3.1 Absolute deductible

If the absolute deductible is applied, the insured pays the first a monetary units of each claim X, and the insurer pays the excess over a, X − a. Then, if an absolute deductible with parameter a ≥ 0 is applied, A and C are defined in Table 1.

Table 1

Definition of A and C if an absolute deductible with parameter a ≥ 0 is applied.

X	A	C
X < a	X	0
X > a	a	X − a

Following [26], we define the s-th partial moment of X about the origin over (0, x₀) as the partial expectation of X^s, $$ H_{X^s}\left(x_0\right)={\int }_0^{x_0}{x}^{s}f\left(x\right)dx$$. Hence, the expectations of A, AX and A² are

From (4) and using (6), (7) and (8), dif(P) is given by

As dif(P) is function of only one variable, in order to find the value of a that maximizes the difference, we substitute the gradient by the derivative with respect to a. The system (5) becomes

Differentiating (6), (7) and (8) with respect to a, we obtain,

Then, using these last derivatives, (10) is

Proposition 3. The function dif(P) attains a global maximum on D.

Proof. The function dif(P) is a continuous and positive function in D = {a|a > 0}} and lim_{a → ∞} dif(P) = lim_{a → 0} dif(P)) = 0. Therefore, (11) has at least one solution and dif(P) attains a global maximum.

We consider two claim amount distributions: exponential and Pareto-Lomax.

Exponential case: If X ∼ exp(γ), γ > 0, the probability density function is given by f(x) = γe ^{− γx}, then

Substituting the previous expressions in (9), dif(P) is

And, substituting (12) and (13) in (11), the equation that allows to obtain the value of a that maximizes dif(P) is

The value of a that fulfills the previous expression, a*, depends on the relationship between E(N) and V(N). So, if E(N) > V(N), a* $$ \in (0,\frac{ln2}{\gamma })\cup (\frac{1}{\gamma },\infty )$$; if E(N) < V[N], $$ a^{*}\in (\frac{ln2}{\gamma },\frac{1}{\gamma })$$, and lastly, when E(N) = V(N), $$ a^{*}=\frac{1}{\gamma }$$. If N is Poisson distributed with parameter λ, the optimal value for a coincides with the mean claim amount, the maximum difference is $$ \frac{2\delta E\left(N\right)}{{\gamma }^{2}e}$$ and hence, for the insured, the best option is to purchase a deductible insurance with a = E(X) and a refundable insurance that covers a.

In Fig 1, we plot dif(P), in the Poisson-exponential case, as a function of a for different values of $$ E\left(X\right)=\frac{1}{\gamma }$$.

Fig 1

dif(F) as a function of a in the Poisson-exponential case for δ = 0.03 and λ = 1 for E(X) = 2, 5, 10, 15.

Pareto-Lomax case: If X ∼ Pareto(θ, ν), the probability density function is given by f(x) = θν^θ(ν + x) ^{− θ−1} and the mean is $$ E\left(X\right)=\frac{\nu }{\theta -1}$$, ν > 0, θ > 2 (so that the variance is finite). Then,

Substituting the previous expressions in (11)

If E(N) > V(N), $$ a^{*}>\frac{\nu }{\theta -2}$$, if E(N) = V(N) (Poisson case) $$ a^{*}=\frac{\nu }{\theta -2}$$, and if E(N) < V(N), $$ a^{*}<\frac{\nu }{\theta -2}$$.

Summarizing the results for the exponential and the Pareto-Lomax cases, in Table 2, the optimal values of a that maximizes dif(P) are included.

Table 2

Optimal a in absolute deductible in the Poisson-exponential and Poisson-Pareto-Lomax cases.

	E(N)<V(N)	E(N) = V(N)	E(N)>V(N)
Exp(γ)	$$	$$	$$
Pareto(θ,ν)	$$	$$	$$

We now consider that N is Poisson distributed with parameter λ, and we do not specify the distribution of X. Now, dif(P) is linear on δ and depends on a, the only parameter of the deductible,

The value of a that maximizes dif(P) is the solution to

The solution to (14) depends only on the distribution of X. Explicit expressions have been obtained in the previous two subsections for the exponential and the Pareto-Lomax distributions. For other claim amount distributions, only numerical solutions can be found. We calculate the optimal value of a and the maximum difference reached for several distributions of the claim amount. The parameters of the distributions are such that the expected value and the variance of the claim amount is the same, so the premium with a zero deductible insurance Π would be the same. The Pareto-Lomax distribution is not included in the comparison because there is no combination of θ and ν that fulfills E(X)² = V(X), as in the exponential case. For the lognormal distribution, with parameters μ and σ, (14) is

If the individual claim amount follows an Inverse-Gaussian distribution, X ∼ IG(μ, λ), μ > 0, λ > 0, the probability distribution function is given by $$ f\left(x\right)=\sqrt{\frac{\lambda }{2\pi x^3}}{e}^{\frac{-\lambda {(x-\mu )}^2}{2{\mu }^{2}x}}$$, the expected value is E(X) = μ and its variance $$ V\left(X\right)=\frac{{\mu }^3}{\lambda }$$. The combination of μ and λ that fulfills E(X)² = V(X) is μ = λ. Then, the optimal value of a is obtained from

In Table 3, the results of the optimal value a* and the maximum value of dif(P) depending on E(N) and δ are presented for an Exponential, a Lognormal and an Inverse-Gaussian distribution.

Table 3

Optimal value of a and difference (a*, dif(a*)) for Exp(γ), Log(μ, σ) and IG(μ, λ) claim amount distributions.

E(X)	V(X)	Exp(γ)	Log(μ, σ)	IG(μ, λ)
0.1	0.01	(0.1, 0.007E(N)δ)	(0.093, 0.001E(N)δ)	(0.102, 0.007E(N)δ)
0.5	0.25	(0.5, 0.184E(N)δ)	(0.468, 0.029E(N)δ)	(0.508, 0.168E(N)δ)
1	1	(1, 0.736E(N)δ)	(0.938, 0.332E(N)δ)	(1.016, 0.672E(N)δ)
2	4	(2, 2.943E(N)δ)	(1.877, 1.621E(N)δ)	(2.033, 2.69E(N)δ)
5	25	(5, 18.394E(N)δ)	(4.692, 9.902E(N)δ)	(5.082, 16.812E(N)δ)
10	100	(10, 73.576E(N)δ)	(9.388, 39.469E(N)δ)	(10.166, 67.248E(N)δ)

Table 3 shows that the maximum difference is proportional to the expected value of N and to the safety loading, which is the highest in the exponential case and the lowest in the Lognormal case. For each value of E(X), the optimal value of a, a*, is the lowest in the Lognormal case and the highest in the Inverse-Gaussian case.

3.2Proportional deductible

In this section, we focus our attention on the proportional deductible. We work with two types of proportional deductible: a first type in which the insured pays a percentage α of each claim, and, a second type in which we include a maximum loss for the insured, B. In a deductible with participation α ∈ (0, 1), A and C are defined in Table 4.

Table 4

Definition of A and C if a deductible with participation α ∈ (0, 1) is applied.

X	A	C
∀X	αX	(1 − α)X

In this case, E(AX) = αE(X²), E(A²) = α² E(X²) and E(A) = αE(X), then

Using (5), the value α that maximizes the difference has to fulfill

Proposition 4. The function dif(P) attains a global maximum in $$ \alpha =\frac{1}{2}$$.

Proof. The function dif(P) is a continuous and positive function in D = {α|α ∈ (0, 1)} and lim_{α → 1} dif(P) = lim_{α → 0} dif(P)) = 0. As the second factor in (16) is always different from 0, the critical point, $$ \alpha =\frac{1}{2}$$, is a global maximum of dif(P).

Independently of the distribution of N and X, $$ \alpha =\frac{1}{2}$$ is the value that maximizes the difference. Then, the maximum value of the difference is $$ \frac{\delta [E\left(N\right)V\left(X\right)+E{\left(X\right)}^{2}V\left(N\right)]}{2}$$. If N ∼ Pois(λ), the maximum difference is $$ \frac{\delta \lambda E\left(X^2\right)}{2}$$.

Now, we generalize the proportional deductible including a maximum loss for the insured. In Table 5, a deductible with participation α ∈ (0, 1) and a maximum loss B > 0 is defined.

Table 5

Definition of A and C if a deductible with participation α ∈ (0, 1) with limit B > 0 is applied.

X	A	C
$$	αX	(1 − α)X
$$	B	X − B

If B tends to infinity, the first type of proportional deductible is obtained as a particular case.

If the deductible with participation α ∈ (0, 1) with limit B > 0 is applied,

From (4), dif(P) can be rewritten as

The deductible depends on two parameters, α and B. The partial derivatives with respect to α and B, from (17), (18) and (19), are

Then, (5) is

The function dif(P) is a continuous and positive function in D = {(α, B)|α ∈ (0, 1), B > 0}. The critical points of the maximization problem are the solutions of the system of Eq (20). If we evaluate the function at the boundaries of D, it is easy to see that lim_{α → 0} dif(P) = lim_{B → 0} dif(P) = 0, lim_{α → 1} dif(P) equals to dif(P) in the absolute deductible case with parameter a = B, and lim_{B → ∞} dif(P) equals to dif(P) in the proportional deductible case. Therefore, we are not able to guarantee the existence of a global maximum for any N and X. Nevertheless, if N is Poisson distributed, we proof (see Proposition 5) that a global maximum does not exist. In this proposition, considering that the insured can choose only one of the parameters that define the deductible while the other is fixed by the insurer, we also obtain marginal maximums.

Proposition 5. If N ∼ Pois(λ), there is no value of (B, α) that maximizes (22), but marginal optimums exist. For a fixed B, $$ {\alpha }^{*}=\frac{1}{2}$$ maximizes the difference. For a fixed $$ \alpha \le \frac{1}{2}$$ the maximum does not exist, and if $$ \alpha >\frac{1}{2}$$ the maximum point, if it exists, fulfills $$ {\int }_{\frac{B}{\alpha }}^{\infty }(x-2B)f\left(x\right)dx=0$$.

Proof. If N ∼ Pois(λ), dif(P) is

From the second equation in (22), knowing that $$ {\int }_0^{\frac{B}{\alpha }}{x}^{2}f\left(x\right)dx\ne 0$$, we obtain $$ \alpha =\frac{1}{2}$$. Substituting this value in the first equation, it is reduced to $$ {\int }_{2B}^{\infty }(x-2B)f\left(x\right)dx=0$$, which is impossible because the integral is always different from 0. Then, there is no value of (B, α) that maximizes (22).

From $$ \frac{\partial dif\left(P\right)}{\partial \alpha }=0$$ (the second equation of (22)), $$ \alpha =\frac{1}{2}$$ is the critical point. As the sign of $$ \frac{{\partial }^{2}dif\left(P\right)}{{\partial }^2\alpha }$$ at $$ \alpha =\frac{1}{2}$$ is negative, the critical point is a maximum.

From $$ \frac{\partial dif\left(P\right)}{\partial B}=0$$ (the first equation of (22)), the value of B that maximizes does not exist because $$ {\int }_{\frac{B}{\alpha }}^{\infty }(x-2B)f\left(x\right)dx$$ is always different from 0 as $$ \frac{B}{\alpha }\ge 2B$$, and if $$ \alpha >\frac{1}{2}$$ the value of B that maximizes has to fulfill $$ {\int }_{\frac{B}{\alpha }}^{\infty }(x-2B)f\left(x\right)dx=0$$.

Poisson-Exponential case: If N ∼ Pois(λ) and X ∼ exp(γ), following Proposition 5, we know that there is no value of (α, B) that maximizes (21), but marginal optimums exist. For a fixed B, $$ {\alpha }^{*}=\frac{1}{2}$$ maximizes the difference and for a fixed α > 0.5, the value $$ B^{*}={\displaystyle \frac{\alpha }{\gamma (2\alpha -1)}}$$ maximizes dif(P). For example, assuming γ = 0.6, δ = 0.03 and λ = 1, dif(P) as a function of α and B is plotted in Fig 2.

Fig 2

dif(F) in the Poisson-exponential case with δ = 0.03, λ = 1, γ = 0.6.

Poisson-Pareto case: If N ∼ Pois(λ) and c for a fixed B, $$ {\alpha }^{*}=\frac{1}{2}$$ maximizes dif(P) and for a fixed $$ \alpha >\frac{1}{2}$$, $$ B^{*}={\displaystyle \frac{\alpha }{2\alpha (\theta -1)-\theta }}$$ also maximizes the difference. For example, if θ = 3, δ = 0.03 and λ = 1, dif(P) is plotted in Fig 3.

Fig 3

dif(F) in the Poisson-Pareto case with δ = 0.03, λ = 1, θ = 3.

Poisson-Lognormal case: If X ∼ LN(μ, σ), for a fixed $$ \alpha >\frac{1}{2}$$, the optimal value of B is the one that fulfils $$ {\int }_{\frac{B}{\alpha }}^{\infty }(x-2B)f\left(x\right)dx=0$$, which can be written as,

In this case, it is not possible to obtain analytical results of the value B that optimizes the difference. Some numerical results are presented in Table 6 for different values of $$ \alpha >\frac{1}{2}$$.

Table 6

Optimal B if the proportional deductible with limit is applied in the Poisson-Lognormal case for $$ \alpha >\frac{1}{2}$$.

α	0.55	0.6	0.65	0.7	0.75	0.8	0.85	0.9	0.95	1
B*	11.79	3.17	2.01	1.59	1.38	1.26 1.18	1.128	1.088	1.058

From Table 6, we see that B* decreases with α and if α = 1, we have in fact an absolute deductible and thus (23) is reduced to (15).

3.3 Mixture of absolute deductible and proportional deductible

In a mixture of absolute deductible with a > 0 and proportional deductible with α ∈ (0, 1), the insured pays the first a monetary units of each claim X, pays a if the claim amount is greater than a and less than $$ \frac{a}{\alpha }$$, and if the claim amount is greater than $$ \frac{a}{\alpha }$$ the proportional deductible is applied and the insured pays a percentage α of the claim amount. A and C are defined in Table 7.

Table 7

Definition of A and C if a mixture of absolute deductible a > 0 and proportional deductible α ∈ (0, 1) is applied.

X	A	C
X < a	X	0
$$	a	X − a
$$	αX	(1 − α)X

We define the different elements needed in the optimization of dif(P):

This deductible depends on two parameters. We calculate the partial derivatives with respect to α and a,

And the system of equations that allows us to obtain the critical points is,

The function dif(P) is a continuous and positive function in D = {(α, a)|α ∈ (0, 1), a > 0}. We evaluate dif(P) at the boundaries of D: lim_{α → 1} dif(P) = lim_{a → ∞} dif(P) = 0, lim_{α → 0} dif(P) equals to dif(P) in the absolute deductible case and lim_{a → 0} dif(P) equals to dif(P) in the proportional deductible case. As in the proportional deductible with limit B, this analysis does not allow us to assure the existence of a global maximum for any N and X. However, if N is Poisson distributed, we proof (see Proposition 5) that a global maximum does not exist and, additionally, we obtain marginal maximums.

Proposition 6. If N ∼ Pois(λ), there is no value of (a, α) that maximizes (25), but marginal optimums exist. For a fixed a, $$ {\alpha }^{*}=\frac{1}{2}$$ maximizes the difference. For a fixed $$ \alpha \ge \frac{1}{2}$$, the maximum does not exist, and if $$ \alpha <\frac{1}{2}$$ the maximum point, if it exists, fulfills $$ {\int }_a^{\frac{a}{\alpha }}(x-2a)f\left(x\right)dx=0$$.

Proof. If N ∼ Pois(λ),

From the second equation of (25), knowing that $$ {\int }_{\frac{a}{\alpha }}^{\infty }{x}^{2}f\left(x\right)dx\ne 0$$, we obtain $$ {\alpha }^{*}=\frac{1}{2}$$. Substituting this value in the first equation, we obtain $$ {\int }_a^{2a}(x-2a)f\left(x\right)dx=0$$, which is impossible because the integral is always different from 0. Then, there is no value of (a, α) that maximizes (25).

From $$ \frac{\partial dif\left(P\right)}{\partial \alpha }=0$$, $$ \alpha =\frac{1}{2}$$ is the critical point. As the sign of $$ \frac{{\partial }^{2}dif\left(P\right)}{{\partial }^2\alpha }$$ at $$ \alpha =\frac{1}{2}$$ is negative, the critical point is a maximum.

From $$ \frac{\partial dif\left(P\right)}{\partial a}=0$$, the first equation of (25), if $$ \alpha \ge \frac{1}{2}$$, the value of a that maximizes does not exist because $$ {\int }_a^{\frac{a}{\alpha }}(x-2a)f\left(x\right)dx$$ is always different from 0. If $$ \alpha <\frac{1}{2}$$ the value of a that maximizes has to fulfill $$ {\int }_a^{\frac{a}{\alpha }}(x-2a)f\left(x\right)dx=0$$.

Poisson-Exponential case: For X ∼ exp(γ), if $$ \alpha >\frac{1}{2}$$, the value of a that optimizes the difference is obtained from the equation $$ \frac{2a-\frac{a}{\alpha }-\frac{1}{\gamma }}{\left(a-\frac{1}{\gamma }\right)}=e^{\left(\frac{1}{\alpha }-1\right)\gamma a}$$ that has no explicit solution.

We can visualize graphically the behavior of dif(P) and its optimal values. For example, if γ = 0.6, δ = 0.03 and λ = 1, dif(P) is plotted in Fig 4.

Fig 4

dif(P) in the Poisson-exponential case with δ = 0.03, λ = 1, γ = 0.6.

The function dif(P) is plotted for different values of α, with δ = 0.03, λ = 1 and γ = 0.6 (Fig 5) and for different values of a with δ = 0.03, λ = 1 and γ = 0.6 (Fig 6).

Fig 5

dif(P) for α = 0.1, 0.2, 0.3, 0.4, 0.5 in the Poisson-exponential case with δ = 0.03, λ = 1, γ = 0.6.

Fig 6

dif(P) for a = 1, 1.2, 1.4, 1.6, 1.8 in the Poisson-exponential case with δ = 0.03, λ = 1, γ = 0.6.

Poisson-Pareto case: If X ∼ Pareto(θ, 1), θ > 2, the value of a that optimizes the difference is obtained from the equation $$ \frac{2a(\gamma -1)-\gamma a-1}{2a(\gamma -1)-\frac{\gamma }{\alpha }a-1}={\left(\frac{\frac{a}{\alpha }+1}{a+1}\right)}^{-\theta }$$.

In Fig 7, a graphical illustration of dif(P) is included for θ = 3, δ = 0.03 and λ = 1.

Fig 7

dif(P) in the Poisson-Pareto case with δ = 0.03, λ = 1, θ = 0.6.

Partial analysis of dif(P) with respect to a and α are plotted in Fig 8 and in Fig 9, respectively.

Fig 8

dif(P) for α = 0.1, 0.2, 0.3, 0.4 in the Poisson-Pareto case with δ = 0.03, λ = 1, θ = 3.

Fig 9

dif(P) for a = 1, 1.2, 1.4, 1.6, 1.8 in the Poisson-Pareto case with δ = 0.03, λ = 1, θ = 3.

In the figures obtained for the exponential case (Figs 5 and 6) and for the Pareto case (Figs 8 and 9), we can observe a similar behavior in the optimization problem with respect to a and α.

Poisson-Lognormal case: If X ∼ LN(μ, σ), in order to obtain the optimal value of a that maximizes dif(P) for values of $$ \alpha \ge \frac{1}{2}$$, the integral $$ {\int }_a^{\frac{a}{\alpha }}(x-2a)f\left(x\right)dx=0$$ can be written as,

Due to the impossibility of analytically solving the previous equation, Table 8 shows some numerical results from which we see that a* decreases with α.

Table 8

Optimal value of a in the Poisson-Lognormal case for $$ \alpha <\frac{1}{2}$$.

α	0.05	0.1	0.15	0.2	0.25	0.3	0.35	0.4	0.45
a*	1.053	1.052	1.042	1.007	0.934	0.81	0.625	0.377	0.096

3.4 All-nothing deductible

In this section, we analyze the all-nothing deductible with participation M > 0. The idea is that if the individual claim amount, X, is less than M, the insurer pays the whole claim, but if X is greater than M, the insurer does not pay anything (see [15]).

In Table 9, A and C are defined.

Table 9

All-nothing deductible with participation M.

X	A	C
X < M	0	X
X > M	X	0

This deductible does not fulfill Proposition 1 because A and C are not commonotic risks. Then, the following results are not focused on obtaining the optimal deductible for the insured because it is not possible to maintain that the function dif(P) is always positive.

If the all-nothing deductible is applied, E(AX) = E(A²) = E(X²) − H_X²(M), and E(A) = E(X) − H_X(M). Hence, the expression for dif(P) is

If N ∼ Pois(λ), then dif(P) = 0 regardless of the claim amount distribution. Then, in the Poisson case, for the insured it is the same to purchase a Zero Deductible Insurance policy or a Deductible Insurance policy together with a Refundable Deductible Insurance policy.

Knowing that E(X) − H_X(M)>0, we can observe in (26) that the sign of dif(P) depends on the sign of V(N) − E(N). Then, dif(P) can be positive or negative. The explanation is that, in this deductible, A and C are not commonotic risks (see [21]), and therefore the initial hypothesis of this paper is not fulfilled. That is to say, in this deductible, Cov(A, C) is not a positive value. In fact, knowing that E(AC) = E(AX) − E(A²) = 0, Cov(A, C) = −E(A)E(C)<0. From (3), the positiveness of Cov(A, C) is a sufficient, but not necessary, condition for the positiveness of dif(P). We are only interested in the situations in which dif(F) is positive, therefore we impose that V(N)>E(N).

In order to obtain the value that maximizes dif(P), we need the following two previous derivatives,

Considering that H_X(M) is a continuous and increasing function in D = {M|M > 0}, with H_X(0) = 0 and lim_{M → ∞} H_X = E(X), we can assure the existence of a critical point. It is easy to see that dif′′(P) is always negative, then the critical point is a maximum.

In the exponential case, (27) is (2γM + 2)e ^{− Mγ} = 1 and for the Pareto(θ, ν), it is

In Table 10, some numerical results for the exponential distribution are obtained.

Table 10

Optimal value of M in the exponential case.

E(X)	1	2	3	4	5	6	10	20
M	1.6783	3.3567	5.035	6.7134	8.3917	10.07	16.783	33.567

If X follows a Pareto(θ, ν) distribution, the optimal values of M are obtained solving numerically (28). The results are shown in Table 11.

Table 11

Optimal value of M in the Pareto case.

ν θ	2.1	2.5	3	3.5	4	5
1	2.1189	1.4176	1	0.771650	0.627942	0.457323
1.5	3.17833	2.12645	1.5	1.15747	0.941913	0.685984
2	4.23778	2.83527	2	1.5433	1.2559	0.914645
2.5	5.29722	3.54409	2.5	1.92912	1.5698	1.14331

If X ∼ LN(μ, σ), then (27) is,