# The generating function

Consider the function $g(z)=\displaystyle e^{\alpha (z-1)}$ where $\alpha$ is a positive constant. The following shows the derivatives of this function.

\displaystyle \begin{aligned}. \ \ \ \ \ \ &g(z)=e^{\alpha (z-1)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g(0)=e^{-\alpha} \\&\text{ } \\&g'(z)=e^{\alpha (z-1)} \ \alpha \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g'(0)=e^{-\alpha} \ \alpha \\&\text{ } \\&g^{(2)}(z)=e^{\alpha (z-1)} \ \alpha^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ g^{(2)}(0)=2! \ \frac{e^{-\alpha} \ \alpha^2}{2!} \\&\text{ } \\&g^{(3)}(z)=e^{\alpha (z-1)} \ \alpha^3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ g^{(3)}(0)=3! \ \frac{e^{-\alpha} \ \alpha^3}{3!} \\&\text{ } \\&\ \ \ \ \ \ \ \ \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdots \\&\text{ } \\&g^{(n)}(z)=e^{\alpha (z-1)} \ \alpha^n \ \ \ \ \ \ \ \ \ \ \ \ \ \ g^{(n)}(0)=n! \ \frac{e^{-\alpha} \ \alpha^n}{n!} \end{aligned}

Note that the derivative of $g(z)$ at each order is a multiple of a Poisson probability. Thus the Poisson distribution is coded by the function $g(z)=\displaystyle e^{\alpha (z-1)}$. Because of this reason, such a function is called a generating function (or probability generating function). This post discusses some basic facts about the generating function (gf) and its cousin, the moment generating function (mgf). One important characteristic is that these functions generate probabilities and moments. Another important characteristic is that there is a one-to-one correspondence between a probability distribution and its generating function and moment generating function, i.e. two random variables with different cumulative distribution functions cannot have the same gf or mgf. In some situations, this fact is useful in working with independent sum of random variables.

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The Generating Function
Suppose that $X$ is a random variable that takes only nonegative integer values with the probability function given by

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$(1) \ \ \ \ \ \ P(X=j)=a_j, \ \ \ \ j=0,1,2,\cdots$

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The idea of the generating function is that we use a power series to capture the entire probability distribution. The following defines the generating function that is associated with the above sequence $a_j$, .

$(2) \ \ \ \ \ \ g(z)=a_0+a_1 \ z+a_2 \ z^2+ \cdots=\sum \limits_{j=0}^\infty a_j \ z^j$

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Since the elements of the sequence $a_j$ are probabilities, we can also call $g(z)$ the generating function of the probability distribution defined by the sequence in $(1)$. The generating function $g(z)$ is defined wherever the power series converges. It is clear that at the minimum, the power series in $(2)$ converges for $\lvert z \lvert \le 1$.

We discuss the following three properties of generating functions:

1. The generating function completely determines the distribution.
2. The moments of the distribution can be derived from the derivatives of the generating function.
3. The generating function of a sum of independent random variables is the product of the individual generating functions.

The Poisson generating function at the beginning of the post is an example demonstrating property 1 (see Example 0 below for the derivation of the generating function). In some cases, the probability distribution of an independent sum can be deduced from the product of the individual generating functions. Some examples are given below.

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Generating Probabilities
We now discuss the property 1 indicated above. To see that $g(z)$ generates the probabilities, let’s look at the derivatives of $g(z)$:

\displaystyle \begin{aligned}(3) \ \ \ \ \ \ &g'(z)=a_1+2 a_2 \ z+3 a_3 \ z^2+\cdots=\sum \limits_{j=1}^\infty j a_j \ z^{j-1} \\&\text{ } \\&g^{(2)}(z)=2 a_2+6 a_3 \ z+ 12 a_4 \ z^2=\sum \limits_{j=2}^\infty j (j-1) a_j \ z^{j-2} \\&\text{ } \\&g^{(3)}(z)=6 a_3+ 24 a_4 \ z+60 a_5 \ z^2=\sum \limits_{j=3}^\infty j (j-1)(j-2) a_j \ z^{j-3} \\&\text{ } \\&\ \ \ \ \ \ \ \ \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdots \\&\text{ } \\&g^{(n)}(z)=\sum \limits_{j=n}^\infty j(j-1) \cdots (j-n+1) a_j \ z^{j-n}=\sum \limits_{j=n}^\infty \binom{j}{n} n! \ a_j \ z^{j-n} \end{aligned}

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By letting $z=0$ above, all the terms vanishes except for the constant term. We have:

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$(4) \ \ \ \ \ \ g^{(n)}(0)=n! \ a_n=n! \ P(X=n)$

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Thus the generating function is a compact way of encoding the probability distribution. The probability distribution determines the generating function as seen in $(2)$. On the other hand, $(3)$ and $(4)$ demonstrate that the generating function also determines the probability distribution.

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Generating Moments
The generating function also determines the moments (property 2 indicated above). For example, we have:

\displaystyle \begin{aligned}(5) \ \ \ \ \ \ &g'(1)=0 \ a_0+a_1+2 a_2+3 a_3+\cdots=\sum \limits_{j=0}^\infty j a_j=E(X) \\&\text{ } \\&g^{(2)}(1)=0 a_0 + 0 a_1+2 a_2+6 a_3+ 12 a_4+\cdots=\sum \limits_{j=0}^\infty j (j-1) a_j=E[X(X-1)] \\&\text{ } \\&E(X)=g'(1) \\&\text{ } \\&E(X^2)=g'(1)+g^{(2)}(1) \end{aligned}

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Note that $g^{(n)}(1)=E[X(X-1) \cdots (X-(n-1))]$. Thus the higher moment $E(X^n)$ can be expressed in terms of $g^{(n)}(1)$ and $g^{(k)}(1)$ where $k.
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More General Definitions
Note that the definition in $(2)$ can also be interpreted as the mathematical expectation of $z^X$, i.e., $g(z)=E(z^X)$. This provides a way to define the generating function for random variables that may take on values outside of the nonnegative integers. The following is a more general definition of the generating function of the random variable $X$, which is defined for all $z$ where the expectation exists.

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$(6) \ \ \ \ \ \ g(z)=E(z^X)$

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The Generating Function of Independent Sum
Let $X_1,X_2,\cdots,X_n$ be independent random variables with generating functions $g_1,g_2,\cdots,g_n$, respectively. Then the generating function of $X_1+X_2+\cdots+X_n$ is given by the product $g_1 \cdot g_2 \cdots g_n$.

Let $g(z)$ be the generating function of the independent sum $X_1+X_2+\cdots+X_n$. The following derives $g(z)$. Note that the general form of generating function $(6)$ is used.

\displaystyle \begin{aligned}(7) \ \ \ \ \ \ g(z)&=E(z^{X_1+\cdots+X_n}) \\&\text{ } \\&=E(z^{X_1} \cdots z^{X_n}) \\&\text{ } \\&=E(z^{X_1}) \cdots E(z^{X_n}) \\&\text{ } \\&=g_1(z) \cdots g_n(z) \end{aligned}

The probability distribution of a random variable is uniquely determined by its generating function. In particular, the generating function $g(z)$ of the independent sum $X_1+X_2+\cdots+X_n$ that is derived in $(7)$ is unique. So if the generating function is of a particular distribution, we can deduce that the distribution of the sum must be of the same distribution. See the examples below.

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Example 0
In this example, we derive the generating function of the Poisson distribution. Based on the definition, we have:

\displaystyle \begin{aligned}. \ \ \ \ \ \ g(z)&=\sum \limits_{j=0}^\infty \frac{e^{-\alpha} \alpha^j}{j!} \ z^j \\&\text{ } \\&=\sum \limits_{j=0}^\infty \frac{e^{-\alpha} (\alpha z)^j}{j!} \\&\text{ } \\&=\frac{e^{-\alpha}}{e^{- \alpha z}} \sum \limits_{j=0}^\infty \frac{e^{-\alpha z} (\alpha z)^j}{j!} \\&\text{ } \\&=e^{\alpha (z-1)} \end{aligned}

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Example 1
Suppose that $X_1,X_2,\cdots,X_n$ are independent random variables where each $X_i$ has a Bernoulli distribution with probability of success $p$. Let $q=1-p$. The following is the generating function for each $X_i$.

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$. \ \ \ \ \ \ g(z)=q+p z$

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Then the generating function of the sum $X=X_1+\cdots+X_n$ is $g(z)^n=(q+p z)^n$. The following is the binomial expansion:

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\displaystyle \begin{aligned}(8) \ \ \ \ \ \ g(z)^n&=(q+p z)^n \\&\text{ } \\&=\sum \limits_{j=0}^n \binom{n}{j} q^{n-j} \ p^j \ z^j \end{aligned}

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By definition $(2)$, the generating function of $X=X_1+\cdots+X_n$ is:

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$(9) \ \ \ \ \ \ g(z)^n=\sum \limits_{j=0}^\infty P(X=j) \ z^j$

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Comparing $(8)$ and $(9)$, we have

$\displaystyle (10) \ \ \ \ \ \ P(X=j)=\left\{\begin{matrix}\displaystyle \binom{n}{j} p^j \ q^{n-j}&\ 0 \le j \le n\\{0}&\ j>n \end{matrix}\right.$

The probability distribution indicated by $(8)$ and $(10)$ is that of a binomial distribution. Since the probability distribution of a random variable is uniquely determined by its generating function, the independent sum of Bernoulli distributions must ave a Binomial distribution.

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Example 2
Suppose that $X_1,X_2,\cdots,X_n$ are independent and have Poisson distributions with parameters $\alpha_1,\alpha_2,\cdots,\alpha_n$, respectively. Then the independent sum $X=X_1+\cdots+X_n$ has a Poisson distribution with parameter $\alpha=\alpha_1+\cdots+\alpha_n$.

Let $g(z)$ be the generating function of $X=X_1+\cdots+X_n$. For each $i$, the generating function of $X_i$ is $g_i(z)=e^{\alpha_i (z-1)}$. The key to the proof is that the product of the $g_i$ has the same general form as the individual $g_i$.

\displaystyle \begin{aligned}(11) \ \ \ \ \ \ g(z)&=g_1(z) \cdots g_n(z) \\&\text{ } \\&=e^{\alpha_1 (z-1)} \cdots e^{\alpha_n (z-1)} \\&\text{ } \\&=e^{(\alpha_1+\cdots+\alpha_n)(z-1)} \end{aligned}

The generating function in $(11)$ is that of a Poisson distribution with mean $\alpha=\alpha_1+\cdots+\alpha_n$. Since the generating function uniquely determines the distribution, we can deduce that the sum $X=X_1+\cdots+X_n$ has a Poisson distribution with parameter $\alpha=\alpha_1+\cdots+\alpha_n$.

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Example 3
In rolling a fair die, let $X$ be the number shown on the up face. The associated generating function is:

$\displaystyle. \ \ \ \ \ \ g(z)=\frac{1}{6}(z+z^2+z^3+z^4+z^5+z^6)=\frac{z(1-z^6)}{6(1-z)}$

The generating function can be further reduced as:

\displaystyle \begin{aligned}. \ \ \ \ \ \ g(z)&=\frac{z(1-z^6)}{6(1-z)} \\&\text{ } \\&=\frac{z(1-z^3)(1+z^3)}{6(1-z)} \\&\text{ } \\&=\frac{z(1-z)(1+z+z^2)(1+z^3)}{6(1-z)} \\&\text{ } \\&=\frac{z(1+z+z^2)(1+z^3)}{6} \end{aligned}

Suppose that we roll the fair dice 4 times. Let $W$ be the sum of the 4 rolls. Then the generating function of $Z$ is

$\displaystyle. \ \ \ \ \ \ g(z)^4=\frac{z^4 (1+z^3)^4 (1+z+z^2)^4}{6^4}$

The random variable $W$ ranges from 4 to 24. Thus the probability function ranges from $P(W=4)$ to $P(W=24)$. To find these probabilities, we simply need to decode the generating function $g(z)^4$. For example, to find $P(W=12)$, we need to find the coefficient of the term $z^{12}$ in the polynomial $g(z)^4$. To help this decoding, we can expand two of the polynomials in $g(z)^4$.

\displaystyle \begin{aligned}. \ \ \ \ \ \ g(z)^4&=\frac{z^4 (1+z^3)^4 (1+z+z^2)^4}{6^4} \\&\text{ } \\&=\frac{z^4 \times A \times B}{6^4} \\&\text{ } \\&A=(1+z^3)^4=1+4z^3+6z^6+4z^9+z^{12} \\&\text{ } \\&B=(1+z+z^2)^4=1+4z+10z^2+16z^3+19z^4+16z^5+10z^6+4z^7+z^8 \end{aligned}

Based on the above polynomials, there are three ways of forming $z^{12}$. They are: $(z^4 \times 1 \times z^8)$, $(z^4 \times 4z^3 \times 16z^5)$, $(z^4 \times 6z^6 \times 10z^2)$. Thus we have:

$\displaystyle. \ \ \ \ \ \ P(W=12)=\frac{1}{6^4}(1+4 \times 16+6 \times 10)=\frac{125}{6^4}$

To find the other probabilities, we can follow the same decoding process.

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Remark
The probability distribution of a random variable is uniquely determined by its generating function. This fundamental property is useful in determining the distribution of an independent sum. The generating function of the independent sum is simply the product of the individual generating functions. If the product is of a certain distributional form (as in Example 1 and Example 2), then we can deduce that the sum must be of the same distribution.

We can also decode the product of generating functions to obtain the probability function of the independent sum (as in Example 3). The method in Example 3 is quite tedious. But one advantage is that it is a “machine process”, a pretty fool proof process that can be performed mechanically.

The machine process is this: Code the individual probability distribution in a generating function $g(z)$. Then raise it to $n$. After performing some manipulation to $g(z)^n$, decode the probabilities from $g(z)^n$.

As long as we can perform the algebraic manipulation carefully and correctly, this process will be sure to provide the probability distribution of an independent sum.

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The Moment Generating Function
The moment generating function of a random variable $X$ is $M_X(t)=E(e^{tX})$ on all real numbers $t$ for which the expected value exists. The moments can be computed more directly using an mgf. From the theory of mathematical analysis, it can be shown that if $M_X(t)$ exists on some interval $-a, then the derivatives of $M_X(t)$ of all orders exist at $t=0$. Furthermore, it can be show that $E(X^n)=M_X^{(n)}(0)$.

Suppose that $g(z)$ is the generating function of a random variable. The following relates the generating function and the moment generating function.

\displaystyle \begin{aligned}. \ \ \ \ \ \ &M_X(t)=g(e^t) \\&\text{ } \\&g(z)=M_X(ln z) \end{aligned}

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Reference

1. Feller W. An Introduction to Probability Theory and Its Applications, Third Edition, John Wiley & Sons, New York, 1968

# The Pareto distribution

This post takes a closer look at the Pareto distribution. A previous post demonstrates that the Pareto distribution is a mixture of exponential distributions with Gamma mixing weights. We now elaborate more on this point. Through looking at various properties of the Pareto distribution, we also demonstrate that the Pareto distribution is a heavy tailed distribution. In insurance applications, heavy-tailed distributions are essential tools for modeling extreme loss, especially for the more risky types of insurance such as medical malpractice insurance. In financial applications, the study of heavy-tailed distributions provides information about the potential for financial fiasco or financial ruin. The Pareto distribution is a great way to open up a discussion on heavy-tailed distribution.

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Update (11/12/2017). This blog post introduces a catalog of many other parametric severity models in addition to Pareto distribution. The link to the catalog is found in that blog post. To go there directly, this is the link.

Update (10/29/2017). This blog post has updated information on Pareto distribution. It also has links to more detailed contents on Pareto distribution in two companion blogs. These links are also given here: more detailed post on Pareto, Pareto Type I and Type II and practice problems on Pareto.

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The continuous random variable $X$ with positive support is said to have the Pareto distribution if its probability density function is given by

$\displaystyle f_X(x)=\frac{\beta \ \alpha^\beta}{(x+\alpha)^{\beta+1}} \ \ \ \ \ x>0$

where $\alpha>0$ and $\beta>0$ are constant. The constant $\alpha$ is the scale parameter and $\beta$ is the shape parameter. The following lists several other distributional quantities of the Pareto distribution, which will be used in the discussion below.

$\displaystyle S_X(x)=\frac{\alpha^\beta}{(x+\alpha)^\beta}=\biggl(\frac{\alpha}{x+\alpha}\biggr)^\beta \ \ \ \ \ \ \ \ \ \text{survival function}$

$\displaystyle F_X(x)=1-\biggl(\frac{\alpha}{x+\alpha}\biggr)^\beta \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{distribution function}$

$\displaystyle E(X)=\frac{\alpha}{\beta-1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{mean},\beta>1$

$\displaystyle E(X^2)=\frac{2 \alpha^2}{(\beta-1)(\beta-2)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{second momemt},\beta>2$

$\displaystyle Var(X)=\frac{\alpha^2 \beta}{(\beta-1)^2(\beta-2)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{variance},\beta>2$

$\displaystyle E(X^k)=\frac{k! \alpha^k}{(\beta-1) \cdots (\beta-k)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{higher moments},\beta>k, \text{ k positive integer}$

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The Pareto Distribution as a Mixture
The Pareto pdf indicated above can be obtained by mixing exponential distributions using Gamma distributions as weights. Suppose that $X$ follows an exponential distribution (conditional on a parameter value $\theta$). The following is the conditional pdf of $X$.

$\displaystyle f_{X \lvert \Theta}(x \lvert \theta)=\theta e^{-\theta x} \ \ \ x>0$

There is uncertainty in the parameter, which can be viewed as a random variable $\Theta$. Suppose that $\Theta$ follows a Gamma distribution with scale parameter $\alpha$ and shape parameter $\beta$. The following is the pdf of $\Theta$.

$\displaystyle f_{\Theta}(\theta)=\frac{\alpha^\beta}{\Gamma(\beta)} \ \theta^{\beta-1} \ e^{-\alpha \theta} \ \ \ \theta>0$

The unconditional pdf of $X$ is the weighted average of the conditional pdfs with the Gamma pdf as weight.

\displaystyle \begin{aligned}f_X(x)&=\int_0^{\infty} f_{X \lvert \Theta}(x \lvert \theta) \ f_\Theta(\theta) \ d \theta \\&=\int_0^{\infty} \biggl[\theta \ e^{-\theta x}\biggr] \ \biggl[\frac{\alpha^\beta}{\Gamma(\beta)} \ \theta^{\beta-1} \ e^{-\alpha \theta}\biggr] \ d \theta \\&=\int_0^{\infty} \frac{\alpha^\beta}{\Gamma(\beta)} \ \theta^\beta \ e^{-\theta(x+\alpha)} \ d \theta \\&=\frac{\alpha^\beta}{\Gamma(\beta)} \frac{\Gamma(\beta+1)}{(x+\alpha)^{\beta+1}} \int_0^{\infty} \frac{(x+\alpha)^{\beta+1}}{\Gamma(\beta+1)} \ \theta^{\beta+1-1} \ e^{-\theta(x+\alpha)} \ d \theta \\&=\frac{\beta \ \alpha^\beta}{(x+\alpha)^{\beta+1}} \end{aligned}

In the following discussion, $X$ will denote the Pareto distribution as defined above. As will be shown below, the exponential distribution is considered a light tailed distribution. Yet mixing exponentials produces the heavy tailed Pareto distribution. Mixture distributions tend to heavy tailed (see [1]). The Pareto distribution is a handy example.

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The Tail Weight of the Pareto Distribution
When a distribution significantly puts more probability on larger values, the distribution is said to be a heavy tailed distribution (or said to have a larger tail weight). According to [1], there are four ways to look for indication that a distribution is heavy tailed.

1. Existence of moments.
2. Speed of decay of the survival function to zero.
3. Hazard rate function.
4. Mean excess loss function.

Existence of moments
Note that the existence of the Pareto higher moments $E(X^k)$ is capped by the shape parameter $\beta$. In particular, the mean $E(X)=\frac{\alpha}{\beta-1}$ does not exist for $\beta \le 1$. If the Pareto distribution is to model a random loss, and if the mean is infinite (when $\beta=1$), the risk is uninsurable! On the other hand, when $\beta=2$, the Pareto variance does not exist. This shows that for a heavy tailed distribution, the variance may not be a good measure of risk.

For a given random variable $Z$, the existence of all moments $E(Z^k)$, for all positive integers $k$, indicates with a light (right) tail for the distribution of $Z$. The existence of positive moments exists only up to a certain value of a positive integer $k$ is an indication that the distribution has a heavy right tail. In contrast, the exponential distribution and the Gamma distribution are considered to have light tails since all moments exist.

The speed of decay of the survival function
The survival function $S_X(x)=P(X>x)$ captures the probability of the tail of a distribution. If a distribution whose survival function decays slowly to zero (equivalently the cdf goes slowly to one), it is another indication that the distribution is heavy tailed.

The following is a comparison of a Pareto survival function and an exponential survival function. The Pareto survival function has parameters ($\alpha=2$ and $\beta=2$). The two survival functions are set to have the same 75th percentile ($x=2$).

$\displaystyle \begin{pmatrix} \text{x}&\text{Pareto }S_X(x)&\text{Exponential }S_Y(x)&\displaystyle \frac{S_X(x)}{S_Y(x)} \\\text{ }&\text{ }&\text{ }&\text{ } \\{2}&0.25&0.25&1 \\{10}&0.027777778&0.000976563&28 \\{20}&0.008264463&9.54 \times 10^{-7}&8666 \\{30}&0.00390625&9.31 \times 10^{-10}&4194304 \\{40}&0.002267574&9.09 \times 10^{-13}&2.49 \times 10^{9} \\{60}&0.001040583&8.67 \times 10^{-19}&1.20 \times 10^{15} \\{80}&0.000594884&8.27 \times 10^{-25}&7.19 \times 10^{20} \\{100}&0.000384468&7.89 \times 10^{-31}&4.87 \times 10^{26} \\{120}&0.000268745&7.52 \times 10^{-37}&3.57 \times 10^{32} \\{140}&0.000198373&7.17 \times 10^{-43}&2.76 \times 10^{38} \\{160}&0.000152416&6.84 \times 10^{-49}&2.23 \times 10^{44} \\{180}&0.000120758&6.53 \times 10^{-55}&1.85 \times 10^{50} \end{pmatrix}$

Note that at the large values, the Pareto right tails retain much more probability. This is also confirmed by the ratio of the two survival functions, with the ratio approaching infinity. If a random loss is a heavy tailed phenomenon that is described by the above Pareto survival function ($\alpha=2$ and $\beta=2$), then the above exponential survival function is woefully inadequate as a model for this phenomenon even though it may be a good model for describing the loss up to the 75th percentile. It is the large right tail that is problematic (and catastrophic)!

Since the Pareto survival function and the exponential survival function have closed forms, We can also look at their ratio.

$\displaystyle \frac{\text{pareto survival}}{\text{exponential survival}}=\frac{\displaystyle \frac{\alpha^\beta}{(x+\alpha)^\beta}}{e^{-\lambda x}}=\frac{\alpha^\beta e^{\lambda x}}{(x+\alpha)^\beta} \longrightarrow \infty \ \text{ as } x \longrightarrow \infty$

In the above ratio, the numerator has an exponential function with a positive quantity in the exponent, while the denominator has a polynomial in $x$. This ratio goes to infinity as $x \rightarrow \infty$.

In general, whenever the ratio of two survival functions diverges to infinity, it is an indication that the distribution in the numerator of the ratio has a heavier tail. When the ratio goes to infinity, the survival function in the numerator is said to decay slowly to zero as compared to the denominator. We have the same conclusion in comparing the Pareto distribution and the Gamma distribution, that the Pareto is heavier in the tails. In comparing the tail weight, it is equivalent to consider the ratio of density functions (due to the L’Hopital’s rule).

$\displaystyle \lim_{x \rightarrow \infty} \frac{S_1(x)}{S_2(x)}=\lim_{x \rightarrow \infty} \frac{S_1^{'}(x)}{S_2^{'}(x)}=\lim_{x \rightarrow \infty} \frac{f_1(x)}{f_2(x)}$

The Hazard Rate Function
The hazard rate function $h_X(x)$ of a random variable $X$ is defined as the ratio of the density function and the survival function.

$\displaystyle h_X(x)=\frac{f_X(x)}{S_X(s)}$

The hazard rate is called the force of mortality in a life contingency context and can be interpreted as the rate that a person aged $x$ will die in the next instant. The hazard rate is called the failure rate in reliability theory and can be interpreted as the rate that a machine will fail at the next instant given that it has been functioning for $x$ units of time. The following is the hazard rate function of the Pareto distribution.

\displaystyle \begin{aligned}h_X(x)&=\frac{f_X(s)}{S_X(x)} \\&=\frac{\beta}{x+\alpha} \end{aligned}

The interesting point is that the Pareto hazard rate function is an decreasing function in $x$. Another indication of heavy tail weight is that the distribution has a decreasing hazard rate function. One key characteristic of hazard rate function is that it can generate the survival function.

$\displaystyle S_X(x)=e^{\displaystyle -\int_0^x h_X(t) \ dt}$

Thus if the hazard rate function is decreasing in $x$, then the survival function will decay more slowly to zero. To see this, let $H_X(x)=\int_0^x h_X(t) \ dt$, which is called the cumulative hazard rate function. As indicated above, the survival function can be generated by $e^{-H_X(x)}$. If $h_X(x)$ is decreasing in $x$, $H_X(x)$ is smaller than $H_Y(x)$ where $h_Y(x)$ is constant in $x$ or increasing in $x$. Consequently $e^{-H_X(x)}$ is decaying to zero much more slowly than $e^{-H_Y(x)}$.

In contrast, the exponential distribution has a constant hazard rate function, making it a medium tailed distribution. As explained above, any distribution having an increasing hazard rate function is a light tailed distribution.

The Mean Excess Loss Function
Suppose that a property owner is exposed to a random loss $Y$. The property owner buys an insurance policy with a deductible $d$ such that the insurer will pay a claim in the amount of $Y-d$ if a loss occurs with $Y>d$. The insuerer will pay nothing if the loss is below the deductible. Whenever a loss is above $d$, what is the average claim the insurer will have to pay? This is one way to look at mean excess loss function, which represents the expected excess loss over a threshold conditional on the event that the threshold has been exceeded.

Given a loss variable $Y$ and given a deductible $d>0$, the mean excess loss function is $e_Y(d)=E(Y-d \lvert X>d)$. For a continuous random variable, it is computed by

$\displaystyle e_Y(d)=\frac{\int_d^{\infty} (y-d) \ f_Y(y) \ dy}{S_Y(d)}$

Applying the technique of integration by parts produces the following formula:

$\displaystyle e_Y(d)=\frac{\int_d^{\infty} S_Y(y) \ dy}{S_Y(d)}$

It turns out that the mean excess loss function is one more way to examine the tail property of a distribution. The following is the mean excess loss function of the Pareto distribution:

$\displaystyle e_X(d)=\frac{d+\alpha}{\beta-1}=\frac{1}{\beta-1} \ d + \frac{\alpha}{\beta-1}$

Note that the Pareto mean excess loss function is a linear increasing function of the deductible $d$. This means that the larger the deductible, the larger the expected claim if such a large loss occurs! If a random loss is modeled by such a distribution, it is a catastrophic risk situation. In general, an increasing mean excess loss function is an indication of a heavy tailed distribution. On the other hand, a decreasing mean excess loss function indicates a light tailed distribution. The exponential distribution has a constant mean excess loss function and is considered a medium tailed distribution.

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The Pareto distribution has many economic applications. Since it is a heavy tailed distribution, it is a good candidate for modeling income above a theoretical value and the distribution of insurance claims above a threshold value.

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Reference

1. Klugman S.A., Panjer H. H., Wilmot G. E. Loss Models, From Data to Decisions, Second Edition., Wiley-Interscience, a John Wiley & Sons, Inc., New York, 2004