Moments

BASIC DISTRIBUTIONAL QUANTITIES

3.1 Moments

There are a variety of interesting calculations that can be done from the models described in Chapter 2. Examples are the average amount paid on a claim that is subject to a deductible or policy limit or the average remaining lifetime of a person age 40.

Definition 3.1 Thekth raw momentof a random variable is the expected(average)value of the𝑘th power of the variable, provided that it exists. It is denoted byE(𝑋^𝑘)or by𝜇^′_𝑘. The first raw moment is called themeanof the random variable and is usually denoted by𝜇.

Note that𝜇 is not related to 𝜇(𝑥), the force of mortality from Definition 2.7. For random variables that take on only nonnegative values (i.e. Pr(𝑋≥0) = 1),𝑘may be any real number. When presenting formulas for calculating this quantity, a distinction between continuous and discrete variables needs to be made. Formulas will be presented for random variables that are either everywhere continuous or everywhere discrete. For mixed models, evaluate the formula by integrating with respect to its density function wherever the random variable is continuous, and by summing with respect to its probability function wherever

LossModels:FromDatatoDecisions,FifthEdition.

StuartA.Klugman,HarryH.Panjer,andGordonE.Willmot.

© 2019 John Wiley & Sons, Inc. Published 2019 by John Wiley & Sons, Inc.

Companion website: www.wiley.com/go/klugman/lossmodels5e

21

the random variable is discrete and adding the results. The formula for the𝑘th raw moment is

𝜇_𝑘^′ =E(𝑋^𝑘) =

∫

∞

−∞

𝑥^𝑘𝑓(𝑥)𝑑𝑥 if the random variable is continuous

=∑

𝑗 𝑥^𝑘_𝑗𝑝(𝑥_𝑗) if the random variable is discrete, (3.1) where the sum is to be taken over all𝑥_𝑗 with positive probability. Finally, note that it is possible that the integral or sum will not converge, in which case the moment is said not to exist.

EXAMPLE 3.1

Determine the first two raw moments for each of the five models.

The subscripts on the random variable𝑋indicate which model is being used.

E(𝑋₁) =

∫

100 0

𝑥(0.01)𝑑𝑥= 50, E(𝑋²₁) =

∫

100 0

𝑥²(0.01)𝑑𝑥= 3,333.33, E(𝑋₂) =

∫

∞ 0

𝑥 3(2,000)³

(𝑥+ 2,000)⁴𝑑𝑥= 1,000, E(𝑋²₂) =

∫

∞ 0

𝑥² 3(2,000)³

(𝑥+ 2,000)⁴𝑑𝑥= 4,000,000,

E(𝑋3) = 0(0.5) + 1(0.25) + 2(0.12) + 3(0.08) + 4(0.05) = 0.93, E(𝑋²₃) = 0(0.5) + 1(0.25) + 4(0.12) + 9(0.08) + 16(0.05) = 2.25, E(𝑋₄) = 0(0.7) +

∫

∞ 0

𝑥(0.000003)𝑒^{−0.00001𝑥}𝑑𝑥= 30,000,

E(𝑋²₄) = 0²(0.7) +

∫

∞ 0

𝑥²(0.000003)𝑒^{−0.00001𝑥}𝑑𝑥= 6,000,000,000, E(𝑋5) =

∫

50 0

𝑥(0.01)𝑑𝑥+

∫

75 50

𝑥(0.02)𝑑𝑥= 43.75, E(𝑋²₅) =

∫

50 0

𝑥²(0.01)𝑑𝑥+

∫

75 50

𝑥²(0.02)𝑑𝑥= 2,395.83.

□ Definition 3.2 Thekth central momentof a random variable is the expected value of the 𝑘th power of the deviation of the variable from its mean. It is denoted byE[(𝑋−𝜇)^𝑘]or by 𝜇_𝑘. The second central moment is usually called thevarianceand denoted𝜎²orVar(𝑋), and its square root,𝜎, is called thestandard deviation. The ratio of the standard deviation to the mean is called thecoefficient of variation. The ratio of the third central moment to the cube of the standard deviation,𝛾1 = 𝜇3∕𝜎³, is called theskewness. The ratio of the

fourth central moment to the fourth power of the standard deviation,𝛾2=𝜇4∕𝜎⁴, is called thekurtosis.¹

The continuous and discrete formulas for calculating central moments are 𝜇_𝑘=E[(𝑋−𝜇)^𝑘]

=∫

∞

−∞

(𝑥−𝜇)^𝑘𝑓(𝑥)𝑑𝑥 if the random variable is continuous

=∑

𝑗

(𝑥_𝑗−𝜇)^𝑘𝑝(𝑥_𝑗) if the random variable is discrete. (3.2) In reality, the integral needs be taken only over those𝑥values where𝑓(𝑥)is positive. The standard deviation is a measure of how much the probability is spread out over the random variable’s possible values. It is measured in the same units as the random variable itself.

The coefficient of variation measures the spread relative to the mean. The skewness is a measure of asymmetry. A symmetric distribution has a skewness of zero, while a positive skewness indicates that probabilities to the right tend to be assigned to values further from the mean than those to the left. The kurtosis measures flatness of the distribution relative to a normal distribution (which has a kurtosis of 3).²Kurtosis values above 3 indicate that (keeping the standard deviation constant), relative to a normal distribution, more probability tends to be at points away from the mean than at points near the mean. The coefficients of variation, skewness, and kurtosis are all dimensionless.

There is a link between raw and central moments. The following equation indicates the connection between second moments. The development uses the continuous version from (3.1) and (3.2), but the result applies to all random variables:

𝜇2=

∫

∞

−∞

(𝑥−𝜇)²𝑓(𝑥)𝑑𝑥=

∫

∞

−∞

(𝑥²− 2𝑥𝜇+𝜇²)𝑓(𝑥)𝑑𝑥

=E(𝑋²) − 2𝜇E(𝑋) +𝜇²=𝜇^′₂−𝜇². (3.3) EXAMPLE 3.2

The density function of the gamma distribution appears to be positively skewed.

Demonstrate that this is true and illustrate with graphs.

From Appendix A, the first three raw moments of the gamma distribution are 𝛼𝜃,𝛼(𝛼+ 1)𝜃², and𝛼(𝛼+ 1)(𝛼+ 2)𝜃³. From (3.3) the variance is𝛼𝜃², and from the solution to Exercise 3.1 the third central moment is2𝛼𝜃³. Therefore, the skewness is2𝛼^−1∕2. Because𝛼must be positive, the skewness is always positive. Also, as𝛼 decreases, the skewness increases.

Consider the following two gamma distributions. One has parameters𝛼 = 0.5 and𝜃 = 100while the other has𝛼= 5and𝜃 = 10. These have the same mean, but their skewness coefficients are 2.83 and 0.89, respectively. Figure 3.1 demonstrates

the difference. □

1It would be more accurate to call these items the “coefficient of skewness” and “coefficient of kurtosis” because there are other quantities that also measure asymmetry and flatness. The simpler expressions are used in this text.

2Because of this, an alternative definition of kurtosis has 3 subtracted from our definition, giving the normal distribution a kurtosis of zero, which can be used as a convenient benchmark.

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0 20 40 60 80 100

x

Density

f(x) g(x)

Figure 3.1 The densities of𝑓(𝑥) ∼gamma(0.5, 100) and𝑔(𝑥) ∼gamma(5, 10).

Finally, when calculating moments, it is possible that the integral or sum will not exist (as is the case for the third and fourth moments for Model 2). For the models that we typically encounter, the integrand and summand are nonnegative, and so failure to exist implies that the limit of the integral or sum is infinity. For an illustration, see Example 3.9.

Definition 3.3 For a given value of𝑑 withPr(𝑋 > 𝑑) > 0, theexcess loss variable is 𝑌^𝑃 =𝑋−𝑑, given that𝑋 > 𝑑. Its expected value,

𝑒_𝑋(𝑑) =𝑒(𝑑) =E(𝑌^𝑃) =E(𝑋−𝑑|𝑋 > 𝑑),

is called the mean excess loss function. Other names for this expectation are mean residual life functionandcomplete expectation of life. When the latter terminology is used, the commonly used symbol is ˚e_𝑑.

This variable could also be called aleft truncated and shifted variable. It is left truncated because any values of𝑋below𝑑 are not observed. It is shifted because𝑑 is subtracted from the remaining values. When𝑋is a payment variable, the mean excess loss is the expected amount paid, given that there has been a payment in excess of a deductible of𝑑.³ When𝑋is the age at death, the mean excess loss is the expected remaining time until death, given that the person is alive at age𝑑. The𝑘th moment of the excess loss variable is determined from

𝑒^𝑘_𝑋(𝑑) = ∫_𝑑^∞(𝑥−𝑑)^𝑘𝑓(𝑥)𝑑𝑥

1 −𝐹(𝑑) if the variable is continuous

=

∑𝑥_𝑗>𝑑(𝑥_𝑗−𝑑)^𝑘𝑝(𝑥_𝑗)

1 −𝐹(𝑑) if the variable is discrete. (3.4) Here, 𝑒^𝑘_𝑋(𝑑) is defined only if the integral or sum converges. There is a particularly convenient formula for calculating the first moment. The development given below is for a continuous random variable, but the result holds for all types of random variables.

3This provides the meaning of the superscript𝑃, indicating that this payment is per payment. It is made to distinguish this variable from𝑌^𝐿, the per-loss variable to be introduced shortly. These two variables are explored in depth in Chapter 8.

The second line is based on an integration by parts, where the antiderivative of 𝑓(𝑥)is taken as−𝑆(𝑥):

𝑒_𝑋(𝑑) = ∫_𝑑^∞(𝑥−𝑑)𝑓(𝑥)𝑑𝑥 1 −𝐹(𝑑)

= −(𝑥−𝑑)𝑆(𝑥)|^∞_𝑑 +∫_𝑑^∞𝑆(𝑥)𝑑𝑥 𝑆(𝑑)

= ∫_𝑑^∞𝑆(𝑥)𝑑𝑥

𝑆(𝑑) . (3.5)

Definition 3.4 Theleft censored and shifted variableis 𝑌^𝐿= (𝑋−𝑑)₊=

{

0, 𝑋≤𝑑, 𝑋−𝑑, 𝑋 > 𝑑.

It is left censored because values below𝑑 are not ignored but are set equal to zero.

There is no standard name or symbol for the moments of this variable. For dollar events, the distinction between the excess loss variable and the left censored and shifted variable is one of per paymentversusper loss. In the per-payment situation, the variable exists only when a payment is made. The per-loss variable takes on the value zero whenever a loss produces no payment. The moments can be calculated from

E[(𝑋−𝑑)^𝑘₊] =

∫

∞

𝑑 (𝑥−𝑑)^𝑘𝑓(𝑥)𝑑𝑥 if the variable is continuous.

= ∑

𝑥𝑗>𝑑

(𝑥_𝑗−𝑑)^𝑘𝑝(𝑥_𝑗) if the variable is discrete. (3.6)

It should be noted that

E[(𝑋−𝑑)^𝑘₊] =𝑒^𝑘(𝑑)[1 −𝐹(𝑑)]. (3.7) EXAMPLE 3.3

Construct graphs to illustrate the difference between the excess loss variable and the left censored and shifted variable.

The two graphs in Figures 3.2 and 3.3 plot the modified variable𝑌 as a function of the unmodified variable𝑋. The only difference is that for𝑋values below 100 the variable is undefined, while for the left censored and shifted variable it is set equal to

zero. □

These concepts are most easily demonstrated with a discrete random variable.

EXAMPLE 3.4

An automobile insurance policy with no coverage modifications has the following possible losses, with probabilities in parentheses: 100 (0.4), 500 (0.2), 1,000 (0.2), 2,500 (0.1), and 10,000 (0.1). Determine the probability mass functions and expected

–50 0 50 100 150 200 250

0 50 100 150 200 250 300

X

Y

Figure 3.2 The excess loss variable.

–50 0 50 100 150 200 250

0 50 100 150 200 250 300

X

Y

Figure 3.3 A left censored and shifted variable.

values for the excess loss and left censored and shifted variables, where the deductible is set at 750.

For the excess loss variable, 750 is subtracted from each possible loss above that value. Thus the possible values for this random variable are 250, 1,750, and 9,250.

The conditional probabilities are obtained by dividing each of the three probabilities by 0.4 (the probability of exceeding the deductible). They are 0.5, 0.25, and 0.25, respectively. The expected value is250(0.5) + 1,750(0.25) + 9,250(0.25) = 2,875.

For the left censored and shifted variable, the probabilities that had been assigned to values below 750 are now assigned to zero. The other probabilities are unchanged, but the values they are assigned to are reduced by the deductible. The probability mass function is 0 (0.6), 250 (0.2), 1,750 (0.1), and 9,750 (0.1). The expected value is0(0.6) + 250(0.2) + 1,750(0.1) + 9,250(0.1) = 1,150. As noted in (3.7), the ratio of the two expected values is the probability of exceeding the deductible.

Another way to understand the difference in these expected values is to consider 10 accidents with losses conforming exactly to the above distribution. Only four of the accidents produce payments, and multiplying by the expected payment per payment gives a total of4(2,875) = 11,500expected to be paid by the company. Or, consider that the 10 accidents each have an expected payment of 1,150 per loss (accident) for a total expected value of 11,500. Therefore, what is important is not the variable being

used but, rather, that it be used appropriately. □

0 50 100 150 200 250

0 50 100 150 200

Loss

Payment

Deductible Limit

Figure 3.4 A limit of 100 plus a deductible of 100 equals full coverage.

The next definition provides a complementary variable to the excess loss variable.

Definition 3.5 Thelimited loss variableis 𝑌 =𝑋∧𝑢=

{𝑋, 𝑋 < 𝑢, 𝑢, 𝑋≥𝑢.

Its expected value, E(𝑋∧𝑢), is called thelimited expected value.

This variable could also be called theright censored variable. It is right censored because values above𝑢are set equal to𝑢. An insurance phenomenon that relates to this variable is the existence of a policy limit that sets a maximum on the benefit to be paid.

Note that(𝑋−𝑑)₊+ (𝑋∧𝑑) =𝑋. That is, buying one insurance policy with a limit of𝑑 and another with a deductible of𝑑is equivalent to buying full coverage. This is illustrated in Figure 3.4.

The most direct formulas for the𝑘th moment of the limited loss variable are E[(𝑋∧𝑢)^𝑘] =

∫

𝑢

−∞

𝑥^𝑘𝑓(𝑥)𝑑𝑥+𝑢^𝑘[1 −𝐹(𝑢)]

if the random variable is continuous.

= ∑

𝑥_𝑗≤𝑢

𝑥^𝑘_𝑗𝑝(𝑥_𝑗) +𝑢^𝑘[1 −𝐹(𝑢)]

if the random variable is discrete. (3.8) Another interesting formula is derived as follows:

E[(𝑋∧𝑢)^𝑘] =

∫

0

−∞

𝑥^𝑘𝑓(𝑥)𝑑𝑥+

∫

𝑢 0

𝑥^𝑘𝑓(𝑥)𝑑𝑥+𝑢^𝑘[1 −𝐹(𝑢)]

=𝑥^𝑘𝐹(𝑥)|⁰_−∞−

∫

0

−∞

𝑘𝑥^𝑘−1𝐹(𝑥)𝑑𝑥

−𝑥^𝑘𝑆(𝑥)|^𝑢₀+

∫

𝑢 0

𝑘𝑥^𝑘−1𝑆(𝑥)𝑑𝑥+𝑢^𝑘𝑆(𝑢)

= −∫

0

−∞

𝑘𝑥^𝑘−1𝐹(𝑥)𝑑𝑥+

∫

𝑢 0

𝑘𝑥^𝑘−1𝑆(𝑥)𝑑𝑥, (3.9)

where the second line uses integration by parts. For𝑘= 1, we have E(𝑋∧𝑢) = −

∫

0

−∞

𝐹(𝑥)𝑑𝑥+

∫

𝑢 0

𝑆(𝑥)𝑑𝑥.

The corresponding formula for discrete random variables is not particularly interesting.

The limited expected value also represents the expected dollar saving per incident when a deductible is imposed. The𝑘th limited moment of many common continuous distributions is presented in Appendix A. Exercise 3.8 asks you to develop a relationship between the three first moments introduced previously.

EXAMPLE 3.5

(Example 3.4 continued) Calculate the probability function and the expected value of the limited loss variable with a limit of 750. Then show that the sum of the expected values of the limited loss and left censored and shifted random variables is equal to the expected value of the original random variable.

All possible values at or above 750 are assigned a value of 750 and their proba- bilities summed. Thus the probability function is 100 (0.4), 500 (0.2), and 750 (0.4), with an expected value of100(0.4) + 500(0.2) + 750(0.4) = 440. The expected value of the original random variable is 100(0.4) + 500(0.2) + 1,000(0.2) + 2,500(0.1) +

10,000(0.1) = 1,590, which is440 + 1,150. □

3.1.1 Exercises

3.1 Develop formulas similar to (3.3) for𝜇₃and𝜇₄.

3.2 Calculate the standard deviation, skewness, and kurtosis for each of the five models.

It may help to note that Model 2 is a Pareto distribution and the density function in the continuous part of Model 4 is an exponential distribution. Formulas that may help with calculations for these models appear in Appendix A.

3.3 (*) A random variable has a mean and a coefficient of variation of 2. The third raw moment is 136. Determine the skewness.

3.4 (*) Determine the skewness of a gamma distribution that has a coefficient of variation of 1.

3.5 Determine the mean excess loss function for Models 1–4. Compare the functions for Models 1, 2, and 4.

3.6 (*) For two random variables,𝑋and𝑌,𝑒_𝑌(30) =𝑒_𝑋(30) + 4. Let𝑋have a uniform distribution on the interval from 0 to 100 and let𝑌 have a uniform distribution on the interval from 0 to𝑤. Determine𝑤.

3.7 (*) A random variable has density function 𝑓(𝑥) = 𝜆⁻¹𝑒^{−𝑥∕𝜆}, 𝑥, 𝜆 > 0. Determine 𝑒(𝜆), the mean excess loss function evaluated at𝜆.

3.8 Show that the following relationship holds:

E(𝑋) =𝑒(𝑑)𝑆(𝑑) +E(𝑋∧𝑑). (3.10) 3.9 Determine the limited expected value function for Models 1–4. Do this using both (3.8) and (3.10). For Models 1 and 2, also obtain the function using (3.9).

3.10 (*) Which of the following statements are true?

(a) The mean excess loss function for an empirical distribution is continuous.

(b) The mean excess loss function for an exponential distribution is constant.

(c) If it exists, the mean excess loss function for a Pareto distribution is decreasing.

3.11 (*) Losses have a Pareto distribution with 𝛼= 0.5and𝜃 = 10,000. Determine the mean excess loss at 10,000.

3.12 Define a right truncated variable and provide a formula for its𝑘th moment.

3.13 (*) The severity distribution of individual claims has pdf 𝑓(𝑥) = 2.5𝑥^−3.5, 𝑥≥1. Determine the coefficient of variation.

3.14 (*) Claim sizes are for 100, 200, 300, 400, or 500. The true probabilities for these values are 0.05, 0.20, 0.50, 0.20, and 0.05, respectively. Determine the skewness and kurtosis for this distribution.

3.15 (*) Losses follow a Pareto distribution with𝛼 >1and𝜃unspecified. Determine the ratio of the mean excess loss function at𝑥= 2𝜃to the mean excess loss function at𝑥=𝜃. 3.16 (*) A random sample of size 10 has two claims of 400, seven claims of 800, and one claim of 1,600. Determine the empirical skewness coefficient for a single claim.