Month: June 2017
Mathematical models for insurance payments – part 2 – ordinary deductible
The post is the second post on models for insurance payments. The focus here is on the insurance payment when the loss is adjusted by a deductible. The first post is on insurance policy with a limit.
Ordinary Deductible
Suppose that the random variable is the size of the loss (if a loss occurs). Under an insurance policy with a deductible , the first dollars of a loss is responsible by the insured and the amount of the loss in excess of is paid by the insurer. Under this policy provision, what is the expected amount of insurance payment for a given loss? The deductible described here is called an ordinary deductible. The insurance payment (per loss) under the presence of an ordinary deductible is denoted by . The idea for this notation is that the insurance payment is as long as is not negative. The following describes the payment rule more explicitly.
The goal is to describe the probability model of the insurance payment . First and foremost, we need to know how much it will cost the insurer on average to pay the insured, which is the mean . The variance of is also important as it is a measure of risk to the insurer. The mean and the variance can be calculated by using the density function of . At times it will be useful to know the the probability density function (PDF) and cumulative distribution function (CDF) of and other distributional quantities.
Moments
First, let’s focus on the moments of . One way to obtain moments of is through the distribution of the unmodified loss . Let , and be the probability density function (PDF), cumulative distribution function (CDF) and the survival function of the random loss, respectively. Then the moments of can be expressed using these distributional quantities of .
The above calculation is based on first principle. When the loss is less than or equal to , the insurance payment is zero. Note that the zero payment does not have to be explicitly stated in the above integrals. When the loss exceeds the deductible , the payment is , the amount of the loss in excess of . Example 1 and Example 2 below demonstrate how this calculation is performed.
Another way to obtain the moments of is obtained by first finding the PDF of . This is discussed in a section below.
Basic Example
Example 1
Suppose that the loss distribution has PDF . Evaluate and . What is the expected amount of the loss eliminated as a result of imposing the ordinary deductible of 2 (eliminated from the insurer’s perspective)?
From Example 1 in the previous post, . Thus when a loss occurs, the expected amount of the loss that is responsible by the insured is
,which is also the expected amount of the loss eliminated (from the insurer’s perspective) as a result of imposing an ordinary deductible.
Example 2
Continue with Example 1. Calculate the variance of the insurance payment .
From Example 2 in the previous post, . This shows that imposing a deductible that is significant enough has a variance reducing effect on the insurance payment. Had the deductible not present, there would be a greater fluctuation in the payment.
Connection with Policy Limit
The calculation in Example 1 and Example 2 are based on first principle. For many parametric families of distributions, the calculation for can be done using formulas. To that end, consider summing the payment and the payment .
Summing up each case produces the original loss . Thus buying a policy with an ordinary deductible along with a policy with a limit of equals to full coverage. From a calculation standpoint, can be computed as follows:
The advantage of using the above idea is that the calculation of is routine for a large number of parametric families of distributions. For examples, most of the distributions in the table in this link have formulas for and . Then subtracting into would give . The following table gives the limited expectation for three distributions, taken from the table in the given link.
Limited Expectation 


Examples are given to demonstrate how these formulas are used.
Example 3
You are given the following:
 Losses follow a Pareto distribution with parameters and .
 The coverage has an ordinary deductible of 100.
For the next loss that will occur, determine the expected amount that will be paid by the insurer.
The following shows the calculation.
As a result of imposing a deductible of 1000, the payment made by the insurer per loss goes from 250 to 173.611. The reduction of payment of = 76.389.
Example 4
You are given the following:
 Losses follow an exponential distribution with mean 2500.
 The coverage has an ordinary deductible of 1000.
Determine the expected amount per loss that is paid by the insurer.
The following shows the calculation.
Note that the reduction of payment by the insurer per loss is , which is the amount per loss that has to be met by the insured.
One observation about exponential loss. When the loss is an exponential distribution, can actually be directly calculated using first principle since the exponential distribution is very tractable mathematically.
Example 5
You are given the following:
 Losses follow a lognormal distribution with parameters and .
 The coverage has an ordinary deductible of 1000.
Determine the expected amount per loss that is paid by the insurer.
The following shows the calculation.
PDF and CDF of Insurance Payment
The preceding section shows that is a great way to evaluate mean claim cost per loss when the loss distribution is from this list of distributions. For the distributions in this list, the limited expectation is readily available. To calculate the higher moments of , it is helpful to know more about its distribution. To that end, we derive its PDF and CDF and the survival function.
Before deriving the PDF and CDF, note that the payment variable is a censored and shifted random variable. It is censored from below and is shifted to the left by the amount of the deductible . As far as payment is concerned, any loss that is below the deductible is considered zero. In effect, any loss in the interval is recorded as zero. The insurance policy only pays for losses that are in excess of the deductible . Hence the positive payments are derived by shifting losses in the interval to the left by . The resulting distribution for the payment is a mixed random variable. It has a point mass at with probability where is the CDF of the loss . On the interval or some appropriate interval , the density curve of is continuous and is the resulting of shifting the density curve of to the left by the amount . Let . Let and be the PDF and CDF of , respectively. The following is the PDF of .
Note the point mass at , which is the probability that the loss is less than or equal to the deductible. The curve is the density curve shifted to the left by the amount . The following is the CDF and the survival function of .
Now that we have a description of the model for , moments and other distributional quantities can be derived. Consider the following two examples.
Example 6
You are given the following:
 Losses follow a uniform distribution on the interval .
 The coverage has an ordinary deductible of 4.
Determine the mean and variance of .
The following shows the PDF, CDF and survival function of the loss .
The point mass is . The following is the PDF .
Note that the PDF over the interval is shifted to the interval . The following gives the CDF and survival function of .
The moments of can be can be obtained by integrating with the PDF .
Note that the variance of the unmodified loss is . In this case, imposing a deductible of 4 reduces the variance by a little more than half.
Example 7
Consider the loss distribution and the coverage in Example 3. Determine the PDF, CDF and the survival function of . Then find the mean and variance of .
The loss has a Pareto distribution with parameters and . The following shows its PDF, CDF and survival function.
The point mass is . The following is the PDF .
Note that the part for is times the Pareto density function for parameters and . Thus the moments of can be obtained by multiplying with the moments of this new Pareto distribution. Since the point mass is for , we can ignore it.
To complete the example, the following gives the CDF and survival function.
Insurance Payment Per Loss versus Per Payment
Note that the is calculated over all losses (whether they are below or above the deductible). The expected value reflects the probability that is assigned to the payment of zero (the point mass), and the probabilities that are applied to the payment over all in the interval (the integral). Thus is the average payment per loss (or over all possible losses). Another average payment to consider is the average payment over all payments, i.e. over all losses larger than the deductible. This topic is continued in the next post.
2017 – Dan Ma
Mathematical models for insurance payments – part 1 – policy limit
Suppose an individual (or an entity) faces a random loss and the loss is modeled by the random variable . The individual purchases an insurance coverage to cover this loss. If the policy pays the loss in full (if a loss occurs), then the expected insurance payment is mean loss . Usually the expected insurance payment is less than due to the presence of policy provisions such as a deductible and/or limit. This post is the first post in a series of posts in discussing the probability models of insurance payments.
Policy Limit
In this post, we assume that the random loss is a continuous random variable taking on the positive real numbers or numbers from an interval such as . The methodology can be adjusted to handle discrete loss variable (mostly replacing integrals with summation).
Suppose that the random variable is the size of the loss (if a loss occurs). Under an insurance policy with a limit , the insurer pays the loss in full if the loss is less than the limit. Furthermore if the loss is or greater, the insurance payment is capped at . Under this policy provision, what is the expected amount of insurance payment if there is a loss? The insurance payment in the presence of a limit is denoted by . This is called the limited loss random variable.
First, let’s consider the mean (called limited expectation). Let , and be the probability density function (PDF), cumulative distribution function (CDF) and the survival function of the random loss, respectively.
The first integral is based on the definition of the limited loss variable, summing up the payments in the interval and in the interval . The second integral is an alternative way to evaluate the first integral (see here for more explanation). The following gives the second moment, which can be used in evaluating the variance of insurance payment.
Once again, the second moment is also expressed using the alternative calculation. The following two examples demonstrate the evaluation of the first two moments of .
Basic Example
Example 1
Suppose that the loss distribution has PDF . Evaluate and
The CDF of the loss is .
The limited expected value is calculated in two ways to demonstrate that the two approaches are equivalent. Without the policy limit, the expected loss is 3.3333. With the benefit payment capped at 4, the expected amount paid by the insurance policy is 2.6133 (per loss). The difference is 3.3333 – 2.6133 = 0.72, which is the expected loss responsible by the insured.
Example 2
Continue with Example 1. Calculate the variance of the insurance payment .
.
To compare, the variance of is as calculated below.
The example demonstrates that policy provisions such as limit has a variance reducing effect. Had the insurer been liable to pay for the entire loss amount, there would be a greater fluctuation in the payment.
A Censored Variable
Mathematically, the random variable is an upper censored random variable. Any realized value of above the limit is known only known as the limit (as far as payment is concerned). On the other hand, applying a limit on the loss turns the continuous random variable into a mixed random variable. In the interval , is continuous while there is a point mass at the point with probability mass .
The policy provisions such as deductible and limit have the effect of turning the unmodified loss variable into censored or truncated variables. As a result, these variables are mixed random variables. The subsequent posts will further demonstrate this point.
Parametric Distributions
There is a vast inventory of parametric distributions that are potential candidates for models of random losses. For example, the table in this link is an inventory of distributions that may be useful in modeling insurance losses. Included in the table are the limited expectations for various distributions. The following table shows three of the distributions. They are listed here primarily because the calculation is tractable and is better suited for demonstration of the concept.
Limited Expectation 


For formulas for the higher moments, refer to the table. This table will be used throughout the discussion (in subsequent blog posts) on estimating insurance payments. Here’s some blog posts on these three distributions – exponential, Pareto and lognormal.
The following example shows how these formulas are used.
Example 3
Evaluate for the following loss distributions.
 Exponential: mean 1000, = 2000.
 Pareto: mean 5, variance 75, = 10.
 Lognormal: mean 177.682811, variance 13680.72152, = 250.
The exponential distribution with mean 1000 has parameter (the scale parameter). The following gives the limited expected value.
The Pareto distribution with mean 5 and variance 75 translates to the parameters and . The following gives the limited expected value.

.
The lognormal distribution with mean 177.682811 and variance 13680.72152 corresponds to and . The following gives the limited expected value.
Comments
The discussion here is just the beginning. The limited expected value is for a simple policy provision, having just one modification of loss. It is a building block for other payments under more complicated insurance payment rules.
2017 – Dan Ma
Mathematical models for insurance payments – part 0 – introduction
Insurance losses depend on two random variables. The first one is the number of losses that will occur in a specified period in connection to an insured (or a group of insureds). This is commonly referred to as the loss frequency or claim frequency. Its probability distribution is called the loss (claim) frequency distribution. The second random variable is the amount of the loss (if a loss has occurred). The amount of loss is usually referred to as the severity and its probability distribution is called the severity distribution. Then putting these two distributions together will lead to the total loss distribution.
In the several posts that follow, the focus is on the severity distribution, i.e. we will focus on the size of the loss or size of the claim. Once the methodology for the severity is discussed in a fair amount of details, we will add claim frequency.
Given that a loss has occurred and that the amount is , the insurance company will make a payment to the insured to cover the loss . The insurance payment is a random quantity since is random. Due to the presence of policy provisions such as deductible and limit, the insurance payment is likely less than . The focus is on determining the distributional quantities of the probability model of the insurance payment to the insured. First and foremost, we would like to calculate the mean and variance of the distribution of payment as well as probability density function and cumulative distribution function and other distributional quantities.
The subsequent post will show, from a mathematical standpoint, the random variable of the insurance payment is a truncated and/or censored variable of the loss due to policy provisions such as deductible and policy limit.
In practice, the mean and variance of the insurance payments are often estimated from historical observations rather than using a hypothesized distribution of losses. However, the parametric distribution approach is a good starting point of the discussion of estimation of insurance losses.
2017 – Dan Ma
Integrating survival function to calculate the mean
For a continuous random variable that are nonnegative in values, the mean is obtained by integrating from to where is the probability density function of . In some cases, this integral is hard (or impossible) to do. In these cases, it may be possible to find the mean and higher moments by integrating the survival functions.
Here’s the usual way to calculate moments.
Moments can be calculated by integrating the survival function.
In fact, the mean and moments of many of the distributions in the Exam C table are calculated using the survival function method, e.g. the Pareto distribution and the exponential distribution.
The integrals in (3) and (4) are derived from (1) and (2) by performing integration by parts. The derivation is not very hard. You will find out that some assumptions are needed to make the integration by parts work. Specifically, the assumption is that the following limit is 0.
The integrals in (3) and (4) works only if this limit converges to zero. Essentially this limit is zero if the expectation exists.
The same idea can be applied to a discrete distribution.
Examples
Example 1
Suppose that the useful life (in months) of a device is modeled by the CDF for . Calculated the expected useful of a brand new device. Calculate the variance of the lifetime of such a device.
Note that the survival function is . The following shows the calculation.
Note that the integral for uses the method of substitution.
Example 2
Suppose that the lifetime of a machine follows a distribution with the following CDF.
Calculate the mean and variance of the lifetime.
The survival function is . The first is to calculate the first 2 moments.
Note that the integral for is manipulated to be in terms of integrals of exponential density functions while the integral for is manipulated to be in terms of integrals of times exponential density functions (each becoming the mean of that exponential distribution).
Example 3
The twoparameter Pareto distribution (Type II Lomax) is discussed in here in a companion blog. Its survival function is
Derive the mean of the Pareto distribution using the survival function approach. What assumption is made on the shape parameter ?
In order for the integral to converge, we need to assume .
The next two examples are left as exercises.
Practice Problems
Practice Problem 1
The survival function for the distribution of a lifetime of a type of electronic devices is .
Calculate the mean and variance of the lifetime of such devices.
Practice Problem 2
The probability that the size of a randomly selected auto collision loss is greater than is .
Calculate the mean and variance of the loss size.
Answers
Practice Problem 1
Practice Problem 2
2017 – Dan Ma
Practice Problem Set 3 – basic lognormal problems
This post has several practice problems to go with this previous discussion on lognormal distribution.
Practice Problem 3A 
The amount of annual losses from an insured follows a lognormal distribution with parameters and = 0.6 and with mode = 2.5. Calculate the mean annual loss for a randomly selected insured. 
Practice Problem 3B 
Claim size for an auto insurance coverage follows a lognormal distribution with mean 149.157 and variance 223.5945. Determine the probability that a randomly selected claim will be greater than 170. 
Practice Problem 3C 
For xray machines produced by a certain manufacturer, the following is known.
Calculate the probability that an xray machine produced by this manufacturer will last at least 12 years. 
Practice Problem 3D 
Claim sizes expressed in US dollars follow a lognormal distribution with parameters = 5 and = 0.25. One Canadian dollar is currently worth $0.75 US dollars. Calculate the 75th percentile of a claim in Canadian dollars. 
Practice Problem 3E 
For a commercial fire coverage, the size of a loss follows a lognormal distribution with parameters = 2.75 and = 0.75. Determine where is the 75th percentile of a loss and is the 25th percentile of a loss. Note that is known as the interquartile range. 
Practice Problem 3F 
Claim sizes in the current year follow a lognormal distribution with = 4.75 and = 0.25. In the next year, all claims are expected to be inflated uniformly by 25%. One claim is expected in the next year for an insured. Determine where is the 80th percentile of the size of this claim and is the 40th percentile of the size of this claim. 
Practice Problem 3G 
Determine the median of the portion of next year’s loss distribution that is above 10. 
Practice Problem 3H 
Losses follow a lognormal distribution with mean 17 and variance 219. Determine the skewness of the loss distribution. 
Practice Problem 3I 
Determine the probability that a loss is less than 5000. 
Practice Problem 3J 
Determine the mean of the losses. 
All normal probabilities are obtained by using the normal distribution table found here.
Problem  Answer 

3A  4.29 
3B  0.0869 
3C  0.2033 
3D  233.9675 
3E  16.39085 
3F  42.5155 
3G  21.143268 
3H  3.271185 
3I  0.7764 
3J  1124.394559 
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2017 – Dan Ma
Basic properties of lognormal distribution
A detailed discussion of the mathematical properties of lognormal distribution is found in this previous post in a companion blog. This post shows how to work basic calculation problems for lognormal distribution. A summary of lognormal distribution is given and is followed by several examples. Practice problems are in the next post.
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Basic Properties
The random variable is said to follow a lognormal distribution with parameters and if follows a normal distribution with mean and variance . Here, is the natural logarithm in base = 2.718281828…. It is difficult (if not impossible) to calculate probabilities by integrating the lognormal density function. Since the lognormal distribution is intimately related to the normal distribution, the basic lognormal calculation is performed by calculating the corresponding normal distribution. The following summary shows how.
In the following points, has a lognormal distribution with parameters and and is the corresponding normal distribution with mean and variance . The notation means raising to the number .
1. Lognormal observations and normal observations 

2. Lognormal CDF and normal CDF 

3. Lognormal density function and normal density function 

4. Lognormal moments and normal moment generating function 

5. Examples of lognormal moments 

6. Lognormal percentiles and normal percentiles 

7. Constant multiple of lognormal distribution 

8. Mode of lognormal distribution 

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Examples
Two examples are given to illustrate the calculation discussed here. The next post has practice problems.
All normal probabilities are obtained by using the normal distribution table found here.
Example 1
Suppose that the random variable has a lognormal distribution with parameters = 1 and = 2. Calculate the following.
 and
 The 67th, 95th and 99th percentiles of .
 Let . Find and
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To find the percentiles, first find the standard normal percentiles, either by using calculator or by looking up a table. Using a standard normal table, we get 0.44 (67th percentile), 1.645 (95th percentile) and 2.33 (99th percentile). The following gives the lognormal percentiles.

(67th percentile)
(95th percentile)
(99th percentile)
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The random variable has a lognormal distribution with parameters and = 2.
Note. One interpretation of is that of inflation, in this case a 10% inflation. For example, let be the size of a randomly selected auto insurance collision claim in the current year. If the claims are expected to increase 10% in the following year, is the the size of a randomly selected claim in the following year.
Example 2
Suppose that the random variable has a lognormal distribution with mean 12.18 and variance 255.02. Calculation the following.
 The skewness and kurtosis of .
First, determine the parameters and by setting up the following equations.
Plug the first equation into the second equation and obtain the equation . Solving for produces = 1. Plug = 1 into the first equation produces = 2. The following gives the desired probability.
To find the skewness and kurtosis, one way is to find the first 4 lognormal moments and then calculate the third standardized moment (skewness) and the fourth standardized moment (kurtosis). To see how this is done, see this previous post. Another is to use the formulas given above.
Example 3
Suppose that the lifetime (in years) of a certain type of machines follows the lognormal distribution described in Example 2. Suppose that you purchased such a machine that is 10year old. What is the probability that it will last another 10 years?
This is a conditional probability since the machine already survived for 10 years already.
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2017 – Dan Ma
Practice Problem Set 2 – finding median losses
This post has two practice problems to find the median of a distribution that models insurance losses.
Practice Problem 2a 
Losses have a distribution with the following density function:
Calculate the median loss amount. 
Practice Problem 2b 
Losses are modeled by a distribution that is a mixture of two exponential distributions, one with mean 6 and the other with mean 12. The weight of each distribution is 50%. Calculate the median loss amount. 
Solutions
Problem 2a

Median
log is logarithm to base = 2.718281828…
Problem 2b

Median
log is logarithm to base = 2.718281828…
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2017 – Dan Ma