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 nx 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 (, 2 ) is a normal random variable with mean 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 denition 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 nh P(Yn = y) = p q , h := npqy + np, h where y = (k np)/ npq and k = 0, 1, , n. [HINT: Use the denition 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) = et , t 0, where E(Xi ) = 1/ is the mean of waiting time. If we redene 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 xn1 ex . (n 1)! Use this fact to show that the exact probability density function of Yn nn1/2 (n 1)! y 1+ n n1 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 dierent 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) = x1 , 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 ex = 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 and 2 ) is heavy-tailed. [HINT: Apply L'Hospitl'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 i1 i Qn (p) = x(i) , for