# Modeling lower-truncated and right-censored insurance claims with an
extension of the MBBEFD class

Selim Gatti111RiskLab, Department of Mathematics, ETH Zurich,
<EMAIL_ADDRESS>Mario V. Wüthrich222RiskLab, Department of
Mathematics, ETH Zurich<EMAIL_ADDRESS>
(Version of October 13, 2023)

###### Abstract

In general insurance, claims are often lower-truncated and right-censored
because insurance contracts may involve deductibles and maximal covers. Most
classical statistical models are not (directly) suited to model lower-
truncated and right-censored claims. A surprisingly flexible family of
distributions that can cope with lower-truncated and right-censored claims is
the class of MBBEFD distributions that originally has been introduced by
Bernegger (1997) for reinsurance pricing, but which has not gained much
attention outside the reinsurance literature. We derive properties of the
class of MBBEFD distributions, and we extend it to a bigger family of
distribution functions suitable for modeling lower-truncated and right-
censored claims. Interestingly, in general insurance, we mainly rely on
unimodal skewed densities, whereas the reinsurance literature typically
proposes monotonically decreasing densities within the MBBEFD class.

Keywords. General insurance claims, lower-truncation, right-censoring, MBBEFD
distribution, unimodal density, skewed density, normalized loss, exposure
curve, Swiss Re exposure curve, Lloyd’s exposure curve.

##  1 Introduction

Insurance contracts in general insurance often involve deductibles $d>0$ and
maximal covers $M>0$. Deductibles are introduced to reduce the number of small
claims which mainly cause administrative expenses but which are not essential
in risk mitigation. Maximal covers are introduced to control the maximal loss
of an insurer. A maximal cover may, e.g., refer to the property value insured
(after subtracting the deductible), or to the maximal insurance coverage
warranted to a liability claim. Denote by $Y^{\dagger}$ the total financial
loss. The insurance claim $Y$ after subtracting the deductible $d>0$ and with
a maximal cover of size $M>0$ is given by

$Y=\min\left\\{(Y^{\dagger}-d)_{+},\,M\right\\}.$ (1.1)

We say, this financial loss is lower-truncated at $d>0$ and right-censored at
$M>0$ (after subtracting the deductible). Statistical modeling of lower-
truncated and right-censored claims is a notoriously difficult problem because
most statistical models have an unbounded support, e.g., the supports of the
gamma and the log-normal distributions are the entire positive real line
${\mathbb{R}}_{+}$. In most cases, this implies that fitting a statistical
model to lower-truncated and right-censored data is not a problem that is
easily analytically tractable. We give an example. We start from a classical
statistical model such as the gamma distribution for the total financial loss
$Y^{\dagger}\sim F_{\Gamma}$, where $F_{\Gamma}$ denotes the gamma
distribution with corresponding gamma density $f_{\Gamma}$ on
${\mathbb{R}}_{+}$. Lower-truncation and right-censoring introduces two
difficulties which are illustrated in Figure 1. First, the lower-truncation
$(Y^{\dagger}-d)_{+}$ of the total financial loss $Y^{\dagger}$ implies, in
general, that the density of the lower-truncated claim is positive in 0, see
Figure 1. In the above mentioned gamma case, this means that the lower-
truncation with $d>0$ leads to a new density given by

$y\geq
0~{}\mapsto~{}\frac{f_{\Gamma}(d+y)}{\int_{d}^{\infty}f_{\Gamma}(z)\,dz}=\frac{f_{\Gamma}(d+y)}{1-F_{\Gamma}(d)}~{}>~{}0.$

Second, right-censoring at $M>0$ of this lower-truncated claim
$(Y^{\dagger}-d)_{+}$ leads to a point mass in $M$, resulting in the density
$f$ of the lower-truncated and right-censored claim $Y$

$y\geq
0~{}\mapsto~{}f(y)=\frac{f_{\Gamma}(d+y)}{1-F_{\Gamma}(d)}\,\mathds{1}_{\\{y<M\\}}+\frac{1-F_{\Gamma}(d+M)}{{1-F_{\Gamma}(d)}}\,\mathds{1}_{\\{y=M\\}},$
(1.2)

where (1.2) is a density w.r.t. the $\sigma$-finite measure being the Lebesgue
measure on $(0,M)$ and having a point mass in $M$; this point mass is not
illustrated in Figure 1, but only the absolutely continuous part on $(0,M)$;
the point mass in $M$ equals one minus the volume of the blue area in Figure
1.

Figure 1: Lower-truncated and right-censored claim with $d=2000$ and $M=5000$.

More generally, for maximum likelihood estimation (MLE) based on lower-
truncated and right-censored claims $Y\in(0,M]$, ${\mathbb{P}}$-a.s., we
consider the log-likelihood function of an unknown parameter $\theta$ given by

$\theta~{}\mapsto~{}\ell_{Y}(\theta)=\log\left(f_{\theta}(Y)\right)\mathds{1}_{\\{Y<M\\}}+\log\left(1-F_{\theta}(M_{-})\right)\mathds{1}_{\\{Y=M\\}},$
(1.3)

assuming that the response variable $Y$ is absolutely continuous on $(0,M)$
with density $f_{\theta}(y)$, having a point mass
$1-F_{\theta}(M_{-})=1-\lim_{y\uparrow M}F_{\theta}(y)$ in $M$, and with
(unknown) model parameter $\theta$. Fitting such a model with MLE can be
difficult because we need an analytically tractable form for both the density
$f_{\theta}(\cdot)$ and its distribution function $F_{\theta}(\cdot)$, see
(1.3). This is not the case, e.g., in the lower-truncated and right-censored
gamma model given in (1.2). Therefore, in such cases, one either needs to rely
on numerical integration of the density (which can be computationally
demanding) or one uses a version of the Expectation-Maximization (EM)
algorithm by interpreting the lower-truncation and right-censoring as a
missing information problem; we refer to Verbelen et al. [16], Fung et al. [5]
and Sections 6.4.2 and 6.4.3 in Wüthrich–Merz [17]. However, also this EM
algorithm approach has its drawbacks as it requires tractability of
conditional tail expectations and reasonable dispersion estimates in multi-
dimensional parameter settings. These two side constraints lead to further
restrictions on the class of solvable models, e.g., these problems can only be
solved for a very small number of models within the class of Tweedie’s models
[15], namely, for the Tweedie’s models stated in Theorem 3 of Blæsild–Jensen
[3]; we also refer to Landsman–Valdez [6]. In a series of papers, Poudyal [9,
10] and Poudyal–Brazauskas [11] consider trimmed and/or winsorized method of
moments estimators for truncated and/or censored data; in statistics,
truncation is also called trimming and censoring winsorizing. In these papers,
trimming and winsorizing is also shown to be a useful method of robustifying
moment estimation under extreme claims.

We take a different approach in this paper to solve the fitting problem of
lower-truncated and right-censored data. In reinsurance, often so-called
MBBEFD exposure curves are used for exposure rating. Those exposure curves
have been introduced by Bernegger [2], and the acronym MBBEFD indicates that
this class includes the Maxwell–Boltzmann (MB), the Bose–Einstein (BE) and the
Fermi–Dirac (FD) distributions; these are well known distributions in
statistical mechanics. These MBBEFD exposure curves are based on the
assumption that there is a maximal cover $M$, and they directly describe
right-censored claims up to this maximal cover. Differentiating twice these
MBBEFD exposure curves provides us with densities being absolutely continuous
on the interval $(0,M)$ and having a point mass in $M$; we refer to formula
(3.7) in Bernegger [2].

The goal of this paper is first to study the properties of these MBBEFD
densities and to extend it to a bigger class of models that will be called the
Bernegger class. Our contribution is to show that the Bernegger class is
rather rich including monotonically decreasing densities, unimodal densities
and monotonically increasing densities, and our extension provides new
families of lower-truncated and right-censored random variables that allow for
skewness in the absolutely continuous part of the distribution. This is of
particular interest because unimodal skewed densities are suited for modeling
lower-truncated and right-censored claims in general insurance since their
empirical density roughly looks like the one given in Figure 1. In particular,
one can recognize the distribution of right-censored exponential random
variables as a special case of the Bernegger class.

Surprisingly, the class of MBBEFD densities of Bernegger [2] has only entered
the reinsurance literature; see, e.g., Parodi–Watson [8], Abramson [1], Riegel
[13], Chapter 21 of Parodi [7], and the R [12] package mbbefd of Dutang et al.
[4, 14]. Popular examples in reinsurance pricing are the so-called Swiss Re
and Lloyd’s exposure curves that are special cases of MBBEFD exposure curves;
see Bernegger [2]. However, in this reinsurance pricing literature, one mainly
focuses on exposure curves and not on the resulting densities nor on their
properties. We show that the most popular choices from reinsurance lead to
monotonically decreasing densities, whereas we are mainly interested into the
unimodal case, as this is the common situation in general insurance pricing,
e.g., in private insurance lines.

Organization. This manuscript is organized as follows. In the next section, we
state the necessary properties that any exposure curve has to fulfill in order
to describe distribution functions allowing to model lower-truncated and
right-censored claims. In Section 3, we start from the MBBEFD class of
distributions of Bernegger [2] by stating some of its properties, and then we
extend it to a richer family of distributions, the Bernegger class. In Section
4, we introduce a subclass of the Bernegger class that incorporates the MBBEFD
class of distributions, whereas in Section 5, we treat another subclass of
distributions that includes the exponential distribution. In Section 6, we use
a real dataset of lower-truncated and right-censored claims in order to
compare the performance of the MBBEFD class of distributions to four other
classes of distributions when trying to fit those models using maximum
likelihood estimation (MLE). Finally, the last section concludes this work.
All mathematical proofs are provided in the appendix.

##  2 Exposure curves and their resulting densities

###  2.1 From Exposure curves to distributions

In reinsurance claims modeling, one often works with exposure curves instead
of distribution functions. Assume we have a positively supported response
variable $Y\sim F_{Y}$, and assume that the maximal possible loss (MPL) is
given by $M>0$, i.e., $0<Y\leq M$, ${\mathbb{P}}$-a.s. We define the
normalized loss $Z=Y/M$. Denote the distribution function of the normalized
loss $Z$ by $F_{Z}$, being supported in $(0,1]$. The exposure curve of a
normalized loss $Z\sim F_{Z}$ is given by

$z~{}\mapsto~{}G(z)=\frac{\int_{0}^{z}1-F_{Z}(s)\,ds}{\int_{0}^{1}1-F_{Z}(s)\,ds}=\frac{\int_{0}^{z}1-F_{Z}(s)\,ds}{{\mathbb{E}}[Z]},$

for $z\in[0,1]$; see Bernegger [2]. This exposure curve $G:[0,1]\to[0,1]$ is
non-decreasing, concave, and satisfies the property $G^{\prime}(0)>0$, as well
as the normalization $G(0)=0$ and $G(1)=1$; for examples see Figure 2 (lhs),
below. In the last integral, a change of variable $s\in[0,1]\mapsto
t=sM\in[0,M]$ gives us

$G(z)=\frac{\int_{0}^{zM}1-F_{Z}(t/M)\,dt}{M{\mathbb{E}}[Z]}=\frac{\int_{0}^{zM}1-F_{Y}(t)\,dt}{{\mathbb{E}}[Y]}=G_{Y}(zM),$

the latter being the exposure curve of $Y$ on $[0,M]$. Thus, we can equally
work with the responses $Y$ and $Z$, but the normalized losses $Z$ will have
the advantage that they live on a common unit interval $[0,1]$. Now, let us
take the opposite view and characterize the distribution function of a random
variable $Z\sim F_{Z}$ obtained from a function $G:[0,1]\rightarrow\mathbb{R}$
satisfying the same properties of an exposure curve. Under the assumptions of
the next theorem, this distribution $F_{Z}$ leads to an absolutely continuous
density on $[0,1)$ and a point mass at 1.

######  Theorem 2.1.

Let $G:[0,1]\rightarrow\mathbb{R}$ be a non-decreasing, concave, and twice
continuously differentiable function with $G(0)=0,\,G(1)=1,\,G^{\prime}(0)>0$.
The function $F_{Z}:[0,1]\rightarrow\mathbb{R}$ defined by

$F_{Z}(z)=\left(1-\frac{G^{\prime}(z)}{G^{\prime}(0)}\right)\mathds{1}_{\\{z<1\\}}+\mathds{1}_{\\{z=1\\}}$
(2.1)

is a distribution function on $[0,1]$. Furthermore, this distribution has as
density

$f_{Z}(z)=-\frac{G^{\prime\prime}(z)}{G^{\prime}(0)},$ (2.2)

for $z\in[0,1)$, and a point mass at 1 given by

$p=\frac{G^{\prime}(1)}{G^{\prime}(0)}.$ (2.3)

Finally, the mean of $Z\sim F_{Z}$ is equal to
${\mathbb{E}}[Z]=1/G^{\prime}(0)$.

The proofs of all statements are given in the appendix. Due to this last
result, functions $G$ satisfying the assumptions of Theorem 2.1 will be called
exposure curves.

######  Definition 2.2.

An exposure curve is a function $G:[0,1]\rightarrow\mathbb{R}$, which is non-
decreasing, concave, and twice continuously differentiable with
$G(0)=0,\,G(1)=1,\,G^{\prime}(0)>0$.

As seen previously, if we start from any such function $G$, we can derive a
distribution whose density is absolutely continuous and of closed form on
$[0,1)$, with a point mass in 1 and a mean that are of closed form too, i.e.,
we have a class of models that has a full tractable mean, density and point
mass, which is suitable to model right-censored claims, and if
$G^{\prime\prime}(0)<0$, we can also consider lower-truncation, see (2.2). It
turns out that a linear combination of exposure curves allows us to define a
mixture of their respective associated distribution functions as the next
result shows.

######  Lemma 2.3.

Let $(\alpha_{i})_{i=1}^{n}$ be non-negative weights adding up to 1 and let
$(G_{i})_{i=1}^{n}$ be exposure curves leading to distributions functions
$(F_{i})_{i=1}^{n}$, densities $(f_{i})_{i=1}^{n}$, and point masses in 1
equal to $(p_{i})_{i=1}^{n}$ as in Theorem 2.1. The convex combination

$G(z)=\sum_{i=1}^{n}\alpha_{i}G_{i}(z),$ (2.4)

for $z\in[0,1]$, is again an exposure curve, allowing to define the
distribution function of a random variable $Z\sim F_{Z}$ given by

$F_{Z}(z)=\sum_{i=1}^{n}w_{i}F_{i}(z),$

for $z\in[0,1]$, and where
$w_{i}=\alpha_{i}G_{i}^{\prime}(0)/\sum_{j=1}^{n}\alpha_{j}G_{j}^{\prime}(0)$
are non-negative weights summing up to 1. In particular, the density of $Z$ on
$[0,1)$ is given by

$f_{Z}(z)=\sum_{i=1}^{n}w_{i}f_{i}(z),$

and the point mass in 1 is equal to

$p=\sum_{i=1}^{n}w_{i}p_{i}.$

Finally, the mean of $Z\sim F_{Z}$ is given by

${\mathbb{E}}[Z]=\sum_{i=1}^{n}w_{i}\,\frac{1}{G_{i}^{\prime}(0)}.$

###  2.2 Flexibility of the point mass in the right-censoring point

The point mass $p$ in 1 is automatically determined by formula (2.3). Often,
one may require more modeling flexibility in the choice of this point mass,
while still retaining the tractability of the density and the mean as it was
shown in Theorem 2.1. A simple way to do so connects to so-called one-inflated
distributions; we refer to Dutang et al. [4]. In the case of Theorem 2.1, this
can easily be achieved. The next corollary shows, how we can obtain them by
looking at the conditional density of the random variable
$Z_{0}\stackrel{{\scriptstyle\rm(d)}}{{=}}Z|_{\\{Z<1\\}}$.

######  Corollary 2.4.

Let $G:[0,1]\rightarrow\mathbb{R}$ be an exposure curve, we receive an
absolutely continuous density on $[0,1)$ given by

$\begin{split}f_{0}(z)=-\frac{1}{1-p}\frac{G^{\prime\prime}(z)}{G^{\prime}(0)}=\frac{G^{\prime\prime}(z)}{G^{\prime}(1)-G^{\prime}(0)}~{}\geq~{}0,\end{split}$

which integrates to 1. This density provides the mean for the random variable
$Z_{0}\sim f_{0}$

${\mathbb{E}}[Z_{0}]=\frac{1}{1-p}({\mathbb{E}}[Z]-p)=\frac{1-G^{\prime}(1)}{G^{\prime}(0)-G^{\prime}(1)}.$

We can now add a point mass $q\in(0,1)$ at 1 to this density. This adds one
more parameter to the model, giving us a mixture distribution between an
absolutely continuous part $f_{0}$ on $[0,1)$ and a point mass in 1. We have
the following corollary.

######  Corollary 2.5.

Let $G:[0,1]\rightarrow\mathbb{R}$ be an exposure curve, the random variable
$Z$ that has an absolutely continuous density on $[0,1)$ given by

$f_{q}(z)=\left(1-q\right)\frac{G^{\prime\prime}(z)}{G^{\prime}(1)-G^{\prime}(0)}~{}\geq~{}0,$

with fixed point mass $q\in(0,1)$ in 1 has expected value

${\mathbb{E}}[Z]=\left(1-q\right)\frac{1-G^{\prime}(1)}{G^{\prime}(0)-G^{\prime}(1)}+q.$

Note that the transformation achieved in Corollary 2.5 can also be obtained
using Lemma 2.3. Indeed, let us consider the function

$\tilde{G}(z)=w\,G(z)+(1-w)z,$

for $z\in[0,1]$, $w\in[0,1]$, and where $G$ is an exposure curve. Then, the
function $\tilde{G}$ is an exposure curve. Using Theorem 2.1, if we denote by
$F$ the distribution function obtained from the exposure curve $G$, by $f$ the
absolutely continuous density on $[0,1)$ and by $p$ the point mass in 1, we
can characterize the distribution function $\tilde{F}$, the absolutely
continuous density on $[0,1)$ $\tilde{f}$, and the point mass $\tilde{p}$
obtained from the exposure curve $\tilde{G}$ using

$\tilde{F}(z)=wF(z)+(1-w)\mathds{1}_{\\{z=1\\}},$

for $z\in[0,1]$,

$\quad\tilde{f}(z)=wf(z)=-w\frac{G^{\prime\prime}(z)}{G^{\prime}(0)},$

for $z\in[0,1)$, and

$\tilde{p}=wp+(1-w).$

In what follows, the goal will be to introduce examples of exposure curves
that are useful to model lower-truncated and right-censored losses. Before
that, we start by studying the explicit family of exposure curves introduced
by Bernegger [2].

##  3 The Bernegger class of distributions

###  3.1 The class of MBBEFD exposure curves and densities

The MBBEFD class of Bernegger [2] selects an explicit family of exposure
curves. This family is parametrized through two parameters $g\geq 1$ and
$b\geq 0$, and is given as follows for $z\in[0,1]$

$G_{b,g}(z)=\left\\{\begin{array}[]{ll}z&\text{ for $g=1$ or $b=0$,}\\\
\frac{\log(1+(g-1)z)}{\log(g)}&\text{ for $g>1$ and $b=1$,}\\\
\frac{1-b^{z}}{1-b}&\text{ for $g>1$ and $bg=1$,}\\\
\frac{\log\left(\frac{(g-1)b+(1-bg)b^{z})}{(1-b)}\right)}{\log(bg)}&\text{ for
$g>1$, $b>0$, $b\neq 1$ and $bg\neq 1$.}\end{array}\right.$ (3.1)

The three cases $bg=1$, $bg>1$ and $bg<1$ give the MB (Maxwell-Boltzmann), the
BE (Bose-Einstein) and the FD (Fermi-Dirac) distributions, respectively. Using
(2.1), we can calculate the distributions for $z\in[0,1)$, see (3.6) in
Bernegger [2],

$F_{b,g}(z)=\left\\{\begin{array}[]{ll}0&\text{ for $g=1$ or $b=0$,}\\\
1-\left(1+(g-1)z\right)^{-1}&\text{ for $g>1$ and $b=1$,}\\\ 1-b^{z}&\text{
for $g>1$ and $bg=1$,}\\\ 1-\frac{1-b}{(g-1)b^{1-z}+(1-bg)}&\text{ for $g>1$,
$b>0$, $b\neq 1$ and $bg\neq 1$.}\end{array}\right.$ (3.2)

We observe that the first case is not of interest because it gives a point
mass of 1 to $z=1$. For this reason, we skip this case in the sequel. On
$z<1$, we can calculate the second derivatives of these exposure curves
$G_{b,g}$. This gives us the densities for $z\in[0,1)$, see (3.7) in Bernegger
[2],

$f_{b,g}(z)=\left\\{\begin{array}[]{ll}(g-1)\left(1+(g-1)z\right)^{-2}&\text{
for $g>1$ and $b=1$,}\\\ -\log(b)b^{z}&\text{ for $g>1$ and $bg=1$,}\\\
\frac{(g-1)(b-1)\log(b)b^{1-z}}{\left((g-1)b^{1-z}+(1-bg)\right)^{2}}&\text{
for $g>1$, $b>0$, $b\neq 1$ and $bg\neq 1$,}\end{array}\right.$ (3.3)

and we have a point mass in $z=1$ given by

$p=\frac{1}{g}~{}\in~{}(0,1).$ (3.4)

Thus, we have an absolutely continuous distribution on $[0,1)$, with a point
mass $p=1/g$ in 1, and, e.g., in the last case of (3.3), we have a strictly
positive density in 0

$f_{b,g}(0)=\frac{(g-1)\log(b)b}{b-1}~{}>~{}0.$

Such a density may therefore come from a lower-truncated claim. Finally, the
mean is given by

${\mathbb{E}}_{b,g}[Z]=\left\\{\begin{array}[]{ll}\frac{\log(g)}{g-1}&\text{
for $g>1$ and $b=1$,}\\\ \frac{b-1}{\log(b)}&\text{ for $g>1$ and $bg=1$,}\\\
\frac{b-1}{\log(b)}\frac{\log(bg)}{bg-1}&\text{ for $g>1$, $b>0$, $b\neq 1$
and $bg\neq 1$.}\end{array}\right.$ (3.5)

We give an example of such an exposure curve that is typically used for
exposure rating in reinsurance.

######  Example 3.1 (Swiss Re and Lloyd’s exposure curves).

Bernegger [2] provides an explicit parametrization for the MBBEFD class which
can be used for reinsurance exposure rating in case of scarce data. Namely,
both parameters $b$ and $g$ are parametrized as a function of a single
parameter $c>0$ as follows

$b=b(c)=\exp\\{3.1-0.15(1+c)c\\}\qquad\text{ and }\qquad
g=g(c)=\exp\\{(0.78+0.12c)c\\}.$ (3.6)

For $c=1.5,2,3,4$, one obtains the Swiss Re exposure curves, and for $c=5$,
the Lloyd’s exposure curve, these are illustrated in Figure 2 (lhs). The
right-hand side of this figure shows the resulting densities
$f_{b(c),g(c)}(\cdot)$, and we remark that all of the resulting MBBEFD
densities are monotonically decreasing on $(0,1)$. $\blacksquare$

Figure 2: Swiss Re and Lloyd’s exposure curves (lhs) and the resulting
densities (rhs).

###  3.2 Properties of the MBBEFD class

Example 3.1 has shown that the exposure curves (3.6) proposed by Bernegger [2]
for reinsurance exposure rating lead to monotonically decreasing densities
$f_{b,g}$, see Figure 2 (rhs). For general insurance pricing, we are rather
interested into unimodal densities similar to Figure 1, because this more
commonly reflects the properties of general insurance claims data.

######  Proposition 3.2.

The density $f_{b,g}$ for $g>1$ and $b>0$ given in (3.3) has the following
properties:

  * •

Case $bg<1$. The density $f_{b,g}$ is

    * –

monotonically decreasing on $[0,1)$ for $(1-bg)/(g-1)\leq b$;

    * –

unimodal on $[0,1)$ for $b<(1-bg)/(g-1)<1$, with a maximum in

$z^{*}=1-\frac{\log\left((1-bg)/(g-1)\right)}{\log(b)}~{}\in~{}(0,1);$ (3.7)

    * –

monotonically increasing on $[0,1)$ for $(1-bg)/(g-1)\geq 1$.

  * •

Case $bg\geq 1$. The density $f_{b,g}$ is monotonically decreasing on $[0,1)$.

This proposition shows that in practical applications in general insurance,
the FD distributions (with $bg<1$) are the most interesting ones, as they can
be unimodal, or either monotonically decreasing or increasing. This excludes
the exposure curves of Example 3.1, as these Swiss Re and Lloyd’s exposure
curves provide us with $bg>1$. Next, we show that the MBBEFD distribution
$F_{b,g}$ for $bg<1$ can be derived from the logistic function (distribution)

$\psi(t)=\frac{e^{t}}{e^{t}+1}~{}\in~{}(0,1),$

for $t\in{\mathbb{R}}$. The logistic function has first derivative (logistic
density)

$\psi^{\prime}(t)=\frac{e^{t}}{(e^{t}+1)^{2}}=\psi(t)\left(1-\psi(t)\right).$

This derivative is symmetric around zero. We have the following result.

######  Proposition 3.3.

Let $b>0$ and $g>1$. For $bg<1$, the MBBEFD density has the functional form,
for $z\in[0,1)$,

$f_{b,g}(z)=(a+1)\log(1/b)\,\psi^{\prime}\left(z\log(1/b)+\log(a)\right),$

where we set

$a=\frac{b(g-1)}{1-bg}\qquad\text{ and }\qquad g=\frac{a+b}{(a+1)b},$ (3.8)

respectively. If $b<(1-bg)/(g-1)<1$, i.e. $b<a<1$, this MBBEFD density is bell
shaped around $z^{*}$ given in (3.7).

For further terminology, we call unimodal symmetric densities, bell shaped
densities. Of course, this includes for example the Gaussian density but also
the logistic density. We conclude that for $bg<1$, the MBBEFD density is the
logistic density $\psi^{\prime}$ on the interval

$\left[\log(a),\,\log(a/b)\right),$

scaled by a constant factor $(1+a)\log(1/b)>0$. It is symmetric around the
mode $z^{*}$, and it decays slower than the Gaussian density. This now shows
why the MBBEFD densities are not sufficient for general insurance claims
modeling, because general insurance claims are typically positively skewed,
which cannot be captured by the logistic density.

###  3.3 Extension of Bernegger’s idea using non-bell shaped densities

We extend the class of bell-shaped MBBEFD densities to a more general class of
exposure curves, which allows, in particular, for skewness in their
corresponding densities. We call this extended family the Bernegger class. In
Section 2.1, we have started from a generic exposure curve $G:[0,1]\to[0,1]$
which is a non-decreasing, concave and twice continuously differentiable
function with the normalization $G(0)=0,\,G(1)=1$ and with $G^{\prime}(0)>0$.
The MBBEFD exposure curve (3.1) can be reparametrized. Indeed, by using (3.8),
we obtain for $z\in[0,1]$

$G_{b,a}(z)=\frac{\log(a+b^{z})-\log(a+1)}{\log(a+b)-\log(a+1)},$ (3.9)

for $b\geq 0$ and $a>-\min(1,b)$; we refer to Section 3.1 of Bernegger [2].
This structure can be used to design exposure curve forms that do not have the
bell-shape property of Proposition 3.3. The function
$a+b^{z}=a+\exp\\{-z\log(1/b)\\}$ is strictly decreasing and convex for
$b\in(0,1)$. We modify the modeling set-up (3.9) as follows. Choose a function
$B:[0,1]\rightarrow{\mathbb{R}}$ that satisfies

$B(z)=h(b(z)),$ (3.10)

for some functions $h$ and $b$, in order to define an exposure curve (under
further assumptions on $h$ and $b$)

$z\mapsto G(z)=\frac{B(z)-B(0)}{B(1)-B(0)},$ (3.11)

which ensures that the normalization property $G(0)=0$ and $G(1)=1$ is
satisfied. The function $h$ will be denoted as the link function, whereas the
function $b$ will be named the inner function, and we notice that Bernegger’s
original choice was $b(z)=a+b^{z}$ for $b\in(0,1)$ and $h(x)=\log(x)$, meaning
that he used a logarithmic linked exposure curve. We will first explore some
examples using the same link function and then introduce the exponentially
linked exposure curves, which use the link function $h(x)=\exp(x)$ and seem to
be more suitable in modeling general insurance claims. We call the class of
distributions induced by exposure curves of the form (3.11) the Bernegger
class.

##  4 Logarithmic linked exposure family

We start by considering logarithmic linked examples of the Bernegger class,
which are obtained by choosing $h(x)=\log(x)$ in (3.10).

######  Proposition 4.1.

Choose a function $b:[0,1]\to(0,\infty)$ with $b(0)\neq b(1)$ that is twice
continuously differentiable and define for $z\in[0,1]$ the function

$G(z)=\frac{\log(b(z))-\log(b(0))}{\log(b(1))-\log(b(0))}.$ (4.1)

The function $G$ is an exposure curve if and only if one of the following
holds:

$b^{\prime}(0)>0,\,b^{\prime}(z)\geq 0\textrm{ and
}b^{\prime\prime}(z)b(z)-b^{\prime}(z)^{2}\leq 0\textrm{ for all }z\in[0,1],$
(4.2)

or

$b^{\prime}(0)<0,\,b^{\prime}(z)\leq 0\textrm{ and
}b^{\prime\prime}(z)b(z)-b^{\prime}(z)^{2}\geq 0\textrm{ for all }z\in[0,1].$
(4.3)

Using Theorem 2.1, one can then derive the distribution function of a random
variable leading to an absolutely continuous density on $[0,1)$ and a point
mass in 1.

######  Corollary 4.2.

Assume that a twice continuously differentiable function
$b:[0,1]\to(0,\infty)$ with $b(0)~{}\neq~{}b(1)$ fulfills condition (4.2) or
(4.3). The exposure curve $G$ defined in (4.1) provides the distribution of a
random variable $Z\sim F_{Z}$

$F_{Z}(z)=\left(1-\frac{b^{\prime}(z)}{b^{\prime}(0)}\frac{b(0)}{b(z)}\right)\mathds{1}_{\\{z<1\\}}+\mathds{1}_{\\{z=1\\}},$
(4.4)

for $z\in[0,1]$, with density for $z\in[0,1)$

$f_{Z}(z)=\frac{b(0)}{-b^{\prime}(0)}\frac{b^{\prime\prime}(z)b(z)-b^{\prime}(z)^{2}}{b(z)^{2}},$

and with point mass in $z=1$ equal to

$p=\frac{b^{\prime}(1)}{b^{\prime}(0)}\frac{b(0)}{b(1)}.$

Moreover, the mean of $Z\sim F_{Z}$ is equal to

${\mathbb{E}}[Z]=\frac{b(0)}{-b^{\prime}(0)}\,\log\left(\frac{b(0)}{b(1)}\right).$

For general insurance pricing, we are interested into unimodal densities and
one is thus interested in the first derivative of $f_{Z}$ in order to
characterize the maximum of the density

$f^{\prime}_{Z}(z)=\frac{b(0)}{-b^{\prime}(0)}\frac{b^{\prime\prime\prime}(z)b(z)^{2}-3b^{\prime\prime}(z)b^{\prime}(z)b(z)+2b^{\prime}(z)^{3}}{b(z)^{3}}.$

Note that this derivative only exists if the third derivative of $b$ exists.

##  5 Exponentially linked exposure family

Next we introduce the exponentially linked exposure family by setting
$h(x)=\exp(x)$ in (3.10).

######  Proposition 5.1.

Choose a function $b:[0,1]\to{\mathbb{R}}$ with $b(0)\neq b(1)$ that is twice
continuously differentiable and define for $z\in[0,1]$ the function

$G(z)=\frac{e^{b(z)}-e^{b(0)}}{e^{b(1)}-e^{b(0)}}.$ (5.1)

The function $G$ is an exposure curve if and only if one of the following
holds:

$b^{\prime}(0)>0,\,b^{\prime}(z)\geq 0\textrm{ and
}b^{\prime\prime}(z)+b^{\prime}(z)^{2}\leq 0\textrm{ for all }z\in[0,1],$
(5.2)

or

$b^{\prime}(0)<0,\,b^{\prime}(z)\leq 0\textrm{ and
}b^{\prime\prime}(z)+b^{\prime}(z)^{2}\geq 0\textrm{ for all }z\in[0,1].$
(5.3)

As for the logarithmic linked exposure curves, one can then derive using
Theorem 2.1 a distribution function leading to an absolutely continuous
density on $[0,1)$ and to a point mass in 1.

######  Corollary 5.2.

Assume that a twice continuously differentiable function
$b:[0,1]\to(0,\infty)$ with $b(0)\neq b(1)$ fulfills condition (5.2) or (5.3).
The exposure curve $G$ defined in (5.1) provides the distribution of a random
variable $Z\sim F_{Z}$

$F_{Z}(z)=\left(1-\frac{e^{b(z)}}{e^{b(0)}}\frac{b^{\prime}(z)}{b^{\prime}(0)}\right)\mathds{1}_{\\{z<1\\}}+\mathds{1}_{\\{z=1\\}},$
(5.4)

for $z\in[0,1]$, with density for $z\in[0,1)$

$f_{Z}(z)=-\frac{e^{b(z)}}{e^{b(0)}}\frac{b^{\prime}(z)^{2}+b^{\prime\prime}(z)}{b^{\prime}(0)},$
(5.5)

and with point mass in $z=1$ equal to

$p=\frac{e^{b(1)}}{e^{b(0)}}\frac{b^{\prime}(1)}{b^{\prime}(0)}.$

The mean of $Z\sim F_{Z}$ is equal to

${\mathbb{E}}[Z]=\frac{e^{b(1)-b(0)}-1}{b^{\prime}(0)}.$

The first derivative of the density $f_{Z}$ given in (5.5) allows to
characterize its extrema and is given by

$f_{Z}^{\prime}(z)=-\frac{e^{b(z)}}{e^{b(0)}}\frac{b^{\prime}(z)^{3}+3b^{\prime}(z)b^{\prime\prime}(z)+b^{\prime\prime\prime}(z)}{b^{\prime}(0)}.$
(5.6)

Note that this derivative only exists if the third derivative of $b$ exists.

######  Example 5.3 (Exponential distribution).

A well-known example of a distribution belonging to the exponentially linked
exposure family arises when choosing $b(z)=-\lambda z$ for $z\in[0,1]$ and
$\lambda>0$. The distribution obtained using Corollary 5.2 is the distribution
of $Z=\min\\{Y,1\\}$, where $Y\sim\textrm{Exp}(\lambda)$. Indeed, the
construction in (5.4) gives as distribution function

$F_{Z}(z)=\left(1-e^{-\lambda
z}\right)\mathds{1}_{\\{z<1\\}}+\mathds{1}_{\\{z=1\\}},$

for $z\in[0,1]$. $\blacksquare$

##  6 Examples

In this section, the goal is to explore some examples belonging to the
logarithmic and the exponentially linked exposure family of the Bernegger
class. These examples will be used to fit general insurance claims data using
the tractability of the models described in Section 2.1. In other words, MLE
can directly be used since the distribution functions as well as their
associated densities are of closed form.

For this, we will vary the choice of the inner function $b(z)$. In the
following, $\theta\in\Theta$ will denote the set of parameters appearing in
the inner function. Using a dataset for $Z$ taking values in $[0,1]$, we will
fit the models with MLE, maximizing the log-likelihood function

$\theta\mapsto\ell_{Z}(\theta)=\log(f^{(\theta)}(Z))\mathds{1}_{\\{Z<1\\}}+\log(p^{(\theta)})\mathds{1}_{\\{Z=1\\}},$
(6.1)

where the absolutely continuous density $f^{(\theta)}$ on $[0,1)$ and the
point mass $p^{(\theta)}$ are obtained as in Section 2.1. We also study the
model of Corollary 2.5, which extends the previous model by adding a flexible
point mass $q$ in 1. Its log-likelihood function is

$\begin{split}(\theta,q)\mapsto\ell_{Z}(\theta,q)&=\log((1-q)f^{(\theta)}_{0}(Z))\mathds{1}_{\\{Z<1\\}}+\log(q)\mathds{1}_{\\{Z=1\\}}\\\
&=\log(f^{(\theta)}_{0}(Z))\mathds{1}_{\\{Z<1\\}}+\log(1-q)\mathds{1}_{\\{Z<1\\}}+\log(q)\mathds{1}_{\\{Z=1\\}}.\end{split}$
(6.2)

The maximization problem in (6.1) will be denoted as the MLE of the standard
problem, whereas the maximization problem in (6.2) will be called the MLE of
the extended problem.

The dataset used in this section consists of real non-life insurance claims
with their associated deductibles and maximal covers. The lower-truncated and
right-censored i.i.d. claims $Z_{i}$ taking values in the interval $[0,1]$ are
obtained using the transformation (1.1) as well as a rescaling with the
maximal cover $M$. In total, we have $n=126\,026$ observations for $Z_{i}$,
with an empirical point mass in $1$ equal to $0.034$ and an empirical mean
equal to $0.339$. The histogram and the empirical density of the claims that
are strictly smaller than $1$ are illustrated in the left and the right plot
of Figure 3, respectively.

Figure 3: Histogram (lhs) and empirical density (rhs) of the claims $Z_{i}$.

These normalized losses $Z_{i}$, along with different models, will be used to
solve the above MLE maximization problems in order to produce the results of
this section. For this, we use the R function optim in order to minimize the
sum of the negative of the log-likelihoods evaluated at $Z_{i}$, after having
possibly transformed our parameters in a way to ensure that they lie in a
suitable open domain, as described below. The first fitted model is the
classical MBBEFD model of Bernegger [2].

###  6.1 The MBBEFD example

We have seen in Section 3.3 that the MBBEFD example belongs to the logarithmic
linked exposure family and is obtained by choosing

$b(z)=a+b^{z},$

for parameters $b\geq 0$ and $a>-\min(1,b)$. Using the parametrization in
(3.8), it is possible to give the conditions under which this class of
distributions leads to unimodal densities on $[0,1)$. Indeed, according to
Proposition 3.2, such unimodal densities are obtained if and only if

$bg<1\qquad\text{ and }\qquad b<\frac{1-bg}{g-1}<1,$

for parameters $g>1$ and $b>0$. Therefore, we use the domain

$\Theta=\Big{\\{}g>1,\max\left(0,\frac{2-g}{g}\right)<b<\frac{1}{2g-1}\Big{\\}},$

in order to find unimodal MLE solutions of the standard and extended
maximization problems described in (6.1) and (6.2). In Figure 4, the empirical
density of the data points strictly smaller than 1 is plotted in blue color.
Since this corresponds to a true density integrating to one, we further show
the conditional density of $Z|_{\\{Z<1\\}}$ of the fitted models, using the
parameters obtained by solving the standard problem (6.1) (in green) and the
extended problem (6.2) (in red). The point mass and the mean are shown in
Table 1, as well as the log-likelihoods of the random variables
$Z|_{\\{Z<1\\}}$ and $Z$, and the AIC scores that are computed using
$\ell_{Z}$. We see that, as expected, the MLE solution of the extended problem
gives better results, even if the densities obtained are not close to the
empirical density. Proposition 3.3 helps us to understand why the fit is not
accurate since the empirical density is skewed to the right, whereas the
MBBEFD class only allows for symmetric densities.

Figure 4: The MBBEFD: densities of the random variable $Z|_{\\{Z<1\\}}$. | Point Mass | Mean | $\ell_{Z|_{\\{Z<1\\}}}$ | $\ell_{Z}$ | AIC
---|---|---|---|---|---
Empirical density (Blue) | 0.034 | 0.339 | - | - | -
MLE of the standard problem (Green) | 0.020 | 0.337 | 32 682 | 13 475 | -26 947
MLE of the extended problem (Red) | 0.034 | 0.336 | 33 257 | 14 587 | -29 168
Table 1: The MBBEFD: results.

###  6.2 Power logarithmic linked exposure example

An example belonging to the logarithmic linked exposure family is obtained by
choosing the power function

$b(z)=\left(1-\frac{z}{\alpha}\right)^{\delta}+a,$

for $\alpha>1$, $\delta>1$ and a sufficiently large parameter $a>0$. This is a
smooth and strictly convex curve on $[0,1]$ with
$b(1)=(1-1/\alpha)^{\delta}+a<1+a=b(0)$. The first and second derivatives for
$z\in[0,1]$ are given by

$b^{\prime}(z)=-\frac{\delta}{\alpha}\left(1-\frac{z}{\alpha}\right)^{\delta-1}~{}<~{}0\qquad\text{
and }\qquad
b^{\prime\prime}(z)=\frac{\delta(\delta-1)}{\alpha^{2}}\left(1-\frac{z}{\alpha}\right)^{\delta-2}~{}>~{}0.$

In order to achieve $b^{\prime\prime}(z)b(z)-b^{\prime}(z)^{2}\geq 0$ for all
$z\in[0,1]$, which is necessary due to Proposition 4.1, the parameter $a$ has
to satisfy

$a>\max_{z\in[0,1]}\left(\frac{b^{\prime}(z)^{2}}{b^{\prime\prime}(z)}-\tilde{b}(z)\right),$
(6.3)

where we set $\tilde{b}(z)=\left(1-\frac{z}{\alpha}\right)^{\delta}$. This
reads in this example

$a>\max_{z\in[0,1]}\left(\frac{1}{\delta-1}\left(1-\frac{z}{\alpha}\right)^{\delta}\right)~{}=~{}\frac{1}{\delta-1}.$
(6.4)

With Corollary 4.2, we then obtain as density for $z\in[0,1)$

$f_{Z}(z)=\frac{a+1}{\alpha}\frac{\left(1-{z}/{\alpha}\right)^{\delta-2}}{\left(a+(1-{z}/{\alpha})^{\delta}\right)^{2}}\left(a(\delta-1)-\left(1-{z}/{\alpha}\right)^{\delta}\right),$
(6.5)

with point mass in $z=1$ equal to

$p=\left(1-\frac{1}{\alpha}\right)^{\delta-1}\frac{a+1}{a+(1-{1}/{\alpha})^{\delta}},$

and mean

${\mathbb{E}}[Z]=\frac{\alpha(a+1)}{\delta}\,\log\left(\frac{a+1}{a+(1-{1}/{\alpha})^{\delta}}\right).$

The derivative of the density $f_{Z}$ is given by

$f^{\prime}_{Z}(z)=-\frac{a+1}{\alpha^{2}}\,\frac{\left[a^{2}(\delta-1)(\delta-2)-a(\delta-1)(\delta+4)(1-{z}/{\alpha})^{\delta}+2(1-{z}/{\alpha})^{2\delta}\right]}{(a+(1-{z}/{\alpha})^{\delta})^{3}\,(1-{z}/{\alpha})^{-\delta+3}}.$
(6.6)

######  Lemma 6.1.

The power logarithmic linked exposure example with the above parameters leads
to a well-defined distribution. The density $f_{Z}$ of this power logarithmic
linked exposure example given in (6.5) can only be unimodal on $[0,1)$ if
$\delta>2$.

Therefore, we use the domain

$\Theta=\Big{\\{}\alpha>1,\delta>2,a>\frac{1}{\delta-1}\Big{\\}},$

in order to find unimodal solutions to the MLE of the standard and extended
maximization problems described in (6.1) and (6.2). The results displayed in
Figure 5 are very similar to the ones obtained for the MBBEFD example. This
can be confirmed by comparing Table 1 and Table 2, where most values coincide
although they are actually different if we look at digits after the decimal
point.

Figure 5: Power logarithmic linked exposure: densities of the random variable $Z|_{\\{Z<1\\}}$. | Point Mass | Mean | $\ell_{Z|_{\\{Z<1\\}}}$ | $\ell_{Z}$ | AIC
---|---|---|---|---|---
Empirical density (Blue) | 0.034 | 0.339 | - | - | -
MLE of the standard problem (Green) | 0.020 | 0.337 | 32 680 | 13 475 | -26 945
MLE of the extended problem (Red) | 0.034 | 0.336 | 33 257 | 14 587 | -29 166
Table 2: Power logarithmic linked exposure example: results.

###  6.3 Sine logarithmic linked exposure example

A third example belonging to the logarithmic linked exposure family is given
by the sine function

$b(z)=\sin(\alpha z+\beta)+a,$

for
$\beta\in\left(-\frac{\pi}{2},0\right),\alpha\in\left(0,\frac{\pi}{2}-\beta\right)$
and $-\sin(\beta)<a<-1/\sin(\beta)$. This is a smooth curve on $[0,1]$ with
$b(0)=\sin(\beta)+a<\sin(\alpha+\beta)+a=b(1)$. The first and second
derivatives for $z\in[0,1]$ are given by

$b^{\prime}(z)=\alpha\cos(\alpha z+\beta)~{}>~{}0\qquad\text{ and }\qquad
b^{\prime\prime}(z)=-\alpha^{2}\sin(\alpha z+\beta).$

Note that in this case, the function $b$ is in general neither concave, nor
convex on the entire interval $[0,1]$. We claim that the condition
$b^{\prime\prime}(z)b(z)-b^{\prime}(z)^{2}\geq 0$ for all $z\in[0,1]$ is
fulfilled, which is necessary in order to obtain a distribution function due
to Proposition 4.1. Moreover, the density for $z\in[0,1)$ reads

$f_{Z}(z)=\frac{\sin(\beta)+a}{\cos(\beta)}\,\frac{\alpha\left[1+a\sin(\alpha
z+\beta)\right]}{\left[\sin(\alpha z+\beta)+a\right]^{2}},$ (6.7)

with point mass in $z=1$

$p=\frac{\cos(\alpha+\beta)}{\cos(\beta)}\,\frac{\sin(\beta)+a}{\sin(\alpha+\beta)+a},$

and mean

${\mathbb{E}}[Z]=-\frac{\sin(\beta)+a}{\alpha\cos(\beta)}\,\log\left(\frac{\sin(\beta)+a}{\sin(\alpha+\beta)+a}\right).$

The derivative of the density $f_{Z}$ is given by

$f^{\prime}_{Z}(z)=-\frac{\sin(\beta)+a}{\alpha\cos(\beta)}\,\frac{\alpha^{3}\cos(\alpha
z+\beta)\left[-a^{2}+a\sin(\alpha z+\beta)+2\right]}{\left[\sin(\alpha
z+\beta)+a\right]^{3}}.$ (6.8)

######  Lemma 6.2.

The sine logarithmic linked exposure example with the above parameters leads
to a well-defined distribution. Moreover, the density $f_{Z}$ given in (6.7)
can only be unimodal on $[0,1)$ if $1\leq a\leq 2$.

Figure 6: Sine logarithmic linked exposure: densities of the random variable $Z|_{\\{Z<1\\}}$. | Point Mass | Mean | $\ell_{Z|_{\\{Z<1\\}}}$ | $\ell_{Z}$ | AIC
---|---|---|---|---|---
Empirical density (Blue) | 0.034 | 0.339 | - | - | -
MLE of the standard problem (Green) | 0.040 | 0.380 | 26 907 | 8 164 | -16 323
MLE of the extended problem (Red) | 0.034 | 0.362 | 30 453 | 11 783 | -23 558
Table 3: Sine logarithmic linked exposure example: results.

We use the domain

$\Theta=\left\\{\beta\in\left(-\frac{\pi}{2},0\right),\alpha\in\left(0,\frac{\pi}{2}-\beta\right),1<a<\min\left(-\frac{1}{\sin(\beta)},2\right)\right\\},$

in order to find unimodal solutions to the standard and extended maximization
problems described in (6.1) and (6.2). Similarly as for the previous examples,
we obtain the results displayed in Figure 6 and Table 3. Although this example
manages to produce rather skewed densities, see Figure 6, the fit does not
seem to be accurate for this data, i.e. it is less accurate than in the first
examples, and with AIC, we clearly prefer the previous models.

###  6.4 Quadratic exponentially linked exposure example

Let us now treat an example belonging to the exponentially linked exposure
family, and given by the quadratic function

$b(z)=\alpha z^{2}+\beta z,$

for $\alpha<0$ and $\beta<-\sqrt{-2\alpha}$. This is a smooth curve on $[0,1]$
with $b(0)=0>\alpha+\beta=b(1)$. The first and second derivatives for
$z\in[0,1]$ are given by

$b^{\prime}(z)=2\alpha z+\beta~{}<~{}0\qquad\text{ and }\qquad
b^{\prime\prime}(z)=2\alpha~{}<~{}0.$

This function $b$ is a smooth and strictly concave curve on $[0,1]$. We claim
that the condition $b^{\prime\prime}(z)+b^{\prime}(z)^{2}\geq 0$ is fulfilled
for all $z\in[0,1]$, which is necessary in order to obtain a distribution
function due to Proposition 5.1. Moreover, the density for $z\in[0,1)$ reads

$f_{Z}(z)=-\frac{e^{\alpha z^{2}+\beta z}\left[4\alpha^{2}z^{2}+4\alpha\beta
z+\beta^{2}+2\alpha\right]}{\beta},$ (6.9)

with point mass in $z=1$

$p=\frac{e^{\alpha+\beta}\,[2\alpha+\beta]}{\beta},$

and mean

${\mathbb{E}}[Z]=\frac{e^{\alpha+\beta}-1}{\beta}.$

The derivative of the density $f_{Z}$ is given by

$f^{\prime}_{Z}(z)=-\frac{e^{\alpha z^{2}+\beta z}\,(2\alpha
z+\beta)\left[(2\alpha z+\beta)^{2}+6\alpha\right]}{\beta}.$ (6.10)

######  Lemma 6.3.

The quadratic logarithmic linked exposure example with the above parameters
leads to a well-defined distribution. Moreover, the density $f_{Z}$ given in
(6.9) is unimodal on $[0,1)$ if and only if
$-\sqrt{-6\alpha}<\beta<-2\alpha-\sqrt{-6\alpha}$.

We use the domain

$\Theta=\Big{\\{}\alpha<0,\beta\in\left(-\sqrt{-6\alpha}\,,\min\left(-2\alpha-\sqrt{-6\alpha},-\sqrt{-2\alpha}\right)\right)\Big{\\}}$

in order to find unimodal solutions to the standard and extended maximization
problems described in (6.1) and (6.2). Similarly as for the previous examples,
we obtain the results shown in Figure 7 and Table 4. Although a new family of
exposure curves is used here, the results are close to the ones obtained with
the power logarithmic linked exposure example or the MBBEFD example. In fact,
in the extended problem, we obtain a slightly better model than in the MBBEFD
class according to AIC. However, looking at Figure 7, this new model is still
not satisfactory for this data.

Figure 7: Quadratic exponentially linked exposure example: densities of the random variable $Z|_{\\{Z<1\\}}$. | Point Mass | Mean | $\ell_{Z|_{\\{Z<1\\}}}$ | $\ell_{Z}$ | AIC
---|---|---|---|---|---
Empirical density (Blue) | 0.034 | 0.339 | - | - | -
MLE of the standard problem (Green) | 0.016 | 0.345 | 32 101 | 12 425 | -24 846
MLE of the extended problem (Red) | 0.034 | 0.341 | 33 340 | 14 726 | -29 447
Table 4: Quadratic exponentially linked exposure example: results.

###  6.5 Power exponentially linked exposure example

Finally, we consider a last example belonging again to the exponentially
linked exposure family. This last example is a bit more difficult in handling,
but it provides the best results for our example. Choose the power function

$b(z)=\epsilon(z+\delta)^{\alpha}-\beta z,$

for $\alpha\in(1,2),\delta>0,\epsilon<0$ and
$\beta>\epsilon\alpha\delta^{\alpha-1}+\sqrt{-\epsilon\alpha(\alpha-1)\delta^{\alpha-2}}$.
This is a smooth curve on $[0,1]$ with
$b(0)=\epsilon\delta^{\alpha}>\epsilon(1+\delta)^{\alpha}-\beta=b(1)$. The
first and second derivatives for $z\in[0,1]$ are given by

$b^{\prime}(z)=\alpha\epsilon(z+\delta)^{\alpha-1}-\beta~{}<~{}0\qquad\text{
and }\qquad
b^{\prime\prime}(z)=\alpha(\alpha-1)\epsilon(z+\delta)^{\alpha-2}~{}<~{}0.$

This function $b$ is a smooth and strictly concave curve on $[0,1]$. We claim
that the condition $b^{\prime\prime}(z)+b^{\prime}(z)^{2}\geq 0$ for all
$z\in[0,1]$ is fulfilled, which is necessary in order to obtain a distribution
function due to Proposition 5.1. Moreover, the density for $z\in[0,1)$ reads
as

$f_{Z}(z)=-\frac{e^{\epsilon(z+\delta)^{\alpha}-\beta
z}\left[(\alpha\epsilon(z+\delta)^{\alpha-1}-\beta)^{2}+\alpha(\alpha-1)\epsilon(z+\delta)^{\alpha-2}\right]}{e^{\epsilon\delta^{\alpha}}\left[\alpha\epsilon\delta^{\alpha-1}-\beta\right]},$

with point mass in $z=1$

$p=\frac{e^{\epsilon(1+\delta)^{\alpha}-\beta}\,[\alpha\epsilon(1+\delta)^{\alpha-1}-\beta]}{e^{\epsilon\delta^{\alpha}}\,[\alpha\epsilon\delta^{\alpha-1}-\beta]},$

and mean

${\mathbb{E}}[Z]=\frac{e^{\epsilon\left[(1+\delta)^{\alpha}-\delta^{\alpha}\right]-\beta}-1}{\alpha\epsilon\delta^{\alpha-1}-\beta}.$

The derivative of the density $f_{Z}$ can be obtained using (5.6). It does
however not allow to characterize the extrema of the density in this example
explicitly. Nevertheless, we observe in Figure 8 that this example contains
unimodal densities.

######  Lemma 6.4.

The power exponentially linked exposure example with the above parameters
leads to a well-defined distribution.

We use the domain

$\Theta=\Big{\\{}\alpha\in(1,2),\delta>0,\epsilon<0,\beta>\epsilon\alpha\delta^{\alpha-1}+\sqrt{-\epsilon\alpha(\alpha-1)\delta^{\alpha-2}}\Big{\\}},$

in order to find solutions to the standard and extended maximization problems
described in (6.1) and (6.2). Similarly as for the previous examples, we
obtain the results shown in Figure 8 and Table 5. This time, the fit seems
much better. This is especially the case for the red curve, which represents
the solution of the flexible maximization problem. The tail behavior on the
right and left sides seem adequate in contrast to the plots presented in the
previous examples. This observation is confirmed by the AIC scores attained
here, which are lower than the AIC scores of all other examples, i.e., we give
clear preference to this last example for this data.

Figure 8: Power exponentially linked exposure example: densities of the random
variable $Z|_{\\{Z<1\\}}$.

| Point Mass | Mean | $\ell_{Z|_{\\{Z<1\\}}}$ | $\ell_{Z}$ | AIC
---|---|---|---|---|---
Empirical density (Blue) | 0.034 | 0.339 | - | - | -
Standard MLE density (Green) | 0.025 | 0.341 | 34 068 | 15 199 | -30 390
Flexible MLE density (Red) | 0.034 | 0.339 | 34 402 | 15 731 | -31 453
Table 5: Power exponentially linked exposure example: results.

##  7 Conclusion

Most classical statistical models are not directly suited to model lower-
truncated and right-censored claims in general insurance since they lead to
problems that are not easily analytically tractable. Bernegger introduced in
[2] the MBBEFD class of distributions that can model such claims using a
distribution function, an absolutely continuous density, and a point mass that
are all of closed form. This class was introduced in the reinsurance
literature, where densities are typically monotonically decreasing.

In general insurance, however, we are mainly interested in unimodal skewed
densities. Therefore, we extended the MBBEFD class to a much bigger family of
distributions that we called the Bernegger class. By starting from the
properties of an exposure curve, we introduced two subfamilies, namely the
logarithmic and exponentially linked exposure curves. Through various
examples, we used the full tractability and flexibility of the Bernegger class
in order to fit parameters to general insurance claims using maximum
likelihood estimations. It turned out that this large class of distributions
contains models allowing to obtain a suitable approximation for the
distribution of the used dataset, and in general, we have a rich family of
unimodal and skewed densities within the Bernegger class. Going forward, it
will be interesting to further characterize and classify the members of the
Bernegger class based on different properties. Of course, this might involve
exploring other link functions. Another next step is to lift these models to
regression models allowing for integrating fixed effects described by
covariates.

## References

  * [1] Abramson, D. (2022). A nonproportional premium rating method for construction risks. North American Actuarial Journal 26/4, 626-645.
  * [2] Bernegger, S. (1997). The Swiss Re exposure curves and the MBBEFD distribution class. ASTIN Bulletin 27/1, 99-111.
  * [3] Blæsild, P., Jensen, J.L. (1985). Saddlepoint formulas for reproductive exponential models. Scandinavian Journal of Statistics 12/3, 193-202.
  * [4] Dutang, C., Gesmann, M., Spedicato, G. (2021). Exposure rating, destruction rate models and the mbbefd package. R Vignettes.
  * [5] Fung, T.C., Badescu, A.L., Lin, X.S. (2022). Fitting censored and truncated regression data using the mixture of experts models. North American Actuarial Journal, 26/4, 496-520.
  * [6] Landsman, Z., Valdez, E.A. (2005). Tail conditional expectation for exponential dispersion models. ASTIN Bulletin 35/1, 189-209.
  * [7] Parodi, P. (2014). Pricing in General Insurance. CRC Press.
  * [8] Parodi, P., Watson, P. (2019). Property graphs - a statistical model for fire and explosion losses based on graph theory. ASTIN Bulletin 49/2, 263-297.
  * [9] Poudyal, C. (2021). Robust estimation of loss models for lognormal insurance payment severity data. ASTIN Bulletin 51/2, 475-507.
  * [10] Poudyal, C. (2021). Truncated, censored, and actuarial payment-type moments for robust fitting of a single-parameter Pareto distribution. Journal of Computational and Applied Mathematics 388, 113310.
  * [11] Poudyal, C., Brazauskas, V. (2022). Robust estimation of loss models for truncated and censored severity data. Variance 15/2.
  * [12] R Core Team (2018). R: A language and environment for statistical computing. R Foundation for Statistical Computing, Vienna, Austria. https://www.R-project.org/
  * [13] Riegel, U. (2010). On fire exposure rating and the impact of the risk profile type. ASTIN Bulletin 40/2, 727-777.
  * [14] Spedicato, G.A., Dutang, C. (2015). The mbbefd package: a package for handling destruction rates in R including MBBEFD distributions. R Vignettes.
  * [15] Tweedie, M.C.K. (1984). An index which distinguishes between some important exponential families. In: Statistics: Applications and New Directions. Ghosh, J.K., Roy, J. (Eds.). Proceeding of the Indian Statistical Golden Jubilee International Conference, Indian Statistical Institute, Calcutta, 579-604.
  * [16] Verbelen, R., Gong, L., Antonio, K., Badescu, A., Lin, S. (2015). Fitting mixtures of Erlangs to censored and truncated data using the EM algorithm. ASTIN Bulletin 45/3, 729-758.
  * [17] Wüthrich, M.V., Merz, M. (2023). Statistical Foundations of Actuarial Learning and its Applications. Springer Actuarial. https://link.springer.com/book/10.1007/978-3-031-12409-9

##  Appendix A Proofs

We prove all statements in this appendix.

Proof of Theorem 2.1. The function $F_{Z}$, as defined in (2.1), is
continuously differentiable on the interval $(0,1)$ by our assumptions on $G$.
This means that a derivative exists, and is equal to

$f_{Z}(z)=F_{Z}^{\prime}(z)=-\frac{G^{\prime\prime}(z)}{G^{\prime}(0)}\geq 0,$

for $z\in[0,1)$. Since $G$ was assumed to be twice continuously
differentiable, non-decreasing and concave, we obtain that $0\leq
G^{\prime}(z)\leq G^{\prime}(0)$ and $f_{Z}(z)\geq 0$ for all $z\in[0,1]$.
This implies

$0=F_{Z}(0)\leq F_{Z}(z)\leq F_{Z}(1)=1,\,\text{ for all }z\in[0,1].$

Since $G^{\prime}$ is continuous by assumption, we conclude from (2.1) that
$F_{Z}$ is right-continuous, and hence, a distribution function on $[0,1]$.
The point mass in 1 is then given by

$p=1-F_{Z}(1_{-})=\frac{G^{\prime}(1)}{G^{\prime}(0)}\in[0,1],$

and the mean of $Z\sim F_{Z}$ is equal to

${\mathbb{E}}[Z]=\int_{0}^{1}1-F_{Z}(s)\,ds=\int_{0}^{1}\frac{G^{\prime}(s)}{G^{\prime}(0)}\,ds=\frac{G(1)-G(0)}{G^{\prime}(0)}=\frac{1}{G^{\prime}(0)}.$

This proves the theorem.

$\Box$

Proof of Lemma 2.3. First, it is clear that the function
$G:[0,1]\rightarrow{\mathbb{R}}$ defined in (2.4) is an exposure curve.
Moreover, we have by (2.1) that the distribution function $F_{Z}$ generated by
$G$ satisfies

$\begin{split}F_{Z}(z)&=\left(1-\frac{G^{\prime}(z)}{G^{\prime}(0)}\right)\mathds{1}_{\\{z<1\\}}+\mathds{1}_{\\{z=1\\}}\\\
&=\left(1-\frac{\sum_{i=1}^{n}\alpha_{i}G_{i}^{\prime}(z)}{\sum_{j=1}^{n}\alpha_{j}G_{j}^{\prime}(0)}\right)\mathds{1}_{\\{z<1\\}}+\mathds{1}_{\\{z=1\\}}\\\
&=\left(1-\sum_{i=1}^{n}\frac{\alpha_{i}G_{i}^{\prime}(0)}{\sum_{j=1}^{n}\alpha_{j}G_{j}^{\prime}(0)}\frac{G_{i}^{\prime}(z)}{G_{i}^{\prime}(0)}\right)\mathds{1}_{\\{z<1\\}}+\mathds{1}_{\\{z=1\\}}\\\
&=\left(1-\sum_{i=1}^{n}w_{i}\frac{G_{i}^{\prime}(z)}{G_{i}^{\prime}(0)}\right)\mathds{1}_{\\{z<1\\}}+\mathds{1}_{\\{z=1\\}}\\\
&=\sum_{i=1}^{n}w_{i}\left(\left(1-\frac{G_{i}^{\prime}(z)}{G_{i}^{\prime}(0)}\right)\mathds{1}_{\\{z<1\\}}+\mathds{1}_{\\{z=1\\}}\right)=\sum_{i=1}^{n}w_{i}F_{i}(z),\end{split}$

for $z\in[0,1]$, and where the elements

$w_{i}=\frac{\alpha_{i}G_{i}^{\prime}(0)}{\sum_{j=1}^{n}\alpha_{j}G_{j}^{\prime}(0)}~{}\geq~{}0$

are weights summing up to 1. The proof then follows similarly to Theorem 2.1.

$\Box$

Proof of Proposition 3.2. We calculate the first derivative for $g>1$, $b>0$,
$b\neq 1$ and $bg\neq 1$

$\displaystyle f^{\prime}_{b,g}(z)$ $\displaystyle=$
$\displaystyle-\frac{(g-1)(b-1)(\log(b))^{2}b^{1-z}}{\left((g-1)b^{1-z}+(1-bg)\right)^{2}}+2\frac{(g-1)^{2}(b-1)(\log(b))^{2}b^{2-2z}}{\left((g-1)b^{1-z}+(1-bg)\right)^{3}}$
(A.1) $\displaystyle=$
$\displaystyle\frac{(g-1)(b-1)(\log(b))^{2}b^{1-z}}{\left((g-1)b^{1-z}+(1-bg)\right)^{3}}\left[(g-1)b^{1-z}-(1-bg)\right]$
$\displaystyle=$
$\displaystyle(g-1)(b-1)\,(\log(b))^{2}b^{1-z}\,\frac{(g-1)b^{1-z}-(1-bg)}{\left((g-1)b^{1-z}+(1-bg)\right)^{3}}.$

This derivative can be zero (for $g>1$, $b>0$, $b\neq 1$) if and only if the
last ratio is zero.

(a) Case $bg<1$. This implies $1-bg>0$ and $b<1$, and the term in front of the
ratio in (A.1) is negative and the denominator in the ratio is positive. In
this case, the derivative is thus positive if
$0~{}<~{}b^{1-z}~{}<~{}(1-bg)/(g-1)$ and negative if $0<(1-bg)/(g-1)<b^{1-z}$.
Since $b^{1-z}$ is increasing in $z$, we have a monotonically increasing
density on $[0,1)$ if $(1-bg)/(g-1)\geq 1$ and we have a monotonically
decreasing density on $[0,1)$ if $(1-bg)/(g-1)\leq b$. For $b<(1-bg)/(g-1)<1$,
we have a unimodal density with critical point

$z^{*}=1-\frac{\log\left((1-bg)/(g-1)\right)}{\log(b)}~{}\in~{}(0,1).$

(b) Case $bg=1$. This implies $b<1$. The density $f_{b,g}(z)=-\log(b)b^{z}$ is
monotonically decreasing in $z$.

(c) Case $bg>1$. This implies $bg-1>0$. which means that the numerator of the
last ratio in (A.1) is thus always positive. We therefore need to analyze for
$b\neq 1$ the ratio

$\frac{b-1}{(g-1)b^{1-z}+(1-bg)},$

in order to determine the sign of the derivative $f^{\prime}_{b,g}$.

(c1) Consider the first case $b<1$. In this case, we have a negative numerator
and $b^{1-z}$ is increasing in $z\in[0,1)$. If $0<b^{1-z}<(bg-1)/(g-1)$, we
have an increasing derivative, and for $0<(bg-1)/(g-1)<b^{1-z}$, we have a
decreasing derivative. Since in this case, $(bg-1)/(g-1)<b<b^{1-z}$ holds for
all $z\in[0,1]$, we have a monotonically decreasing density.

(c2) Consider the second case $b>1$. In this case, we have a positive
numerator and $b^{1-z}$ is decreasing in $z\in[0,1)$. If
$0<b^{1-z}<(bg-1)/(g-1)$, we have a decreasing derivative, and for
$0<(bg-1)/(g-1)<b^{1-z}$, we have an increasing derivative. Since in this
case, we have $(bg-1)/(g-1)>b>b^{1-z}$ for all $z\in[0,1]$, the density is
decreasing.

Finally, the case $b=1$ also immediately follows. This completes the proof.
$\Box$

Proof of Proposition 3.3. For $bg<1$, we have the MBBEFD density

$\displaystyle f_{b,g}(z)$ $\displaystyle=$
$\displaystyle\frac{(g-1)(b-1)\log(b)b^{1-z}}{\left((g-1)b^{1-z}+(1-bg)\right)^{2}}~{}=~{}\frac{\frac{b(g-1)(b-1)\log(b)}{(1-bg)^{2}}b^{-z}}{\left(\frac{b(g-1)}{1-bg}b^{-z}+1\right)^{2}}$
$\displaystyle=$
$\displaystyle\frac{(b-1)\log(b)}{1-bg}\,\frac{\exp\left\\{z\log(1/b)+\log\left(\frac{b(g-1)}{1-bg}\right)\right\\}}{\left(\exp\left\\{z\log(1/b)+\log\left(\frac{b(g-1)}{1-bg}\right)\right\\}+1\right)^{2}}.$

This proves the first claim. For the second claim, we remark that the function
$t\mapsto\psi^{\prime}(t)$ is symmetric around the origin $t=0$. In view of
our claim, there is $z=z^{*}\in(0,1)$ such that
$z\log(1/b)+\log(b(g-1)/(1-bg))=0$ if and only if $b<(1-bg)/(g-1)<1$. Using
reparametrization (3.8) completes the proof. $\Box$

Proof of Proposition 4.1. First the function $G(z)$ is well-defined for all
$z\in[0,1]$ due to the assumption that the function $b$ maps to the interval
$(0,\infty)$ and that $b(0)\neq b(1)$. We compute the first and second
derivative of $G$ and obtain

$\displaystyle G^{\prime}(0)$ $\displaystyle=$
$\displaystyle\frac{b^{\prime}(0)}{b(0)}\,\frac{1}{\log(b(1))-\log(b(0))}~{}>~{}0,$
$\displaystyle G^{\prime}(z)$ $\displaystyle=$
$\displaystyle\frac{b^{\prime}(z)}{b(z)}\,\frac{1}{\log(b(1))-\log(b(0))}~{}\geq~{}0,$
$\displaystyle G^{\prime\prime}(z)$ $\displaystyle=$
$\displaystyle\frac{b^{\prime\prime}(z)b(z)-b^{\prime}(z)^{2}}{b(z)^{2}}\,\frac{1}{\log(b(1))-\log(b(0))}~{}\leq~{}0,$

where the inequalities hold if and only if $b^{\prime}(0)>0$,
$b^{\prime}(z)\geq 0$ and $b^{\prime\prime}(z)b(z)-b^{\prime}(z)^{2}\leq 0$
for all $z\in[0,1]$, or $b^{\prime}(0)<0$, $b^{\prime}(z)\leq 0$ and
$b^{\prime\prime}(z)b(z)-b^{\prime}(z)^{2}\geq 0$ for all $z\in[0,1]$. $\Box$

Proof of Corollary 4.2. According to Theorem 2.1, the function $F_{Z}$ defined
in (4.4) satisfies for $z\in[0,1]$

$\begin{split}F_{Z}(z)&=\left(1-\frac{G^{\prime}(z)}{G^{\prime}(0)}\right)\mathds{1}_{\\{z<1\\}}+\mathds{1}_{\\{z=1\\}}\\\
&=\left(1-\frac{b^{\prime}(z)}{b^{\prime}(0)}\,\frac{b(0)}{b(z)}\right)\mathds{1}_{\\{z<1\\}}+\mathds{1}_{\\{z=1\\}}.\end{split}$

The remaining statements then follow.

$\Box$

Proof of Proposition 5.1. First the function $G(z)$ is well-defined for all
$z\in[0,1]$ due to the assumption that $b(0)\neq b(1)$. We compute now the
first and second derivative of $G$ and obtain

$\displaystyle G^{\prime}(0)$ $\displaystyle=$
$\displaystyle\frac{e^{b(0)}}{e^{b(1)}-e^{b(0)}}\,b^{\prime}(0)~{}>~{}0,$
$\displaystyle G^{\prime}(z)$ $\displaystyle=$
$\displaystyle\frac{e^{b(z)}}{e^{b(1)}-e^{b(0)}}\,b^{\prime}(z)~{}\geq~{}0,$
$\displaystyle G^{\prime\prime}(z)$ $\displaystyle=$
$\displaystyle\frac{e^{b(z)}}{e^{b(1)}-e^{b(0)}}[b^{\prime}(z)^{2}+b^{\prime\prime}(z)]~{}\leq~{}0,$

where the inequalities hold if and only if $b^{\prime}(0)>0$,
$b^{\prime}(z)\geq 0$ and $b^{\prime}(z)^{2}+b^{\prime\prime}(z)\leq 0$ for
all $z\in[0,1]$, or $b^{\prime}(0)<0$, $b^{\prime}(z)\leq 0$ and
$b^{\prime}(z)^{2}+b^{\prime\prime}(z)\geq 0$ for all $z\in[0,1]$.

$\Box$

Proof of Corollary 5.2. According to Theorem 2.1, the function $F_{Z}$ defined
in (5.4) satisfies for $z\in[0,1]$

$\begin{split}F_{Z}(z)&=\left(1-\frac{G^{\prime}(z)}{G^{\prime}(0)}\right)\mathds{1}_{\\{z<1\\}}+\mathds{1}_{\\{z=1\\}}\\\
&=\left(1-\frac{e^{b(z)}}{e^{b(0)}}\frac{b^{\prime}(z)}{b^{\prime}(0)}\right)\mathds{1}_{\\{z<1\\}}+\mathds{1}_{\\{z=1\\}}.\end{split}$

The remaining statements then follow.

$\Box$

Proof of Lemma 6.1. The power logarithmic linked exposure example leads to a
well-defined distribution due to (6.3) and Propositon 4.1. Set
$y=(1-z/\alpha)^{\delta}\in((1-1/\alpha)^{\delta},1]$ for $z\in[0,1)$. The
derivative (6.6) is zero only if

$a^{2}(\delta-1)(\delta-2)-a(\delta-1)(\delta+4)y+2y^{2}=0.$

If there exist real-valued solutions, they are given by

$\displaystyle y_{\pm}$ $\displaystyle=$ $\displaystyle
a\,\frac{(\delta-1)(\delta+4)\pm\sqrt{(\delta-1)^{2}(\delta+4)^{2}-8(\delta-1)(\delta-2)}}{4}.$
(A.2)

We start with the case $\delta\in(1,2]$. In that case, we have

$(\delta-1)(\delta+4)\leq\sqrt{(\delta-1)^{2}(\delta+4)^{2}-8(\delta-1)(\delta-2)}.$

This implies for the smaller solution $y_{-}\leq 0$, thus, this solution
provides $y_{-}\not\in((1-1/\alpha)^{\delta},1]$. For the bigger solution, we
have, using (6.4) in the second step,

$y_{+}\geq a\,\frac{(\delta-1)(\delta+4)}{4}>\frac{\delta+4}{4}>1.$ (A.3)

Thus, also this second solution provides
$y_{+}\not\in((1-1/\alpha)^{\delta},1]$, therefore the density is monotone for
$\delta\in(1,2]$.

Next, we analyze the case $\delta>2$. The term under the square root in (A.2)
is given by

$\displaystyle(\delta-1)^{2}(\delta+4)^{2}-8(\delta-1)(\delta-2)$
$\displaystyle=$
$\displaystyle(\delta-1)\left[(\delta-1)(\delta+4)^{2}-8(\delta-2)\right]$
$\displaystyle=$
$\displaystyle(\delta-1)\left[(\delta-1)(\delta^{2}+8\delta+8)+8\right]~{}>~{}0.$

Thus, there are two real-valued solutions to (A.2). The bigger solution
$y_{+}$ also provides (A.3), and henceforth,
$y_{+}\not\in((1-1/\alpha)^{\delta},1]$. Therefore, we can focus on the
smaller solution $y_{-}$. It is given by

$\displaystyle y_{-}$ $\displaystyle=$ $\displaystyle
a(\delta-1)\left[(\delta/4+1)-\sqrt{(\delta/4+1)^{2}-\frac{\delta-2}{2\delta-2}}\right].$

The square bracket is in $[0,0.1]$, therefore there are $\alpha>1$ and
$a>1/(\delta-1)$ such that $y_{-}\in((1-1/\alpha)^{\delta},1]$. In particular,
there is a critical point $z^{*}\in(0,1]$ with $f_{Z}(z^{*})=0$ in these cases
and we easily see from (6.6) that $z^{*}$ is a maximum. This does however not
hold for any $\alpha>1$ and $a>1/(\delta-1)$. $\Box$

Proof of Lemma 6.2. In order to prove that the sine logarithmic linked
exposure example leads to a well-defined distribution function, we need to
show that the function $h$ defined through
$h(z)=b^{\prime\prime}(z)b(z)-b^{\prime}(z)^{2}$ satisfies $h(z)\leq 0$ for
all $z\in[0,1]$ according to Proposition 4.1. For this, it suffices to show
that $h(0)\leq 0$ and $h^{\prime}(z)\leq 0$ for any $0\leq z\leq 1$. This
indeed holds since

$\begin{split}h(0)&=b^{\prime\prime}(0)b(0)-b^{\prime}(0)^{2}\\\
&=-\alpha^{2}\sin(\beta)\Big{[}\sin(\beta)+a\Big{]}-\Big{[}\alpha\cos(\beta)\Big{]}^{2}\\\
&=-\alpha^{2}[1+a\sin(\beta)]~{}<~{}0,\end{split}$

where the last inequality is due to $a<-\frac{1}{\sin(\beta)}$. Furthermore,
let $z\in[0,1]$, then

$\begin{split}h^{\prime}(z)&=b^{\prime\prime\prime}(z)b(z)-b^{\prime}(z)b^{\prime\prime}(z)\\\
&=-\alpha^{3}\cos(\alpha z+\beta)\Big{[}\sin(\alpha
z+\beta)+a\Big{]}-\alpha\cos(\alpha z+\beta)(-\alpha^{2}\sin(\alpha
z+\beta))\\\ &=-a\alpha^{3}\cos(\alpha z+\beta)~{}<~{}0,\end{split}$

since $a>0$, $\alpha>0$ and $\alpha z+\beta\in(-\frac{\pi}{2},\frac{\pi}{2})$
for any $z\in[0,1]$. From (6.8), we see that this derivative has either no
root or a unique root $z^{*}\in[0,1]$ satisfying

$z^{*}=\frac{1}{\alpha}\left[\sin^{-1}\Big{(}\frac{a^{2}-2}{a}\Big{)}-\beta\right],$
(A.4)

where the inverse function $\sin^{-1}(\cdot)$ is chosen to map to the interval
$[-\frac{\pi}{2},\frac{\pi}{2}]$. This root can only exist if $-2\leq a\leq-1$
or $1\leq a\leq 2$. Since $a$ was assumed to be positive, this means that the
density $f_{Z}$ can only be unimodal when $1\leq a\leq 2$. Note that this
condition is not sufficient since we do not a priori obtain that there is a
root $z^{*}$ lying in the interval $(0,1)$. However, in the case where this
root $z^{*}\in(0,1)$ exists, it is clear from (6.8) that it corresponds to a
maximum of the density.

$\Box$

Proof of Lemma 6.3. The function $b$ of this example is strictly decreasing,
which means that the necessary condition in order to obtain a well-defined
distribution reads $b^{\prime\prime}(z)+b^{\prime}(z)^{2}\geq 0$ for all
$z\in[0,1]$ according to Proposition 5.1. This condition holds due to the
assumption $\beta<-\sqrt{-2\alpha}$. Furthermore, the roots of the derivative
of the density in (6.10) are given by solving

$2\alpha z+\beta=0\quad\textrm{or}\quad(2\alpha z+\beta)^{2}+6\alpha=0.$

Note that by the conditions imposed on the parameters, we have $2\alpha
z+\beta<0$ for all $z\in[0,1]$. This means that the only extrema in the
interval $[0,1]$ can lie at

$z_{-}=\frac{-\beta-\sqrt{-6\alpha}}{2\alpha}\quad\textrm{and}\quad
z_{+}=\frac{-\beta+\sqrt{-6\alpha}}{2\alpha}.$

Furthermore, since $\alpha<0$ and $\beta<0$, we obtain $z_{+}<0$. This means
that the density can have at most one extremum $z^{*}$ lying between 0 and 1,
and we have that $z^{*}\in(0,1)\iff 0>-\beta-\sqrt{-6\alpha}>2\alpha$, which
is equivalent to $-\sqrt{-6\alpha}<\beta<-2\alpha-\sqrt{-6\alpha}$. Note
finally that this extremum $z^{*}\in(0,1)$ corresponds to the maximum of the
density due to (6.10).

$\Box$

Proof of Lemma 6.4. Since the function $b$ of this example is strictly
decreasing, the necessary condition in order to obtain a well-defined
distribution reads $b^{\prime\prime}(z)+b^{\prime}(z)^{2}\geq 0$ for all
$z\in[0,1]$ according to Proposition 5.1. This condition holds due to the
assumptions made on the different parameters. Indeed, these assumptions imply
that $b^{\prime}(z)$ is decreasing, while $b^{\prime\prime}(z)$ is increasing
for all $z\in[0,1]$. This means in particular that
$b^{\prime\prime}(z)+b^{\prime}(z)^{2}$ is increasing for all $z\in[0,1]$. In
order to prove the above necessary condition, it thus suffices to show that

$b^{\prime\prime}(0)+b^{\prime}(0)^{2}>0.$

Given that $\alpha>0$, $\delta>0$ and $\epsilon<0$, this is equivalent to

$\alpha(\alpha-1)\epsilon\delta^{\alpha-2}+(\alpha\epsilon\delta^{\alpha-1}-\beta)^{2}>0\iff\left|\alpha\epsilon\delta^{\alpha-1}-\beta\right|>\sqrt{-\alpha(\alpha-1)\epsilon\delta^{\alpha-2}}.$

Let us first assume that $\beta>\alpha\epsilon\delta^{\alpha-1}$, which is
necessary in order to obtain $b^{\prime}(z)<0$ for all $z\in[0,1]$. This
implies that the interior of the absolute value is negative, we can thus write

$\begin{split}b^{\prime\prime}(0)+b^{\prime}(0)^{2}>0&\iff\beta-\alpha\epsilon\delta^{\alpha-1}>\sqrt{-\alpha(\alpha-1)\epsilon\delta^{\alpha-2}}\\\
&\iff\beta>\alpha\epsilon\delta^{\alpha-1}+\sqrt{-\alpha(\alpha-1)\epsilon\delta^{\alpha-2}}.\end{split}$

The parameter $\beta$ needs thus to satisfy precisely this last condition in
order to obtain a well-defined distribution function.

$\Box$