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(solution) Hi I would appreciate any help with exercises 3.2, 3.4, and 3.5.

Hi I would appreciate any help with exercises 3.2, 3.4, and 3.5. Thanks!

CASE STUDY: HEAVY-TAILED DISTRIBUTION AND

REINSURANCE RATE-MAKING

The purpose of this case study is to give a brief introduction to a heavytailed distribution and its distinct behaviors in contrast with familiar lighttailed distributions in standard texts. You will learn about QQ-plot, which is

a popular tool for checking goodness-of-?t for a particular statistical model.

You will also work on a real-life application of heavy-tailed distributions in

reinsurance rate-making.

Reinsurance is a very important component of the global ?nancial market.

It allows insurers to take on risks that they would otherwise not be able

to. Did you know that NASA buys insurance contracts for every rocket it

launches and every satellite and probe it sends to the outer space? These

equipments are so expensive that typical insurers would not be able to cover

on their own. Therefore, they can go to the reinsurance market, slide up

the coverage and transfer partial coverage to reinsurers that exceed their

?nancial capabilities. By the end of this case study, you will be able to learn

basic principles of pricing an reinsurance contract.

Learning Objectives:

? Visualize the concepts in central limit theorem;

? Identify cases where classical central limit theorem does not apply;

? Reinforce the concept of cumulative distribution function;

? Understand why and how QQ-plot works for the assessment of goodnessof-?t;

? Reinforce the concepts of conditional probability and conditional

expectation;

? Apply basic integration technique to compute mean excess function;

? Learn about behaviors of a heavy-tailed distribution;

? Learn how to use order statistics to estimate quantiles and mean

excess function;

? Develop intuition behind point estimators.

1. Central limit theorem

This section is to provide visualization of central limit theorem which governs most of familiar distributions introduced in the ?rst course of probability. We provide examples on both discrete random variable and continuous

random variable.

Example 1.1. (Bernoulli random variables) Suppose we intend to test the

fairness of a coin, i.e. whether the coin has equal chance of landing on a

head or a tail. We can do so by counting the number of heads in a sequence

1 2 CASE STUDY: REINSURANCE of coin tosses. The number of heads in each toss is a Bernoulli random

variable, denoted by X1 . Let p be the probability of a head and q = 1 ? p

be the probability of a tail. Note that these parameters are unknown before

our experiments. Then its probability mass function is given by

p,

q, P(X1 = x) = x = 1,

x = 0. We let Xk be the number of heads in the k-th coin toss, k = 1, 2, · · · , n.

Then we count the total number of heads after n coin tosses.

n Sn := Xk .

k=1 Then it is easy to show that Sn is a binomial random variable with parameters n and p and its probability mass function is given by

P(Sn = x) = n x n?x

p q

,

x x = 0, 1, · · · , n. For example, suppose that we have an unfair coin with p = 0.3, which

means it has only 30% chance of landing on a head. Figure 1 shows the

probability mass functions of the number of heads, Sn , where the number of

coin tosses n = 1, 2, 3, 10, 20, 50. Since p < 1/2, we are more likely to see a

smaller number of heads that that of tails in any given n tosses. In general,

the probabiltiy mass function of Sn tends to skew towards to the right.

However, as one can see in the later graphs in Figure 1, the probability mass

function becomes more and more symmetric1 as n gets bigger and bigger.

This phenonmenon is present for any p ? (0, 1), no matter how extreme is

p. Why is this happening? The answer lies in the ?invisible hands? of a

governing law known as the Central Limit Theorem, which we have already

learned in class.

Let us consider the sample average

1

X n := Sn .

n

Since the expectation of the sample average

E(X n ) = 1

n n E(Xk ) = p,

k=1 the sample mean provides an unbiased estimator of the unknown parameter

of fairness p.

Exercise 1.1. The Central Limit Theorem tells us that the estimator X n

is asymptotically normal, i.e.

Yn := ? Xn ? p

n ?

? N (0, 1),

pq 1Note, however, this is not to suggest that the coin becomes fair. CASE STUDY: REINSURANCE 3 Figure 1. Probability mass function of Sn

where N (µ, ? 2 ) is a normal random variable with mean µ and variance ? 2

and ??? refers to convergence in distribution. Explain why the estimator

X n behaves roughly like N (p, pq/n). [HINT: Noting that Yn is approximately N (0, 1), you can reformulate the given equation to express X n as a

function of Yn . What kind of function do you get? Combine these two facts

to determine the distribution of X n , including E(X n ) and Var(X n ).]

As the sample size n gets big, the variance is so small that the sample

average gives very good estimate of the actual parameter p. That is why in

practice we use the value of X n as an estimate, despite the fact that it is in

fact a random variable.

Exercise 1.2. What is the exact distribution of X n ? [HINT: Use the

de?nition of X n as a function of Sn , and apply the probability mass function

of Sn (which you should know) to derive the probability mass function of

X n .]

Let us verify numerically the conclusion of central limit theorem. Similar

to what you showed in Exercise 1.2, one can show that the exact probability

mass function of Yn is given by

n h n?h

?

P(Yn = y) =

p q

,

h := npqy + np,

h

?

where y = (k ? np)/ npq and k = 0, 1, · · · , n. [HINT: Use the de?nition

of X n given earlier to write Yn as a function of Sn . Using the probability

mass function of Sn , to derive the probability mass function of Yn .] 4 CASE STUDY: REINSURANCE We can draw graphs of the probability mass functions and see how they

converge to a normal distribution as n increases. Figure 2 below is an illustration of the central limiting theorem. The blue bars visually depict

how a point mass function of a binomial random variable behave over an

interval. The red dashed lines indicate the normal density function. From

left to right, top to bottom we have the densities for binomial random variables with sample size n=1,2,5,20,100,1000 respectively, with probability of

success being once again 30%. Figure 2. Convolution of binomial probability mass converging to normal density

Example 1.2. (Exponential random variables) There is empirical evidence

to show that the inter-arrival times of 911 calls are generally exponentially

distributed. As a 911 operator, you might be interested in the average

waiting period. Denote the time between the (i ? 1)-th and the i-th calls by

Xi . The probability density function is given by

f (t) = ?e??t , t ? 0, where E(Xi ) = 1/? is the mean of waiting time. If we rede?ne X n , Yn , and

Sn according to this new random variable, then the Central Limit Theorem

tells us that

?

Yn := n(?X n ? 1) ? N (0, 1). CASE STUDY: REINSURANCE 5 In this case, we can also observe how the actual probability density of Yn

converges to a normal distribution.

Exercise 1.3. It can be shown using integration by parts that the probability density function of Sn is given by

?n

xn?1 e??x .

(n ? 1)!

Use this fact to show that the exact probability density function of Yn

nn?1/2

(n ? 1)! y

1+ ?

n n?1 y

exp ?n 1 + ?

n , ?

y > ? n. [HINT: As in Exercise 1.2, write Yn as a function of Sn , then use the

probability density function of Sn to determine the probability density function of Yn .]

Figure 3 is a visual illustration of the probability density function of Yn

for various choices of n. We can see how the probability density function

for Yn converges to the standard normal density function. Again the red

dashed lines is the standard normal density function while the blue lines are

the densities for Yn , given above, for n = 1, 2, 3, 5, 10, 100.

Exercise 1.4. Use python to create graphs indicating how the density function of the scaled sum of exponential random variables Yn converges to the

standard normal density by the Central Limit Theorem. It does not need

to look exactly like Figure 3 but it must at least contain the normal density

as well as the density for Yn for n = 1, 2, 10.

In general, central limit theorems deal with the sum Sn := X1 + X2 + · · · +

Xn and tries to ?nd constants an > 0 and bn such that

1

(Sn ? bn )

Yn :=

an

tends in distribution to a non-degenerate distribution. In the classical case

in the textbook, an = Var(Sn ) and bn = E(Sn ). Once the limit is known,

it can be used to approximate the otherwise cumbersome distribution of

Yn . That limit is typically the normal distribution, except when Xi ?s possesses a too heavy tail, in which case a stable distribution appears as a limit.

The discussion of a stable distribution is beyond the scope of this course.

However, we can visualize how the extremes produced by heavy-tailed distributions will corrupt the average so that an asymptotic behavior di?erent

from the normal behavior is obtained.

Example 1.3. (Pareto random variables) Consider the strict Pareto random

variable whose density is given by

f (x) = ?x???1 , x > 1, where ? is a positive number, called the Pareto index. For the purpose of

this case study, there are a few properties that we want to pay attention to. 6 CASE STUDY: REINSURANCE Figure 3. Convolution of exponential densities converging

to normal density

(1) Pareto does not always have ?nite mean or variance.

Exercise 1.5. Calculate the mean and variance of a strict Pareto

random variable. Are there any values of ? for which either does not

take a ?nite value?

(2) Pareto is related to exponential in much the same way lognormal is

related to normal.

Exercise 1.6. Show that ln X is exponentially distributed with

mean 1/? if X is a strict Pareto random variable.

(3) Pareto is a heavy-tailed distribution.

Exercise 1.7. A random variable X is said to have a heavy tail if

e??x

= 0,

x?? F (x)

lim for all ? > 0, where F (x) := P(X > x). Even though the tail function F decays

to zero as x goes to ?, it never goes down as fast as the tail of

any exponential random variable. In other words, the tail of this

distribution is heavier than any exponential tail. CASE STUDY: REINSURANCE 7 Show that a strict Pareto random variable is heavy-tailed while

neither an exponential (with any rate ?) nor a normal random variable (with and mean and variance µ and ? 2 ) is heavy-tailed. [HINT:

Apply L?Hospit?l?s rule]

o

We shall demonstrate in the next section with the visual aid of QQ-plot

that the sample average X n for Pareto random variables will not behave like

a normal random variable no matter how large n is.

2. QQ-plot

Quantile-quantile plot, also known as QQ-plot for short, is a visual tool to

check if a proposed model provides a plausible ?t to the distribution of the

random variable at hand. We start by explaining and illustrating the idea

of QQ-plot for the exponential model. Consider the standard exponential

distribution

F1 (x) := 1 ? exp(?x),

x>0

as the standard example from the class of distributions with general survival

function

F ? (x) := 1 ? F? (x) = exp(??x),

x > 0.

Suppose we have real data x1 , x2 , · · · , xn at our disposal, which we suspect

are realizations of a exponentially distributed random variable Exp(?) for

some unknown parameter ? > 0. It is important to note that this parameter

value can be considered as a nuisance parameter since its value is not our

main point of interest at this point. We can arrange these observations from

the smallest to the largest. Denote the i-th smallest observation by x(i) .

The quantile function for the exponential function has the form

1

p ? (0, 1).

Q? (p) = ? ln(1 ? p),

?

Hence, there exists a simple linear relation between the quantiles of any exponential distribution and the corresponding standard exponential quantiles

1

Q? (p) = Q1 (p),

p ? (0, 1).

?

Note that we do not know the exact distribution of the unknown random

variable, let alone the quantile. Nevertheless, we can replace the unknown

?

quantile Q? by the empirical distribution Qn where

i?1

i

?

Qn (p) = x(i) ,

for

<p? .

n

n

In a rectangular coordinate system, we plot the points with values

?

(? ln(1 ? p), Qn (p))

for several values of p ? (0, 1). A typical choice of values of p is given by

p? 1 ? 1/2 2 ? 1/2

n ? 1 ? 1/2

1/2

,

,··· ,

,1 ?

n

n

n

n . 8 CASE STUDY: REINSURANCE In other words, p = (j ? 1/2)/n for j = 1, 2, . . . , n. We then expect that a

?

straight line pattern will appear in the scatter plot if Qn (p) indeed resembles

Q? (p), in other words, if the exponential model provides a plausible statistical ?t for the given data set. When a straight line pattern is obtained, the

slope of a ?tted line can be used as an estimate of the parameter 1/?. Figure 4. Scatter plot of insurance claims data

Example 2.1. For a practical example of the QQ-plot, we investigate the

insurance claim data from a reinsurance company. The data set contains automobile claims from 1988 until 2001, which are at least as large as 1, 200, 000

euro. The original claim numbers were corrected for in?ation. This data set

contains n = 371 observations and a scatterplot can be seen in Figure 4.

The ultimate goal of this case study is to provide the participating reinsurance companies with an objective statistical analysis in order to assist

in the pricing of an unlimited excess-of-loss reinsurance to be discussed in

the next section. We follow the above-mentioned procedure to create an

exponential QQ-plot in Figure 5, where the horizontal axis represents the

quantiles of standard exponential and the vertical axis represents the empirical quantiles of claim size data. In this plot, we included the lowest 300

observations of insurance claims. If we imagine a curve connecting all points

in the plot, it appears that the curve would bend downwards and exhibit

a concave pattern. The downward curve indicates that the claim size distribution has a heavier tail than expected from an exponential distribution,

because the incremental increases in the quantile values appear to be smaller

than those of a true exponential distribution. In other words, the empirical survival function would decay slower than that of the survival function CASE STUDY: REINSURANCE 9 Figure 5. Exponential QQ-plot of data

of an exponential random variable. (Recall the de?nition of heavy-tailed

distribution in Exercise 1.7.)

Exercise 2.1. Take a logarithmic transform of the largest 270 observations

in the data set and construct an exponential QQ-plot. Observe that the

new plot is more or less linear except for the largest few points, indicating

a reasonable ?t of the Pareto distribution to the tail of claim sizes.

QQ-plots can be used in cases more general than the exponential distribution discussed above. In fact, they can be used to assess the ?t of any

statistical model. For example, we can consider the standard normal random

variable with cumulative distribution function

x

y2

1

? exp ?

dy.

?(x) :=

2

2?

??

We shall denote its inverse function

??1 (p) = inf{x ? R : ?(x) ? p}.

Most computational platforms have built-in algorithms to compute the distribution function ? and its inverse ??1 in a very e?cient manner. https:

//en.wikipedia.org/wiki/Error_function#Inverse_functions

Exercise 2.2. Show that the quantile function of a normal random variable

with mean µ and variance ? is given by

µ + ???1 (p). 10 CASE STUDY: REINSURANCE Figure 6. Normal QQ-plot

Similar to the exponential case, if we create a QQ-plot with values

?

(??1 (p), Qn (p))

for various values of p ? (0, 1) and the data {x1 , x2 , · · · , xn } is generated by

any normal random variable, then there should be a straight line pattern on

the QQ-plot.

Example 2.2. Consider 100 copies of independent and identically distributed Pareto random variables with ? = 3/2. We create a random sample

by simulations of the sample average of the random variables. Figure 6 depicts a normal QQ-plot to show that the sample average is far from a normal

distribution. This illustrates numerically that the classical central limit theorem does not apply. Furthermore, we observe that the imaginary curve

connecting all dots would bend upward on both the left tail and the right

tail, indicating the distribution of the sum of i.i.d. Therefore, the Pareto

random variables has a lighter left and heavier right tails than those of a

normal distribution. Exercise 2.3. Give a justi?cation why the QQ-plot in Example 2.2 fails to

exhibit a pattern of normality even with a relatively large sample size of 100.

[HINT: Consider the requirements for applying the central limit theorem.] CASE STUDY: REINSURANCE 11 3. Mean excess function and reinsurance

Reinsurance is an insurance policy purchased by an insurance company

from one or more other insurance companies, known as the ?reinsurer?, as

a means of risk management. It is a very common market practice when

insurance companies undertake high risk pro?les with potential catastrophic

losses. A reinsurance agreement details the conditions upon which the reinsurer would pay a share of the claims incurred by the insurer and the reinsurer is paid a ?reinsurance premium? by the insurer, which issues insurance

policies to its own policyholders. A diagram of cash ?ows among the participants in an insurance market can be found in Figure 7.

A common form of reinsurance is the excess of loss (XL) reinsurance,

where the insurer covers insurance claims from policyholders up to the maximum of its retention level and any amount beyond the retention will be

reimbursed by the reinsurer. For example, an insurance company might

insure commercial property risks with policy limits up to $10 million, and

then buy reinsurance of $5 million in excess of $5 million. In this case a loss

of $6 million on that policy will result in the recovery of $1 million from the

reinsurer.

Policyholder (P) Insurer (I) Reinsurer (R) (?) Pay insurance premium to I

(+) Receive claim from I (+) Receive premium from P

(?) Pay claims to P

(?) Pay reinsurance premiums

to R

(+) Recoup claims in excess of

retention from R (+) Receive reinsurance premiums from I

(?) Pay claims above retention

to I Figure 7. Participants in an insurance market

The modeling of the XL reinsurance relies on an important mathematical

concept, called mean excess function. Suppose a ceding insurer enters into

an XL treaty with a retention level t. Let X be a random variable governing

the size of a particular policyholder?s claim. After claim investigation, the

ceding insurer will make the payment and the reinsurer has to pay X ? t if

X > t. Pricing actuaries from reinsurance companies would want to know

the average cost of such claims, which is theoretically determined by the

mean excess function

e(t) = E(X ? t|X > t) = E(X|X > t) ? t.

Exercise 3.1. Show that the mean excess function of an exponential random variable with mean 1/? is given by

e(t) = 1

,

? t > 0. 12 CASE STUDY: REINSURANCE [HINT: Note that fX|X>t (x), the probability density function of X given

that X > t, can be expressed as fX|X>t (x) = fX (x)/P (X > t) if X > t

(and 0 otherwise), where fX (x) is the PDF of random variable X. Apply the

de?nition of expectation, using this distribution, to compute E(X ? t|X >

t)].

Exercise 3.2. Show that the mean excess function of a strict Pareto random

variable with the Pareto index ? is given by

t

e(t) =

,

t > 1.

??1

[HINT: See the hint for the previous exercise].

Observe that the mean excess function can also be written as2

E[XI(X > t)]

? t,

e(t) =

E[I(X > t)]

where I(A) = 1 if the event A is true and 0 otherwise. In practice, we

replace the theoretical mean by its empirical counterpart. Given the sample

data x1 , x2 , · · · , xn , the mean excess function is estimated by

en (t) =

? n

i=1 xi I(xi > t)

n

i=1 I(xi > t) ? t. Often the empirical function en is evaluated at t = x(n?k) , the (k + 1)?

n

th largest observation. Then the numerator equals

i=1 xi I(xi > t) =

k

j=1 x(n?j+1) , while the number of xi larger than t equals k. The estimates

of the mean excesses are then given by

(1) ek := en (x(n?k) ) =

? 1

k k x(n?j+1) ? x(n?k) .

j=1 Consider an XL reinsurance contract with a retention level R. The reinsurer is obligated to pay for the claim amount in excess over the limit R.

The fair net premium3 is given by

(2) ?(R) = E[(X ? R)+ ], where (x)+ = max{x, 0}. Equivalently, we obtain

(3) ?(R) = e(R)F (R). Since reinsurance contracts are meant to transfer risks of catastrophic

losses, claims of small and medium sizes provide no useful information for the

valuation of reinsurance contracts. Let us consider a reinsurance contract

with various retention levels, for example R = 5, 000, 000 euro, which is

2To see this result, note that XI(X > t) is a function of random variable X, such that XI(X > t) = X if X > t (and 0 otherwise). Hence, we can compute E[XI(X > t)] as we

do any other function of X

3Net premium refers to the pure cost of insurance coverage. It is used in contrast with

gross premium, which includes commission, policy expenses and other administrative costs. CASE STUDY: REINSURANCE 13 typically used in practice. Observe that only 12 observations are larger

than that level in the given data set. We use two methods in this case to

determine the net premium.

Exercise 3.3. (Estimator #1: non-parametric) The simplest way to estimate the net premium ?(R) is to use an empirical estimator of (2) given

by

n

? 1 (R) := 1

?

(xi ? R)+ .

n

i=1 Develop a computer algorithm in Python to estimate the net premiums for

various retention levels in Table 1.

Exercise 3.4. (Estimator #2: non-parametric) Another way to determine

the net premium is to make use of the identity (3)

?

?

e

?2 (R) := F n (R)?n (R),

?

where F n (R) is some estimator of the tail probability F (R) = P(X > R).

If R is ?xed at one of the sample points, that is, R = x(n?k) , the nonparametric estimator is given by

k

?

?2 (X(n?k) ) = ek .

n

If R is not ?xed at one of the sample points, then we have to introduce an

estimator for F (R). There are many di?erent ways of de?ning an estimator.

For example, one can estimate F (R) as

1

?

F n (R) :=

n n I(xi > R).

i=1 ?

Substituting this back into our estimator of ?2 (R), we ?nd

en (R)

?

?

?2 (R) :=

n n I(xi > R).

i=1 ?

Show that, if we de?ne ?2 (R) in this way, then this estimator for the net

?

premium is the same as ?1 . [HINT: You should be able to demonstrate

that these are equal by algebraic manipulation.]

Note that in the previous two pricing models we did not use any assumption of a parametric model. As much as we prefer simplicity, we should also

keep in mind a general rule from statistical theory that statistical estimators

do not provide accurate results when the size of sample data for estimation

is very small. When R = 5, 000, 000 euros, only 12 observations enter into

?

?

the calculations of ?1 and ?2 . It is essentially a waste of information from

other data points below 5 million eurors. 14 CASE STUDY: REINSURANCE We learned from Exercise 2.1 that Pareto distribution provides a good

?t for the large observations in the data set. Hence, we should take advantage of this extra information extracted from the data set. One should

keep in mind, however, that the strictly Pareto ?ts the large observations

well but not necessarily for small data points. It indicates that the data

set can be modeled by a Pareto-type distribution, whose tail behaves like

a Pareto distribution. Without getting into too much technical details regarding Pareto-type distributions, we consider a scaled Pareto distribution

for our analysis. Suppose the large observations all came from a random

variable X, whose survival probability function is given by

F (x) = x

C ?? , x > C, for some large number C > 0. It means that X is C times a strictly Pareto

random variable. Hence it is easy to show that the mean excess function

of the scaled Pareto random variable remains the same as in Exercise 3.2.

This indicates that we can get around without estimating the constant C

when using estimates of the mean excess function.

Hence it remains to estimate the unknown parameter ?, whose reciprocal,

? := 1/?, is known as the extreme value index, in the extreme value theory.

A well-known estimator of the extreme value index, called the Hill estimator,

was proposed in Hill [1]. The intuition behind this estimator can be easily

explained. It follows from Exercise 1.6 and Exercise 3.2 that the mean excess

function of the logarithmic of the strict Pareto random variable (and that

of the scaled Pareto random variable) is 1/?. According to (1), it would be

natural to consider

?k,n :=

? 1

k k ln x(n?j+1) ? ln x(n?k) .

j=1 This is the Hill estimator. It enjoys a high degree of popularity thanks to

some nice theoretical properties, which we do not intend to discuss in this

class. For example, it has been proven that the Hill estimator is a consistent estimator for ?. (https://en.wikipedia.org/wiki/Consistent_

estimator) One should note that the Hill estimator is based on k-largest

ob...

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