In this section we evaluate the impacts of coverage modifications: a) deductibles, b) policy limit, c) coinsurance and inflation on insurer’s costs.

3.4.1 Policy Deductibles

Under an ordinary deductible policy, the insured (policyholder) agrees to cover a fixed amount of an insurance claim before the insurer starts to pay. This fixed expense paid out of pocket is called the deductible and often denoted by $d$. The insurer is responsible for covering the loss $X$ less the deductible $d$. Depending on the agreement, the deductible may apply to each covered loss or to a defined benefit period (month, year, etc.)

Deductibles eliminate a large number of small claims, reduce costs of handling and processing these claims, reduce premiums for the policyholders and reduce moral hazard. Moral hazard occurs when the insured takes more risks, increasing the chances of loss due to perils insured against, knowing that the insurer will incur the cost (e.g. a policyholder with collision insurance may be encouraged to drive recklessly). The larger the deductible, the less the insured pays in premiums for an insurance policy.

Let $X$ denote the loss incurred to the insured and $Y$ denote the amount of paid claim by the insurer. Speaking of the benefit paid to the policyholder, we differentiate between two variables: The payment per loss and the payment per payment. The payment per loss variable, denoted by $Y^{L}$, includes losses for which a payment is made as well as losses less than the deductible and hence is defined as
$$Y^{L} = left( X – d right)_{+}
= left{ begin{array}{cc}
0 & X lt d, \
X – d & X > d
end{array} right. .$$
$Y^{L}$ is often referred to as left censored and shifted variable because the values below $d$ are not ignored and all losses are shifted by a value $d$.

On the other hand, the payment per payment variable, denoted by $Y^{P}$, is not defined when there is no payment and only includes losses for which a payment is made. The variable is defined as
$$Y^{P} = left{ begin{matrix}
text{Undefined} & X le d \
X – d & X > d
end{matrix} right. $$
$Y^{P}$ is often referred to as left truncated and shifted variable or excess loss variable because the claims smaller than $d$ are not reported and values above $d$ are shifted by $d$.

Even when the distribution of $X$ is continuous, the distribution of $Y^{L}$ is partly discrete and partly continuous. The discrete part of the distribution is concentrated at $Y = 0$ (when $X leq d$) and the continuous part is spread over the interval $Y > 0$ (when $X > d$). For the discrete part, the probability that no payment is made is the probability that losses fall below the deductible; that is,
$$Prleft( Y^{L} = 0 right) = Prleft( X leq d right) = F_{X}left( d right).$$ Using the transformation $Y^{L} = X – d$ for the continuous part of the
distribution, we can find the probability density function of $Y^{L}$ given by
$$f_{Y^{L}}left( y right) = left{ begin{matrix}
F_{X}left( d right) & y = 0, \
f_{X}left( y + d right) & y > 0
end{matrix} right. $$

We can see that the payment per payment variable is the payment per loss variable conditioned on the loss exceeding the deductible; that is, $Y^{P} = left. Y^{L} right|X > d$. Hence, the probability density function of $Y^{P}$ is given by
$$f_{Y^{P}}left( y right) = frac{f_{X}left( y + d right)}{1 – F_{X}left( d right)},$$
for $y > 0$. Accordingly, the distribution functions of $Y^{L}$and $Y^{P}$ are given by
$$F_{Y^{L}}left( y right) = left{ begin{matrix}
F_{X}left( d right) & y = 0, \
F_{X}left( y + d right) & y > 0. \
end{matrix} right. $$
and
$$F_{Y^{P}}left( y right) = frac{F_{X}left( y + d right) – F_{X}left( d right)}{1 – F_{X}left( d right)},$$
for $y > 0$, respectively.

The raw moments of $Y^{L}$ and $Y^{P}$ can be found directly using the probability density function of $X$ as follows
$$Eleftlbrack left( Y^{L} right)^{k} rightrbrack = int_{d}^{infty}left( x – d right)^{k}f_{X}left( x right)dx ,$$
and
$$Eleftlbrack left( Y^{P} right)^{k} rightrbrack = frac{int_{d}^{infty}left( x – d right)^{k}f_{X}left( x right) dx }{{1 – F}_{X}left( d right)} = frac{Eleftlbrack left( Y^{L} right)^{k} rightrbrack}{{1 – F}_{X}left( d right)},$$
respectively.

We have seen that the deductible $d$ imposed on an insurance policy is the amount of loss that has to be paid out of pocket before the insurer makes any payment. The deductible $d$ imposed on an insurance policy reduces the insurer’s payment. The loss elimination ratio (*LER*) is the percentage decrease in the expected payment of the insurer as a result of imposing the deductible. *LER* is defined as
$$LER = frac{Eleft( X right) – Eleft( Y^{L} right)}{Eleft( X right)}.$$

A little less common type of policy deductible is the franchise deductible. The Franchise deductible will apply to the policy in the same way as ordinary deductible except that when the loss exceeds the deductible $d$, the full loss is covered by the insurer. The payment per loss and payment per payment variables are defined as
$$Y^{L} = left{ begin{matrix}
0 & X leq d, \
X & X > d, \
end{matrix} right. $$
and
$$Y^{P} = left{ begin{matrix}
text{Undefined} & X leq d, \
X & X > d, \
end{matrix} right. $$
respectively.

Example 3.15 (SOA) A claim severity distribution is exponential with mean 1000. An insurance company will pay the amount of each claim in excess of a deductible of 100. Calculate the variance of the amount paid by the insurance company for one claim, including the possibility that the amount paid is 0.
Solution

Example 3.16 (SOA) For an insurance:

Losses have a density function
$$f_{X}left( x right) = left{ begin{matrix}
0.02x & 0 lt x lt 10, \
0 & text{elsewhere.} \
end{matrix} right. $$
The insurance has an ordinary deductible of 4 per loss.
$Y^{P}$ is the claim payment per payment random variable.

Calculate $Eleft( Y^{P} right)$.
Solution

Example 3.17 (SOA) You are given:

Losses follow an exponential distribution with the same mean in all
years.
The loss elimination ratio this year is 70%.
The ordinary deductible for the coming year is 4/3 of the current
deductible.

Compute the loss elimination ratio for the coming year.
Solution

3.4.2 Policy Limits

Under a limited policy, the insurer is responsible for covering the actual loss $X$ up to the limit of its coverage. This fixed limit of coverage is called the policy limit and often denoted by $u$. If the loss exceeds the policy limit, the difference $X – u$ has to be paid by the policyholder. While a higher policy limit means a higher payout to the insured, it is associated with a higher premium.

Let $X$ denote the loss incurred to the insured and $Y$ denote the amount of paid claim by the insurer. Then $Y$ is defined as
$$Y = X land u = left{ begin{matrix}
X & X leq u, \
u & X > u. \
end{matrix} right. $$
It can be seen that the distinction between $Y^{L}$ and $Y^{P}$ is not needed under limited policy as the insurer will always make a payment.

Even when the distribution of $X$ is continuous, the distribution of $Y$ is partly discrete and partly continuous. The discrete part of the distribution is concentrated at $Y = u$ (when $X > u$), while the continuous part is spread over the interval $Y lt u$ (when $X leq u$). For the discrete part, the probability that the benefit paid is $u$, is the probability that the loss exceeds the policy limit $u$; that is,
$$Pr left( Y = u right) = Pr left( X > u right) = {1 – F}_{X}left( u right).$$
For the continuous part of the distribution $Y = X$, hence the probability density function of $Y$ is given by
$$f_{Y}left( y right) = left{ begin{matrix}
f_{X}left( y right) & 0 lt y lt u, \
1 – F_{X}left( u right) & y = u. \
end{matrix} right. $$
Accordingly, the distribution function of $Y$ is given by
$$F_{Y}left( y right) = left{ begin{matrix}
F_{X}left( x right) & 0 lt y lt u, \
1 & y geq u. \
end{matrix} right. $$
The raw moments of $Y$ can be found directly using the probability density function of $X$ as follows
$$Eleft( Y^{k} right) = Eleftlbrack left( X land u right)^{k} rightrbrack = int_{0}^{u}x^{k}f_{X}left( x right)dx + int_{u}^{infty}{u^{k}f_{X}left( x right)} dx \ int_{0}^{u}x^{k}f_{X}left( x right)dx + u^{k}leftlbrack {1 – F}_{X}left( u right) rightrbrack dx.$$

Example 3.18 (SOA) Under a group insurance policy, an insurer agrees to pay 100% of the medical bills incurred during the year by employees of a small company, up to a maximum total of one million dollars. The total amount of bills incurred, $X$, has probability density function
$$f_{X}left( x right) = left{ begin{matrix}
frac{xleft( 4 – x right)}{9} & 0 lt x lt 3, \
0 & text{elsewhere.} \
end{matrix} right. $$
where $x$ is measured in millions. Calculate the total amount, in millions of dollars, the insurer would expect to pay under this policy.
Solution

3.4.3 Coinsurance

As we have seen in Section 3.4.1, the amount of loss retained by the policyholder can be losses up to the deductible $d$. The retained loss can also be a percentage of the claim. The percentage $alpha$, often referred to as the coinsurance factor, is the percentage of claim the insurance company is required to cover. If the policy is subject to an ordinary deductible and policy limit, coinsurance refers to the percentage of claim the insurer is required to cover, after imposing the ordinary deductible and policy limit. The payment per loss variable, $Y^{L}$, is defined as
$$Y^{L} = left{ begin{matrix}
0 & X leq d, \
alphaleft( X – d right) & d lt X leq u, \
alphaleft( u – d right) & X > u. \
end{matrix} right. $$
The policy limit (the maximum amount paid by the insurer) in this case is $alphaleft( u – d right)$, while $u$ is the maximum covered loss.

The $k$-th moment of $Y^{L}$ is given by
$$Eleftlbrack left( Y^{L} right)^{k} rightrbrack = int_{d}^{u}leftlbrack alphaleft( x – d right) rightrbrack^{k}f_{X}left( x right)dx + int_{u}^{infty}leftlbrack alphaleft( u – d right) rightrbrack^{k}f_{X}left( x right) dx .$$

A growth factor $left( 1 + r right)$ may be applied to $X$ resulting in an inflated loss random variable $left( 1 + r right)X$ (the prespecified d and u remain unchanged). The resulting per loss variable can be written as
$$Y^{L} = left{ begin{matrix}
0 & X leq frac{d}{1 + r}, \
alphaleftlbrack left( 1 + r right)X – d rightrbrack & frac{d}{1 + r} lt X leq frac{u}{1 + r}, \
alphaleft( u – d right) & X > frac{u}{1 + r}. \
end{matrix} right. $$
The first and second moments of $Y^{L}$ can be expressed as
$$Eleft( Y^{L} right) = alphaleft( 1 + r right)leftlbrack Eleft( X land frac{u}{1 + r} right) – Eleft( X land frac{d}{1 + r} right) rightrbrack,$$ and
$$Eleftlbrack left( Y^{L} right)^{2}
rightrbrack = alpha^{2}left( 1 + r right)^{2} left{ Eleftlbrack left( X land frac{u}{1 + r} right)^{2} rightrbrack – Eleftlbrack left( X land frac{d}{1 + r} right)^{2} rightrbrack right. \
left. – 2left( frac{d}{1 + r} right)leftlbrack Eleft( X land frac{u}{1 + r} right) – Eleft( X land frac{d}{1 + r} right) rightrbrack right} ,$$ respectively.

The formulae given for the first and second moments of $Y^{L}$ are general. Under full coverage, $alpha = 1$, $r = 0$, $u = infty$, $d = 0$ and $Eleft( Y^{L} right)$ reduces to $Eleft( X right)$. If only an ordinary deductible is imposed, $alpha = 1$, $r = 0$, $u = infty$ and $Eleft( Y^{L} right)$ reduces to $Eleft( X right) – Eleft( X land d right)$. If only a policy limit is imposed $alpha = 1$, $r = 0$, $d = 0$ and $Eleft( Y^{L} right)$ reduces to $Eleft( X land u right)$.

Example 3.19 (SOA) The ground up loss random variable for a health insurance policy in 2006 is modeled with *X*, an exponential distribution with mean 1000. An insurance policy pays the loss above an ordinary deductible of 100, with a maximum annual payment of 500. The ground up loss random variable is expected to be 5% larger in 2007, but the insurance in 2007 has the same deductible and maximum payment as in 2006. Find the percentage increase in the expected cost per payment from 2006 to 2007.
Solution

3.4.4 Reinsurance

In Section 3.4.1 we introduced the policy deductible, which is a contractual arrangement under which an insured transfers part of the risk by securing coverage from an insurer in return for an insurance premium. Under that policy, when the loss exceeds the deductible, the insurer is not required to pay until the insured has paid the fixed deductible. We now introduce reinsurance, a mechanism of insurance for insurance companies. Reinsurance is a contractual arrangement under which an insurer transfers part of the underlying insured risk by securing coverage from another insurer (referred to as a reinsurer) in return for a reinsurance premium. Although reinsurance involves a relationship between three parties: the original insured, the insurer (often referred to as cedent or cedant) and the reinsurer, the parties of the reinsurance agreement are only the primary insurer and the reinsurer. There is no contractual agreement between the original insured and the reinsurer. The reinsurer is not required to pay under the reinsurance contract until the insurer has paid a loss to its original insured. The amount retained by the primary insurer in the reinsurance agreement (the reinsurance deductible) is called retention.

Reinsurance arrangements allow insurers with limited financial resources to increase the capacity to write insurance and meet client requests for larger insurance coverage while reducing the impact of potential losses and protecting the insurance company against catastrophic losses. Reinsurance also allows the primary insurer to benefit from underwriting skills, expertize and proficient complex claim file handling of the larger reinsurance companies.

Example 3.20 (SOA) In 2005 a risk has a two-parameter Pareto distribution with $alpha = 2$ and $theta = 3000$. In 2006 losses inflate by 20%. Insurance on the risk has a deductible of 600 in each year. $P_{i}$, the premium in year $i$, equals 1.2 times expected claims. The risk is reinsured with a deductible that stays the same in each year. $R_{i}$, the reinsurance premium in year $i$, equals 1.1 times the expected reinsured claims. $frac{R_{2005}}{P_{2005} = 0.55}$. Calculate $frac{R_{2006}}{P_{2006}}$.
Solution

Let us use the following notation:

$X_{i}:$ The risk in year $i$
$Y_{i}:$ The insured claim in year $i$
$P_{i}:$ The insurance premium in year $i$
$Y_{i}^{R}:$ The reinsured claim in year $i$
$R_{i}:$ The reinsurance premium in year $i$
$d:$ The insurance deductible in year $i$ (the insurance deductible is fixed each year, equal to 600)
$d^{R}:$ The reinsurance deductible or retention in year $i$ (the reinsurance deductible is fixed each year, but unknown) where $i = 2005, 2006$

$$Y_{i} = left{ begin{matrix}
0 & X_{i} leq 600 \
X_{i} – 600 & X_{i} > 600 \
end{matrix} right. $$
where $i = 2005, 2006$

$$X_{2005}sim Paleft( 2,3000 right)$$

$$Eleft( Y_{2005} right) = Eleft( X_{2005} – 600 right)_{+} = Eleft( X_{2005} right) – Eleft( X_{2005} land 600 right)$$

$= 3000 – 3000left( 1 – frac{3000}{3600} right) = 2500$

$$P_{2005} = 1.2Eleft( Y_{2005} right) = 3000$$

Since $X_{2006} = 1.2X_{2005}$ and Pareto is a scale distribution with scale parameter $theta$, then $X_{2006}sim Paleft( 2,3600 right)$

$$Eleft( Y_{2006} right) = Eleft( X_{2006} – 600 right)_{+} = Eleft( X_{2006} right) – Eleft( X_{2006} land 600 right)$$

$= 3600 – 3600left( 1 – frac{3600}{4200} right) = 3085.714$

$$P_{2006} = 1.2Eleft( Y_{2006} right) = 3702.857$$

$$Y_{i}^{R} = left{ begin{matrix}
0 & X_{i} – 600 leq d^{R} \
X_{i} – 600 – d^{R} & X_{i} – 600 > d^{R} \
end{matrix} right. $$

Since $frac{R_{2005}}{P_{2005}} = 0.55$, then $R_{2005} = 3000 times 0.55 = 1650$

Since $R_{2005} = 1.1Eleft( Y_{2005}^{R} right)$, then $Eleft( Y_{2005}^{R} right) = frac{1650}{1.1} = 1500$

$$Eleft( Y_{2005}^{R} right) = Eleft( X_{2005} – 600 – d^{R} right)_{+} = Eleft( X_{2005} right) – Eleft( X_{2005} land left( 600 + d^{R} right) right)$$

$= 3000 – 3000left( 1 – frac{3000}{3600 + d^{R}} right) = 1500 Rightarrow d^{R} = 2400$

$$Eleft( Y_{2006}^{R} right) = Eleft( X_{2006} – 600 – d^{R} right)_{+} = Eleft( X_{2006} – 3000 right)_{+} = Eleft( X_{2006} right) – Eleft( X_{2006} land 3000 right)$$

$= 3600 – 3600left( 1 – frac{3600}{6600} right) = 1963.636$

$$R_{2006} = 1.1Eleft( Y_{2006}^{R} right) = 1.1 times 1963.636 = 2160$$

Therefore $frac{R_{2006}}{P_{2006}} = frac{2160}{3702.857} = 0.583$

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