\xi e^{-bt})^{-q/2}\,h_{t}\, e^{-h_{t}z}z^{q/2}I_{q}(2h_{t}e^{-bt/2}\sqrt{\xi z}), $$

(D.15)

where $h_{t}=\frac{2b}{s^{2}(1-e^{-bt})}$, $q=\frac{2\theta}{\sigma^{2}}-1$ and $I_{q}$ is the modified Bessel function of the first type of order $q$. For $(s,\sigma)$ near $(\sigma_{0},\sigma_{0})$, we prove that

$$\begin{aligned} \frac{\partial}{\partial s}\mathbb{E}_{\xi}^{\mathbb{Q}}[(f/\pi_{s})(Z_{T})] &=\frac{\partial}{\partial s}\int_{0}^{\infty}(f/\pi_{s})(z)\,\ell(z;T,s)\, dz \\ &=\int_{0}^{\infty}f(z)\,\frac{\partial}{\partial s} \frac{\ell (z;T,s)}{\pi_{s}(z)}\,dz. \end{aligned}$$

(D.16)

To prove the interchangeability of the differentiation and the integration in the second equality, it suffices to show that for $(s,\sigma)$ near $(\sigma_{0},\sigma_{0})$ and for all $z>0$,

$$ \left|f(z)\,\frac{\partial}{\partial s} \frac{\ell(z;t,s)}{\pi _{s}(z)}\right|\leq Ce^{-\frac{1}{\sigma_{0}^{2}}\sqrt{a^{2}+2\sigma_{0}^{2}}\,z}=: G(z) $$

for a positive constant $C$ because the function $G(z)$ is integrable over $(0,\infty)$. Let us estimate how fast the function $|f(z)\,\frac{\partial}{\partial s} (\frac{\ell(z;t,s)}{\pi_{s}(z)})|$ grows as $z \to\infty$ by considering the growth rates of $f(z)$, $\frac{1}{\pi_{s}(z)}$, $\frac{\partial }{\partial s}\frac{1}{\pi_{s}(z)}$, $\ell(z;t,s)$ and each term of

$$ \begin{aligned} \frac{\partial}{\partial s}\,\ell(z;t,s)&=\frac{\,2\,}{s}h_{t}\xi e^{-bt}\ell(z;t,s)-\frac{\,2\,}{s}\ell(z;t,s)+\frac{\,2\,}{s}zh_{t}\ell (z;t,s)\\ &\phantom{=:}-\frac{\,2\,}{s}e^{-h_{t}\xi e^{-bt}}\xi ^{(-q+1)/2}e^{(q-1)bt/2}\,h_{t}^{2}\,e^{-h_{t}z}z^{(q+1)/2}(I_{q-1}+I_{q+1}). \end{aligned} $$

Given $\sigma>0$ and large $t>0$, for $s$ near $\sigma_{0}$, each term of $\frac{\partial}{\partial s} \ell(z;t,s)$ is dominated by one of

$$ \ell(z;t,s), \quad z\ell(z;t,s), \quad z^{q}e^{-h_{t}z+2h_{t}e^{-bt/2}\sqrt {\xi z}}, \quad z^{q+1}e^{-h_{t}z+2h_{t}e^{-bt/2}\sqrt{\xi z}} $$

up to constant multiples. We have used the upper bound of $I_{q}$ given in (D.3). The growth of each term is essentially dominated by $e^{-\frac {2b}{s^{2}}z}$ because $\frac{2b}{s^{2}}< h_{t}$. Thus the growth rate of $|\frac{\partial}{\partial s}(\ell(z;t,s)/\pi_{s} (z))|$ is less than or equal to that of $e^{(\eta(s)-\frac{2b}{s^{2}})z}$. Because the growth of $f(z)$ is at most polynomial, it follows that the growth rate of $|f(z)\frac{\partial}{\partial s}(\ell(z;t,s)/\pi_{s} (z))|$ is less than or equal to that of $e^{(\eta(s)-\frac{2b}{s^{2}})z}$. For $(s,\sigma)$ near $(\sigma_{0},\sigma_{0})$, the exponent of $e^{(\eta(s)-\frac{2b}{s^{2}})z}$ satisfies

$$ \eta(s)-\frac{2b}{s^{2}}=\frac{\sqrt{a^{2}+2s^{2}}-a}{s^{2}}-\frac{2\sqrt {a^{2}+2\sigma^{2}}}{s^{2}}< -\frac{\sqrt{a^{2}+2\sigma_{0}^{2}}}{\sigma_{0}^{2}}, $$

(D.17)

which is the desired inequality. Since (D.16) holds, we can directly derive that the partial derivative $\frac{\partial}{\partial s}\mathbb{E}_{q(s)}^{\mathbb{Q}}[(G_{s}/\Pi _{s})(U_{T})]=\frac{\partial}{\partial s}\mathbb{E}_{\xi}^{\mathbb{Q}}[(f/\pi _{s})(Z_{T})]$ exists and is continuous in $(s,\sigma)$ in a neighborhood of $(\sigma_{0},\sigma_{0})$.

Step (iii). Finally, we demonstrate that

$$ \int_{0}^{\infty}f(z)\,\frac{\partial}{\partial s} \bigg| _{s=\sigma} \frac {\ell(z;T,s)}{\pi_{s}(z)}\,dz $$

converges to a finite constant as $T\rightarrow\infty$. This can be proved by the dominated convergence theorem and by estimating how fast the function $|\frac{\partial}{\partial s}(\ell(z;t,s)/\pi_{s} (z))|$ grows as $T\to \infty$ in the same manner as above. This completes the proof. □

We now prove (iii) in Theorem 4.13. For simplicity, we omit the variable $s$ in the following notations; so

$$ \pi=\pi_{s}, \quad\Pi=\Pi_{s}, \quad G=G_{s}, \quad q=q(s) $$

for the functions defined in (D.14).

Proposition D.4

Fix $\sigma_{0} > 0$. The partial derivative $\frac{\partial}{\partial\sigma}\mathbb{E}_{q}^{\mathbb{Q}}[(G/\Pi)(U_{T})]$ is continuous in $(s,\sigma)$ in a neighborhood of $(\sigma_{0},\sigma_{0})$.

Proof

We only sketch the main idea because the proof is similar to that of Proposition D.3. Define the process $Z=(Z_{t})_{t\geq0}$ by $Z_{t}=s^{2}U_{t}^{2}/4$ so that we obtain $\mathbb{E}_{q}^{\mathbb{Q}}[(G/\Pi)(U_{T})] =\mathbb{E}_{\xi}^{\mathbb{Q}}[(f/\pi)(Z_{T})]$. Consider the density function $\ell=\ell(z;t)$ of $Z_{t}$ given in (D.15) as

$$ \ell(z;t)=e^{-h_{t}\xi e^{-bt}}(\xi e^{-bt})^{-q/2}\,h_{t}\, e^{-h_{t}z}z^{q/2}I_{q}(2h_{t}e^{-bt/2}\sqrt{\xi z}), $$

where $h_{t}=\frac{2b}{s^{2}(1-e^{-bt})}$ and $q=\frac{2\theta}{\sigma^{2}}-1$. For $(s,\sigma)$ near $(\sigma_{0},\sigma_{0})$, we prove that

$$\begin{aligned} \frac{\partial}{\partial\sigma}\mathbb{E}_{\xi}^{\mathbb{Q}}[(f/\pi)(Z_{T})] &=\frac{\partial}{\partial\sigma}\int_{0}^{\infty}(f/\pi)(z)\,\ell(z;T)\, dz \\ &=\int_{0}^{\infty}(f/\pi)(z)\,\frac{\partial}{\partial\sigma}\ell(z;T)\,dz. \end{aligned}$$

(D.18)

To prove the interchangeability of differentiation and integration in the above equality, it suffices to show that for $(s,\sigma)$ near $(\sigma_{0},\sigma_{0})$ and for all $z>0$,

$$ \left|(f/\pi)(z)\frac{\partial}{\partial\sigma}\ell(z;t)\right|\leq Ce^{-\frac{1}{\sigma_{0}^{2}}\sqrt{a^{2}+2\sigma_{0}^{2}}\,z}=: G(z) $$

for a positive constant $C$ because the function $G(z)$ is integrable over $(0,\infty)$. Let us estimate how fast the function $|(f/\pi)(z)\frac{\partial}{\partial\sigma}\ell(z;t)|$ grows as $z \to\infty$. Given $\sigma>0$ and large $t>0$, for $s$ near $\sigma_{0}$, each term of $\frac{\partial}{\partial\sigma}\ell(z;t)$ is dominated by one of

$$\begin{aligned} &\ell(z;t), \quad z\ell(z;t), \quad\ln(z)\ell(z;t),\quad z^{q/2}e^{-h_{t}z+2h_{t}e^{-bt/2}\sqrt{\xi z}}, \\ &z^{q+1}e^{-h_{t}z+2h_{t}e^{-bt/2}\sqrt{\xi z}} \end{aligned}$$

(D.19)

up to constant multiples. In the calculation of $\frac{\partial}{\partial\sigma}\ell(z;t)$, we use the upper bound of $I_{q}$ given in (D.3) and the equality

$$ \frac{\partial}{\partial q}I_{q}(z)=I_{q}(z)\ln(z/2)+\frac{\Gamma '(q+1/2)}{\Gamma(q+1/2)}I_{q}(z)+\int_{0}^{\pi}\big(e^{z\cos u}\sin^{2q}u\ln (\sin^{2} u)\big)\,du. $$

Let $x=\sin^{2}u$ for $u\in[0,\pi]$. Then for $x\in[0,1]$, it is easy to check that the range of $x^{q}\ln x$ is $[-1/(qe),0]$. Thus we get

$$ -\frac{\pi}{q} e^{z-1} \leq\int_{0}^{\pi}\big(e^{z\cos u}\sin^{2q}u\ln(\sin^{2} u)\big)\,du\leq 0. $$

The growth of each term in (D.19) is essentially dominated by $e^{-\frac{2b}{s^{2}}z}$ up to polynomial multiples. Thus the growth rate of $|(f/\pi)(z)\frac{\partial}{\partial\sigma}\ell (z;t)|$ is less than or equal to that of $e^{(\eta(s)-\frac {2b}{s^{2}})z}$ up to polynomial multiples since the growth of $f(z)$ is at most polynomial. From the argument in (D.17), we obtain the desired result. Since (D.18) holds, we can directly derive that the partial derivative $\frac{\partial}{\partial\sigma}\mathbb {E}_{q}^{\mathbb{Q}}[(G/\Pi)(U_{T})]=\frac{\partial}{\partial\sigma}\mathbb {E}_{\xi}^{\mathbb{Q}}[(f/\pi)(Z_{T})]$ is continuous in $(s,\sigma)$ in a neighborhood of $(\sigma_{0},\sigma_{0})$. □

Appendix E: The quadratic-term structure model

5.1 E.1 Hansen–Scheinkman decomposition

First, observe that $(b,\sigma,r,f)$ and $\xi$ satisfy Assumptions 4.1 and 4.2 (that is, Assumptions 1.1–1.3 and 2.1–2.4). Assumptions 1.1 and 2.1–2.3 can be confirmed from [23, Sect. 6.2], and the other conditions are trivial. The notations $X$, $\mathcal{P}$, ℒ, $M$, ℚ, $(\lambda,\phi)$, $\varphi$, $\nu$ are self-explanatory. The recurrent eigenpair is given in (5.4). The ℚ-dynamics of $X$ is

$$ dX_{t}=\big(b-au+(A-2aV)X_{t}\big)\,dt+\sigma\,dW_{t}, $$

where $W$ is a ℚ-Brownian motion.

5.2 E.2 Sensitivity of $\xi$

We want to find the long-term sensitivity of the expectation $p_{T}$ with respect to the initial value $\xi$. The aim is to show that $\lim_{T\rightarrow\infty} \nabla_{\xi}\ln p_{T}=\frac{\nabla_{\xi}\,\phi(\xi )}{\phi(\xi)} =-u-2V\xi$ by applying Proposition 3.2. The first variation process $Y$ is given by $dY_{t}\!=(A\!-2aV)Y_{t}\,dt$ with $Y_{0}=I_{d}$, where $I_{d}$ is the $d\times d$ identity matrix. It follows that

$$ \mathbb{E}_{\xi}^{\mathbb{Q}}[|\!|Y_{T}|\!|^{2}]=|\!|Y_{T}|\!|^{2}=|\! |e^{(A-2aV)T}|\!|^{2}. $$

Because all eigenvalues of $A-2aV$ have negative real parts, it follows that $\mathbb{E}_{\xi}^{\mathbb{Q}}[|\!|Y_{T}|\!|^{2}]$ is bounded uniformly in $T$ on $[0,\infty)$. This gives the desired result.

5.3 E.3 Sensitivity of $b$

We perform a sensitivity analysis of the expectation $p_{T}$ with respect to the drift coefficients $b=(b_{1},b_{2},\dots,b_{d})^{\top}$. Fix $i=1,2,\dots,d$. The parameter $b_{i}$ can be regarded as a perturbation parameter. The goal is to show by applying Corollary 4.11 that $\lim_{T\rightarrow\infty}\frac{1}{T}\frac{\partial}{\partial b_{i}}\ln p_{T}=-\frac{\partial\lambda}{\partial b_{i}}$. Assumption 4.6 is easy to confirm from (5.4) and the fact that $f$ is a bounded function with bounded support. We now apply Theorem 4.8. Define

$$ k(x)=\sigma^{-1}b-\sigma^{\top}u+(\sigma^{-1}A-2\sigma^{\top}V)x $$

(E.1)

and let $g(x)=C$ be a constant function for a sufficiently large $C>0$ such that $|\frac{\partial}{\partial b_{i}}k(x)| \leq|(\sigma^{-1})_{i}|< C=g(x)$ for $i=1,2,\dots,d$, where $(\sigma^{-1})_{i}$ is the $i$th column of $\sigma^{-1}$. Then the functions $k$ and $g$ satisfy (4.6) and (4.7) in Sect. 4.1. Because $g$ is a constant function, (i) and (ii) of Theorem 4.8 are trivially satisfied with $p=q=2$. We now consider (iii) of Theorem 4.8. As a function of two variables $(x,b_{i})$, we write the function $\phi (x)$ as $\phi(x,b_{i})$. Since $f$ has bounded support, we choose a compact set $K$ such that $\mbox{supp}(f)\subseteq K$. For a bounded open neighborhood $I\subseteq\mathbb{R}$ of 0, define $\overline{I}_{b_{i}}:=\{b_{i}+r\in\mathbb{R}:r\in\overline{I}\}$. Since $\phi$ is a positive and continuous function in the two variables $(x,b_{i})$, its reciprocal $1/\phi$ has a positive maximum on the compact set $K\times\overline{I}_{b_{i}}$. We define

$$ M:=\max_{(x,z)\in K\times\overline{I}_{b_{i}} }\frac{1}{\phi(x,z)}, \qquad\psi(x):=Mf(x). $$

(E.2)

Then (4.8) is satisfied. With this function $\psi$, it is easy to check (iii) because $f$ is a bounded function and supp$(f)\subseteq K$.

5.4 E.4 Sensitivity of $A$

We investigate the long-term sensitivity of the expectation $p_{T}$ with respect to the matrix $A=(A_{ij})_{1\leq i,j\leq d}$. The parameter $A_{ij}$ can be regarded as a perturbation parameter. The goal is to show that $\lim_{T\rightarrow\infty}\frac{1}{T}\frac{\partial}{\partial A_{ij}}\ln p_{T}=-\frac{\partial\lambda}{\partial A_{ij}}$ by applying Corollary 4.11. For condition (i) in Theorem 4.3, it is sufficient to check that $V$ (and hence $u$) is continuously differentiable in $A_{ij}$. Here, the continuous differentiability of a matrix and a vector means that all components are continuously differentiable. The continuous differentiability of $V$ is from [25, Eq. (2.5)] and [24, Theorem 3.1]. Condition (ii) in Theorem 4.3 is easy to check because $f$ is a bounded function with bounded support.

We now apply Theorem 4.10. From the definition of $k$ in (E.1), because $V$ and $u$ are continuously differentiable in $A_{ij}$, there exist sufficiently large constants $c_{1}$ and $c_{2}$ such that

$$ \left|\frac{\partial}{\partial A_{ij}}k(x)\right|\leq c_{1}+c_{2}|x|=:g(x). $$

Then the functions $k$ and $g$ satisfy (4.6) and (4.7) in Sect. 4.1. To check condition (i) in Theorem 4.10, it suffices to show that there exists a positive $\epsilon_{0}$ such that $\mathbb{E}_{\xi}^{\mathbb{Q}}[e^{\epsilon_{0}|X_{T}|^{2}}]$ is bounded uniformly in $T$ on $[0,\infty)$. Consider the density function of $X_{T}$, which is a multivariate normal random variable. We have

$$ \mathbb{E}_{\xi}^{\mathbb{Q}}\big[e^{\epsilon_{0}|X_{T}|^{2}}\big] =\frac{1}{\sqrt{(2\pi)^{d}\det\Sigma_{T}}}\int_{\mathbb{R}^{d}} e^{\epsilon_{0} |z|^{2}-\frac{1}{2}(z-\mu_{T})^{\top}\Sigma_{T}^{-1}(z-\mu_{T})}\,dz, $$

where $\mu_{T}$ and $\Sigma_{T}$ are the mean vector and the covariance matrix of $X_{T}$, respectively. Observe the exponent $\epsilon_{0} |z|^{2}-\frac{1}{2}(z-\mu_{T})^{\top}\Sigma_{T}^{-1}(z-\mu_{T})$ of the integrand. Under the recurrent eigenmeasure ℚ, because Assumption 2.3 is satisfied, the distribution of $X_{T}$ converges to an invariant distribution which is a non-degenerate multivariate normal distribution. Let $\Sigma_{\infty}$ be the covariance matrix of the invariant distribution. Choose $\epsilon_{0}$ less than the smallest eigenvalue of $\Sigma_{\infty}^{-1}$; then the above integral converges to a constant as $T\rightarrow\infty$, which implies condition (i). Condition (ii) in Theorem 4.10 can also be checked by the method in (E.2).

5.5 E.5 Sensitivity of $\sigma$

This section investigates the sensitivity of the expectation $p_{T}$ with respect to the volatility matrix $\sigma=(\sigma_{ij})_{1\leq i,j\leq d}$. Assume that $f$ is continuously differentiable with compact support. It can be shown that $\lim_{T\rightarrow\infty}\frac{1}{T}\frac{\partial}{\partial\sigma _{i}}\ln p_{T}=-\frac{\,\partial\lambda\,}{\partial\sigma_{i}}$ by using Theorems 4.3 and 4.16. We check only the hypothesis of Theorem 4.16 because the other conditions are easy to prove. The corresponding variation process $Z=(Z_{t})_{t\geq0}$ is given by

$$ dZ_{t}=(A-2aV)Z_{t}\,dt+\sigma\,dW_{t}. $$

It follows that $\mathbb{E}_{\xi}^{\mathbb{Q}}[|Z_{T}|]$ is convergent as $T\rightarrow\infty$ because the process $Z$ is an Ornstein–Uhlenbeck (OU) process and all eigenvalues of $A-2aV$ have negative real parts. This gives the desired result.

Appendix F: The $3/2$ model

The aim of this section is to prove the sensitivities discussed in Sect. 6.2. For the sensitivity of $\xi$, (6.3) is obtained by showing $\lim_{T\rightarrow\infty}\frac{\partial}{\partial\xi}\ln q_{T} = \frac{\phi'(\xi)}{\phi(\xi)} =-\ell\xi^{-1}$, where $q_{T}$ is the expectation in (6.1). The reciprocal $Y:=1/X$ satisfies

$$ dY_{t}=\big(a+\sigma^{2}(\ell+1)-\theta Y_{t}\big)\,dt-\sigma\sqrt{Y_{t}}\,dW_{t}, $$

which is a CIR model, and therefore we can use the results of Sect. D.2. By Theorem 3.1, it is enough to show that the expectation $\mathbb{E}_{\xi}^{\mathbb{Q}}[(f/\phi)(X_{T})]=\mathbb{E}_{\xi}^{\mathbb {Q}}[Y_{T}^{-\alpha\beta-\ell}]$ is continuously differentiable in $\xi$ and that

$$ \lim_{T\rightarrow\infty}\frac{\partial}{\partial\xi}\mathbb{E}_{\xi }^{\mathbb{Q}}[(f/\phi)(X_{T})] =\lim_{T\rightarrow\infty}\frac{\partial}{\partial\xi}\mathbb{E}_{\xi }^{\mathbb{Q}}[Y_{T}^{-\alpha\beta-\ell}]=0. $$

This can be proved by the method in (D.5).

For the sensitivity of $\theta$, Corollary 4.11 with Theorem 4.8 will be used to show $\lim_{T\rightarrow\infty}\frac{1}{T}\frac{\partial}{\partial\theta }\ln p_{T}=-\frac{\partial\lambda}{\partial\theta} =-\ell$. We only show the conditions of Theorem 4.8 because the other conditions are easily checked. From $k(x)=\frac{\theta}{\sigma\sqrt{x}}-(\frac{\,a\,}{\sigma}+\sigma\ell )\sqrt{x}$, we define $g(x):=\frac{1}{\sigma\sqrt{x}}$. Condition (i) is evident since $1/X$ is a CIR process. Consider (ii) and (iii) with $q=1+\epsilon$ for a sufficiently small $\epsilon>0$. Observe that for any $n\in\mathbb{N}$, the expectation $\mathbb{E}_{\xi}^{\mathbb{Q}} [(1/X_{T})^{n}]$ converges to a constant as $T\rightarrow\infty$ since $1/X$ is a CIR process. This proves (ii) in Theorem 4.8. For (iii), since $f$ and $\phi$ are independent of the parameter $\theta$, we define $\psi(x)$ as $f(x)/\phi(x)$. For a sufficiently small positive number $\epsilon$, it is easy to show that the expectation $\mathbb{E}_{\xi}^{\mathbb{Q}}[\psi^{1+\epsilon}(X_{T})]=\mathbb{E}_{\xi}^{\mathbb{Q}}[X_{T}^{(1+\epsilon)(\alpha\beta+\ell)}]$ converges as $T\rightarrow\infty$ by considering the density function of the CIR process $1/X$.

For the sensitivity of $a$, the goal is to prove (6.4) by using Corollary 4.11 and Theorem 4.8. We only check condition (ii) in Theorem 4.3 because the other conditions are easy to prove. Define $Y:=1/X$ so that $Y$ is a CIR process. Let $h(y;t)$ be the density function of $Y_{t}$. We temporarily employ a new parameter $b$ to distinguish the parameter $a$ in $f/\phi$ from the parameter $a$ in the drift of $X$. Define

$$ \ell_{b}:=\sqrt{\bigg(\frac{1}{2}+\frac{b}{\sigma^{2}}\bigg)^{2}+\alpha\beta (\beta-1)}-\bigg(\frac{1}{2}+\frac{b}{\sigma^{2}}\bigg), \qquad\pi _{b}(x):=x^{-\ell_{b}}, $$

so that $\ell_{a}=\ell$ and $\pi_{a}(x)=x^{-\ell}=\phi(x)$ for the constant $\ell$ in (6.2). First we show that the partial derivative $\frac{\partial}{\partial b}\mathbb{E}_{\xi}^{\mathbb{Q}}[(f/\pi_{b})(X_{T})]$ exists and that

$$ \frac{\partial}{\partial b}\mathbb{E}_{\xi}^{\mathbb{Q}}[(f/\pi_{b})(X_{T})] =\mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[\frac{\partial}{\partial b}(f/\pi _{b})(X_{T})\bigg]. $$

The proof is obtained from Theorem G.1 by defining the dominating function $g$ as $g(x)=c_{1}x^{\alpha\beta+\ell+1}+ c_{2}x^{\alpha \beta+\ell-1}$ for sufficiently large constants $c_{1}$ and $c_{2}$, since

$$ \left|\frac{\partial}{\partial b}(f/\pi_{b})(x)\right|=\left|\frac {\partial\ell_{b}}{\partial b}x^{\alpha\beta+\ell_{b}}\ln x\right|\leq c_{1}x^{\alpha\beta+\ell+1}+ c_{2}x^{\alpha\beta+\ell-1}=g(x) $$

for all $b$ in a small open neighborhood of $a$. The expectation

$$ \mathbb{E}_{\xi}^{\mathbb{Q}}[g(X_{T})]=c_{1}\mathbb{E}_{\xi}^{\mathbb {Q}}[Y_{t}^{-\alpha\beta-\ell-1}]+c_{2}\mathbb{E}_{\xi}^{\mathbb {Q}}[Y_{t}^{-\alpha\beta-\ell+1}] $$

is finite when $\frac{a}{\sigma^{2}}+1-\alpha\beta>0$ because the growth of the density $h(y,t)$ is dominated by $e^{-\frac{2\theta}{\sigma^{2} }y}$ as $y\rightarrow\infty$ and is dominated by $y^{\frac{2a}{\sigma ^{2}}+2\ell+1}$ as $y\rightarrow0+$. The joint continuity of $\frac{\partial}{\partial b}\mathbb{E}_{\xi}^{\mathbb{Q}}[(f/\pi_{b})(X_{T})]$ in the two variables $(b,a)$ can be obtained from the joint continuity of $h(y;T)$ and the equality

$$\begin{aligned} \frac{\partial}{\partial b}\mathbb{E}_{\xi}^{\mathbb{Q}}[(f/\pi_{b})(X_{T})] &=\mathbb{E}_{\xi}^{\mathbb{Q}}\bigg[\frac{\partial}{\partial b}(f/\pi _{b})(X_{T})\bigg] =\frac{\partial\ell_{b}}{\partial b}\mathbb{E}_{\xi}^{\mathbb{Q}}[X_{T}^{\alpha \beta+\ell_{b}}\ln X_{T}]\\ &=-\frac{\partial\ell_{b}}{\partial b}\mathbb{E}_{\xi}^{\mathbb{Q}}[Y_{T}^{-\alpha\beta-\ell_{b}}\ln Y_{T}]\\ &=-\frac{\partial\ell_{b}}{\partial b}\int_{0}^{\infty}h(y;T) y^{-\alpha\beta -\ell_{b}}\ln y \,dy. \end{aligned}$$

It is easy to check that $\mathbb{E}_{\xi}^{\mathbb{Q}}[g(X_{T})]$ is convergent as $T\to\infty$, which gives (4.3).

For the sensitivity of $\sigma$, consider a quadruple $(\delta,1,R,F)$ and an initial value $\zeta$ defined by the Lamperti transformation in Sect. 4.2.1. Defining $u(x) =\frac{2}{\sigma\sqrt{x}}$ for $x>0$, we get

$$\begin{aligned} \delta(u)&=\bigg(\frac{2a}{\sigma^{2}}+2\ell+\frac{3}{2}\bigg)\frac {1}{u}-\frac{\theta u}{2} ,\quad R(u)=2\alpha\beta(1-\beta)u^{2},\\ F(u)&=(\sigma u/2)^{-2\alpha\beta}, \quad\zeta=\frac{2}{\sigma\sqrt{\xi}}. \end{aligned}$$

By Proposition 4.12, the quadruple $(\delta,1,R,F)$ and $\zeta$ satisfy Assumptions 4.1 and 4.2 because the quadruple $(b,\sigma,r,f)$ and the initial value $\xi$ also satisfy them. One can show (6.5) by using Theorem 4.13. Conditions (i) and (iii) can be proved by the methods in Propositions D.3 and D.4, and condition (ii) can be shown by Theorem 4.8 when $\frac{a}{\sigma^{2}}+1-\alpha\beta>0$ by applying the same method as used in the sensitivity analysis of $a$.

Appendix G: Perturbation of payoff function

In this section, we are interested in the partial derivative $\frac {\partial}{\partial\epsilon}\mathbb{E}[h_{\epsilon}(X)]$ of an expectation. The interchangeability of differentiation and expectation is an important issue in this paper. The following theorem is a well known fact, and it is noteworthy because this theorem is useful for checking condition (ii) in Theorem 4.3 when the initial value $\xi$ is not perturbed.

Theorem G.1

Let $X$ be a random variable and $h_{\epsilon}(x)$ a function of two variables $(\epsilon,x)$ on $I\times\mathbb{R}^{d}$, where $I$ is an open neighborhood of 0. Assume that $\mathbb{E}[h_{\epsilon}(X)]<\infty$ for each $\epsilon$ in $I$ and that $h_{\epsilon}(x)$ is continuously differentiable in $\epsilon$ on $I$ for each $x$. Suppose that there exists a positive function $g$, which is called a dominating function, such that $\mathbb{E}[g(X)]<\infty$ and

$$ \left|\frac{\partial}{\partial\epsilon}\,h_{\epsilon}(x)\right|\leq g(x)\qquad\mbox{on } I\times\mathbb{R}^{d}. $$

Then the expectation $\mathbb{E}[h_{\epsilon}(X)]$ is continuously differentiable in $\epsilon$ on $I$, and

$$ \frac{\partial}{\partial\epsilon}\mathbb{E}[h_{\epsilon}(X)]= \mathbb{E}\bigg[\frac{\partial}{\partial\epsilon} h_{\epsilon}(X)\bigg]. $$