Category Archives: Deductibles

Frequency with Respect to Exposure and Coverage Modifications

In a portfolio of risks, there are two types of modifications which can influence the frequency distribution of payments.

  1. Exposure Modification (not in syllabus) — increasing or decreasing the number of risks or time periods of coverage in the portfolio
  2. Coverage Modification — applying limits, deductible or adjusting for inflation in each individual risk
EXPOSURE MODIFICATION
If there is an exposure modification, you would adjust the frequency distribution by scaling the appropriate parameter to reflect the change in exposure.  The following list provides the appropriate parameter to adjust for each distribution:
  1. Poisson:  \lambda
  2. Negative Binomial:  r
    (Geometric is a Negative Binomial with r = 1)
  3. Binomial:  m
    (Only valid if the new value remains an integer)
Example 1:  You own a portfolio of 10 risks.  You model the frequency of claims with a negative binomial having parameters r = 2 and \beta = 0.5.  The number of risks in your portfolio increases to 15.  What are the parameters for the new distribution?
Answer:  The frequency distribution now has parameters r = 3 and \beta = 0.5  Note that since the mean and variance are r\beta and r\beta(1+\beta) respectively, the new mean and variance are multiplied by 1.5.
COVERAGE MODIFICATION
Coverage modifications shift, censor, or scale the individual risks, usually in the presence of a deductible or claim limit, and they change the conditions that trigger a payment.  For example, adding a deductible d is considered a coverage modification and this changes the condition for payment because any losses below the deductible do not qualify for payment.  If the risk is represented by random variable X, then adding a deductible would change the random variable to (X-d)_+. Scaling in the presence of a deductible or claim limit also affects the frequency distribution.  The following lists parameters affected by coverage modifications:
  1. Poisson:  \lambda
  2. Negative Binomial:  \beta
  3. Binomial:  q
  4. Geometric:  \beta
These parameters are scaled by the probability that a payment occurs.
Example 2:  The frequency of loss is modeled as a Poisson distribution with parameter \lambda = 5.  A deductible is imposed so that only 80% of losses result in payments.  What is the new distribution?
Answer:  It is Poisson with \lambda = 0.8(5) = 4.
Example 3:  The frequency of payment N is modeled as a negative binomial with parameters r = 3 and \beta = 0.5.  Losses X are pareto distributed with parameters \alpha = 2 and \theta = 100.  The deductible is changed from d=10 to d=15.  What are the new parameters in the frequency distribution?
Answer:  Firstly, N is the frequency of payment.  So it reflects the current deductible.  If you wanted the distribution of N without deductible, you would divide \beta by \Pr{(X>10)}.  Now to find the distribution of N with the deductible of 15, you multiply \beta by Pr(X>15).  To summarize:
\begin{array}{rll} \beta_{new} &=& \displaystyle \beta_{old}\times\frac{\Pr(X>15)}{\Pr(X>10)} \\ \\ &=& \displaystyle 0.5\times \frac{0.75614367}{0.82644628} \\ \\ &=& 0.45746692 \end{array}
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Other Coverage Modifications

Coinsurance \alpha is the fraction of losses covered by the policy.  For example, \alpha = 0.8 means if a loss is incurred, 80% will be paid by the insurance company.  A claims limit u is the maximum amount that will be paid.  The order in which coinsurance, claims limits, and deductibles is applied to a loss is important and will be specified by the problem.  The expected payment per loss when all three are present in a policy is given by

E\left[Y\right] = \alpha \left[E\left[X\wedge u\right] - E\left[X \wedge d\right]\right]

where Y is the payment variable and X is the original loss variable.  The second moment is given by

E\left[Y^2\right] = \alpha^2\left(E\left[(X\wedge u)^2\right] - E\left[(X \wedge d)^2\right]-2d\left(E\left[X \wedge u\right]-E\left[X \wedge d\right]\right)\right)

The second moment can be used to find the variance of payment per loss.  If inflation r is present, multiply the second moment by (1+r)^2 and divide u and d by (1+r).   For payment per payments, divide the expected values by P(X>d) or 1-F(d).

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The Loss Elimination Ratio

If you impose a deductible d on an insurance policy that you’ve written, what fraction of expected losses do you eliminate from your expected liability?  This is measured by the Loss Elimination Ratio LER(d).

\displaystyle LER(d) = \frac{E\left[X \wedge d\right]}{E\left[X\right]}

Definitions:

  1. Ordinary deductible d— The payment made by the writer of the policy is the loss X minus the deductible d.  If the loss is less than d, then nothing is paid.
  2. Franchise deductible d_f—  The payment made by the writer of the policy is the complete amount of the loss X if X is greater than d_f.
A common type of question considers what happens to LER if an inflation rate r increases the amount of all losses, but the deductible remains unadjusted.  Let X be the loss variable.  Then Y=(1+r)X is the inflation adjusted loss variable.  If losses Y are subject to deductible d, then
\begin{array}{rll} \displaystyle LER_Y(d) &=& \frac{E\left[(1+r)X\wedge d\right]}{E\left[(1+r)X\right]} \\ \\ \displaystyle &=&\frac{(1+r)E\left[X\wedge \frac{d}{1+r}\right]}{(1+r)E\left[X\right]} \\ \\ &=& \frac{E\left[X \wedge \frac{d}{1+r}\right]}{E\left[X\right]}\end{array}
Memorize:
\displaystyle E\left[X \wedge d\right] = \int_0^d{x f(x) dx} + d\left(1-F(x)\right)

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Expected Values for Insurance

Before I begin, please note: I hated this chapter.  If there are any errors please let me know asap!

A deductible d is an amount that is subtracted from an insurance claim.  If you have a $500 deductible on your car insurance, your insurance company will only pay damages incurred beyond $500.  We are interested in the following random variables: (X - d)_+ and (X\wedge d).

Definitions:

  1. Payment per Loss: (X-d)_+ = \left\{ \begin{array}{ll} X-d &\mbox{ if } X>d \\ 0 &\mbox{ otherwise} \end{array} \right.
  2. Limited Payment per Loss:  (X\wedge d) = \left\{ \begin{array}{ll} d &\mbox{ if } X>d \\ X &\mbox{ if } 0<X<d \\ 0 &\mbox{ otherwise} \end{array} \right.
Expected Values:
  1. \begin{array}{rll} E[(X-d)_+] &=& \displaystyle \int_{d}^{\infty}{(x-d)f(x)dx} \\ \\ &=& \displaystyle \int_{d}^{\infty}{S(x)dx} \end{array}
     
  2. \begin{array}{rll} E[(X\wedge d)] &=& \displaystyle \int_{0}^{d}{xf(x)dx +dS(x)} \\ \\ &=& \displaystyle \int_{0}^{d}{S(x)dx} \end{array}
We may also be interested in the payment per loss, given payment is incurred (payment per payment) X-d|X>d.
By definition:
E[X-d|X>d] = \displaystyle \frac{E[(X-d)_+]}{P(X>d)}
Since actuaries like to make things more complicated than they really are, we have special names for this expected value.  It is denoted by e_X(d) and is called mean excess loss in P&C insurance and \displaystyle {\mathop{e}\limits^{\circ}}_d is called mean residual life in life insurance.  Weishaus simplifies the notation by using the P&C notation without the random variable subscript.  I’ll use the same.
Memorize!
  1. For an exponential distribution,
    e(d) = \theta
  2. For a Pareto distribution,
    e(d) = \displaystyle \frac{\theta +d}{\alpha - 1}
  3. For a single parameter Pareto distribution,
    e(d) = \displaystyle \frac{d}{\alpha - 1}
Useful Relationships:
  1. \begin{array}{rll} E[X] &=& E[X\wedge d] + E[(X-d)_+] \\ &=& E[X\wedge d] + e(d)[1-F(d)] \end{array}
Actuary Speak (important for problem comprehension):
  1. The random variable (X-d)_+ is said to be shifted by d and censored.
  2. e(d) is called mean excess loss or mean residual life.
  3. The random variable X\wedge d can be called limited expected value, payment per loss with claims limit, and amount not paid due to deductible.  d can be called a claims limit or deductible depending on how it is used in the problem.
  4. If data is given for X with observed values and number of observations or probabilities, the data is called the empirical distribution.  Sometimes empirical distributions may be given for a problem, but you are still asked to assume an parametric distribution for X.

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