Abstract
Hougaard processes, which include gamma and inverse Gaussian processes as special cases, as well as the moments of the corresponding first-passage-time (FPT) distributions are commonly used in many applications. Because the density function of a Hougaard process involves an intractable infinite series, the Birnbaum–Saunders (BS) distribution is often used to approximate its FPT distribution. This article derives the finite moments of FPT distributions based on Hougaard processes and provides a theoretical justification for BS approximation in terms of convergence rates. Further, we show that the first moment of the FPT distribution for a Hougaard process approximated by the BS distribution is larger and provide a sharp upper bound for the difference using an exponential integral. The conditions for convergence coincidentally elucidate the classical convergence results of Hougaard distributions. Some numerical examples are proposed to support the validity and precision of the theoretical results.
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Acknowledgements
This work by Peng was partially supported by the Ministry of Science and Technology (MOST-109-2118-M-001-009-MY3) and Academia Sinica (AS-CDA-107-M09) of Taiwan, Republic of China. Professor Fan was supported by the Ministry of Science and Technology (MOST-109-2118-M-008-004-MY3) of Taiwan, Republic of China. The authors are grateful to the editor-in-chief, associate editor and referees for their insightful suggestions and constructive comments which greatly improved the quality of the article.
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Appendix
Appendix
Lemma 5.1
For \(c \in {\mathbb {R}}{\setminus } \{0\} \) and \(0< \alpha < 1\), the real part of \((1+ c \imath )^\alpha -1\) is positive, i.e., \({{\textrm{Re}}}{(1+c \imath )^\alpha } > 1\).
Proof
By Euler formula and De Moivre formula, there is a unique principal argument \(\vartheta = \arctan \left( c \right) \in (-\pi /2,\pi /2]\) such that \((1+ c \imath )^\alpha = (1+c^2)^\frac{\alpha }{2} \left( \cos (\alpha \vartheta ) + \imath \sin (\alpha \vartheta ) \right) \). Let \(g( \alpha ) = \ln \left( {{\textrm{Re}}}(1+c\imath )^\alpha \right) = \frac{\alpha }{2} \ln (1+c^2) + \ln \left( \cos (\alpha \vartheta )\right) \) for \(\alpha \in [0, 1]\). Hence, it is easy to check \(g(0) = g(1) = 0\) and \({ \text{ d}^2 }g(\alpha )/{\text{ d } \alpha ^2} = - \vartheta ^2 \sec (\alpha \vartheta )^2 < 0\), indicating that \(g(\alpha )\) is concave. Thus, we have \(\ln \left( {{\textrm{Re}}}(1+c\imath )^\alpha \right) = g(\alpha ) > \min \{ g(0), g(1) \} = 0\) for \(\alpha \in (0, 1)\), which is equivalent to \({{\textrm{Re}}}{(1+c \imath )^\alpha }>1\). \(\square \)
Lemma 5.2
-
(i)
For \(n \in {\mathbb {N}}, s > 0\) and \(0< \alpha < 1\),
$$\begin{aligned}&\frac{ (1 -\exp (\alpha \pi \imath )s^\alpha )^{-n} - (1 -\exp (-\alpha \pi \imath ) s^\alpha )^{-n} }{2 \imath } \nonumber \\&\quad = \frac{\sin \left( n \arctan \left( \frac{\sin (\alpha \pi ) s^\alpha }{1- \cos (\alpha \pi ) s^\alpha } \right) \right) }{ \left( 1-2\cos (\alpha \pi )s^\alpha + s^{2\alpha }\right) ^\frac{n}{2} } {{\textrm{sgn}}} \left( \left( 1 - \cos (\alpha \pi ) s^\alpha \right) ^n \right) . \end{aligned}$$(13) -
(ii)
When \(\alpha = 0\), the above equation reduces to
$$\begin{aligned}{} & {} \frac{ (-\pi \imath - \ln s)^{-n} - ( \pi \imath - \ln s )^{-n} }{2 \imath } \\{} & {} \quad = -\frac{\sin \left( n \arctan \left( \pi / \ln s \right) \right) }{ \left( \pi ^2 + (\ln s)^2 \right) ^\frac{n}{2} } {{\textrm{sgn}}} \left( \left( 1 - s \right) ^n \right) . \end{aligned}$$
Proof
(i) By the identity \(\exp ( \alpha \pi \imath ) = \cos (\alpha \pi ) + \sin (\alpha \pi ) \imath \) and binomial Theorem, we have
where \(\lfloor \cdot \rfloor \) denotes the floor function. Let \(x = 1 - \cos (\alpha \pi ) s^\alpha \) and \(y = \sin (\alpha \pi ) s^\alpha >0\). Use the multi-angle formula to get
Substituting the above identity to the numerator of (14) gives the desired result. (ii) For \(\alpha = 0\), multiplying \(\alpha ^n\) on both sides in (13) and applying L’Hôpital rule, the result follows. \(\square \)
Lemma 5.3
-
(i)
For \(0<\alpha <1\),
$$\begin{aligned}&\int _{0}^{\infty } \frac{ 1 }{ s + 1} \frac{ \sin \left( n \arctan \left( \frac{\sin (\alpha \pi ) s^\alpha }{1 - \cos (\alpha \pi ) s^\alpha } \right) \right) }{ ( 1 - 2 \cos (\alpha \pi ) s^\alpha + s^{2\alpha } )^\frac{n}{2} }\nonumber \\&\quad {{\textrm{sgn}}} \left( \left( 1 - \cos (\alpha \pi ) s^\alpha \right) ^n \right) {{\textrm{d}}} s \nonumber \\&\quad = \pi {{\textrm{Res}}}\left( \frac{(1-s^\alpha )^{-n}}{1-s},s=1 \right) . \end{aligned}$$(15) -
(ii)
For \(\alpha = 0\),
$$\begin{aligned}{} & {} \int _{0}^{\infty } \frac{ -1 }{ s + 1} \frac{ \sin \left( n \arctan \left( \pi / \ln s \right) \right) }{ ( \pi ^2 + (\ln s)^2 )^\frac{n}{2} } {{\textrm{sgn}}} \left( \left( 1 - s \right) ^n \right) {{\textrm{d}}} s \\{} & {} \quad = \pi {{\textrm{Res}}}\left( \frac{(- \ln s)^{-n}}{1-s},s=1 \right) . \end{aligned}$$
Proof
(i) Define
Then \(s=1\) is a single pole of q(s) with order \(n+1\). Consider the following closed keyhole contour IJKLI plotted in Fig. 5.
We show that
and
By the Euler formula, we have
Then the result follows by Lemma 5.2(i) and the Cauchy residue Theorem. (ii) For \(\alpha = 0\), multiplying \(\alpha ^n\) on both sides in (15) and applying L’Hôpital rule, the result follows. \(\square \)
Proof of Theorem 2.2
The following auxiliary Lemmas 5.4–5.5 are needed to prove Theorem 2.2.
Lemma 5.4
Let \(n, \ell , m_\ell \in {\mathbb {N}}\), \(h(x) = (1 - b \imath x)^\alpha \) for \( 0< \alpha < 1\), \(b > 0\) and \(x \in {\mathbb {C}}\). Then
Proof
Define
and \({\mathscr {E}}_i^n = \bigcup _{1 \le w_1< w_2< \cdots < w_i \le n} \bigcap _{j = 1}^i {\mathscr {D}}^n_{w_j}\) for \(i = 1,2, \ldots , n-1\).
By the general Leibniz rule, we have
Since
for each \(j=1,2,\ldots , n\),
Then applying the exclusion-inclusion principle, we obtain
where the last equality is due to the generalized Chu-Vandermonde identity (see Graham et al. 1994, page 248). By substituting the above identity into (16), we get the desired result. \(\square \)
Lemma 5.5
(Gould 1978)
-
(i)
If \(f: {\mathbb {R}} \rightarrow {\mathbb {R}}\) is a function which nth derivative exists, then we get
$$\begin{aligned} f^{(n)}(x) = \lim \limits _{h \rightarrow 0} \frac{1}{h^n} \sum \limits _{k=0}^n (-1)^{n-k} {n \atopwithdelims ()k} f\left( x + k h \right) . \end{aligned}$$ -
(ii)
For \(n \in {\mathbb {N}}\), we have
$$\begin{aligned} \sum \limits _{k=0}^n (-1)^k {n \atopwithdelims ()k} k^n = \Gamma (n+1). \end{aligned}$$
Now, we return to the proof of Theorem 2.2.
Proof
(i)-(a) Since the integrand, \(\frac{1-\exp \left( -\omega \imath s\right) }{s(1-h(s))^n}\), has a pole of order n at \(s=0\), we have
where \(h(s) = (1 - b \imath s)^\alpha \) is defined in Lemma 5.4. By Lemma 5.5 (i), the Eq. (17) can be expressed as
To calculate the limit in (18) by L’Hôpital rule, taking the derivative \(n^2-1\) times of the denominator of (18) and using Lemma 5.5(i), we get
where the last identity holds according to Lemma 5.5(ii). Taking the derivative \(n^2-1\) times of the numerator of (18) and using the general Leibniz rule, we have
where
Using Lemma 5.4 with
we get
By combining (19) and (20), we obtain the desired result of the residue.
(i)-(b) The derivation of the residue can be followed by similar procedure. (ii) The results can be established by L’Hôpital rule directly. This completes the proof. \(\square \)
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Peng, CY., Dong, YS. & Fan, TH. The first-passage-time moments for the Hougaard process and its Birnbaum–Saunders approximation. Stat Comput 33, 59 (2023). https://doi.org/10.1007/s11222-023-10235-1
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DOI: https://doi.org/10.1007/s11222-023-10235-1