2. Some preliminaries

In this section, we recall some known results on the expected present value of dividend payments before ruin and the expected discounted penalty functions. Throughout this paper, for any function $f$ defined on the positive real line, we denote its Fourier transform and Laplace transform by

$$\mathcal{F}f(s)=\int_0^\infty e^{isx} f(x)\,dx,\quad \mathcal{L}f(s)=\int_0^\infty e^{{-}sx}f(x)\,dx,$$

where $s$ is such that the above integrals are well defined.

Note that $U^\infty _t$ is a spectrally negative Lévy process, and its Laplace exponent is defined by

(2.1)\begin{equation} \psi_U(s)=\frac{1}{t}\ln E[ e^{sU^\infty_t}]=\frac{\sigma^2}{2}s^2+cs-\lambda(1-\mathcal{L}f_X(s)). \end{equation}

For each $q\geq 0$, let

$$\rho_q=\sup\{s\geq 0: \psi_U(s)=q\}$$

be the right inverse of $\psi _U$. For the special case $q=\delta$, we shall put $\rho =\rho _\delta$ for simplicity.

For each $q, x\geq 0$, let $W_q(x)$ denote the $q$-scale function associated with the process $U^\infty _t$, which is a strictly increasing and continuous function with Laplace transform given by

(2.2)\begin{equation} \int_0^\infty e^{{-}sx} W_q(x)\,dx=\frac{1}{\psi_U(s)-q},\quad \text{for}\ s>\rho_q. \end{equation}

We can extend $W_q$ to the whole real line by setting $W_q(x)=0$ for $x<0$. The $q$-scale function and its various extensions have many applications in the fields related to the spectrally negative Lévy processes. For a thorough introduction to the $q$-scale function and some of its applications, we refer the interested readers to the survey Kuznetsov et al. [Reference Kuznetsov, Kyprianou and Rivero13].

In our applications, it is useful to introduce the following auxiliary function,

(2.3)\begin{equation} h(x):=\frac{\delta}{\psi_U'(\rho)} e^{\rho x}-\delta W_\delta (x),\quad x\in \mathbb{R}. \end{equation}

It follows from Zhang and Cui [Reference Zhang and Cui30] that $h(x)$ is the probability density function of the random variable $U^\infty _0-U^\infty _{\mathbf {e}_\delta }$, where $\mathbf {e}_\delta$, independent of $U^\infty$, denotes an exponential random variable with rate $\delta >0$.

For the expected present value of total dividend payments before ruin, it follows from Albrecher and Gerber [Reference Albrecher and Gerber1] that

(2.4)\begin{equation} V(u;b)=\frac{W_\delta(u)}{W_\delta^{ '}(b)},\quad 0\leq u\leq b. \end{equation}

Then, by formula (2.9), in Xie and Zhang [Reference Xie and Zhang27], we obtain

(2.5)\begin{equation} V(u;b) =\frac{{\delta}/{\psi_U'(\rho)} -e^{-\rho u}h_+(u)}{{\delta\rho e^{\rho(b-u)}}/{\psi_U'(\rho)}-e^{-\rho u}g_{+}(b)},\quad 0\leq u\leq b, \end{equation}

where

$$h_+(x)=\left\{\begin{array}{ll} h(x), & x\geq 0,\\ 0, & x< 0, \end{array}\right. \quad g_+(x)=\left\{\begin{array}{ll} h_+'(x), & x\geq 0,\\ 0, & x<0. \end{array}\right.$$

Let $\tau _\infty =\inf \{t\geq 0: U^\infty _t\leq 0\}$ denote the ruin time of the model $U^\infty$, and accordingly define

\begin{align*} \phi_d(u;\infty)& =E[e^{-\delta\tau_\infty} I(U^\infty_{\tau_\infty}=0, \tau_\infty<\infty)| U_0^\infty=u],\quad u\geq 0,\\ \phi_c(u;\infty)& =E[e^{-\delta\tau_\infty} w(|U^\infty_{\tau_\infty}|)I(U^\infty_{\tau_\infty}<0, \tau_\infty<\infty)| U_0^\infty=u],\quad u\geq 0, \end{align*}

to be the Laplace transform of ruin time when ruin is caused by the Brownian motion, and the expected discounted penalty function when ruin is due to a claim. For the expected discounted penalty functions $\phi _d(u;b)$ and $\phi _c(u;b)$, they can be expressed via the dividends-penalty identities as follows,

(2.6)\begin{align} & \phi_d(u;b)=\phi_d(u;\infty)-V(u;b)\phi_d'(b;\infty),\quad 0\leq u\leq b, \end{align}

(2.7)\begin{align} & \phi_c(u;b)=\phi_c(u;\infty)-V(u;b)\phi_c'(b;\infty),\quad 0\leq u\leq b. \end{align}

For example, see Lin et al. [Reference Lin, Willmot and Drekic17].

3. The COS approximation

In this section, we shall use the COS method to approximate $V(u;b)$, $\phi _d(u;b)$ and $\phi _c(u;b)$. Recall that for an integrable function $f$ with a finite support $[a_1, a_2]$, it has the following cosine series expansion,

(3.1)\begin{equation} f(x)=\mathop{{\sum}'}_{k=0}^{\infty}A_{f,k} \cos\left(k\pi\frac{x-a_1}{a_2-a_1}\right), \end{equation}

where $\mathop {{\sum }'}$ denotes a summation with its first term weighted by a half, and the cosine coefficients are given by

(3.2)\begin{equation} A_{f,k}=\frac{2}{a_2-a_1}\text{Re}\left\{ \int_{a_1}^{a_2} f(x)\exp\left(ik\pi\frac{x-a_1}{a_2-a_1}\right)dx \right\},\quad k=0,1,2,\ldots, \end{equation}

where $\text {Re}(\cdot )$ means taking real part and $i=\sqrt {-1}$ is the imaginary unit.

If $f$ is an integrable function supported on the positive real line $[0, \infty )$, we can expand $f$ on a closed interval $[0, a]$, and the COS method suggests that, for large enough $a$, the COS coefficients can be approximated as follows,

(3.3)\begin{equation} A_{f,k}\approx B_{f,k}:=\frac{2}{a}\text{Re}\left\{\mathcal{F}f \left(\frac{k\pi}{a}\right) \right\},\quad k=0,1,2,\ldots. \end{equation}

Hence, we have

(3.4)\begin{equation} f(x) \approx \tilde{f}(x):= \mathop{{\sum}'}_{k=0}^K B_{f,k} \cos\left(k\pi\frac{x-a_1}{a_2-a_1}\right),\quad a_1\leq x\leq a_2, \end{equation}

where $K$ is a large integer applied to truncate the infinite series.

The following lemma is proved in Xie and Zhang [Reference Xie and Zhang27], which gives the approximation error for the COS method.

Lemma 1. For real-valued integrable function $f$ supported on $[0, \infty )$, suppose that $|f'(0+)|<\infty$, $|f'(a)|<\infty$, and $\int _0^\infty |f''(y)|\,dy<\infty$. Then, for some positive constants $c_1$ and $c_2$, we have

(3.5)\begin{equation} \sup_{x\in [0, a]}|f(x)-\tilde{f}(x)|\leq c_1 aK^{{-}1}+c_2Ka^{{-}1}\int_a^\infty |f(y)|\,dy. \end{equation}

Remark 2. Usually, the decay rate of the COS coefficients depends heavily on the smoothness of the objective function $f$. If $f$ is infinitely times differentiable, the series $\{A_{f,k}\}$ will show exponential convergence; otherwise, it will yield algebraic convergence. See, for example, Fang and Oosterlee [Reference Fang and Oosterlee9]. In our paper, note that the function $f$ that we approximate usually has support $[0, \infty )$, then using integration by parts we can find that

$$A_{f,k}=\frac{2a}{k^2\pi^2}\left\{ f'(a)\cos(k\pi)-f'(0+)-\int_0^a f''(x)\cos\left( k\pi\frac{x}{a} \right)dx\right\},$$

which yields that $A_{f,k}=O(k^{-2})$ for fixed $a$. Furthermore, if $f$ is $2n$ times differentiable and $f^{(2n)}$ is integrable, and $f^{(j)}(0+)=0$ for $j=1,3,\ldots, 2n-1$, then repeatedly using integration by parts we can prove that

$$A_{f,k}=O\left( a^{{-}1}\sum_{j=1}^{n} \left(\frac{a}{k}\right)^{2j} f^{(2j-1)}(a) +\frac{a^{2n-1}}{k^{2n}} \right).$$

Hence, if the derivatives $f^{(1)}, f^{(3)}, \ldots, f^{(2n-1)}$ all have exponential decay rate, we can first choose large $a$ to ignore the term $a^{-1}\sum _{j=1}^{n} ({a}/{k})^{2j} f^{(2j-1)}(a)$, then for such $a$, $A_{f,k}=O(k^{-2n})$, which can further lead to faster convergence for the COS approximation.

In order to use the COS method to approximate $V(u;b)$, we should first approximate $h_+$ and $g_+$. It follows from Xie and Zhang [Reference Xie and Zhang27] that

(3.6)\begin{equation} \mathcal{L}h_+(s)=\frac{\delta}{\psi_U'(\rho)}\cdot \frac{1}{s-\rho}-\frac{\delta}{\psi_U(s)-\delta},\quad s\neq \rho, \end{equation}

and

(3.7)\begin{equation} \mathcal{F}h_{+}(s)=\mathcal{L}h_+({-}is)=\frac{\delta}{\psi_U'(\rho)}\cdot \frac{1}{-is-\rho}-\frac{\delta}{\psi_U({-}is)-\delta},\quad s\in \mathbb{R}. \end{equation}

Since $W_\delta (0)=0$ as $\sigma >0$, we have from (2.3) that $h_+(0)= {\delta } / {\psi _U'(\rho )}$, and this yields

(3.8)\begin{equation} \mathcal{L}g_+(s)=s\mathcal{L}h_+(s)-h_+(0) =\frac{\delta}{\psi_U'(\rho)}\cdot \frac{\rho}{s-\rho}-\frac{\delta s}{\psi_U(s)-\delta},\quad s\neq \rho, \end{equation}

and

(3.9)\begin{equation} \mathcal{F}g_+(s)=\mathcal{L}g_+({-}is)=\frac{\delta}{\psi_U'(\rho)}\cdot \frac{\rho}{-is-\rho}+\frac{i\delta s}{\psi_U({-}is)-\delta},\quad s\in \mathbb{R}. \end{equation}

Now using the closed-form Fourier transforms $\mathcal {F}h_+(s)$ and $\mathcal {F}g_+(s)$, the functions $h_+$ and $g_+$ can be approximated by the COS method as follows,

(3.10)\begin{equation} h_+(x) \approx \tilde{h}_{+}(x):= \mathop{{\sum}'}_{k=0}^K B_{h_+,k} \cos\left(k\pi\frac{x}{a}\right),\quad 0\leq x\leq a, \end{equation}

and

(3.11)\begin{equation} g_+(x) \approx \tilde{g}_{+}(x):= \mathop{{\sum}'}_{k=0}^K B_{g_+,k} \cos\left(k\pi\frac{x}{a}\right),\quad 0\leq x\leq a, \end{equation}

where the COS coefficients are given by

(3.12)\begin{equation} B_{h_+,k}:=\frac{2}{a}\text{Re}\left\{\mathcal{F}h_+ \left(\frac{k\pi}{a}\right) \right\},\quad B_{g_+,k}:=\frac{2}{a}\text{Re}\left\{\mathcal{F}g_+ \left(\frac{k\pi}{a}\right) \right\},\quad k=0,1,\ldots. \end{equation}

As for the expected present value of dividend payments before ruin, using formula (2.5) we can approximate it as follows,

(3.13)\begin{equation} V(u;b)\approx \tilde{V}(u;b): =\frac{{\delta}/{\psi_U'(\rho)} -e^{-\rho u}\tilde{h}_{+}(u)}{{\delta\rho e^{\rho(b-u)}}/{\psi_U'(\rho)}-e^{-\rho u}\tilde{g}_{+}(b)},\quad 0\leq u\leq b, \end{equation}

where we take $a>b$.

Remark 4. Suppose that $|h'(0+)|<\infty,\ |h''(0+)|<\infty$, $|h'(a)|<\infty$, $|h''(a)|<\infty$ and

$$\int_0^\infty |h''(y)|\,dy<\infty,\quad \int_0^\infty |h'''(y)|\,dy<\infty.$$

Furthermore, let $C$ be a positive generic constant that may vary at different steps. Then by Lemma 1 we can obtain

(3.14)\begin{equation} |\tilde{V}(u;b)-V(u;b)|\leq C\cdot \mathcal{E}_h(a,K), \end{equation}

where

(3.15)\begin{equation} \mathcal{E}_h(a,K)=aK^{{-}1}+Ka^{{-}1}\int_a^\infty h(y)\,dy+ Ka^{{-}1}\int_a^\infty |h'(y)|\,dy. \end{equation}

Next, we consider how to use the COS method to approximate the expected discounted penalty functions $\phi _d(u;b)$ and $\phi _c(u;b)$. To this end, we use the dividends-penalty identities given in (2.6) and (2.7). It is easily seen that we should use the COS method to approximate $\phi _{d}(u;\infty ), \phi _c(u;\infty )$ and their first-order derivatives.

For $\phi _d(u;\infty )$, by Zhang [Reference Zhang29], we know that its Laplace transform and Fourier transform are given by

(3.16)\begin{equation} \mathcal{L}\phi_d(s;\infty)=\frac{ ({\sigma^2}/{2})(s-\rho) }{\psi_U(s)-\delta},\quad \mathcal{F}\phi_d(s;\infty)=\frac{ ({\sigma^2}/{2})({-}is-\rho) }{\psi_U({-}is)-\delta}. \end{equation}

Then, the Laplace transform of the derivative $\phi _d'(u;\infty )$ is given by

(3.17)\begin{equation} \mathcal{L}\phi_d'(s;\infty)=s\mathcal{L}\phi_d(s;\infty)-\phi_d(0;\infty)= \frac{\delta-(c+{\sigma^2}/{2}\rho)s+\lambda[ 1-\mathcal{L}f_X(s) ] }{\psi_U(s)-\delta }, \end{equation}

from which we obtain

(3.18)\begin{equation} \mathcal{F}\phi_d'(s;\infty)=\mathcal{L}\phi_d'({-}is;\infty)=\frac{\delta+(c+({\sigma^2}/{2})\rho)is+\lambda[ 1-\mathcal{F}f_X(s) ] }{\psi_U({-}is)-\delta }. \end{equation}

For $\phi _c(u;\infty )$, its Laplace transform and Fourier transform are given by

(3.19)\begin{equation} \mathcal{L}\phi_c(s;\infty)=\frac{ \lambda [ \mathcal{L}\omega(\rho)-\mathcal{L}\omega(s) ] }{\psi_U(s)-\delta},\quad \mathcal{F}\phi_c(s;\infty)=\frac{ \lambda[\mathcal{L}\omega(\rho)-\mathcal{F}\omega(s)] }{\psi_U({-}is)-\delta}, \end{equation}

where $\omega (u)=\int _u^\infty w(x-u)f_X(x)\,dx$, $u\geq 0$. Furthermore, since $\phi _c(0;\infty )=0$, we have

(3.20)\begin{equation} \mathcal{L}\phi_c'(s;\infty)=s\mathcal{L}\phi_c(s;\infty) =\frac{\lambda s[\mathcal{L}\omega(\rho)-\mathcal{L}\omega(s)] }{\psi_U(s)-\delta } \end{equation}

and

(3.21)\begin{equation} \mathcal{F}\phi_c'(s;\infty)=\mathcal{L}\phi_c'({-}is;\infty) =\frac{-\lambda is[\mathcal{L}\omega(\rho)-\mathcal{F}\omega(s)] }{\psi_U({-}is)-\delta }. \end{equation}

Now, using the COS method, we have

(3.22)\begin{align} \phi_d(u;\infty) \approx \tilde{\phi}_{d}(u;\infty):= \mathop{{\sum}'}_{k=0}^K B_{\phi_d,k} \cos\left(k\pi\frac{u}{a}\right),\quad 0\leq u\leq a, \end{align}

(3.23)\begin{align} \phi_c(u;\infty) \approx \tilde{\phi}_{c}(u;\infty):= \mathop{{\sum}'}_{k=0}^K B_{\phi_c,k} \cos\left(k\pi\frac{u}{a}\right),\quad 0\leq u\leq a, \end{align}

(3.24)\begin{align} \phi_d'(u;\infty) \approx \tilde{\phi}_{d}'(u;\infty):= \mathop{{\sum}'}_{k=0}^K B_{\phi_d',k} \cos\left(k\pi\frac{u}{a}\right),\quad 0\leq u\leq a, \end{align}

(3.25)\begin{align} \phi_c'(u;\infty) \approx \tilde{\phi}_{c}'(u;\infty):= \mathop{{\sum}'}_{k=0}^K B_{\phi_c',k} \cos\left(k\pi\frac{u}{a}\right),\quad 0\leq u\leq a, \end{align}

where, for $k=0,1,\ldots$, the COS coefficients are given by

(3.26)\begin{align} B_{\phi_d,k}:=\frac{2}{a}\text{Re}\left\{\mathcal{F}\phi_d \left(\frac{k\pi}{a};\infty\right) \right\},\quad B_{\phi_c,k}:=\frac{2}{a}\text{Re}\left\{\mathcal{F}\phi_c \left(\frac{k\pi}{a};\infty\right) \right\}, \end{align}

(3.27)\begin{align} B_{\phi_d',k}:=\frac{2}{a}\text{Re}\left\{\mathcal{F}\phi_d' \left(\frac{k\pi}{a};\infty\right) \right\},\quad B_{\phi_c',k}:=\frac{2}{a}\text{Re}\left\{\mathcal{F}\phi_c' \left(\frac{k\pi}{a};\infty\right) \right\}. \end{align}

Finally, by the dividends-penalty identities (2.6) and (2.7), we obtain

(3.28)\begin{equation} \tilde{\phi}_d(u;b)= \tilde{\phi}_d(u;\infty)-\tilde{V}(u;b)\tilde{\phi}_d'(b;\infty),\quad 0\leq u\leq b, \end{equation}

and

(3.29)\begin{equation} \tilde{\phi}_c(u;b)= \tilde{\phi}_c(u;\infty)-\tilde{V}(u;b)\tilde{\phi}_c'(b;\infty),\quad 0\leq u\leq b. \end{equation}

Remark 5. Again, the approximating error can be obtained by Lemma 1. Suppose all the conditions in Remark 4 hold true. Furthermore, suppose that $|\phi _d'(0+;\infty )|<\infty,\ |\phi _d''(0+;\infty )|<\infty$, $|\phi _d'(a;\infty )|<\infty$, $|\phi _d''(a;\infty )|<\infty$ and

$$\int_0^\infty |\phi_d''(u;\infty)|\,du<\infty,\quad \int_0^\infty |\phi_d'''(u;\infty)|\,du<\infty,$$

then we can obtain from Lemma 1 and Remark 4 that

(3.30)\begin{equation} |\tilde{\phi}_d(u;b)-\phi_d(u;b)|\leq C\cdot (\mathcal{E}_{\phi_d}(a,K)+\mathcal{E}_{h}(a,K)), \end{equation}

where $\mathcal {E}_h(a,K)$ is defined in (3.15), and

(3.31)\begin{equation} \mathcal{E}_{\phi_d}(a,K)=aK^{{-}1}+Ka^{{-}1}\int_a^\infty \phi_d(u;\infty)\,du+Ka^{{-}1}\int_a^\infty |\phi_d'(u;\infty)|\,du. \end{equation}

Similarly, suppose that $|\phi _c'(0+;\infty )|<\infty,\ |\phi _c''(0+;\infty )|<\infty$, $|\phi _c'(a;\infty )|<\infty$, $|\phi _c''(a;\infty )|<\infty$ and

$$\int_0^\infty |\phi_c''(u;\infty)|du<\infty,\ \ \int_0^\infty |\phi_c'''(u;\infty)|\,du<\infty.$$

Then, we can obtain

(3.32)\begin{equation} |\tilde{\phi}_c(u;b)-\phi_c(u;b)|\leq C\cdot (\mathcal{E}_{\phi_c}(a,K)+\mathcal{E}_{h}(a,K)), \end{equation}

where

(3.33)\begin{equation} \mathcal{E}_{\phi_c}(a,K)=aK^{{-}1}+Ka^{{-}1}\int_a^\infty \phi_c(u;\infty)\,du+Ka^{{-}1}\int_a^\infty |\phi_c'(u;\infty)|\,du. \end{equation}

4. The estimation method

In this section, we show how to estimate $V(u;b)$, $\phi _d(u;b)$ and $\phi _c(u;b)$ from a random sample on the surplus process. We assume that the premium rate $c$ is known, but the Poisson intensity $\lambda$, the claim size density function $f_X$ and the diffusion parameter $\sigma$ are all unknown.

Suppose that the surplus process can be observed in the long time interval $[0, T]$, and the following random sample on the claim number and individual claim sizes during $[0, T]$ is available,

$$\{n_T,\ X_1,\ldots, X_{n_T}\},$$

where $n_T$ denotes the number of claims received up to time $T$ and it is assumed to be strictly positive w.l.o.g. In addition, suppose that the surplus process $U^b_t$, the aggregate claims process $S_t$ and the aggregate dividends process $D(t)$ can be observed at a sequence of discrete time points so that the following sample is available

$$\{(U^b_{k\Delta}, S_{k\Delta}, D(k\Delta)):\ k=0, 1,2,\ldots, n\},$$

where $\Delta >0$ is a fixed sampling step satisfying $n\Delta =T$.

First, we use formula (3.13) to propose an estimator for the expected present value of dividend payments before ruin. It is easily seen that we need to estimate the following quantities,

$$\rho,\ \ \psi_U',\ \ \tilde{h}_+,\ \ \tilde{g}_+.$$

Since $\rho$ is the positive root of equation $\psi _U(s)=\delta$, we should estimate the Laplace exponent $\psi _U(s)$ by formula (2.1). We estimate the Poisson intensity by $\hat {\lambda }={n_T}/{T}$, and estimate the Laplace transform $\mathcal {L}f_X(s)$ by $\widehat {\mathcal {L}f}_X(s) =({1}/{n_T})\sum _{j=1}^{n_T} e^{-sX_j}$. It remains to estimate the diffusion parameter $\sigma$. Since $U^b_t=U^\infty _t-D(t)$, then we have

$$\sigma B_t=U^\infty_t-u-ct+S_t=U^b_t+D(t)+S_t-u-ct.$$

For $k=1,\ldots, n$, set

$$Z_k=[U^b_{k\Delta}-U^b_{(k-1)\Delta}]+[D(k\Delta)-D((k-1)\Delta)]+[S_{k\Delta}-S_{(k-1)\Delta}]-c\Delta,$$

which are available since the premium rate $c$ is known. Now, we estimate $\sigma ^2$ by

$$\hat{\sigma}^2=\frac{1}{n\Delta}\sum_{k=1}^nZ_k^2.$$

It is easily seen that $\hat {\lambda }, \widehat {\mathcal {L}f}_X(s)$ and $\hat {\sigma }^2$ are all unbiased estimators. Now using these estimators, we can estimate the Laplace exponent $\psi _U(s)$ and its derivative $\psi _U'(s)$ as follows,

\begin{align*} \hat{\psi}_U(s)& =\frac{\hat{\sigma}^2}{2}s^2+cs-\hat{\lambda} (1-\widehat{\mathcal{L}f}_X(s)),\\ \hat{\psi}'_U(s)& =\hat{\sigma}^2s+c-\frac{1}{T}\sum_{j=1}^{n_T} X_j e^{{-}sX_j}. \end{align*}

Since $\rho$ is the positive root of equation $\psi _U(s)=\delta$, we define its estimator, denoted by $\hat {\rho }$, to be the positive root of the following estimating equation

(4.1)\begin{equation} \hat{\psi}_U(s)=\delta. \end{equation}

Then, we can estimate $\psi _U'(\rho )$ by $\hat {\psi }_U'(\hat {\rho })$.

In order to estimate $\tilde {h}_+$ and $\tilde {g}_+$, we should first estimate the Fourier transforms ${\mathcal {F}h}_+(s)$ and $\mathcal {F}g_+(s)$. It follows from (3.7) and (3.9) that they can be estimated by the following estimators

(4.2)\begin{align} \widehat{\mathcal{F}h}_{+}(s)& =\frac{\delta}{\hat{\psi}_U'(\hat{\rho})}\cdot \frac{1}{-is-\hat{\rho}}-\frac{\delta}{\hat{\psi}_U({-}is)-\delta},\quad s\in \mathbb{R}, \end{align}

(4.3)\begin{align} \widehat{\mathcal{F}g}_+(s)& =\frac{\delta}{\hat{\psi}_U'(\hat{\rho})}\cdot \frac{\hat{\rho}}{-is-\hat{\rho}}+\frac{i\delta s}{\hat{\psi}_U({-}is)-\delta},\quad s\in \mathbb{R}. \end{align}

Then, using formulas (3.10) and (3.11), we propose the following estimators for $\tilde {h}_+$ and $\tilde {g}_+$,

(4.4)\begin{equation} *\hat{h}_{+}(x):= \mathop{{\sum}'}_{k=0}^K \hat{B}_{h_+,k} \cos\left(k\pi\frac{x}{a}\right),\quad 0\leq x\leq a, \end{equation}

and

(4.5)\begin{equation} \hat{g}_{+}(x):= \mathop{{\sum}'}_{k=0}^K \hat{B}_{g_+,k} \cos\left(k\pi\frac{x}{a}\right),\quad 0\leq x\leq a, \end{equation}

where the COS coefficients are given by

(4.6)\begin{equation} \hat{B}_{h_+,k}:=\frac{2}{a}\text{Re}\left\{\widehat{\mathcal{F}h}_+ \left(\frac{k\pi}{a}\right) \right\},\quad \hat{B}_{g_+,k}:=\frac{2}{a}\text{Re}\left\{\widehat{\mathcal{F}g}_+ \left(\frac{k\pi}{a}\right) \right\},\quad k=0,1,\ldots. \end{equation}

Now, using formula (3.13), we obtain the following estimator of $V(u;b)$,

(4.7)\begin{equation} \hat{V}(u;b): =\frac{{\delta}/{\hat{\psi}_U'(\hat{\rho})} -e^{-\hat{\rho} u}\hat{h}_{+}(u)}{{\delta\hat{\rho} e^{\hat{\rho}(b-u)}}/{\hat{\psi}_U'(\hat{\rho})}-e^{-\hat{\rho} u}\hat{g}_{+}(b)},\quad 0\leq u\leq b< a. \end{equation}

Next, we consider how to estimate the expected discounted penalty function $\phi _d(u;b)$. To this end, we shall use the dividends-penalty identity (2.6). Note that the Fourier transform of $\phi _d(u;\infty )$ and its derivative can be estimated by

(4.8)\begin{equation} \widehat{\mathcal{F}\phi}_d(s;\infty)=\frac{ ({\hat{\sigma}^2}/{2})({-}is-\hat{\rho}) }{\hat{\psi}_U({-}is)-\delta}, \quad \widehat{\mathcal{F}\phi}_d'(s;\infty)= \frac{\delta+(c+({\hat{\sigma}^2}/{2})\hat{\rho})is+\hat{\lambda}[ 1-\widehat{\mathcal{F}f}_X(s) ] }{\hat{\psi}_U({-}is)-\delta }. \end{equation}

So we can estimate $\phi _d(u;\infty )$ and $\phi _d'(u;\infty )$ as follows,

(4.9)\begin{equation} \hat{\phi}_{d}(u;\infty):= \mathop{{\sum}'}_{k=0}^K \hat{B}_{\phi_d,k} \cos\left(k\pi\frac{u}{a}\right),\quad \hat{\phi}_{d}'(u;\infty):= \mathop{{\sum}'}_{k=0}^K \hat{B}_{\phi_d',k} \cos\left(k\pi\frac{u}{a}\right),\quad 0\leq u\leq a, \end{equation}

where, for $k=0,1,\ldots,$

(4.10)\begin{equation} \hat{B}_{\phi_d,k}:=\frac{2}{a}\text{Re}\left\{\widehat{\mathcal{F}\phi}_d \left(\frac{k\pi}{a};\infty\right) \right\},\quad \hat{B}_{\phi_d',k}:=\frac{2}{a}\text{Re}\left\{\widehat{\mathcal{F}\phi}_d' \left(\frac{k\pi}{a};\infty\right) \right\}. \end{equation}

By the dividends-penalty identity (2.6), we propose the following estimator for $\phi _d(u;b)$,

(4.11)\begin{equation} \hat{\phi}_d(u;b)= \hat{\phi}_d(u;\infty)-\hat{V}(u;b)\hat{\phi}_d'(b;\infty),\quad 0\leq u\leq b. \end{equation}

Finally, we estimate the expected discounted penalty function $\phi _c(u;b)$. It is easily seen that

$$\mathcal{L}\omega(s)=\int_0^\infty \int_0^x e^{{-}su}w(x-u) \,du f_X(x)\,dx,\quad \mathcal{F}\omega(s)=\int_0^\infty\int_0^x e^{isu} w(x-u)\,du f_X(x)\,dx,$$

from which we obtain the following estimators for $\mathcal {L}\omega (\rho )$ and $\mathcal {F}\omega (s)$,

(4.12)\begin{equation} \widehat{\mathcal{L}\omega}(\hat{\rho})=\frac{1}{n_T}\sum_{j=1}^{n_T} \int_0^{X_j} e^{-\hat{\rho}u} w(X_j-u)\,du,\quad \widehat{\mathcal{F}\omega}(s) =\frac{1}{n_T}\sum_{j=1}^{n_T} \int_0^{X_j} e^{isu} w(X_j-u)\,du. \end{equation}

Now the Fourier transforms $\mathcal {F}\phi _c(s;\infty )$ and $\mathcal {F}\phi _c'(s;\infty )$ can be estimated by

(4.13)\begin{equation} \widehat{\mathcal{F}\phi}_c(s;\infty)=\frac{ \hat{\lambda}[\widehat{\mathcal{L}\omega}(\hat{\rho})- \widehat{\mathcal{F}\omega}(s)] }{\hat{\psi}_U({-}is)-\delta},\quad \widehat{\mathcal{F}\phi}_c'(s;\infty) =\frac{-\hat{\lambda} is[\widehat{\mathcal{L}\omega}(\hat{\rho})-\widehat{\mathcal{F}\omega}(s)] }{\hat{\psi}_U({-}is)-\delta }. \end{equation}

Then $\phi _c(u;\infty )$ and $\phi _c'(u;\infty )$ are estimated as follows,

(4.14)\begin{equation} \hat{\phi}_{c}(u;\infty):= \mathop{{\sum}'}_{k=0}^K \hat{B}_{\phi_c,k} \cos\left(k\pi\frac{u}{a}\right),\quad \hat{\phi}_{c}'(u;\infty):= \mathop{{\sum}'}_{k=0}^K \hat{B}_{\phi_c',k} \cos\left(k\pi\frac{u}{a}\right),\quad 0\leq u\leq a, \end{equation}

where, for $k=0,1,\ldots,$

(4.15)\begin{equation} \hat{B}_{\phi_c,k}:=\frac{2}{a}\text{Re}\left\{\widehat{\mathcal{F}\phi}_c \left(\frac{k\pi}{a};\infty\right) \right\},\quad \hat{B}_{\phi_c',k}:=\frac{2}{a}\text{Re}\left\{\widehat{\mathcal{F}\phi}_c' \left(\frac{k\pi}{a};\infty\right) \right\}. \end{equation}

By the dividends-penalty identity (2.7), we propose the following estimator for $\phi _c(u;b)$,

(4.16)\begin{equation} \hat{\phi}_c(u;b)= \hat{\phi}_c(u;\infty)-\hat{V}(u;b)\hat{\phi}_c'(b;\infty),\quad 0\leq u\leq b< a. \end{equation}

5. Consistency properties

In this section, we derive the consistency properties of our estimators when the observation interval $[0, T]$ is very large. First, it is easily seen that

$$\hat{\lambda}-\lambda=O_p(T^{{-}1/2}),\quad \hat{\sigma}^2-\sigma^2=O_p(T^{{-}1/2}).$$

Here, the notation $O_p(1)$ denotes a sequence that is bounded in probability; or more generally in our paper, for a given sequence of $\{R_{T,K,a}\}$, $O_p(R_{T,K,a})$ denotes a sequence that is bounded in probability at rate $R_{T,K,a}$. The following result on the estimator $\hat {\rho }$ is also well known. See, for example, Zhang [Reference Zhang29].

Suppose the conditions in Lemma 2 hold true. For the estimator $\hat {\psi }_U'(\hat {\rho })$, note that

(5.1)\begin{equation} \hat{\psi}_U'(\hat{\rho})-\psi_U'(\rho)= (\hat{\sigma}^2\hat{\rho}-\sigma^2\rho) -\left( \frac{1}{T}\sum_{j=1}^{n_T} X_j e^{-\hat{\rho} X_j} -\lambda E[ X e^{-\rho X}] \right). \end{equation}

Lemma 2 and $\hat {\sigma }^2-\sigma ^2=O_p(T^{-1/2})$ imply that $(\hat {\sigma }^2\hat {\rho }-\sigma ^2\rho )=O_p(T^{-1/2})$. By Proposition 4 in Xie and Zhang [Reference Xie and Zhang27], we obtain

$$\frac{1}{T}\sum_{j=1}^{n_T} X_j e^{-\hat{\rho} X_j} -\lambda E[ X e^{-\rho X}]=O_p(T^{{-}1/2}).$$

Hence, it follows from (5.1) that

(5.2)\begin{equation} \hat{\psi}_U'(\hat{\rho})-\psi_U'(\rho)=O_p(T^{{-}1/2}). \end{equation}

Recall that $\rho$ is the root of equation $\psi _U(s)=\delta$, then we have

(5.3)\begin{align} \psi_U({-}is)-\delta& =\psi_U({-}is)-\psi_U(\rho)\nonumber\\ & =({-}is-\rho)\left( \frac{\sigma^2}{2}({-}is+\rho)+c-\lambda \frac{\mathcal{L}f_X({-}is)-\mathcal{L}f_X(\rho) }{ -is-\rho} \right)\nonumber\\ & =({-}is-\rho)l(s), \end{align}

where $l(s)= ({\sigma ^2}/{2})(-is+\rho )+c+\lambda \int _0^\infty e^{isx}\int _0^x e^{-(\rho +is)y}\,dyf_X(x)\,dx$. Note that

(5.4)\begin{align} |l(s)|& \geq \left|\frac{\sigma^2}{2}({-}is+\rho)+c\right|-\left|\lambda \int_0^\infty e^{isx}\int_0^x e^{-(\rho+is)y}\,dyf_X(x)\,dx\right|\nonumber\\ & \geq \left|\frac{\sigma^2}{2}({-}is+\rho)+c\right|-\lambda EX \end{align}

(5.5)\begin{align} & \geq \frac{\sigma^2}{2}\rho+c-\lambda EX>c-\lambda EX>0. \end{align}

Similarly, since $\hat {\rho }$ is the root of equation $\hat {\psi }_U(s)=\delta$, we have

(5.6)\begin{align} \hat{\psi}_U({-}is)-\delta& =({-}is-\hat{\rho})\left( \frac{\hat{\sigma}^2}{2}({-}is+\hat{\rho})+c-\hat{\lambda} \frac{\widehat{\mathcal{L}f}_X({-}is)-\widehat{\mathcal{L}f}_X(\hat{\rho}) }{ -is-\hat{\rho}}\right)\nonumber\\ & =({-}is-\hat{\rho})\hat{l}(s), \end{align}

where $\hat {l}(s)= ({\hat {\sigma }^2}/{2})(-is+\hat {\rho })+c +({1}/{T})\sum _{j=1}^{n_T} e^{is X_j} \int _0^{X_j} e^{-(\hat {\rho }+is)y}\,dy$. Furthermore, note that

(5.7)\begin{align} \hat{l}(s)& \geq \left|\frac{\hat{\sigma}^2}{2}({-}is+\hat{\rho})+c\right| -\left|\frac{1}{T}\sum_{j=1}^{n_T} e^{is X_j} \int_0^{X_j} e^{-(\hat{\rho}+is)y}\,dy\right|\nonumber\\ & \geq \left|\frac{\hat{\sigma}^2}{2}({-}is+\hat{\rho})+c\right| -\frac{1}{T}\sum_{j=1}^{n_T} X_j \end{align}

(5.8)\begin{align} & \geq \frac{\hat{\sigma}^2}{2}\hat{\rho}+c-\frac{1}{T}\sum_{j=1}^{n_T} X_j>c-\frac{1}{T}\sum_{j=1}^{n_T} X_j, \end{align}

which together with (5.6) gives

(5.9)\begin{equation} |\hat{\psi}_U({-}is)-\delta| \geq |is+\hat{\rho}|\cdot \left( c -\frac{1}{T}\sum_{j=1}^{n_T} X_j\right). \end{equation}

The following lemma shows the uniform convergence of ${1}/{(\hat {\psi }_U(-is)-\delta )}$ and ${s}/{(\hat {\psi }_U(-is)-\delta )}$, which plays an important role in studying the convergence of our estimators.

Lemma 3. Suppose that $c>\lambda EX$, $EX^2<\infty$ and $a=o(K)$. Then, we have

(5.10)\begin{equation} \sup_{s\in \mathcal{S}} \left|\frac{1}{\hat{\psi}_U({-}is)-\delta} -\frac{1}{\psi_U({-}is)-\delta} \right|=O_p\left( \sqrt{\frac{\log(K/a) }{T } }\right) \end{equation}

and

(5.11)\begin{equation} \sup_{s\in \mathcal{S}} \left|\frac{s}{\hat{\psi}_U({-}is)-\delta} -\frac{s}{\psi_U({-}is)-\delta} \right|=O_p\left( \sqrt{\frac{\log(K/a) }{T } }\right). \end{equation}

Proof. First, we study the uniform convergence of ${1}/{(\hat {\psi }_U(-is)-\delta )}$. By (5.3) and (5.5), we have for any real number $s$,

(5.12)\begin{equation} |\psi_U({-}is)-\delta| \geq |is+\rho|\cdot [c-\lambda EX], \end{equation}

and by (5.6) and (5.8) we have for any real number $s$,

(5.13)\begin{equation} |\hat{\psi}_U({-}is)-\delta| \geq |is+\hat{\rho}|\cdot \left( c -\frac{1}{T}\sum_{j=1}^{n_T} X_j \right). \end{equation}

Recall the condition $c>\lambda EX$. In the remainder of this proof, suppose that $c -({1}/{T})\sum _{j=1}^{n_T} X_j>0$, since $({1}/{T})\sum _{j=1}^{n_T} X_j$ converges to $\lambda EX$ a.s.

Now using (5.12) and (5.13) and the following result

(5.14)\begin{align} \psi_U({-}is)-\hat{\psi}_U({-}is)& =\left(\frac{\sigma^2}{2}-\frac{\hat{\sigma}^2}{2}\right) ({-}is)^2-(\lambda-\hat{\lambda})+\lambda\mathcal{L}f_X({-}is)-\hat{\lambda} \widehat{\mathcal{L}f}_X({-}is)\nonumber\\ & =\frac{1}{2}(\hat{\sigma}^2-\sigma^2)s^2+(\hat{\lambda}-\lambda) -\left( \frac{1}{T}\sum_{j=1}^{n_T} e^{isX_j} -\lambda Ee^{isX} \right), \end{align}

we can obtain

(5.15)\begin{align} & \sup_{s\in \mathcal{S}} \left|\frac{1}{\hat{\psi}_U({-}is)-\delta} -\frac{1}{\psi_U({-}is)-\delta} \right|\nonumber\\ & \quad =\sup_{s\in \mathcal{S}} \frac{ |\psi_U({-}is)-\hat{\psi}_U({-}is)| }{ |\psi_U({-}is)-\delta |\cdot | \hat{\psi}_U({-}is)-\delta | } \nonumber\\ & \quad \leq \sup_{s\in\mathcal{S}} \frac{|\frac{1}{2}(\hat{\sigma}^2-\sigma^2)s^2+(\hat{\lambda}-\lambda) -( ({1}/{T})\sum_{j=1}^{n_T} e^{isX_j} -\lambda Ee^{isX} )|}{|is+\rho|\cdot |is+\hat{\rho}|\cdot\left( c-\lambda EX \right)\cdot ( c -({1}/{T})\sum_{j=1}^{n_T} X_j) }\nonumber\\ & \quad \leq \sup_{s\in\mathcal{S}} \frac{\left|\frac{1}{2}(\hat{\sigma}^2-\sigma^2)s^2\right|}{|is+\rho|\cdot |is+\hat{\rho}|\cdot\left( c-\lambda EX \right)\cdot ( c -({1}/{T})\sum_{j=1}^{n_T} X_j) }\nonumber\\ & \qquad + \sup_{s\in\mathcal{S}} \frac{|(\hat{\lambda}-\lambda) -( ({1}/{T})\sum_{j=1}^{n_T} e^{isX_j} -\lambda Ee^{isX} )|}{|is+\rho|\cdot |is+\hat{\rho}|\cdot\left( c-\lambda EX \right)\cdot (c -({1}/{T})\sum_{j=1}^{n_T} X_j ) }\nonumber\\ & \quad \leq \frac{\frac{1}{2}|\hat{\sigma}^2-\sigma^2|}{\left( c-\lambda EX \right)\cdot ( c -({1}/{T})\sum_{j=1}^{n_T} X_j) }+ \frac{ \sup_{s\in\mathcal{S}}|(\hat{\lambda}-\lambda) -( ({1}/{T})\sum_{j=1}^{n_T} e^{isX_j} -\lambda Ee^{isX} )|}{\rho\hat{\rho}\cdot \left( c-\lambda EX \right)\cdot ( c -({1}/{T})\sum_{j=1}^{n_T} X_j ) }. \end{align}

The convergence rate $\hat {\sigma }^2-\sigma ^2=O_p(T^{-1/2})$ implies that

(5.16)\begin{equation} \frac{\frac{1}{2}|\hat{\sigma}^2-\sigma^2|}{\left( c-\lambda EX \right)\cdot ( c -({1}/{T})\sum_{j=1}^{n_T} X_j ) }=O_p(T^{{-}1/2}). \end{equation}

It follows from Lemma 3 in Xie and Zhang [Reference Xie and Zhang27] that, for $a=o(K)$,

(5.17)\begin{equation} \sup_{s\in \mathcal{S}}\left|(\hat{\lambda}-\lambda) -\left( \frac{1}{T}\sum_{j=1}^{n_T} e^{isX_j} -\lambda Ee^{isX} \right)\right|=O_{p}\left(\sqrt{\frac{\log(K/a)}{T}}\right), \end{equation}

leading to

$$\frac{\sup_{s\in\mathcal{S}}|(\hat{\lambda}-\lambda) -( ({1}/{T})\sum_{j=1}^{n_T} e^{isX_j} -\lambda Ee^{isX})|}{\rho\hat{\rho}\cdot ( c-\lambda EX )\cdot ( c -({1}/{T})\sum_{j=1}^{n_T} X_j) }=O_{p}\left(\sqrt{\frac{\log(K/a)}{T}}\right),$$

which together with (5.15), (5.16) yields (5.10).

Next, we prove (5.11). By (5.3) and (5.6), we have

(5.18)\begin{align} \frac{s}{\hat{\psi}_U({-}is)-\delta} -\frac{s}{\psi_U({-}is)-\delta}&=\frac{ s [\psi_U({-}is)-\hat{\psi}_U({-}is)] }{ [\psi_U({-}is)-\delta]\cdot [\hat{\psi}_U({-}is)-\delta] }\nonumber\\ & =\frac{ s(\frac{1}{2}(\hat{\sigma}^2-\sigma^2)s^2+(\hat{\lambda}-\lambda) -( ({1}/{T})\sum_{j=1}^{n_T} e^{isX_j} -\lambda Ee^{isX} )) }{({-}is-\rho)({-}is-\hat{\rho})l(s)\hat{l}(s) }, \end{align}

which gives

(5.19)\begin{align} & \sup_{s\in\mathcal{S}}\left|\frac{s}{\hat{\psi}_U({-}is)-\delta} -\frac{s}{\psi_U({-}is)-\delta} \right|\nonumber\\ & \quad \leq \sup_{s\in \mathcal{S}}\left| \frac{\frac{1}{2}(\hat{\sigma}^2-\sigma^2) s^3 }{ ({-}is-\rho)({-}is-\hat{\rho}) l(s)\hat{l}(s) } \right|+\sup_{s\in\mathcal{S}}\left|\frac{ s((\hat{\lambda}-\lambda) -( ({1}/{T})\sum_{j=1}^{n_T} e^{isX_j} -\lambda Ee^{isX} )) }{({-}is-\rho)({-}is-\hat{\rho})l(s)\hat{l}(s) }\right|\nonumber\\ & \quad \leq \frac{1}{2}\cdot \sup_{s\in \mathcal{S}} \left| \frac{(\hat{\sigma}^2-\sigma^2)s}{l(s)\hat{l}(s)} \right|+\frac{1}{\rho}\cdot \sup_{s\in\mathcal{S}}\left|\frac{(\hat{\lambda}-\lambda) -( ({1}/{T})\sum_{j=1}^{n_T} e^{isX_j} -\lambda Ee^{isX} ) }{l(s)\hat{l}(s) }\right|. \end{align}

It follows from (5.4) that

$$\sup_{s\in \mathcal{S}} | s/l(s)| \leq \sup_{s\in\mathcal{S}} \left|\frac{s}{|(\sigma^2/{2})({-}is+\rho)+c|-\lambda EX}\right|\leq C,$$

which together with (5.8) and the convergence rate $\hat {\sigma }^2-\sigma ^2=O_p(T^{-1/2})$ gives

(5.20)\begin{equation} \sup_{s\in \mathcal{S}} \left| \frac{(\hat{\sigma}^2-\sigma^2)s}{l(s)\hat{l}(s)} \right|\leq \frac{C}{c-({1}/{T})\sum_{j=1}^{n_T} X_j} |\hat{\sigma}^2-\sigma^2|=O_p(T^{{-}1/2}). \end{equation}

By (5.5), (5.8) and (5.17), we obtain

(5.21)\begin{align} & \sup_{s\in\mathcal{S}}\left|\frac{(\hat{\lambda}-\lambda) -( ({1}/{T})\sum_{j=1}^{n_T} e^{isX_j} -\lambda Ee^{isX} ) }{l(s)\hat{l}(s) }\right|\nonumber\\ & \quad \leq \frac{ \sup_{s\in\mathcal{S}}|(\hat{\lambda}-\lambda) -( ({1}/{T})\sum_{j=1}^{n_T} e^{isX_j} -\lambda Ee^{isX} ) | }{ (c-\lambda EX)\cdot (c-({1}/{T})\sum_{j=1}^{n_T} X_j) }=O_{p}\left(\sqrt{\frac{\log(K/a)}{T}}\right). \end{align}

By combining (5.19)–(5.21), we obtain (5.11).

In order to derive the estimation error of $\hat {V}(u;b)$, we should first consider the estimation errors of $\hat {h}_+$ and $\hat {g}_+$. We have the following results.

Proposition 1. Suppose that $c>\lambda EX$, $EX^2<\infty$ and $a=o(K)$. Then, we have

(5.22)\begin{equation} \sup_{0\leq x\leq a} | \hat{h}_{+}(x)-\tilde{h}_{+}(x)| =O_p\left((K/a) \sqrt{\frac{\log(K/a) }{T } }\right) \end{equation}

and

(5.23)\begin{equation} \sup_{0\leq x\leq a} | \hat{g}_{+}(x)-\tilde{g}_{+}(x)| =O_p\left((K/a) \sqrt{\frac{\log(K/a) }{T } }\right). \end{equation}

Proof. We only prove (5.22), since (5.23) can be proved similarly by using (5.11). First, it is easily seen that

(5.24)\begin{equation} \sup_{0\leq x\leq a} | \hat{h}_{+}(x)-\tilde{h}_{+}(x)|\leq \sum_{k=0}^K |\hat{B}_{h_+,k}-B_{h_+,k}|\leq \frac{2(K+1)}{a}\sup_{s\in\mathcal{S}}|\widehat{\mathcal{F}h}_+ (s)-{\mathcal{F}h}_+ (s)|, \end{equation}

where $\mathcal {S}=\{k\pi /a: k=0,1,\ldots, K\}$. For each $s\in \mathcal {S}$, we have

$$\widehat{\mathcal{F}h}_{+}(s)-{\mathcal{F}h}_{+}(s) =I_h(s)+II_h(s),$$

where

$$I_h(s)=\left(\frac{\delta}{\hat{\psi}_U'(\hat{\rho})}\cdot \frac{1}{-is-\hat{\rho}}- \frac{\delta}{{\psi}_U'({\rho})}\cdot \frac{1}{-is-{\rho}}\right), \quad II_h(s)=\left(\frac{\delta}{{\psi}_U({-}is)-\delta} -\frac{\delta}{\hat{\psi}_U({-}is)-\delta}\right).$$

By Lemma 2 and (5.2), we can easily prove that

$$\sup_{s\in \mathcal{S}} |I_h(s)|=O_p(T^{-{1}/{2}}).$$

By Lemma 3, we have

$$\sup_{s\in \mathcal{S}}|II_h(s)|= \delta\cdot \sup_{s\in \mathcal{S}} \left|\frac{1}{\hat{\psi}_U({-}is)-\delta} -\frac{1}{\psi_U({-}is)-\delta} \right|=O_p\left( \sqrt{\frac{\log(K/a) }{T } }\right).$$

Hence, we have

$$\sup_{s\in\mathcal{S}}|\widehat{\mathcal{F}h}_+ (s)-{\mathcal{F}h}_+ (s)|\leq \sup_{s\in \mathcal{S}}|I_h(s)|+ \sup_{s\in \mathcal{S}}|II_h(s)|=O_{p}\left(\sqrt{\frac{\log(K/a)}{T}}\right).$$

which together with (5.24) completes the proof.

By the convergence rate given in (5.2), Lemma 2 and Proposition 1, we can easily obtain the following result.

Proposition 2. Suppose that $c>\lambda EX$, $EX^2<\infty$ and $a=o(K)$. Then, we have

(5.25)\begin{equation} \hat{V}(u;b)-\tilde{V}(u;b) =O_p\left((K/a) \sqrt{\frac{\log(K/a) }{T } }\right). \end{equation}

Next, we derive the convergence rate for the estimators of the expected discounted penalty functions. It follows from formulas (4.11) and (4.16) that we should first study the consistency properties of $\hat {\phi }_d(u;\infty )$ and $\hat {\phi }_d'(u;\infty )$.

Proposition 3. Suppose that $c>\lambda EX$, $EX^2<\infty$ and $a=o(K)$. Then, we have

(5.26)\begin{align} \sup_{0\leq u\leq a}|\hat{\phi}_d(u;\infty)-\tilde{\phi}_d(u;\infty) | =O_p\left((K/a) \sqrt{\frac{\log(K/a) }{T } }\right), \end{align}

(5.27)\begin{align} \sup_{0\leq u\leq a}|\hat{\phi}_d'(u;\infty)-\tilde{\phi}_d'(u;\infty) | =O_p\left((K/a) \sqrt{\frac{\log(K/a) }{T } }\right). \end{align}

Proof. First, we prove (5.26). By formulas (3.22) and (4.9), we can obtain

(5.28)\begin{align} \sup_{0\leq u\leq a}|\hat{\phi}_d(u;\infty)-\tilde{\phi}_d(u;\infty) | & \leq \sum_{k=0}^K |\hat{B}_{\phi_d,k}-B_{\phi_d,k}|\nonumber\\ & \leq \frac{2(K+1)}{a}\cdot \sup_{s\in\mathcal{S}}|\widehat{\mathcal{F}\phi}_d (s;\infty)-{\mathcal{F}\phi}_d (s;\infty)|. \end{align}

By (3.19) and (4.8), we obtain

(5.29)\begin{align} & \sup_{s\in \mathcal{S}}\left|\widehat{\mathcal{F}\phi}_d (s;\infty)-{\mathcal{F}\phi}_d (s;\infty)\right|\nonumber\\ & \quad =\sup_{s\in \mathcal{S}}\left|(\frac{ {\hat{\sigma}^2}/{2})({-}is-\hat{\rho}) }{\hat{\psi}_U({-}is)-\delta}-\frac{ ({\sigma^2}/{2})({-}is-\rho) }{\psi_U({-}is)-\delta}\right|\nonumber\\ & \quad \leq \sup_{s\in \mathcal{S}}\left|\frac{ ({\hat{\sigma}^2}/{2})s }{\hat{\psi}_U({-}is)-\delta}-\frac{ ({\sigma^2}/{2})s }{\psi_U({-}is)-\delta}\right| +\sup_{s\in \mathcal{S}}\left|\frac{ {(\hat{\sigma}^2}/{2})\hat{\rho} }{\hat{\psi}_U({-}is)-\delta}-\frac{ ({\sigma^2}/{2})\rho }{\psi_U({-}is)-\delta}\right|. \end{align}

Furthermore, by Lemma 3, we have

(5.30)\begin{align} & \sup_{s\in \mathcal{S}}\left|\frac{ ({\hat{\sigma}^2}/{2})s }{\hat{\psi}_U({-}is)-\delta}-\frac{ ({\sigma^2}/{2})s }{\psi_U({-}is)-\delta}\right|\nonumber\\ & \quad\leq \frac{\hat{\sigma}^2}{2}\cdot \sup_{s\in \mathcal{S}}\left|\frac{s }{\hat{\psi}_U({-}is)-\delta}-\frac{s }{\psi_U({-}is)-\delta}\right|+ \frac{|\hat{\sigma}^2-\sigma^2|}{2} \cdot \sup_{s\in \mathcal{S}}\left| \frac{s}{\psi_U({-}is)-\delta} \right|\nonumber\\ & \quad\leq \frac{\hat{\sigma}^2}{2}\cdot \sup_{s\in \mathcal{S}}\left|\frac{s }{\hat{\psi}_U({-}is)-\delta}-\frac{s }{\psi_U({-}is)-\delta}\right|+ \frac{|\hat{\sigma}^2-\sigma^2|}{2} \cdot \sup_{s\in \mathcal{S}}\left| \frac{s}{({-}is-\rho)(c-\lambda EX)} \right|\nonumber\\ & \quad=O_p(1)\cdot O_p\left(\sqrt{\frac{\log(K/a) }{T } }\right)+O_p(T^{{-}1/2})\nonumber\\ & \quad=O_p\left(\sqrt{\frac{\log(K/a) }{T } }\right). \end{align}

Similarly, we have

(5.31)\begin{align} & \sup_{s\in \mathcal{S}}\left|\frac{ ({\hat{\sigma}^2}/{2})\hat{\rho} }{\hat{\psi}_U({-}is)-\delta}-\frac{ ({\sigma^2}/{2})\rho }{\psi_U({-}is)-\delta}\right|\nonumber\\ & \quad \leq \frac{\hat{\sigma}^2\hat{\rho}}{2}\cdot \sup_{s\in \mathcal{S}}\left|\frac{1 }{\hat{\psi}_U({-}is)-\delta}-\frac{1 }{\psi_U({-}is)-\delta}\right|+ \frac{|\hat{\sigma}^2\hat{\rho}-\sigma^2\rho|}{2} \cdot \sup_{s\in \mathcal{S}}\left| \frac{1}{({-}is-\rho)(c-\lambda EX)} \right| \nonumber\\ & \quad =O_p(1)\cdot O_p\left(\sqrt{\frac{\log(K/a) }{T } }\right)+O_p(T^{{-}1/2})\nonumber\\ & \quad = O_p\left(\sqrt{\frac{\log(K/a) }{T } }\right). \end{align}

Hence, (5.29) gives

$$\sup_{s\in \mathcal{S}}|\widehat{\mathcal{F}\phi}_d (s;\infty)-{\mathcal{F}\phi}_d (s;\infty)|= O_p\left(\sqrt{\frac{\log(K/a) }{T } }\right),$$

which together with (5.28) yields (5.26).

Next, we prove (5.27). Again, we can obtain

(5.32)\begin{equation} \sup_{0\leq u\leq a}|\hat{\phi}_d'(u;\infty)-\tilde{\phi}_d'(u;\infty) | \leq \frac{2(K+1)}{a}\cdot \sup_{s\in\mathcal{S}}|\widehat{\mathcal{F}\phi}_d' (s;\infty)-{\mathcal{F}\phi}_d' (s;\infty)|, \end{equation}

where

\begin{align*} \widehat{\mathcal{F}\phi}_d'(s;\infty)-{\mathcal{F}\phi}_d'(s;\infty)& = \frac{\delta+(c+{\hat{\sigma}^2}/{2})is+\hat{\lambda}[ 1-\widehat{\mathcal{F}f}_X(s) ] }{\hat{\psi}_U({-}is)-\delta }\\ & \quad -\frac{\delta+(c+{{\sigma}^2}/{2})is+{\lambda}[ 1-{\mathcal{F}f}_X(s) ] }{{\psi}_U({-}is)-\delta }. \end{align*}

Furthermore, we have

\begin{align*} \sup_{s\in \mathcal{S}}\left| \widehat{\mathcal{F}\phi}_d'(s;\infty)-{\mathcal{F}\phi}_d'(s;\infty)\right| & \leq \sup_{s\in \mathcal{S}}\left| \frac{\delta+\hat{\lambda}[ 1-\widehat{\mathcal{F}f}_X(s) ] }{\hat{\psi}_U({-}is)-\delta } -\frac{\delta+{\lambda}[ 1-{\mathcal{F}f}_X(s) ] }{{\psi}_U({-}is)-\delta } \right|\\ & \quad +\sup_{s\in \mathcal{S}}\left| \frac{(c+{\hat{\sigma}^2}/{2})is }{\hat{\psi}_U({-}is)-\delta } -\frac{(c+{{\sigma}^2}/{2})is }{{\psi}_U({-}is)-\delta }\right|. \end{align*}

By Lemma 3 and (5.17) and using the same arguments in (5.30) and (5.31), we can obtain

$$\sup_{s\in \mathcal{S}}\left| \frac{\delta+\hat{\lambda}[ 1-\widehat{\mathcal{F}f}_X(s) ] }{\hat{\psi}_U({-}is)-\delta } -\frac{\delta+{\lambda}[ 1-{\mathcal{F}f}_X(s) ] }{{\psi}_U({-}is)-\delta } \right|=O_p\left(\sqrt{\frac{\log(K/a) }{T } }\right)$$

and

$$\sup_{s\in \mathcal{S}}\left| \frac{(c+{\hat{\sigma}^2}/{2})is }{\hat{\psi}_U({-}is)-\delta } -\frac{(c+{{\sigma}^2}/{2})is }{{\psi}_U({-}is)-\delta }\right| =O_p\left(\sqrt{\frac{\log(K/a) }{T } }\right).$$

Hence, we obtain

$$\sup_{s\in \mathcal{S}}\left| \widehat{\mathcal{F}\phi}_d'(s;\infty)-{\mathcal{F}\phi}_d'(s;\infty)\right| =O_p\left(\sqrt{\frac{\log(K/a) }{T } }\right),$$

which together with (5.32) gives (5.27).

Next, we derive the convergence rates for $\hat {\phi }_c(u;\infty )$ and $\hat {\phi }_c'(u;\infty )$.

Proposition 4. Suppose that $c>\lambda EX$, $EX^4<\infty$, $a=o(K)$, and

$$E\left(\int_0^X w(X-x)\,dx\right)<\infty,\quad E\left(\int_0^X x w(X-x)\,dx\right)^2<\infty.$$

Then, we have

(5.33)\begin{align} & \sup_{0\leq u\leq a}|\hat{\phi}_c(u;\infty)-\tilde{\phi}_c(u;\infty) | =O_p\left((K/a) \sqrt{\frac{\log(K/a) }{T } }\right), \end{align}

(5.34)\begin{align} & \sup_{0\leq u\leq a}|\hat{\phi}_c'(u;\infty)-\tilde{\phi}_c'(u;\infty) | =O_p\left((K/a) \sqrt{\frac{\log(K/a) }{T } }\right). \end{align}

Proof. First, we prove (5.33). It is easily obtained that

(5.35)\begin{equation} \sup_{0\leq u\leq a}|\hat{\phi}_c(u;\infty)-\tilde{\phi}_c(u;\infty) | \leq \frac{2(K+1)}{a}\cdot \sup_{s\in\mathcal{S}}|\widehat{\mathcal{F}\phi}_c (s;\infty)-{\mathcal{F}\phi}_c (s;\infty)|, \end{equation}

where

\begin{align*} & \sup_{s\in\mathcal{S}}\left|\widehat{\mathcal{F}\phi}_c(s;\infty) -{\mathcal{F}\phi}_c(s;\infty)\right|\\ & \quad =\sup_{s\in \mathcal{S}}\left|\frac{ \hat{\lambda}[\widehat{\mathcal{L}\omega}(\hat{\rho})- \widehat{\mathcal{F}\omega}(s)] }{\hat{\psi}_U({-}is)-\delta} -\frac{ {\lambda}[{\mathcal{L}\omega}({\rho})- {\mathcal{F}\omega}(s)] }{{\psi}_U({-}is)-\delta}\right|\nonumber\\ & \quad \leq \sup_{s\in \mathcal{S}}\left|\frac{ \hat{\lambda}\widehat{\mathcal{L}\omega}(\hat{\rho}) }{\hat{\psi}_U({-}is)-\delta} -\frac{ {\lambda}{\mathcal{L}\omega}({\rho})}{{\psi}_U({-}is)-\delta}\right| +\sup_{s\in \mathcal{S}}\left|\frac{ \hat{\lambda} \widehat{\mathcal{F}\omega}(s) }{\hat{\psi}_U({-}is)-\delta} -\frac{ {\lambda} {\mathcal{F}\omega}(s) }{{\psi}_U({-}is)-\delta}\right|. \end{align*}

Furthermore, by Xie and Zhang [Reference Xie and Zhang27], we have

$$\hat{\lambda}\widehat{\mathcal{L}\omega}(\hat{\rho}) -{\lambda}{\mathcal{L}\omega}(\rho) =O_p(T^{{-}1/2}),\quad \sup_{s\in \mathcal{S}} |\hat{\lambda} \widehat{\mathcal{F}\omega}(s) - {\lambda} {\mathcal{F}\omega}(s) |=O_p\left(\sqrt{\frac{\log(K/a) }{T } }\right).$$

Then, by Lemma 3, we can obtain

$$\sup_{s\in\mathcal{S}}|\widehat{\mathcal{F}\phi}_c(s;\infty) -{\mathcal{F}\phi}_c(s;\infty)|=O_p\left(\sqrt{\frac{\log(K/a) }{T } }\right),$$

which together with (5.35) gives (5.33). The proof of (5.34) is similar, and we omit the detailed arguments.

Now by combining the results in Propositions 2–4, we can obtain the convergence rates of $\hat {\phi }_d(u;b)$ and $\hat {\phi }_c(u;b)$.

Proposition 5. Suppose that $c>\lambda EX$, $EX^2<\infty$ and $a=o(K)$. Then, we have

(5.36)\begin{equation} \hat{\phi}_d(u;b)-\tilde{\phi}_d(u;b) =O_p\left((K/a) \sqrt{\frac{\log(K/a) }{T } }\right). \end{equation}

Proposition 6. Suppose that $c>\lambda EX$, $EX^4<\infty$, $a=o(K)$, and

$$E\left(\int_0^X w(X-x)\,dx\right)<\infty,\quad E\left(\int_0^X x w(X-x)\,dx\right)^2<\infty.$$

Then, we have

(5.37)\begin{equation} |\hat{\phi}_c(u;b)-\tilde{\phi}_c(u;b) | =O_p\left((K/a) \sqrt{\frac{\log(K/a) }{T } }\right). \end{equation}

6. Numerical results

In this section, we present some numerical results to illustrate the performance of our method. All computations are performed in MATLAB on a desktop computer with Intel(R) Core(TM) i7-6700 CPU, 3.4 GHz and a RAM of 8 GB. In this part, we set $c=4$, $\lambda =3$, $\sigma =1$, $\delta =0.1$ and $b=30$. We consider the following three claim size density functions,

1. 1. Exp(1): $f_X(x)=e^{-x},\ x>0$;

2. 2. Erlang(2,2): $f_X(x)=4xe^{-2x},\ x>0$;

3. 3. ${\rm Gamma}(\frac {3}{2},\frac {3}{2})$: $f_X(x)=\frac {3\sqrt {6x}e^{-{3x}/{2}}}{2\sqrt {\pi }},\ x>0$.

The choice of the truncation parameter $a$ is selected according to the following rule (see [Reference Fang and Oosterlee9]):

(6.1)\begin{equation} a=\kappa_1+L\sqrt{\kappa_2+\sqrt{\kappa_4}}, \end{equation}

where

(6.2)\begin{equation} \kappa_j=\frac{d^j}{ds^j}\log\left(\int e^{sx}f(x)\,dx\right)\Big|_{s=0}= \left.\frac{d^j}{ds^j}\log\mathcal{F}f({-}is)\right|_{s=0}, \end{equation}

and $f$ is a probability density function. As in Xie and Zhang [Reference Xie and Zhang27], we set $L=10$. When computing $\tilde {h}_{+}$ and $\tilde {g}_{+}$, we take $\mathcal {F}f=\mathcal {F}{h}_{+}$; when computing $\tilde {\phi }_{d}$ and $\tilde {\phi }_{d}'$, we take $\mathcal {F}f=\mathcal {F}{\phi }_{d}$; when computing $\tilde {\phi }_{c}$ and $\tilde {\phi }_{c}'$, we take $\mathcal {F}f=\mathcal {F}{\phi }_{c}$.

For the expected present value of total dividend payments before ruin, we easily obtain the explicit formula by (2.4) when $f_X$ is exponential or Erlang(2,2). When $f_X$ is ${\rm Gamma}(\frac {3}{2},\frac {3}{2})$, in order to provide a benchmark, we compute the reference values by the COS method with truncation parameter $K=2^{13}$. For the expected discounted penalty function, we set $w \equiv 1$, then $\phi _d(u;b)$ is the Laplace transform of the time of ruin by oscillation, and $\phi _c(u;b)$ is the Laplace transform of the time of ruin due to a claim. When $f_X$ is exponential or Erlang(2,2), we can use Laplace inversion transform to compute the explicit formulas by (3.16) and (3.19) for $\phi _d(u;b)$ and $\phi _c(u;b)$, respectively. When $f_X$ is ${\rm Gamma}(\frac {3}{2},\frac {3}{2})$, we also compute the reference values by the COS method with truncation parameter $K=2^{13}$.

First, we study the approximation performance of the COS method. In order to measure the accuracy of our method, we consider average absolute errors for $\tilde {V}(u;b)$, $\tilde {\phi }_d(u;b)$ and $\tilde {\phi }_c(u;b)$, respectively, which are defined by

$$\frac{1}{\# \mathcal{U}}\sum_{u\in \mathcal{U}}|\tilde{V}(u;b)-V^{{\rm ref}}(u;b)| ,\quad \frac{1}{\#\mathcal{U}'}\sum_{u\in \mathcal{U}'}|\tilde{\phi}_d(u;b)-\phi_d^{{\rm ref}}(u;b) |$$

and

$$\frac{1}{\#\mathcal{U}''}\sum_{u\in \mathcal{U}''}|\tilde{\phi}_c(u;b)-\phi_c^{{\rm ref}}(u;b) |,$$

where $V^{{\rm ref}}(u;b)$, $\phi _d^{{\rm ref}}(u;b)$ and $\phi _c^{{\rm ref}}(u;b)$ are the reference values. We take $\mathcal {U}=\{0.1, 0.2, \ldots, 30\}$, $\mathcal {U}'=\{0.1, 0.2,\ldots, 5\}$ and $\mathcal {U}''=\{0.1, 0.2,\ldots, 15\}$, since $\phi _d(u;b)$ is very small for $u>5$, and $\phi _c(u;b)$ is very close to zero for $u>15$. The average absolute errors are reported in Table 1, where we consider the truncation parameter $K=2^p$ for $p=6, 7, 8, 9, 10$. All results in the following tables are rounded at the sixth decimal place. In Table 1, it can be easily seen that the absolute errors decrease with $p$ (or equivalently $K$) for $\tilde {V}(u;b)$, $\tilde {\phi }_d(u;b)$ and $\tilde {\phi }_c(u;b)$, which means that the large truncation parameter can reduce the error in our examples.

Table 1. Average absolute errors by the COS method.

Next, we consider the effectiveness of our statistical estimation method, based on a random sample on the individual claim sizes $\{X_1, \ldots, X_{n_T} \}$ and claim number $n_T$. For the parameter $a$, we consider the following empirical version of the cumulant-based method,

(6.3)\begin{equation} a=\hat{\kappa}_1+L\sqrt{\hat{\kappa}_2+\sqrt{\hat{\kappa}_4} }, \end{equation}

where

(6.4)\begin{equation} \hat{\kappa}_j= \left.\frac{d^j}{ds^j}\log\widehat{\mathcal{F}f}({-}is)\right|_{s=0},\quad j=1,2,4. \end{equation}

When computing $\hat {h}_+$ and $\hat {g}_+$, we take $\widehat {\mathcal {F}f}=\widehat {\mathcal {F}h_+}$; when computing $\hat {\phi }_d$ and $\hat {\phi }_{d}'$, we take $\widehat {\mathcal {F}f}=\widehat {\mathcal {F}\phi _d}$; when computing $\hat {\phi }_c$ and $\hat {\phi }_{c}'$, we take $\widehat {\mathcal {F}f}=\widehat {\mathcal {F}\phi _c}$. Again, we set $L=10$.

Set the truncation parameter $K=2^{10}$ and the inter-observation interval $\Delta =1$. We perform $200$ experiments and consider average absolute errors for $\hat {V}(u;b)$, $\hat {\phi }_d(u;b)$ and $\hat {\phi }_c(u;b)$, respectively, which are defined by

$$\frac{1}{\# \mathcal{U}}\sum_{u\in \mathcal{U}}\frac{1}{200}\sum_{j=1}^{200}|\hat{V}_j(u;b)-V^{{\rm ref}}(u;b)| ,\quad \frac{1}{\#\mathcal{U}'}\sum_{u\in \mathcal{U}'}\frac{1}{200}\sum_{j=1}^{200}|\hat{\phi}_{d,j}(u;b)-\phi_d^{{\rm ref}}(u;b) |$$

and

$$\frac{1}{\#\mathcal{U}''}\sum_{u\in \mathcal{U}''}\frac{1}{200}\sum_{j=1}^{200}|\hat{\phi}_{c,j}(u;b)-\phi_c^{{\rm ref}}(u;b) |,$$

where $\hat {V}_j(u;b)$, $\hat {\phi }_{d,j}(u;b)$ and $\hat {\phi }_{c,j}(u;b)$ denote the $j$th experiment values of $\hat {V}(u;b)$, $\hat {\phi }_d(u;b)$ and $\hat {\phi }_c(u;b)$, respectively.

In Table 2, we list the empirical average absolute errors of the estimator for the three functions. For the observation interval $[0, T]$, we take $T=1000\times 2^q$ for $q=0$, 1, 2, 3, 4, 5. It is obvious that the estimation errors are also decreasing with $q$ (or equivalently the time $T$), that is, the approximation results become closer to reference values, which means that our estimator performs better when the sample sizes become large.

Table 2. Empirical average absolute errors by the COS method.

In order to show the stability of our method, we plot $200$ estimated curves (green curves) and the true curves (bold blue curves) on the same picture. For the expected present value of total dividends before ruin with the exponential claim size density function, we plot the curves of $\hat {V}_j(u;b)$ as functions of $u$ when $T=1000\times 2^1$, $1000\times 2^3$ and $1000\times 2^5$in Figure 1. It is obvious that all the lines increase with $u$ for all the three models, which is consistent with the actual situation that the expected present value of total dividends before ruin increases with the initial surplus. Besides, as $T$ increases, the estimator $\hat {V}(u;b)$ tends to be stable and converges to ${V}(u;b)$ .

Figure 1. Beams for estimating $V(u;b)$: 200 estimators in green, and the true in bold blue. (a) $q=1$; (b) $q=3$ and (c) $q=5$.

The functions $\phi _d(u;b)$, $\phi _c(u;b)$ and $\phi (u;b)$ are also investigated for Erlang(2,2) claim size density function. In Figure 2, we find that $\phi _d(u;b)$ is a decreasing function of the initial surplus $u$, which means that ruin is more likely to happen caused by oscillation when $u$ is small. At the same time, we can also observe that as $T$ increases, the estimator $\hat {\phi }_d(u;b)$ tends to be stable and converges to $\phi _d(u;b)$. As for $\hat {\phi }_c(u;b)$ and $\hat {\phi }(u;b)$, we plot the corresponding curves in Figures 3 and 4. Again, we can observe that the estimation becomes better as $T$ becomes larger.

Figure 2. Beams for estimating $\phi _d(u;b)$: 200 estimators in green, and the true in bold blue. (a) $q=1$; (b) $q=3$ and (c) $q=5$.

Figure 3. Beams for estimating $\phi _c(u;b)$: 200 estimators in green, and the true in bold blue. (a) $q=1$; (b) $q=3$ and (c) $q=5$.

Figure 4. Beams for estimating $\phi (u;b)$: 200 estimators in green, and the true in bold blue. (a) $q=1$; (b) $q=3$ and (c) $q=5$.

Finally, we investigate the effectiveness of our method when further using spectral filters to accelerate the decay rate of the COS coefficients. Recall that a real and symmetric function $\Gamma (s)$ is a filter of order $\upsilon$ if: (i) $\Gamma (0)=1$, $\Gamma ^{(l)}(0)=0$, $1\leq l\leq \upsilon -1$; (ii) $\Gamma (s)=0$ for $|s|\geq 1$; (iii) $\Gamma (s)\in C^{\upsilon -1}$, $s\in \mathbb {R}$, where in particular $\Gamma ^{(l)}(\pm 1)=0$ for $0\leq l\leq \upsilon -1$. For a function $f$, its filter-COS approximation is simply defined by

$$\tilde{f}^{{\rm filter}}(x)=\mathop{{\sum}'}_{k=0}^K \Gamma(k/K) B_{f,k} \cos\left(k\pi\frac{x-a_1}{a_2-a_1}\right),\quad a_1\leq x\leq a_2.$$

In the following experiments, we select an exponential filter $\Gamma (s)=\exp (-\nu s^{\upsilon })$ with $\upsilon =6$ and $\nu =-\log \epsilon$, where $\epsilon$ represents the machine epsilon so that $\Gamma (1)=\epsilon \approx 0$. We take $\epsilon =10^{-6}$. In Table 3, we list the absolute errors of the three functions by using the filter-COS method. Similarly, we can find that the calculation errors decrease with the truncation parameter $K$. Compared with Table 1, we can find that the errors calculated by the filter-COS method are much smaller than the errors obtained without filter. In particular, we observed that for the $\tilde {V}(u;b)$ and $\tilde {\phi }_c(u;b)$ functions, the errors are significantly improved with the addition of filtering. We also list the empirical errors of the estimator for the three functions by the filter-COS method in Table 4, and almost all errors obtained by using the filter-COS method are also reduced compared with Table 2. In summary, spectral filters can accelerate the convergence rate.

Table 3. Average absolute errors by the filter-COS method.

Table 4. Empirical average absolute errors by the filter-COS method.