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Convergence rates of large-time sensitivities with the Hansen–Scheinkman decomposition

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Abstract

This paper investigates the large-time asymptotic behavior of the sensitivities of cash flows. In quantitative finance, the price of a cash flow is expressed in terms of a pricing operator of a Markov diffusion process. We study the extent to which the pricing operator is affected by small changes of the underlying Markov diffusion. The main idea is a partial differential equation (PDE) representation of the pricing operator by incorporating the Hansen–Scheinkman decomposition method. The sensitivities of the cash flows and their large-time convergence rates can be represented via simple expressions in terms of eigenvalues and eigenfunctions of the pricing operator. Furthermore, compared to the work of Park (Finance Stoch 4:773–825, 2018), more detailed convergence rates are provided. In addition, we discuss the application of our results to three practical problems: utility maximization, entropic risk measures, and bond prices. Finally, as examples, explicit results for several market models such as the Cox–Ingersoll–Ross (CIR) model, 3/2 model and constant elasticity of variance (CEV) model are presented.

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Acknowledgements

The author sincerely appreciates the valuable suggestions received from the Editor-in-Chief, the Associate Editor and the anonymous referees for their helpful comments and insights, which have greatly improved the quality of the paper. This research was supported by the National Research Foundation of Korea (NRF) grants funded by the Ministry of Science and ICT (Nos. 2018R1C1B5085491, 2017R1A5A1015626 and 2021R1C1C1011675) and the Ministry of Education (No. 2019R1A6A1A10073437) through Basic Science Research Program.

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Appendices

A Proofs of the theorems in Sect. 3.1

We provide the proof of Theorem 1.

Proof

By applying the Feynman–Kac formula to Eq. (5), it follows that

$$\begin{aligned} -f_t+\frac{1}{2}\sigma ^2(x)f_{xx}+\kappa (x)f_x=0,\;f(0,x)=(h/\phi )(x). \end{aligned}$$

Observe that f is \(C^{1,2}\) from Assumption 5. Since every coefficient is continuously differentiable in x in the above PDE, the partial derivative \(f_x\) is also \(C^{1,2}\) by [6, Section 3.5]. Differentiating in x,  we get

$$\begin{aligned} -f_{xt}+\frac{1}{2}\sigma ^2(x)f_{xxx}+(\kappa +\sigma '\sigma )(x)f_{xx}+\kappa '(x)f_x=0,\;f_x(0,x)=(h/\phi )'(x). \end{aligned}$$
(45)

Meanwhile, since \((\kappa +\sigma '\sigma ,\sigma ,-\kappa ',(h/\phi )')\) satisfies Assumptions 15 on the consistent probability space \((\varOmega ,{\mathcal {F}},({\mathcal {F}}_t)_{t\ge 0},(\hat{{\mathbb {P}}}_t)_{t\ge 0})\) having Brownian motion \(({\hat{B}}_t)_{t\ge 0},\) we can construct the corresponding objects

$$\begin{aligned} {\hat{X}},\,\hat{{\mathcal {P}}},\,({\hat{\lambda }},{\hat{\phi }}),\,{\hat{M}},\,{\hat{f}},\,(\tilde{{\mathbb {P}}}_t)_{t\ge 0},({\tilde{B}}_t)_{t\ge 0} \end{aligned}$$

and these notations are self-explanatory. Since the process \({\hat{X}}\) satisfies

$$\begin{aligned} d{\hat{X}}_t=(\kappa +\sigma '\sigma )({\hat{X}}_t)\,dt+\sigma ({\hat{X}}_t)\,d{\hat{B}}_t, \end{aligned}$$

by applying the Itô formula to \(f_x(T-t,{\hat{X}}_t)e^{\int _0^t \kappa '(X_s)\,ds},\) we have

$$\begin{aligned} \begin{aligned}&d\big (f_x(T-t,{\hat{X}}_t)e^{\int _0^t \kappa '({\hat{X}}_s)\,ds}\big )\\ =&\;e^{\int _0^t \kappa '({\hat{X}}_s)\,ds}\Big (-f_{xt}+\frac{1}{2}\sigma ^2({\hat{X}}_t)f_{xxx}+(\kappa +\sigma '\sigma )({\hat{X}}_t)f_{xx}+\kappa '({\hat{X}}_s)f_x\Big )\,dt +e^{\int _0^t \kappa '({\hat{X}}_s)\,ds}\sigma ({\hat{X}}_t) f_{xx}\,d{\hat{B}}_t\\ =&\;e^{\int _0^t \kappa '({\hat{X}}_s)\,ds}\sigma ({\hat{X}}_t) f_{xx}\,d{\hat{B}}_t. \end{aligned} \end{aligned}$$
(46)

Thus, for each \(T>0\), the process \((f_x(T-t,{\hat{X}}_t)e^{\int _0^t \kappa '({\hat{X}}_s)\,ds})_{0\le t\le T}\) is a local martingale under the probability measure \(\hat{{\mathbb {P}}}_T.\)

Suppose that \(f_x\) satisfies

$$\begin{aligned} f_x(T,x)={\mathbb {E}}^{\hat{{\mathbb {P}}}}[ e^{\int _0^T \kappa '({\hat{X}}_s)\,ds}(h/\phi )'({\hat{X}}_T)|{\hat{X}}_0=x]. \end{aligned}$$

Note that this equality holds if \((f_x(T-t,{\hat{X}}_t)e^{\int _0^t \kappa '({\hat{X}}_s)\,ds})_{0\le t\le T}\) is a martingale since

$$\begin{aligned} f_x(T,x)={\mathbb {E}}^{\hat{{\mathbb {P}}}}[ e^{\int _0^T \kappa '({\hat{X}}_s)\,ds}f_x(0,{\hat{X}}_T)|{\hat{X}}_0=x]={\mathbb {E}}^{\hat{{\mathbb {P}}}}[ e^{\int _0^T \kappa '({\hat{X}}_s)\,ds}(h/\phi )'({\hat{X}}_T)|{\hat{X}}_0=x]. \end{aligned}$$
(47)

Replacing T by t,  it follows that

$$\begin{aligned} f_x(t,x)={\mathbb {E}}^{\hat{{\mathbb {P}}}}\big [e^{\int _0^t\kappa '({\hat{X}}_s)\,ds}(h/\phi )'({\hat{X}}_t)\big |{\hat{X}}_0=x\big ]=\hat{{\mathcal {P}}}_t(h/\phi )'(x). \end{aligned}$$

Now, we apply the Hansen–Scheinkman decomposition. Since \(({\hat{\lambda }},{\hat{\phi }})\) is the recurrent eigenpair and

$$\begin{aligned} {\hat{f}}(t,x)={\mathbb {E}}_x^{\hat{{\mathbb {P}}}}[{\hat{M}}_t(h/\phi )'({\hat{X}}_t)]={\mathbb {E}}_x^{\tilde{{\mathbb {P}}}}[((h/\phi )'/{\hat{\phi }})({\hat{X}}_t)] \end{aligned}$$

is the remainder function, we have

$$\begin{aligned} \begin{aligned} f_x(t,x)&={\hat{f}}(t,x)e^{-{\hat{\lambda }} t}{\hat{\phi }}(x). \end{aligned} \end{aligned}$$
(48)

Since f(tx) and \({\hat{f}}(t,x)\) converge to nonzero constants as \(t\rightarrow \infty ,\) Eq. (6) implies that

$$\begin{aligned} \left| \frac{\partial _\xi p_T}{p_T} -\frac{\phi '(\xi )}{\phi (\xi )}\right| =\left| \frac{f_x(T,\xi )}{f(T,\xi )}\right| =\left| \frac{{\hat{f}}(T,\xi )}{f(T,\xi )}\right| e^{-{\hat{\lambda }} T}{\hat{\phi }}(\xi )\le ce^{-{\hat{\lambda }} T} \end{aligned}$$
(49)

for some positive constant c,  which is dependent on \(\xi \) but independent of T. \(\square \)

Now we describe the proof of Proposition 1.

Proof

Recall that \((\varOmega ,{\mathcal {F}},({\mathcal {F}}_t)_{t\ge 0},(\hat{{\mathbb {P}}}_t)_{t\ge 0})\) is a consistent probability space having Brownian motion \(({\hat{B}}_t)_{t\ge 0}\) and the process \(({\hat{X}}_t)_{t\ge 0}\) satisfies Eq. (7). One can show that

$$\begin{aligned} \frac{\phi ({\hat{X}}_0)}{\phi ({\hat{X}}_t)}e^{\int _0^t\frac{\hat{{\mathcal {L}}}\phi ({\hat{X}}_s)}{\phi ({\hat{X}}_s)}\,ds},\;0\le t\le T \end{aligned}$$

is a \(\hat{{\mathbb {P}}}_T\)-local martingale by applying the Itô formula and checking that the dt-term vanishes. Since this process is assumed to be a martingale, we can define a new measure \({\mathbb {Q}}_T\) on \({\mathcal {F}}_T\) as

$$\begin{aligned} \frac{d{\mathbb {Q}}_T}{d\hat{{\mathbb {P}}}_T}=\frac{\phi ({\hat{X}}_0)}{\phi ({\hat{X}}_T)}e^{\int _0^T\frac{\hat{{\mathcal {L}}}\phi ({\hat{X}}_s)}{\phi ({\hat{X}}_s)}\,ds}. \end{aligned}$$

It is easy to check that the family \(({\mathbb {Q}}_t)_{t>0}\) of probability measures is consistent and the process

$$\begin{aligned} B_t^{\mathbb {Q}}:={\hat{B}}_t+(\sigma \phi '/\phi )({\hat{X}}_t),\;t\ge 0 \end{aligned}$$

is a Brownian motion on the consistent probability space \((\varOmega ,{\mathcal {F}},({\mathcal {F}}_t)_{t\ge 0},({\mathbb {Q}}_t)_{t\ge 0}).\) The process \({\hat{X}}\) satisfies

$$\begin{aligned} \begin{aligned} d{\hat{X}}_t&=(\kappa +\sigma '\sigma )({\hat{X}}_t)\,dt+\sigma ({\hat{X}}_t)\,d{\hat{B}}_t\\&=(\kappa +\sigma '\sigma -\sigma ^2\phi '/\phi )({\hat{X}}_t)\,dt+\sigma ({\hat{X}}_t)\,d{B}_t^{{\mathbb {Q}}} . \end{aligned} \end{aligned}$$

Note that \(\hat{{\mathcal {L}}}\) is the generator of \({\hat{X}}\) under the family of probability measures \(({\mathbb {Q}}_t)_{t\ge 0}.\)

Define

$$\begin{aligned} g(t,x):=f_x(t,x)\phi (x). \end{aligned}$$

Then, Eq. (45) gives

$$\begin{aligned} \begin{aligned}&-g_t+\frac{1}{2}\sigma ^2(x)g_{xx}+(\kappa +\sigma '\sigma -\sigma ^2\phi '/\phi )(x)g_x+(\kappa '-{(\hat{{\mathcal {L}}}\phi )}/{\phi })(x)g=0\\&g(0,x)=(h/\phi )'(x)\phi (x)=(h'-h\phi '/\phi )(x). \end{aligned} \end{aligned}$$

The Feynman–Kac formula (Remark 4) states that

$$\begin{aligned} \begin{aligned} f_x(t,x)\phi (x)=g(t,x)&={\mathbb {E}}^{{{\mathbb {Q}}}}\big [ e^{\int _0^t(\kappa '-\frac{\hat{{\mathcal {L}}}\phi }{\phi })({\hat{X}}_s)\,ds}\phi ({\hat{X}}_{t})(h/\phi )'({\hat{X}}_t)\big |{\hat{X}}_0=x\big ]\\&={\mathbb {E}}^{{{\mathbb {Q}}}}\Big [ e^{-\int _0^t\frac{\hat{{\mathcal {L}}}\phi }{\phi }({\hat{X}}_s)\,ds}\,\frac{\phi ({\hat{X}}_{t})}{\phi ({\hat{X}}_{0})} e^{\int _0^t\kappa '({\hat{X}}_s)\,ds}(h/\phi )'({\hat{X}}_t)\Big |{\hat{X}}_0=x\Big ]\phi (x)\\&={\mathbb {E}}^{{{\mathbb {Q}}}}\Big [ \frac{d\hat{{\mathbb {P}}}_t}{d{\mathbb {Q}}_t} e^{\int _0^t\kappa '({\hat{X}}_s)\,ds}(h/\phi )'({\hat{X}}_t)\Big |{\hat{X}}_0=x\Big ]\phi (x)\\&={\mathbb {E}}^{\hat{{\mathbb {P}}}}\big [ e^{\int _0^t\kappa '({\hat{X}}_s)\,ds}(h/\phi )'({\hat{X}}_t)\big |{\hat{X}}_0=x\big ]\phi (x), \end{aligned} \end{aligned}$$

which implies that

$$\begin{aligned} f_x(t,x)={\mathbb {E}}^{\hat{{\mathbb {P}}}}\big [e^{\int _0^t\kappa '({\hat{X}}_s)\,ds}(h/\phi )'({\hat{X}}_t)\big |{\hat{X}}_0=x\big ]. \end{aligned}$$

From the time-homogeneous Markov property,

$$\begin{aligned} \begin{aligned} f_x(T-t,x)&={\mathbb {E}}^{\hat{{\mathbb {P}}}}\big [e^{\int _0^{T-t}\kappa '({\hat{X}}_s)\,ds}(h/\phi )'({\hat{X}}_{T-t})\big |{\hat{X}}_0=x\big ]\\&={\mathbb {E}}^{\hat{{\mathbb {P}}}}\big [e^{\int _t^T\kappa '({\hat{X}}_s)\,ds}(h/\phi )'({\hat{X}}_T)\big |{\hat{X}}_t=x\big ] \end{aligned} \end{aligned}$$

so

$$\begin{aligned} f_x(T-t,{\hat{X}}_t)e^{\int _0^t \kappa '({\hat{X}}_s)\,ds} ={\mathbb {E}}^{\hat{{\mathbb {P}}}}\big [e^{\int _0^T\kappa '({\hat{X}}_s)\,ds}(h/\phi )'({\hat{X}}_T)\big |{\mathcal {F}}_t\big ]. \end{aligned}$$

In conclusion, the process \((f_x(T-t,{\hat{X}}_t)e^{\int _0^t \kappa '({\hat{X}}_s)\,ds})_{0\le t\le T}\) is a martingale. \(\square \)

The proof of Remark 5 is as follows.

Proof

Define an operator \(\overline{{\mathcal {L}}}\) as

$$\begin{aligned} \begin{aligned} \overline{{\mathcal {L}}}h(x)&=\hat{{\mathcal {L}}}h(x)-\frac{\hat{{\mathcal {L}}}\phi (x)}{\phi (x)}h(x)\\&=\frac{1}{2}\sigma ^2(x)h''(x)+(\kappa +\sigma '\sigma -\sigma ^2\phi '/\phi )(x)h'(x)-\frac{\hat{{\mathcal {L}}}\phi (x)}{\phi (x)}h(x),\;h\in C^2({\mathcal {D}}) \end{aligned} \end{aligned}$$

for \(\hat{{\mathcal {L}}}\) in Eq. (12). It is clear that \(\overline{{\mathcal {L}}}\phi =0.\) By Theorem 5.1.8 of [24], Eq. (13) is a martingale under the probability measure \(\hat{{\mathbb {P}}}_T\) if and only if

$$\begin{aligned} \begin{aligned}&\int _a^{x_0} \frac{1}{\phi ^2(x)}e^{-\int _{x_0}^x\frac{2}{\sigma ^2(s)}(\kappa +\sigma '\sigma -\sigma ^2\frac{\phi '}{\phi })(s)\,ds}\int _x^{x_0}\frac{\phi ^2(y)}{\sigma ^2(y)} e^{\int _{x_0}^y\frac{2}{\sigma ^2(s)}(\kappa +\sigma '\sigma -\sigma ^2\frac{\phi '}{\phi })(s)\,ds}\,dy\,dx =\infty ,\\&\int _{x_0}^b \frac{1}{\phi ^2(x)}e^{-\int _{x_0}^x\frac{2}{\sigma ^2(s)}(\kappa +\sigma '\sigma -\sigma ^2\frac{\phi '}{\phi })(s)\,ds}\int _{x_0}^{x} \frac{\phi ^2(y)}{\sigma ^2(y)} e^{\int _{x_0}^y\frac{2}{\sigma ^2(s)}(\kappa +\sigma '\sigma -\sigma ^2\frac{\phi '}{\phi })(s)\,ds}\,dy\,dx =\infty , \end{aligned} \end{aligned}$$

where \({\mathcal {D}}=(a,b)\) and \(x_0\) is any reference point in (ab). By simplifying these equations, we obtain the desired result. \(\square \)

The proof of Theorem 2 is as follows.

Proof

Using the same argument as that presented in the proof of Theorem 1, it can be shown that two processes \((f_x(T-t,{\hat{X}}_t)e^{\int _0^t \kappa '({\hat{X}}_s)\,ds})_{0\le t\le T}\) and \(({\hat{f}}_x(T-t,{\tilde{X}}_t)e^{\int _0^t \gamma '({\tilde{X}}_s)\,ds})_{0\le t\le T}\) are local martingales under the probability measures \(\hat{{\mathbb {P}}}_T\) and \(\tilde{{\mathbb {P}}}_T,\) respectively. Thus, we omit the proof.

Now, assume that two functions \(f_x\) and \({\hat{f}}_x\) satisfy Eq. (15). If the two local martingales above are martingales, then Eq. (15) holds by the same argument as that in Eq. (47). Since \((\kappa +\sigma '\sigma ,\sigma ,-\kappa ',(h/\phi )')\) satisfies Assumptions 15 on \((\varOmega ,{\mathcal {F}},({\mathcal {F}}_t),(\hat{{\mathbb {P}}}_t)_{t\ge 0})\) having Brownian motion \(({\hat{B}}_t)_{t\ge 0},\) we can construct the corresponding objects

$$\begin{aligned} {\hat{X}},\,\hat{{\mathcal {P}}},\,({\hat{\lambda }},{\hat{\phi }}),\,{\hat{M}},\,{\hat{f}},\,(\tilde{{\mathbb {P}}}_t)_{t\ge 0},({\tilde{B}}_t)_{t\ge 0} \end{aligned}$$

and these notations are self-explanatory. The remainder function

$$\begin{aligned} {\hat{f}}(t,x)={\mathbb {E}}_x^{\tilde{{\mathbb {P}}}}[((h/\phi )'/{\hat{\phi }})({\hat{X}}_t)] \end{aligned}$$
(50)

satisfies \(f_x(t,x)={\hat{f}}(t,x)e^{-{\hat{\lambda }} t}{\hat{\phi }}(x)\) by Eq. (48), and the dynamics of \({\hat{X}}\) is

$$\begin{aligned} \begin{aligned} d{\hat{X}}_t&=(\kappa +\sigma '\sigma )({\hat{X}}_t)\,dt+\sigma ({\hat{X}}_t)\,d{\hat{B}}_t\\&=\gamma ({\hat{X}}_t)\,dt+\sigma ({\hat{X}}_t)\,d{\tilde{B}}_t \end{aligned} \end{aligned}$$

for \(\gamma =\kappa +\sigma '\sigma +\sigma ^2{\hat{\phi }}'/{\hat{\phi }}.\) Applying the Feynman–Kac formula to Eq. (50), we have

$$\begin{aligned} -{\hat{f}}_t+\frac{1}{2}\sigma ^2(x){\hat{f}}_{xx}+\gamma (x){\hat{f}}_x=0,\;{\hat{f}}(0,x)=((h/\phi )'/{\hat{\phi }})(x). \end{aligned}$$

Observe that the function \({\hat{f}}\) is \(C^{1,2}\) by Assumption 5. Since every coefficient is continuously differentiable in x in the above PDE, the partial derivative \({\hat{f}}_x\) is also \(C^{1,2}\) by [6, Section 3.5]. Differentiating in x,  we get

$$\begin{aligned} -{\hat{f}}_{xt}+\frac{1}{2}\sigma ^2(x){\hat{f}}_{xxx}+(\gamma +\sigma '\sigma )(x){\hat{f}}_{xx}+\gamma '(x){\hat{f}}_x=0,\;{\hat{f}}_x(0,x)=((h/\phi )'/{\hat{\phi }})'(x). \end{aligned}$$

Since \((\gamma +\sigma '\sigma ,\sigma ,-\gamma ',((h/\phi )'/{\hat{\phi }})')\) satisfies Assumptions 15, we can construct the corresponding objects

$$\begin{aligned} {\tilde{X}},\,\tilde{{\mathcal {P}}},\,({\tilde{\lambda }},{\tilde{\phi }}),\,{\tilde{M}},\,{\tilde{f}},\,(\overline{{\mathbb {P}}}_t)_{t\ge 0},\,({\overline{B}}_t)_{t\ge 0}. \end{aligned}$$

Note that the process \({\tilde{X}}\) satisfies

$$\begin{aligned} d{\tilde{X}}_t=(\gamma +\sigma '\sigma )({\tilde{X}}_t)\,dt+\sigma ({\tilde{X}}_t)\,d{\tilde{B}}_t. \end{aligned}$$

The process \(({\hat{f}}_x(T-t,{\tilde{X}}_t)e^{\int _0^t \gamma '({\tilde{X}}_s)\,ds})_{0\le t\le T}\) is assumed to be a martingale; thus,

$$\begin{aligned} {\hat{f}}_x(t,x)={\mathbb {E}}_x^{\tilde{{\mathbb {P}}}}[e^{\int _0^t\gamma '({\tilde{X}}_s)\,ds}((h/\phi )'/{\hat{\phi }})'({\tilde{X}}_t)]=\tilde{{\mathcal {P}}}_T((h/\phi )'/{\hat{\phi }})'(x). \end{aligned}$$

We now apply the Hansen–Scheinkman decomposition. Since \(({\tilde{\lambda }},{\tilde{\phi }})\) is the recurrent eigenpair and

$$\begin{aligned} {\tilde{f}}(t,x)={\mathbb {E}}_x^{\tilde{{\mathbb {P}}}}[{\tilde{M}}_T(((h/\phi )'/{\hat{\phi }})'/{\tilde{\phi }})({\tilde{X}}_t)] ={\mathbb {E}}_x^{\overline{{\mathbb {P}}}}[(((h/\phi )'/{\hat{\phi }})'/{\tilde{\phi }})({\tilde{X}}_t)] \end{aligned}$$

is the remainder function, it follows by Eq. (6) that

$$\begin{aligned} \begin{aligned} {\hat{f}}_x(t,x)&={\tilde{f}}(t,x)e^{-{\tilde{\lambda }} t}{\tilde{\phi }}(x). \end{aligned} \end{aligned}$$

Meanwhile, direct calculation gives

$$\begin{aligned} \begin{aligned} \frac{{\partial _{\xi \xi }} p_T}{p_T}-\Big (\frac{\partial _\xi p_T}{p_T}\Big )^2-\frac{\phi ''(\xi )}{\phi (\xi )}+\Big (\frac{\phi '(\xi )}{\phi (\xi )}\Big )^2 =\frac{f_{xx}(T,\xi )}{f(T,\xi )}-\Big (\frac{f_x(T,\xi )}{f(T,\xi )}\Big )^2. \end{aligned} \end{aligned}$$
(51)

We estimate the two terms on the right-hand side. Eq. (49) states that

$$\begin{aligned} \Big (\frac{f_x(T,\xi )}{f(T,\xi )}\Big )^2\le c_1e^{-2{\hat{\lambda }} T} \end{aligned}$$
(52)

for some positive constant \(c_1.\) To estimate the term \(\frac{f_{xx}(T,\xi )}{f(T,\xi )}\) on the right-hand side of Eq. (51), observe that Eq. (48) gives

$$\begin{aligned} \begin{aligned} f_{xx}(t,x)&= {\hat{f}}_x(t,x)e^{-{\hat{\lambda }} t}{\hat{\phi }}(x)+ {\hat{f}}(t,x)e^{-{\hat{\lambda }} t}{\hat{\phi }}'(x) ={\hat{f}}_x(t,x)e^{-{\hat{\lambda }} t}{\hat{\phi }}(x)+\frac{{\hat{\phi }}'(x)}{{\hat{\phi }}(x)}f_x(t,x). \end{aligned} \end{aligned}$$

Combined with \(\frac{f_x(T,\xi )}{f(T,\xi )}=\frac{\partial _\xi p_T}{p_T}-\frac{\phi '(\xi )}{\phi (\xi )},\) Eq. (51) becomes

$$\begin{aligned} \begin{aligned}&\,\frac{{\partial _{\xi \xi }} p_T}{p_T}-\Big (\frac{\partial _\xi p_T}{p_T}\Big )^2-\frac{\phi ''(\xi )}{\phi (\xi )}+\Big (\frac{\phi '(\xi )}{\phi (\xi )}\Big )^2 -\frac{{\hat{\phi }}'(\xi )}{{\hat{\phi }}(\xi )} \frac{\partial _\xi p_T}{p_T} +\frac{{\hat{\phi }}'(\xi )}{{\hat{\phi }}(\xi )} \frac{\phi '(\xi )}{\phi (\xi )}\\ =&\,\frac{{\hat{f}}_x(T,\xi ){\hat{\phi }}(\xi )e^{-{\hat{\lambda }} T}}{f(T,\xi )}-\Big (\frac{f_\xi (T,\xi )}{f(T,\xi )}\Big )^2\\ =&\,\frac{{\tilde{f}}(T,\xi )e^{-({\tilde{\lambda }}+{\hat{\lambda }}) T}{\tilde{\phi }}(\xi ){\hat{\phi }}(\xi )}{f(T,\xi )}-\Big (\frac{f_\xi (T,\xi )}{f(T,\xi )}\Big )^2. \end{aligned} \end{aligned}$$

Since \(f(T,\xi )\) and \({\tilde{f}}(T,\xi )\) converge to nonzero constants as \(T\rightarrow \infty ,\) combined with Eq. (52), we conclude that

$$\begin{aligned} \left| \frac{{\partial _{\xi \xi }} p_T}{p_T}-\Big (\frac{\partial _\xi p_T}{p_T}\Big )^2-\frac{\phi ''(\xi )}{\phi (\xi )}+\Big (\frac{\phi '(\xi )}{\phi (\xi )}\Big )^2 -\frac{{\hat{\phi }}'(\xi )}{{\hat{\phi }}(\xi )} \frac{\partial _\xi p_T}{p_T} +\frac{{\hat{\phi }}'(\xi )}{{\hat{\phi }}(\xi )} \frac{\phi '(\xi )}{\phi (\xi )}\right| \le c(e^{-{\tilde{\lambda }} T}+e^{-{\hat{\lambda }} T})e^{-{\hat{\lambda }} T} \end{aligned}$$

for some positive constant c,  which is dependent on \(\xi \) but independent of T. In particular, by using Eq. (49), we have

$$\begin{aligned} \left| \frac{{\partial _{\xi \xi }} p_T}{p_T}-\frac{\phi ''(\xi )}{\phi (\xi )} \right| \le c'e^{-{\hat{\lambda }} T} \end{aligned}$$

for some positive constant \(c',\) which is dependent on \(\xi \) but independent of T. \(\square \)

B Proofs of the theorems in Sect. 3.2

The proof of Theorem 3 is as follows.

Proof

Applying the Feynman–Kac formula to Eq. (17), we get

$$\begin{aligned} -f_t^{(\epsilon )}+\frac{1}{2}\sigma ^{2}(x)f_{xx}^{(\epsilon )}+\kappa ^{(\epsilon )}(x)f_x^{(\epsilon )}=0,\;f^{(\epsilon )}(0,x)=(h^{(\epsilon )}/\phi ^{(\epsilon )})(x) \end{aligned}$$

for \(\kappa ^{(\epsilon )}=b^{(\epsilon )}+\sigma ^{2}\phi _x^{(\epsilon )}/\phi ^{(\epsilon )}.\) We differentiate this PDE in \(\epsilon \) and evaluate it at \(\epsilon =0\) (Assumptions 6 and 8). Then

$$\begin{aligned} -f_{\epsilon t}^{(0)}+\frac{1}{2}\sigma ^2(x)f_{\epsilon xx}^{(0)}+\kappa (x)f_{\epsilon x}^{(0)}+\ell (x)f_x^{(0)}=0,\;f_\epsilon ^{(0)}(0,x)=\partial _\epsilon |_{\epsilon =0}(h^{(\epsilon )}/\phi ^{(\epsilon )})(x). \end{aligned}$$

By applying the Itô formula to the process \((f_{\epsilon }^{(0)}(T-t,X_t)+\int _0^t\ell (X_s)f_x(T-s,X_s)\,ds)_{0\le t\le T},\) it follows that

$$\begin{aligned} \begin{aligned}&d\Big (f_{\epsilon }^{(0)}(T-t,{X}_t)+\int _0^t\ell ({X}_s)f_x(T-s,{X_s})\,ds\Big )\\ =&\;\big (-f_{\epsilon t}^{(0)}+\frac{1}{2}\sigma ^2(X_t)f_{\epsilon xx}^{(0)}+\kappa ({X}_t)f_{\epsilon x}^{(0)} +\ell ({X}_t)f_x\big )\,dt+\sigma ({X}_t)f_{\epsilon x}^{(0)} \,d{\hat{B}}_t=\sigma ({X}_t)f_{\epsilon x}^{(0)} \,d{\hat{B}}_t. \end{aligned} \end{aligned}$$
(53)

Thus, the process \((f_{\epsilon }^{(0)}(T-t,X_t)+\int _0^t\ell (X_s)f_x(T-s,X_s)\,ds)_{0\le t\le T}\) is a local martingale under the probability measure \(\hat{{\mathbb {P}}}_T.\)

Now, assume that

$$\begin{aligned} \begin{aligned} f_{\epsilon }^{(0)}(T,\xi ) ={\mathbb {E}}_\xi ^{\hat{{\mathbb {P}}}}\Big [\int _0^T\ell (X_t) f_x(T-t,X_t)\,dt+\partial _\epsilon |_{\epsilon =0}(h^{(\epsilon )}/\phi ^{(\epsilon )})(X_T)\Big ]. \end{aligned} \end{aligned}$$

This equality also holds if the process \((f_{\epsilon }^{(0)}(T-t,X_t)+\int _0^t\ell (X_s)f_x(T-s,X_s)\,ds)_{0\le t\le T}\) is a martingale because

$$\begin{aligned} \begin{aligned} f_{\epsilon }^{(0)}(T,\xi )&={\mathbb {E}}_\xi ^{\hat{{\mathbb {P}}}}\Big [\int _0^T\ell (X_t) f_x(T-t,X_t)\,dt+f_{\epsilon }^{(0)}(0,X_T)\Big ]\\&={\mathbb {E}}_\xi ^{\hat{{\mathbb {P}}}}\Big [\int _0^T\ell (X_t) f_x(T-t,X_t)\,dt+\partial _\epsilon |_{\epsilon =0}(h^{(\epsilon )}/\phi ^{(\epsilon )})(X_T)\Big ]. \end{aligned} \end{aligned}$$

Since the two expectations \({\mathbb {E}}_\xi ^{\hat{{\mathbb {P}}}}[\int _0^T\ell (X_t) f_x(T-t,X_t)\,dt]\) and \({\mathbb {E}}_\xi ^{\hat{{\mathbb {P}}}}[\partial _\epsilon |_{\epsilon =0}(h^{(\epsilon )}/\phi ^{(\epsilon )})(X_T)]\) are uniformly bounded in T on \([0,\infty )\), the function \(f_\epsilon ^{(0)}(T,\xi )\) is also uniformly bounded in T on \([0,\infty ).\) Using Eq. (18), we obtain the desired result. \(\square \)

We now provide the proof of Theorem 4.

Proof

Applying the Feynman–Kac formula to Eq. (17), we get

$$\begin{aligned} -f_t^{(\epsilon )}+\frac{1}{2}\sigma ^{(\epsilon )2}(x)f_{xx}^{(\epsilon )}+\kappa ^{(\epsilon )}(x)f_x^{(\epsilon )}=0,\;f^{(\epsilon )}(0,x)=(h^{(\epsilon )}/\phi ^{(\epsilon )})(x) \end{aligned}$$

for \(\kappa ^{(\epsilon )}=b^{(\epsilon )}+\sigma ^{(\epsilon )2}\phi _x^{(\epsilon )}/\phi ^{(\epsilon )}.\) We differentiate this PDE in \(\epsilon \) and evaluate it at \(\epsilon =0,\) then

$$\begin{aligned}&-f_{\epsilon t}^{(0)}+\frac{1}{2}\sigma ^{2}(x)f_{\epsilon xx}^{(0)}+\kappa (x)f_{\epsilon x}^{(0)}+(\sigma \varSigma )(x)f_{xx}^{(0)} +\ell (x)f_x^{(0)}=0,\;f_\epsilon ^{(0)}(0,x)\\&\quad =\partial _\epsilon |_{\epsilon =0}(h^{(\epsilon )}/\phi ^{(\epsilon )})(x). \end{aligned}$$

From the Itô formula, one can show that the process

$$\begin{aligned} \Big (f_{\epsilon }^{(0)}(T-t,X_t)+\int _0^t((\sigma \varSigma )(X_s)f_{xx}(T-s,X_s)+\ell (X_s)f_x(T-s,X_s))\,ds\Big )_{0\le t\le T} \end{aligned}$$

is a local martingale under the probability measure \(\hat{{\mathbb {P}}}_T\) by checking that the dt term vanishes.

Suppose that \(f_\epsilon \) satisfies

$$\begin{aligned} \begin{aligned}&f_{\epsilon }^{(0)}(T,\xi ) ={\mathbb {E}}_\xi ^{\hat{{\mathbb {P}}}}\Big [\int _0^T((\sigma \varSigma )(X_t) f_{xx}(T-t,X_t)\\&\quad +\ell (X_t) f_x(T-t,X_t))\,dt\Big ]+{\mathbb {E}}_\xi ^{\hat{{\mathbb {P}}}}[\partial _\epsilon |_{\epsilon =0}(h^{(\epsilon )}/\phi ^{(\epsilon )})(X_T)] . \end{aligned} \end{aligned}$$

This equality holds if the local martingale above is a martingale. Since the three expectations \({\mathbb {E}}_\xi ^{\hat{{\mathbb {P}}}}[\int _0^T(\sigma \varSigma )(X_t) f_{xx}(T-t,X_t)\,dt],\) \({\mathbb {E}}_\xi ^{\hat{{\mathbb {P}}}}[\int _0^T\ell (X_t) f_x(T-t,X_t)\,dt]\) and \({\mathbb {E}}_\xi ^{\hat{{\mathbb {P}}}}[\partial _\epsilon |_{\epsilon =0}(h^{(\epsilon )}/\phi ^{(\epsilon )})(X_T)]\) are uniformly bounded in T on \([0,\infty )\), the function \(f_\epsilon ^{(0)}(T,\xi )\) is also uniformly bounded in T on \([0,\infty ).\) Using Eq. (18), we obtain the desired result. \(\square \)

C CIR model

In this appendix, we provide proofs for the asymptotics presented in Sect. 5.1. Only the first-order sensitivity of \(\xi \) and the sensitivity of b are analyzed here. One can find the sensitivities of the other parameters in [22].

Let \((\varOmega ,{\mathcal {F}},({\mathcal {F}}_t)_{t\ge 0},({\mathbb {P}}_t)_{t\ge 0})\) be a consistent probability space that has a one-dimensional Brownian motion \(B=(B_t)_{t\ge 0}.\) The filtration \(({\mathcal {F}}_t)_{t\ge 0}\) is the completed filtration generated by B. The CIR model is a process given as a solution of

$$\begin{aligned} dX_t=(b-aX_t)\,dt+\sigma \sqrt{X_t}\,dB_t,\;X_0=\xi \end{aligned}$$

for \(a,\sigma ,\xi >0\) and \(2b>\sigma ^2.\) For \(q>0\) and a nonzero nonnegative function h with polynomial growth, we define

$$\begin{aligned} p_T={\mathbb {E}}^{\mathbb {P}}[e^{-q\int _0^TX_s\,ds} h(X_T)]. \end{aligned}$$

It can be shown that the quadruple of functions

$$\begin{aligned} (b-ax,\sigma \sqrt{x},qx,h(x)),\;x>0 \end{aligned}$$

satisfies Assumptions 15. The recurrent eigenpair is

$$\begin{aligned} (\lambda ,\phi (x)):=(b\eta , e^{-\eta x}) \end{aligned}$$

where

$$\begin{aligned} \alpha :=\sqrt{a^2+2q\sigma ^2},\; \eta :=\frac{\alpha -a}{\sigma ^2}. \end{aligned}$$

Under the consistent family of recurrent eigen-measures \((\hat{{\mathbb {P}}}_t)_{t\ge 0},\) the process

$$\begin{aligned} {\hat{B}}_t=\sigma \eta \int _0^t\sqrt{X_s}\,ds+B_t,\;t\ge 0 \end{aligned}$$

is a Brownian motion, and X follows

$$\begin{aligned} dX_t= (b-\alpha X_t)\,dt+\sigma \sqrt{X_t}\,d{\hat{B}}_t. \end{aligned}$$

Using the Hansen–Scheinkman decomposition, we have

$$\begin{aligned} \begin{aligned} p_T&={\mathbb {E}}_\xi ^{\mathbb {P}}[e^{-q\int _0^TX_s\,ds} h(X_T)] ={\mathbb {E}}_\xi ^{\hat{{\mathbb {P}}}}[h(X_T)e^{\eta X_T}] \,e^{-\eta \xi }\,e^{-\lambda T}. \end{aligned} \end{aligned}$$
(54)

For \(t\in [0,\infty )\) and \(x>0,\) the remainder function is

$$\begin{aligned} f(t,x)={\mathbb {E}}_x^{\hat{{\mathbb {P}}}}[h(X_t)e^{\eta X_t}] \end{aligned}$$
(55)

such that \(p_T=f(T,\xi )e^{-\eta \xi }e^{-\lambda T}.\) For nonzero and nonnegative h with polynomial growth, it is easy to show that \(f(T,\xi )\) converges to a positive constant as \(T\rightarrow \infty \) by using Lemma 1. It is also easy to check that f is \(C^{1,2}\) by considering the density function of \(X_t.\) We will investigate the behavior of the function \(f(T,\xi )\) by expressing it as a solution of a second-order differential equation. Using the Feynman–Kac formula, the function f satisfies

$$\begin{aligned} -f_t+\frac{1}{2}\sigma ^2xf_{xx}+(b-\alpha x)f_x=0,\;f(0,x)=h(x)e^{\eta x}. \end{aligned}$$
(56)

Lemma 1

Let \({\hat{B}}\) be a Brownian motion on the consistent probability space \((\varOmega ,{\mathcal {F}},({\mathcal {F}}_t)_{t\ge 0},(\hat{{\mathbb {P}}}_t)_{t\ge 0}).\) Suppose that X is a solution of

$$\begin{aligned} dX_t= (b-\alpha X_t)\,dt+\sigma \sqrt{X_t}\,d{\hat{B}}_t,\;X_0=\xi \end{aligned}$$

where \(\alpha ,\sigma ,\xi >0\) and \(2b>\sigma ^2.\) Then, for \(\beta <2\alpha /\sigma ^2\), we have

$$\begin{aligned} {\mathbb {E}}^{\hat{{\mathbb {P}}}}[e^{\beta X_T}]=\Big (\frac{1}{1-\beta c(T)}\Big )^{2b/\sigma ^2}e^{\frac{\beta }{1-\beta c(T)}e^{-\alpha T}\xi } \end{aligned}$$

where \(c(T):=\sigma ^2 (1-e^{-\alpha T})/2\alpha .\) Thus, in this case,

$$\begin{aligned} {\mathbb {E}}^{\hat{{\mathbb {P}}}}[e^{\beta X_T}]\le \Big (\frac{1}{1-\beta \sigma ^2/2\alpha }\Big )^{2b/\sigma ^2}e^{\frac{\beta }{1-\beta \sigma ^2/2\alpha }e^{-\alpha T}\xi }, \end{aligned}$$

and

$$\begin{aligned} \lim _{T\rightarrow \infty }{\mathbb {E}}^{\hat{{\mathbb {P}}}}[e^{\beta X_T}]=\Big (\frac{1}{1-\beta \sigma ^2/2\alpha }\Big )^{2b/\sigma ^2}. \end{aligned}$$

Refer to [13, Corollary 6.3.4.4] for the proof.

1.1 C.1 First-order sensitivity of \(\xi \)

We estimate the large-time asymptotic behavior of the first-order sensitivity of \(p_T\) with respect to the initial value \(\xi .\) In this section, assume that h is continuously differentiable and that h and \(h'\) have polynomial growth. From Eq. (54), it follows that

$$\begin{aligned} \begin{aligned} \frac{\partial _\xi p_T}{p_T} =\frac{ f_x(T,\xi )}{f(T,\xi )} -\eta . \end{aligned} \end{aligned}$$

The function \(f_x(t,x)\) satisfies

$$\begin{aligned} -f_{xt}+\frac{1}{2}\sigma ^2x f_{xxx}+\Big (b+\frac{1}{2}\sigma ^2-\alpha x\Big )f_{xx} -\alpha f_x=0,\;f_x(0,x)=(h'(x)+\eta h(x))e^{\eta x}, \end{aligned}$$
(57)

which is obtained from Eq. (56) by differentiating in x. Note that since f is \(C^{1,2}\) and every coefficient is continuously differentiable in x in Eq. (56), the function f is thrice continuously differentiable in x. It is easy to show that the quadruple of functions

$$\begin{aligned} (b+\sigma ^2/2-\alpha x,\sigma \sqrt{x},\alpha , (h'+\eta h)e^{\eta x}),\;x>0 \end{aligned}$$

satisfies Assumptions 15. The corresponding process \({\hat{X}}\) is the solution of

$$\begin{aligned} d{\hat{X}}_t=(b+\sigma ^2/2-\alpha {\hat{X}}_t)\,dt+\sigma {{\hat{X}}_t}^{1/2}\,d{\hat{B}}_t. \end{aligned}$$

Lemma 2

The remainder function f satisfies

$$\begin{aligned} f_x(t,x)={\mathbb {E}}^{\hat{{\mathbb {P}}}}[(h'({\hat{X}}_t)+\eta h({\hat{X}}_t))e^{\eta {\hat{X}}_t}|{\hat{X}}_0=x]e^{-\alpha t} \end{aligned}$$
(58)

for \(x>0\) and \(t\ge 0.\)

Proof

Define \(g(t,x):=f_x(t,x)\phi (x).\) Eq. (57) gives

$$\begin{aligned} \begin{aligned}&-g_t+\frac{1}{2}\sigma ^2xg_{xx}+(b+\sigma ^2/2-a x)g_x+\Big (-\frac{1}{2}(\alpha +a)\eta x+(b+\sigma ^2/2)\eta -\alpha \Big )g=0\\&g(0,x)=h'(x)+\eta h(x). \end{aligned} \end{aligned}$$

Consider a consistent family \(({\mathbb {Q}}_t)_{t\ge 0}\) of probability measures where each \({\mathbb {Q}}_t\) is a probability measure on \({\mathcal {F}}_t\) defined as

$$\begin{aligned} \frac{d{\mathbb {Q}}_t}{d\hat{{\mathbb {P}}}_t}=\frac{\phi ({\hat{X}}_0)}{\phi ({\hat{X}}_t)}e^{\int _0^t\frac{\hat{{\mathcal {L}}}\phi ({\hat{X}}_s)}{\phi ({\hat{X}}_s)}\,ds}=e^{\eta ({\hat{X}}_t-x)+\int _0^t\frac{1}{2}(\alpha +a)\eta {\hat{X}}_s -(b+\sigma ^2/2)\eta \,ds}=e^{-\frac{1}{2}\sigma ^2\eta ^2\int _0^t{\hat{X}}_s\,ds+\sigma \eta \int _0^t {\hat{X}}_s^{1/2}\,d{\hat{B}}_s}. \end{aligned}$$

It is easy to check that \({\hat{X}}\) satisfies

$$\begin{aligned} \begin{aligned} d{\hat{X}}_t&=(b+\sigma ^2/2-\alpha {\hat{X}}_t)\,dt+\sigma {{\hat{X}}_t}^{1/2}\,d{\hat{B}}_t\\&=(b+\sigma ^2/2-a{\hat{X}}_t)\,dt+\sigma {{\hat{X}}_t}^{1/2}\,d{B}_t^{{\mathbb {Q}}} \end{aligned} \end{aligned}$$

for a \(({\mathbb {Q}}_t)_{t\ge 0}\)-Brownian motion \((B_t^{{\mathbb {Q}}})_{t\ge 0}.\) By [24, Theorem 5.1.8], since this process does not reach the boundaries under the consistent family of probability measures \(({\mathbb {Q}}_t)_{t\ge 0},\) the \({\mathbb {Q}}_T\)-local martingale \((e^{-\frac{1}{2}\sigma ^2\eta ^2\int _0^t{\hat{X}}_s\,ds+\sigma \eta \int _0^t {\hat{X}}_s^{1/2}\,d{\hat{B}}_s})_{0\le t\le T}\) is a \({\mathbb {Q}}_T\)-martingale. Observe that the operator \(\hat{{\mathcal {L}}}\) given as

$$\begin{aligned} \hat{{\mathcal {L}}}f=\frac{1}{2}\sigma ^2xf''(x) +(b+\sigma ^2/2-ax)f'(x) \end{aligned}$$

is the infinitesimal generator of \({\hat{X}}\) under the consistent family of probability measures \(({\mathbb {Q}}_t)_{T\ge 0},\) and for \(\phi (x)=e^{-\eta x},\) we get

$$\begin{aligned} \hat{{\mathcal {L}}}\phi (x)=\big (\frac{1}{2}(\alpha +a)\eta x -(b+\sigma ^2/2)\eta \big )e^{-\eta x}. \end{aligned}$$

The Feynman–Kac formula (Remark 4 or Proposition 1) yields

$$\begin{aligned} \begin{aligned} f_x(t,x)\phi (x)=g(t,x)&={\mathbb {E}}^{{{\mathbb {Q}}}}\big [ e^{\int _0^t-\frac{1}{2}(\alpha +a)\eta {\hat{X}}_s+(b+\sigma ^2/2)\eta -\alpha \,ds}(h'({\hat{X}}_t)+\eta h({\hat{X}}_t))\big |{\hat{X}}_0=x\big ]\\&={\mathbb {E}}^{{{\mathbb {Q}}}}\Big [ e^{-\int _0^t\frac{\hat{{\mathcal {L}}}\phi }{\phi }({\hat{X}}_s)\,ds}\,\frac{\phi ({\hat{X}}_{t})}{\phi ({\hat{X}}_{0})} e^{\int _0^t-\alpha \,ds}(h'({\hat{X}}_t)+\eta h({\hat{X}}_t))e^{\eta {\hat{X}}_t}\Big |{\hat{X}}_0=x\Big ]\phi (x)\\&={\mathbb {E}}^{{{\mathbb {Q}}}}\Big [ \frac{d\hat{{\mathbb {P}}}_t}{d{\mathbb {Q}}_t} e^{-\alpha t}(h'({\hat{X}}_t)+\eta h({\hat{X}}_t))e^{\eta {\hat{X}}_t}\Big |{\hat{X}}_0=x\Big ]\phi (x)\\&={\mathbb {E}}^{\hat{{\mathbb {P}}}}\big [ (h'({\hat{X}}_t)+\eta h({\hat{X}}_t))e^{\eta {\hat{X}}_t}\big |{\hat{X}}_0=x\big ]\phi (x)e^{-\alpha t}, \end{aligned} \end{aligned}$$

which implies that

$$\begin{aligned} f_x(t,x)={\mathbb {E}}^{\hat{{\mathbb {P}}}}\big [(h'({\hat{X}}_t)+\eta h({\hat{X}}_t))e^{\eta {\hat{X}}_t}\big |{\hat{X}}_0=x\big ]e^{-\alpha t}. \end{aligned}$$

Condition (iv) of Proposition 1 can be confirmed from Lemma 1 and the density function of \(X_t,\) and the other conditions are trivial. \(\square \)

Using Eq. (58), we can obtain the large-time behavior of \(\partial _\xi p_T.\) Since h and \(h'\) have polynomial growth, for \(\eta<\beta <2\alpha /\sigma ^2,\) there is a positive constant \(c_0=c_0(\beta )\) such that \(|h'(x)+\eta h(x)|e^{\eta x}\le c_0 e^{\beta x}\) for \(x>0.\) From Lemma 1, we have

$$\begin{aligned} \begin{aligned} |f_x(t,x)|e^{\alpha t}\le {\mathbb {E}}^{\hat{{\mathbb {P}}}}\Big [|h'({\hat{X}}_t)+\eta h({\hat{X}}_t)|e^{\eta {\hat{X}}_t}\Big |{\hat{X}}_0=x\Big ]&\le c_0{\mathbb {E}}^{\hat{{\mathbb {P}}}}[e^{\beta {\hat{X}}_t}|{\hat{X}}_0=x]\\&\le c_0\Big (\frac{1}{1-\beta \sigma ^2/2\alpha }\Big )^{1+2b/\sigma ^2}e^{\frac{\beta }{1-\beta \sigma ^2/2\alpha }e^{-\alpha t}x} \le c_1 \end{aligned} \end{aligned}$$
(59)

for some positive constant \(c_1\) that depends on x but does not depend on t. It follows that

$$\begin{aligned} |f_x(t,x)|\le c_1e^{-\alpha t}. \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned} \left| \frac{\partial _\xi p_T}{p_T}+\eta \right| =\left| \frac{ f_x(T,\xi )}{f(T,\xi )} \right| \le c_2 e^{-\alpha T} \end{aligned} \end{aligned}$$

for some positive constant \(c_2.\) This gives the desired result.

1.2 C.2 Sensitivity of b

We investigate the large-time asymptotic behavior of the sensitivity with respect to the parameter b. In this section, assume that h is continuously differentiable and that h and \(h'\) have polynomial growth. Using Eq. (54) and Eq. (55), since \(\eta \) is independent of b and \(\lambda =b\eta ,\) we have

$$\begin{aligned} \frac{\partial _b p_T}{p_T}=\frac{f_b(T,\xi )}{f(T,\xi )}-\eta T. \end{aligned}$$

It can easily be shown that f is continuously differentiable in b by considering the density function of \(X_t\) or by using [21, Theorem 4.8]. We focus on the large-time behavior of \(f_b(T,\xi ).\) Differentiate Eq. (56) in b,  then

$$\begin{aligned} -f_{bt}+\frac{1}{2}\sigma ^2xf_{bxx}+(b-\alpha x)f_{bx}+f_x=0,\;f_b(0,x)=0. \end{aligned}$$

From the Feynman–Kac formula, one can show that

$$\begin{aligned} f_b(t,x) ={\mathbb {E}}_x^{\hat{{\mathbb {P}}}}\Big [\int _0^tf_x(t-s,X_s)\,ds\Big ] \end{aligned}$$

by the same method used in the proof of Lemma 2.

We can estimate the expectation on the right-hand side by using the same method in Section C.1. For \(\eta<\beta <\alpha /\sigma ^2,\) Eq. (58) and Eq. (59) imply that

$$\begin{aligned} \begin{aligned} |f_x(t-s,x)|&\le c_0\Big (\frac{1}{1-\beta \sigma ^2/2\alpha }\Big )^{1+2b/\sigma ^2}e^{\frac{\beta }{1-\beta \sigma ^2/2\alpha }e^{-\alpha (t-s)}x} e^{-\alpha (t-s)} \le c_1e^{\gamma x} e^{-\alpha (t-s)} \end{aligned} \end{aligned}$$

where

$$\begin{aligned} c_1:=c_0\Big (\frac{1}{1-\beta \sigma ^2/2\alpha }\Big )^{1+2b/\sigma ^2},\;\gamma :=\frac{\beta }{1-\beta \sigma ^2/2\alpha }. \end{aligned}$$

Then,

$$\begin{aligned} |f_b(t,x)| \le {\mathbb {E}}_x^{\hat{{\mathbb {P}}}}\Big [\int _0^t|f_x(t-s,X_s)|\,ds\Big ] = \int _0^t{\mathbb {E}}_x^{\hat{{\mathbb {P}}}}|f_x(t-s,X_s)|\,ds \le c_1\int _0^te^{-\alpha (t-s)} {\mathbb {E}}_x^{\hat{{\mathbb {P}}}}[e^{\gamma X_s}]\,ds. \end{aligned}$$

Since \(\gamma <2\alpha /\sigma ^2,\) by Lemma 1, the expectation \({\mathbb {E}}_x^{\hat{{\mathbb {P}}}}[e^{\gamma X_s}]\) is bounded in s on \([0,\infty ).\) Thus, there is a positive constant \(c_2\) such that \({\mathbb {E}}_x^{\hat{{\mathbb {P}}}}[e^{\gamma X_s}] \le c_2 \alpha /c_1,\) which gives

$$\begin{aligned} |f_b(t,x)| \le c_2 \alpha \int _0^te^{-\alpha (t-s)}\,ds=c_2(1-e^{-\alpha t}). \end{aligned}$$

Since \(f(T,\xi )\) converges to a positive constant as \(T\rightarrow \infty ,\) we conclude that

$$\begin{aligned} \left| \frac{1}{T}\frac{\partial _b p_T}{p_T}+\eta \right| =\frac{1}{T}\left| \frac{f_b(T,\xi )}{f(T,\xi )}\right| \le \frac{c_3}{T} \end{aligned}$$

for some positive constant \(c_3.\)

D 3/2 model

In this appendix, we provide proofs for the asymptotics presented in Sect. 5.2. Only the sensitivities of \(\xi \) and \(\sigma \) are analyzed here. One can find the sensitivities of the other parameters in [22].

The 3/2 model is a process given as a solution of

$$\begin{aligned} dX_t=(b-aX_t)X_t\,dt+\sigma {X_t}^{3/2}\,dB_t,\;X_0=\xi \end{aligned}$$

for \(b,\sigma ,\xi >0\) and \(a>-\sigma ^2/2.\) For \(q>0\) and a nonzero, nonnegative Borel function h with at most linear growth, we define

$$\begin{aligned} p_T={\mathbb {E}}^{\mathbb {P}}[e^{-q\int _0^TX_s\,ds} h(X_T)]. \end{aligned}$$

It can be shown that the quadruple of functions

$$\begin{aligned} ((b-ax)x,\sigma x^{3/2},qx,h),\;x>0 \end{aligned}$$

satisfies Assumptions 15. The recurrent eigenpair is

$$\begin{aligned} (\lambda ,\phi (x)):=(b\eta , x^{-\eta }) \end{aligned}$$

where

$$\begin{aligned} \eta :=\frac{\sqrt{(a+\sigma ^2/2)^2+2q\sigma ^2}-(a+\sigma ^2/2)}{\sigma ^2}. \end{aligned}$$

Under the consistent family of recurrent eigen-measures \((\hat{{\mathbb {P}}}_t)_{t\ge 0},\) the process

$$\begin{aligned} {\hat{B}}_t=\sigma \eta \int _0^t\sqrt{X_s}\,ds+B_t,\;t\ge 0 \end{aligned}$$

is a Brownian motion, and X follows

$$\begin{aligned} \begin{aligned} dX_t=(b-\alpha X_t)X_t\,dt+\sigma {X_t}^{3/2}\,d{\hat{B}}_t \end{aligned} \end{aligned}$$

where \(\alpha :=a+\sigma ^2\eta .\)

Using this consistent family of recurrent eigen-measures, we have the Hansen–Scheinkman decomposition

$$\begin{aligned} \begin{aligned} p_T&={\mathbb {E}}_\xi ^{\mathbb {P}}[e^{-q\int _0^TX_s\,ds} h(X_T)] ={\mathbb {E}}_\xi ^{\hat{{\mathbb {P}}}}[h(X_T)X_T^{\eta }] \,\xi ^{-\eta }\,e^{-\lambda T}. \end{aligned} \end{aligned}$$
(60)

For \(t\in [0,\infty )\) and \(x>0,\) we define

$$\begin{aligned} f(t,x)={\mathbb {E}}_x^{\hat{{\mathbb {P}}}}[h(X_t)X_t^{\eta }], \end{aligned}$$

then

$$\begin{aligned} p_T=f(T,\xi )\phi (\xi )e^{-\lambda T}=f(T,\xi )\xi ^{-\eta }e^{-\lambda T}. \end{aligned}$$
(61)

For a nonzero, nonnegative Borel function h with at most linear growth, it is easy to show that \(f(T,\xi ) \) converges to a positive constant as \(T\rightarrow \infty \) by Lemma 3. We will investigate the large-time behavior of the function \(f(T,\xi )\) by expressing the function as a solution of a second-order differential equation. By the Feynman–Kac formula, f satisfies

$$\begin{aligned} -f_t+\frac{1}{2}\sigma ^2x^3f_{xx}+(b-\alpha x)xf_x=0,\;f(0,x)=h(x)x^{\eta }. \end{aligned}$$
(62)

Lemma 3

Let \({\hat{B}}=({\hat{B}}_t)_{t\ge 0}\) be a Brownian motion on the consistent probability space \((\varOmega ,{\mathcal {F}},({\mathcal {F}}_t)_{t\ge 0},(\hat{{\mathbb {P}}}_t)_{t\ge 0}).\) Suppose that X is a solution of

$$\begin{aligned} dX_t=(b-\alpha X_t)X_t\,dt+\sigma {X_t}^{3/2}\,d{\hat{B}}_t,\;X_0=\xi \end{aligned}$$

where \(\alpha ,b,\sigma ,\xi >0.\) Then, for \(A<\frac{2\alpha }{\sigma ^2}+2,\) we have

$$\begin{aligned} \begin{aligned} {\mathbb {E}}_\xi (X_T^A)&=\frac{\varGamma (\frac{2\alpha }{\sigma ^2}+2-A)}{\varGamma (\frac{2\alpha }{\sigma ^2}+2)}\Big (\frac{2b}{\sigma ^2}\frac{1}{1-e^{-bT}}\Big )^{A} F\Big (A,\frac{2\alpha }{\sigma ^2}+2,-\frac{2b}{\sigma ^2}\frac{1}{(e^{bT}-1)\xi }\Big ), \end{aligned} \end{aligned}$$

and the expectation \({\mathbb {E}}_\xi (X_T^A)\) converges to

$$\begin{aligned} \frac{\varGamma (\frac{2\alpha }{\sigma ^2}+2-A)}{\varGamma (\frac{2\alpha }{\sigma ^2}+2)}\Big (\frac{2b}{\sigma ^2}\Big )^{A} \end{aligned}$$

as \(T\rightarrow \infty \) where F is the confluent hypergeometric function. Moreover, if \(0<A<\frac{2\alpha }{\sigma ^2}+2,\) then the map \(H:[0,\infty )\times (0,\infty )\rightarrow {\mathbb {R}}\) defined by \(H(t,x)={\mathbb {E}}_x(X_t^A)\) is uniformly bounded on the domain \([0,\infty )\times (0,\infty ).\)

Proof

Define a process Y as \(Y=1/X,\) then

$$\begin{aligned} dY_t=(\theta -bY_t)\,dt-\sigma \sqrt{Y_t}\,d{\hat{B}}_t,\;Y_0=\zeta , \end{aligned}$$

where \(\theta :=\alpha +\sigma ^2\) and \(\zeta :=1/\xi .\) Since \(\theta >\sigma ^2/2,\) the Feller condition is satisfied. From [12, Theorem 3.1] or [3, Section 3], for \(A<\frac{2\theta }{\sigma ^2}=\frac{2\alpha }{\sigma ^2}+2,\) we have

$$\begin{aligned} \begin{aligned} {\mathbb {E}}(X_t^A)={\mathbb {E}}(Y_t^{-A})&=\frac{\varGamma (\frac{2\theta }{\sigma ^2}-A)}{\varGamma (\frac{2\theta }{\sigma ^2})}\Big (\frac{2b}{\sigma ^2}\frac{1}{1-e^{-bt}}\Big )^{A} F\Big (A,\frac{2\theta }{\sigma ^2},-\frac{2b}{\sigma ^2}\frac{\zeta }{e^{bt}-1}\Big ). \end{aligned} \end{aligned}$$

Since the confluent hypergeometric function F satisfies

$$\begin{aligned} \lim _{t\rightarrow \infty }F\left( A,\frac{2\theta }{\sigma ^2},-\frac{2b}{\sigma ^2}\frac{\zeta }{e^{bt}-1}\right) =1, \end{aligned}$$

we obtain the desired result. Moreover, if A and \(\frac{2\alpha }{\sigma ^2}+2\) are positive, the function \(z\mapsto F(A,\frac{2\alpha }{\sigma ^2}+2,z)\) is uniformly bounded in z on \((0,\infty ),\) which results directly from [1, 13.1.5 on page 504]. This means that the map H is uniformly bounded on the domain \([0,\infty )\times (0,\infty ).\) \(\square \)

1.1 D.1 First-order sensitivity of \(\xi \)

We estimate the large-time asymptotic behavior of the first-order sensitivity of \(p_T\) with respect to the initial value \(\xi .\) In this section, assume that h is continuously differentiable and that h and \(h'\) have at most linear growth and are nonnegative. (We are mainly interested in the case \(h=1\).) From Eq. (60), it follows that

$$\begin{aligned} \begin{aligned} \frac{\partial _\xi p_T}{p_T} =\frac{ f_x(T,\xi )}{f(T,\xi )} -\frac{\,\eta \,}{\xi }. \end{aligned} \end{aligned}$$

We focus on the term \(f_x(T,\xi ).\) Since f is \(C^{1,2}\) and every coefficient is continuously differentiable in x in Eq. (62), the function f is thrice continuously differentiable in x. This gives

$$\begin{aligned} -f_{xt}+\frac{1}{2}\sigma ^2x^3f_{xxx}+(b-(\alpha -\frac{3}{2}\sigma ^2)x)xf_{xx}+(b-2\alpha x)f_x=0,\;f_x(0,x)=(xh'(x)+\eta h(x))x^{\eta -1}. \end{aligned}$$

It is easy to show that the quadruple of functions

$$\begin{aligned} ((b-(\alpha -\frac{3}{2}\sigma ^2)x)x,\sigma {x}^{3/2},-b+2\alpha x, (xh'(x)+\eta h(x))x^{\eta -1}),\;x>0 \end{aligned}$$

satisfies Assumptions 15. The corresponding process \({\hat{X}}\) is the solution of

$$\begin{aligned} d{\hat{X}}_t=(b-(\alpha -\frac{3}{2}\sigma ^2){\hat{X}}_t){\hat{X}}_t\,dt+\sigma {\hat{X}}_t^{3/2}\,d{\hat{B}}_t. \end{aligned}$$

To analyze \(f_x(t,x),\) we apply the Hansen–Scheinkman decomposition. By the same method used in Lemma 2, one can show that the remainder function f satisfies

$$\begin{aligned} f_x(t,x)={\mathbb {E}}^{\hat{{\mathbb {P}}}}[({\hat{X}}_th'({\hat{X}}_t)+\eta h({\hat{X}}_t)){\hat{X}}_t^{\eta -1}e^{-2\alpha \int _0^t{\hat{X}}_s\,ds}|{\hat{X}}_0=x] e^{bt}. \end{aligned}$$
(63)

It can be shown that the pair \((2b,x^{-2})\) is the recurrent eigenpair by considering the generator of \({\hat{X}}\) with killing rate \(2\alpha {\hat{X}}_t,\) which is given as

$$\begin{aligned} {\mathcal {L}}\psi (x):=\frac{1}{2}\sigma ^2x^3\psi ''(x)+(b-(\alpha -\frac{3}{2}\sigma ^2)x)x\psi '(x)-2\alpha x\psi (x). \end{aligned}$$

Consider a consistent family \((\tilde{{\mathbb {P}}}_t)_{t\ge 0}\) of probability measures where each \(\tilde{{\mathbb {P}}}_t\) is defined on \({\mathcal {F}}_t\) as

$$\begin{aligned} \frac{d\tilde{{\mathbb {P}}}_t}{d\hat{{\mathbb {P}}}_t} =e^{2bt-2\alpha \int _{0}^{t}{\hat{X}}_s\,ds}\,\frac{{\hat{X}}_0^2}{{\hat{X}}_t^2}. \end{aligned}$$

Then,

$$\begin{aligned} {\tilde{B}}_t:=2\sigma \int _0^t{\hat{X}}_s^{1/2}\,ds+{\hat{B}}_t,\;t\ge 0 \end{aligned}$$

is a \((\tilde{{\mathbb {P}}}_t)_{t\ge 0}\)-Brownian motion, and \({\hat{X}}\) follows

$$\begin{aligned} d{\hat{X}}_t=(b-(\alpha +\frac{1}{2}\sigma ^2){\hat{X}}_t){\hat{X}}_t\,dt+\sigma {\hat{X}}_t^{3/2}\,d{\tilde{B}}_t. \end{aligned}$$

Using this consistent family of probability measures, we have

$$\begin{aligned} \begin{aligned} f_x(t,x)&={\mathbb {E}}^{\hat{{\mathbb {P}}}}[({\hat{X}}_th'({\hat{X}}_t)+\eta h({\hat{X}}_t)){\hat{X}}_t^{\eta -1}e^{-2\alpha \int _0^t{\hat{X}}_s\,ds}|{\hat{X}}_0=x] e^{bt}\\&={\mathbb {E}}^{\tilde{{\mathbb {P}}}}[({\hat{X}}_th'({\hat{X}}_t)+\eta h({\hat{X}}_t)){\hat{X}}_t^{\eta +1}|{\hat{X}}_0=x] e^{-bt}x^{-2}. \end{aligned} \end{aligned}$$
(64)

We can obtain the large-time behavior of \(\partial _\xi p_T\) by using Eq. (64). Since h and \(h'\) have at most linear growth, there is a positive constant \(c_0\) such that \(|{\hat{X}}_th'({\hat{X}}_t)+\eta h({\hat{X}}_t)|{\hat{X}}_t^{\eta +1}\le c_0{\hat{X}}_t^{\eta +3}.\) By Lemma 3, the expectation

$$\begin{aligned} {\mathbb {E}}^{\tilde{{\mathbb {P}}}}\big [|{\hat{X}}_th'({\hat{X}}_t)+\eta h({\hat{X}}_t)|{\hat{X}}_t^{\eta +1}\big |{\hat{X}}_0=x\big ] \end{aligned}$$
(65)

is uniformly bounded in (tx) since the constant A in the lemma satisfies \(A=\eta +3<\frac{2\alpha }{\sigma ^2}+3.\) Therefore,

$$\begin{aligned} \begin{aligned} \left| \frac{\partial _\xi p_T}{p_T}+\frac{\,\eta \,}{\xi }\right| =\left| \frac{ f_\xi (T,\xi )}{f(T,\xi )} \right| \le c_1 e^{-bT} \end{aligned} \end{aligned}$$

for some positive constant \(c_1.\) This gives the desired result.

1.2 D.2 Second-order sensitivity of \(\xi \)

We investigate the large-time asymptotic behavior of the second-order sensitivity with respect to the initial value \(\xi .\) In this section, assume that h is twice continuously differentiable, h and \(h'\) have at most linear growth and \(h''\) is bounded, moreover h\(h',\) \(h''\) are nonnegative. (We are mainly interested in the case \(h=1\).) From Eq. (60),

$$\begin{aligned} \begin{aligned} \frac{{\partial _{\xi \xi }} p_T}{p_T} =\frac{ f_{xx}(T,\xi )}{f(T,\xi )} -\frac{2\eta }{\xi } \frac{f_x(T,\xi )}{f(T,\xi )} +\frac{\eta (\eta +1)}{\xi ^2}. \end{aligned} \end{aligned}$$

Since we already estimated the large-time asymptotic behavior of \(f_x(T,\xi ),\) we focus on the second-order derivative \(f_{xx}(T,\xi ).\) Define

$$\begin{aligned} {\hat{f}}(t,x):={\mathbb {E}}^{\tilde{{\mathbb {P}}}}[({\hat{X}}_th'({\hat{X}}_t)+\eta h({\hat{X}}_t)){\hat{X}}_t^{\eta +1}|{\hat{X}}_0=x], \end{aligned}$$

then Eq. (64) gives \(f_x(t,x)={\hat{f}}(t,x)e^{-bt}x^{-2}.\) Thus,

$$\begin{aligned} f_{xx}(t,x)={\hat{f}}_x(t,x)e^{-bt}x^{-2}-2{\hat{f}}(t,x)e^{-bt}x^{-3}. \end{aligned}$$
(66)

We need to estimate the large-time behavior of \({\hat{f}}_x(t,x).\) The Feynman–Kac formula states that

$$\begin{aligned} -{\hat{f}}_t+\frac{1}{2}\sigma ^2x^3{\hat{f}}_{xx}+(b-(\alpha +\frac{1}{2}\sigma ^2)x)x{\hat{f}}_x=0,\;{\hat{f}}(0,x)=(h'(x)x+\eta h(x))x^{\eta +1}. \end{aligned}$$

We differentiate this equation in x,  then

$$\begin{aligned} \begin{aligned}&-{\hat{f}}_{xt}+\frac{1}{2}\sigma ^2x^3{\hat{f}}_{xxx}+(b-(\alpha -\sigma ^2)x)x{\hat{f}}_{xx}+(b-(2\alpha +\sigma ^2)x){\hat{f}}_{x}=0,\\&{\hat{f}}_x(0,x)=(h''(x)x^2+2(\eta +1) x h'(x)+\eta (\eta +1)h(x))x^{\eta }. \end{aligned} \end{aligned}$$

It is easy to show that the quadruple of functions

$$\begin{aligned} \Big ((b-(\alpha -\sigma ^2)x)x,\sigma {x}^{3/2},-b+(2\alpha +\sigma ^2) x, (h''(x)x^2+2(\eta +1) x h'(x)+\eta (\eta +1)h(x))x^{\eta }\Big ),\,x>0 \end{aligned}$$

satisfies Assumptions 15. The corresponding process \({\tilde{X}}\) is the solution of

$$\begin{aligned} d{\tilde{X}}_t=(b-(\alpha -\sigma ^2){\tilde{X}}_t){\tilde{X}}_t\,dt+\sigma {\tilde{X}}_t^{3/2}\,d{\tilde{B}}_t. \end{aligned}$$

We observe that

$$\begin{aligned} {\hat{f}}_x(t,x)={\mathbb {E}}^{\tilde{{\mathbb {P}}}}[(h''({\tilde{X}}_t){\tilde{X}}_t^2 +2(\eta +1) {\tilde{X}}_th'({\tilde{X}}_t)+\eta (\eta +1)h({\tilde{X}}_t)){\tilde{X}}_t^{\eta }e^{-(2\alpha +\sigma ^2)\int _0^t{\tilde{X}}_s\,ds}|{\tilde{X}}_0=x]e^{bt}. \end{aligned}$$

This can be obtained by the same argument used in the derivation of Eq. (63) by considering \(Y=1/X.\)

To analyze the expectation above, we apply the Hansen–Scheinkman decomposition. It can be shown that the pair \((2b,x^{-2})\) is the recurrent eigenpair by considering the generator of \({\tilde{X}}\) with killing rate \((2\alpha +\sigma ^2){\tilde{X}}_t,\) which is given as

$$\begin{aligned} {\mathcal {L}}\psi (x):=\frac{1}{2}\sigma ^2x^3\psi ''(x)+(b-(\alpha -\sigma ^2)x)x\psi '(x)-(2\alpha +\sigma ^2)x\psi (x). \end{aligned}$$

Consider a consistent family \((\bar{{\mathbb {P}}}_t)_{t\ge 0}\) of probability measures where each \(\bar{{\mathbb {P}}}_t\) is defined on \({\mathcal {F}}_t\) as

$$\begin{aligned} \frac{d\bar{{\mathbb {P}}}_t}{d\tilde{{\mathbb {P}}}_t} =e^{2bt-(2\alpha +\sigma ^2)\int _{0}^{t}{\tilde{X}}_s\,ds}\,\frac{{\tilde{X}}_0^2}{{\tilde{X}}_t^2}. \end{aligned}$$

Then,

$$\begin{aligned} {\bar{B}}_t:=2\sigma \int _0^t{\tilde{X}}_s^{1/2}\,ds+{\tilde{B}}_t,\;t\ge 0 \end{aligned}$$

is a \((\bar{{\mathbb {P}}}_t)_{t\ge 0}\)-Brownian motion, and \({\tilde{X}}\) follows

$$\begin{aligned} d{\tilde{X}}_t=(b-(\alpha +\sigma ^2){\tilde{X}}_t){\tilde{X}}_t\,dt+\sigma {\tilde{X}}_t^{3/2}\,d{\bar{B}}_t. \end{aligned}$$

Using this consistent family of probability measures \((\bar{{\mathbb {P}}}_t)_{t\ge 0},\) we have

$$\begin{aligned} \begin{aligned} {\hat{f}}_x(t,x)&={\mathbb {E}}^{\tilde{{\mathbb {P}}}}[(h''({\tilde{X}}_t){\tilde{X}}_t^2 +2(\eta +1) {\tilde{X}}_th'({\tilde{X}}_t)+\eta (\eta +1)h({\tilde{X}}_t)){\tilde{X}}_t^{\eta }e^{-(2\alpha +\sigma ^2)\int _0^t{\tilde{X}}_s\,ds}|{\tilde{X}}_0=x]e^{bt}\\&={\mathbb {E}}^{\bar{{\mathbb {P}}}}[(h''({\tilde{X}}_t){\tilde{X}}_t^2 +2(\eta +1) {\tilde{X}}_th'({\tilde{X}}_t)+\eta (\eta +1)h({\tilde{X}}_t)){\tilde{X}}_t^{\eta +2}|{\tilde{X}}_0=x]e^{-bt}x^{-2}. \end{aligned} \end{aligned}$$
(67)

Since h and \(h'\) have at most linear growth and \(h''\) is bounded, by Lemma 3, the expectation

$$\begin{aligned} {\mathbb {E}}^{\bar{{\mathbb {P}}}}\big [|h''({\tilde{X}}_t){\tilde{X}}_t^2 +2(\eta +1) {\tilde{X}}_th'({\tilde{X}}_t)+\eta (\eta +1)h({\tilde{X}}_t)|{\tilde{X}}_t^{\eta +2}\big |{\tilde{X}}_0=x\big ] \end{aligned}$$

converges to a positive constant as \(t\rightarrow \infty \) since the constant A in the lemma satisfies \(A=\eta +4<\frac{2\alpha }{\sigma ^2}+4.\) Thus,

$$\begin{aligned} \begin{aligned} |f_{xx}(t,x)|&=|{\hat{f}}_x(t,x)e^{-bt}x^{-2}-2g(t,x)e^{-bt}x^{-3}|\\&\le |{\hat{f}}_x(t,x)|e^{-bt}x^{-2}+2|g(t,x)|e^{-bt}x^{-3}\\&\le c_1e^{-2bt}x^{-4}+c_1e^{-bt}x^{-3}\\&\le c_2 e^{-bt} \end{aligned} \end{aligned}$$

for some positive constants \(c_1\) and \(c_2,\) which are independent of t. From Eq.(65), we conclude that

$$\begin{aligned} \begin{aligned} \left| \frac{{\partial _{\xi \xi }} p_T}{p_T} -\frac{\eta (\eta +1)}{\xi ^2}\right| =\left| \frac{ f_{xx}(T,\xi )}{f(T,\xi )} -\frac{2\eta }{\xi } \frac{f_x(T,\xi )}{f(T,\xi )} \right| \le \left| \frac{ f_{xx}(T,\xi )}{f(T,\xi )}\right| +\frac{2\eta }{\xi }\left| \frac{f_x(T,\xi )}{f(T,\xi )} \right| \le c_3e^{-bT} \end{aligned} \end{aligned}$$

for some positive constant \(c_3.\) This gives the convergence rate of the second-order sensitivity. We can also provide a higher-order convergence rate of the second-order sensitivity as follows. From Eq. (66) and \(p_T=f(T,\xi )e^{\lambda T}\phi (\xi )\) presented in Eq. (61), it follows that

$$\begin{aligned} \begin{aligned} \frac{\partial _{\xi \xi }p_T}{p_T}-\Big (\frac{\partial _\xi p_T}{p_T}\Big )^2+\frac{2}{\xi }\Big (\frac{\partial _\xi p_T}{p_T}-\frac{\phi '(\xi )}{\phi (\xi )}\Big )-\frac{\phi ''(\xi )}{\phi (\xi )}+\Big (\frac{\phi '(\xi )}{\phi (\xi )}\Big )^2&=\frac{g_x(T,\xi )}{g(T,\xi )}\frac{f_x(T,\xi )}{f(T,\xi )}-\Big (\frac{f_x(T,\xi )}{f(T,\xi )}\Big )^2. \end{aligned} \end{aligned}$$

Using \(\phi (\xi )=\xi ^{-\eta },\) we have

$$\begin{aligned} \begin{aligned} \left| \frac{\partial _{\xi \xi }p_T}{p_T}-\Big (\frac{\partial _\xi p_T}{p_T}\Big )^2+\frac{2}{\xi }\frac{\partial _\xi p_T}{p_T}+\frac{\eta }{\xi ^2}\right|&\le \left| \frac{g_x(T,\xi )}{g(T,\xi )}\frac{f_x(T,\xi )}{f(T,\xi )}\right| +\Big (\frac{f_x(T,\xi )}{f(T,\xi )}\Big )^2\le c_4e^{-2bT} \end{aligned} \end{aligned}$$

for some positive constant \(c_4.\) For the last inequality, we used Eq.(65) and Eq. (67).

1.3 D.3 Sensitivity of \(\sigma \)

We study the large-time asymptotic behavior of the sensitivity of \(p_T\) with respect to the parameter \(\sigma .\) In this section, assume that h is nonzero, nonnegative and twice continuously differentiable. Moreover, assume that h and \(h'\) have at most linear growth and \(h''\) is bounded. From the decomposition in Eq. (60), it follows that

$$\begin{aligned} \frac{\partial _\sigma p_T}{p_T}=\frac{f_\sigma (T,\xi )}{f(T,\xi )} -(\partial _\sigma \eta ) \ln \xi -T\partial _\sigma \lambda . \end{aligned}$$

It can be easily shown that f is continuously differentiable in \(\sigma \) by considering the density function of \(X_t\) or by using [21, Theorem 4.13]. We focus on the large-time behavior of \(f_\sigma (T,\xi ).\)

We differentiate Eq. (62) in \(\sigma ,\) then

$$\begin{aligned} -f_{\sigma t}+\frac{1}{2}\sigma ^2x^3f_{\sigma xx}+(b-\alpha x)xf_{\sigma x}+\sigma x^3f_{xx}-(\partial _\sigma \alpha )x^2f_x=0,\;f(0,x)=h(x)x^{\eta }(\ln x)\partial _\sigma \eta . \end{aligned}$$

It follows that

$$\begin{aligned} \begin{aligned} f_\sigma (t,x)&={\mathbb {E}}_x^{\hat{{\mathbb {P}}}}\Big [\sigma \int _0^tX_s^3f_{xx}(s,X_s)\,ds-(\partial _\sigma \alpha )\int _0^tX_s^2f_x(s,X_s)\,ds+h(X_t)X_t^\eta (\ln X_t) \partial _\sigma \eta \Big ]\\&={\mathbb {E}}_x^{\hat{{\mathbb {P}}}}\Big [\sigma \int _0^tX_s^3f_{xx}(s,X_s)\,ds\Big ]+{\mathbb {E}}_x^{\hat{{\mathbb {P}}}}\Big [-(\partial _\sigma \alpha )\int _0^tX_s^2f_x(s,X_s)\,ds+h(X_t)X_t^\eta (\ln X_t) \partial _\sigma \eta \Big ]. \end{aligned} \end{aligned}$$

We claim that the two expectations on the right-hand side are bounded in t on \([0,\infty ).\) It can be easily checked that the second expectation is bounded in t on \([0,\infty ).\) To estimate the first expectation, observe that

$$\begin{aligned} f_{xx}(t,x)=g_x(t,x)e^{-bt}x^{-2}-2g(t,x)e^{-bt}x^{-3}, \end{aligned}$$

which is presented in Eq. (66). It follows that

$$\begin{aligned} {\mathbb {E}}_x^{\hat{{\mathbb {P}}}}\Big [\int _0^tX_s^3f_{xx}(s,X_s)\,ds\Big ]={\mathbb {E}}_x^{\hat{{\mathbb {P}}}}\Big [\int _0^tX_sg_x(s,X_s)e^{-bs}\,ds\Big ]-2{\mathbb {E}}_x^{\hat{{\mathbb {P}}}}\Big [\int _0^tg(s,X_s)e^{-bs}\,ds\Big ]. \end{aligned}$$

There is a positive constant \(c_1\) that is independent of t and x such that

$$\begin{aligned} |g(t,x)|\le {\mathbb {E}}^{\tilde{{\mathbb {P}}}}\big [|Y_th'(Y_t)+\eta h(Y_t)|Y_t^{\eta +1}\big |Y_0=x\big ]\le c_1. \end{aligned}$$

Thus, \({\mathbb {E}}_x^{\hat{{\mathbb {P}}}}[\int _0^tg(s,X_s)e^{-bs}\,ds]\) is bounded in t on \([0,\infty ).\) From Eq. (67),

$$\begin{aligned} \begin{aligned} |g_x(t,x)|&\le {\mathbb {E}}^{\bar{{\mathbb {P}}}}\big [|h''(Z_t)Z_t^2 +2(\eta +1) Z_th'(Z_t)+\eta (\eta +1)h(Z_t)|Z_t^{\eta +2}\big |Z_0=x\big ]e^{-bt}x^{-2}\\&\le c_2{\mathbb {E}}^{\bar{{\mathbb {P}}}}[Z_t^{\eta +4}\big |Z_0=x]e^{-bt}x^{-2} \end{aligned} \end{aligned}$$

for some positive constant \(c_2.\) By Lemma 3, the expectation \({\mathbb {E}}^{\bar{{\mathbb {P}}}}[Z_t^{\eta +4}\big |Z_0=x]\) is uniformly bounded in (tx) on the domain \([0,\infty )\times (0,\infty ),\) which implies that

$$\begin{aligned} |g_x(t,x)| \le c_3e^{-bt}x^{-2} \end{aligned}$$

for some positive constant \(c_3.\) Thus,

$$\begin{aligned} {\mathbb {E}}_x^{\hat{{\mathbb {P}}}}\Big [\int _0^tX_s|g_x(s,X_s)|e^{-bs}\,ds\Big ]\le c_3{\mathbb {E}}_x^{\hat{{\mathbb {P}}}}\Big [\int _0^tX_s^{-1}e^{-2bs}\,ds\Big ]=c_3 \int _0^t{\mathbb {E}}_x^{\hat{{\mathbb {P}}}}[X_s^{-1}]e^{-2bs}\,ds. \end{aligned}$$

By Lemma 3, the expectation \({\mathbb {E}}_x^{\hat{{\mathbb {P}}}}[X_s^{-1}]\) is bounded in s on \([0,\infty )\) since the expectation converges to a positive constant, which gives the desired result. In conclusion,

$$\begin{aligned} \left| \frac{1}{T}\frac{\partial _\sigma p_T}{p_T}+\partial _\sigma \lambda \right| =\frac{1}{T}\left| \frac{f_\sigma (T,\xi )}{f(T,\xi )}-(\partial _\sigma \eta ) \ln \xi \right| \le \frac{c_4}{T} \end{aligned}$$

for some positive constant \(c_4.\) Furthermore, direct calculation gives

$$\begin{aligned} \partial _\sigma \lambda = \frac{b(a+{\sigma ^2}/{2}+2 q-\sqrt{(a+{\sigma ^2}/{2})^2+2 q\sigma ^2})}{\sigma \sqrt{(a+{\sigma ^2}/{2})^2+2 q\sigma ^2}} -\frac{2b}{\sigma ^3}(\sqrt{(a+{\sigma ^2}/{2})^2+2 q \sigma ^2}-a- {\sigma ^2}/{2}). \end{aligned}$$

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Park, H. Convergence rates of large-time sensitivities with the Hansen–Scheinkman decomposition. Math Finan Econ 16, 1–50 (2022). https://doi.org/10.1007/s11579-021-00300-6

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