{"id":5904,"date":"2016-12-27T08:41:01","date_gmt":"2016-12-27T14:41:01","guid":{"rendered":"http:\/\/www.ssc.wisc.edu\/~jfrees\/?page_id=5904"},"modified":"2016-12-27T09:40:53","modified_gmt":"2016-12-27T15:40:53","slug":"test-of-markdown","status":"publish","type":"page","link":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/testing-wordpress-features\/test-of-markdown\/","title":{"rendered":"Test of Markdown"},"content":{"rendered":"

Chapter 3: Modeling loss severity<\/strong><\/p>\n

Zeinab Amin<\/strong><\/p>\n

September 15, 2016<\/strong><\/p>\n

3.1 Chapter preview<\/strong><\/p>\n

The traditional loss distribution approach to modeling aggregate losses
\nstarts by separately fitting a frequency distribution to the number of
\nlosses and a severity distribution to the size of losses. The estimated
\naggregate loss distribution combines the loss frequency distribution and
\nthe loss severity distribution by convolution. Discrete distributions
\noften referred to as counting or frequency distributions were used in
\nChapter 2 to describe the number of events such as number of accidents
\nto the driver or number of claims to the insurer. Lifetimes, asset
\nvalues, losses and claim sizes are usually modeled as continuous random
\nvariables and as such are modeled using continuous distributions, often
\nreferred to as loss or severity distributions. Mixture distributions are
\nused to model phenomenon investigated in a heterogeneous population,
\nsuch as modelling more than one type of claims in liability insurance
\n(small frequent claims and large relatively rare claims). In this
\nchapter we explore the use of continuous as well as mixture
\ndistributions to model the random size of loss. We present key
\nattributes that characterize continuous models and means of creating new
\ndistributions from existing ones. In this chapter we explore the effect
\nof\u00a0coverage modifications, which change the conditions that trigger a
\npayment, such as applying deductibles, limits, or adjusting for
\ninflation, on the distribution of individual loss amounts.<\/p>\n

3.2 Continuous distributions for loss severity<\/strong><\/p>\n

3.2.1 Basic distributional quantities<\/strong><\/p>\n

In this section we calculate the basic distributional quantities:
\nmoments, percentiles and generating functions.<\/p>\n

Moments<\/strong><\/p>\n

Let \(X\) be a continuous random variable with probability density
\nfunction (f_{X}(x)), the k<\/em>-th raw moment of $X$, denoted by
\n$$\\mu_{k}^{‘}$$, is the expected value of the k<\/em>-th power of $X$,
\nprovided it exists. The first raw moment (\\mu_{1}^{‘}) is the mean of
\n$X$ usually denoted by (\\mu ). The formula for $\\mu_{k}^{‘}$ is given as<\/p>\n

$$\\mu_{k}^{‘} = E\\left( X^{k} \\right) = \\int_{0}^{\\infty}{x^{k}f_{X}\\left( x \\right)\\text{dx}}$$.<\/p>\n

The support of the random variable $X$ is assumed to be nonnegative
\nsince actuarial phenomena are rarely negative.<\/p>\n

The k<\/em>-th central moment of $X$, denoted by $\\mu_{k}$, is the expected
\nvalue of the k<\/em>-th power of the deviation of $X$ from its mean $\\mu$.
\nThe formula for $\\mu_{k}$ is given as<\/p>\n

$\\mu_{k} = E\\left\\lbrack {(X – \\mu)}^{k} \\right\\rbrack = \\int_{0}^{\\infty}{\\left( x – \\mu \\right)^{k}f_{X}\\left( x \\right)\\text{dx}}$.<\/p>\n

The second central moment $\\mu_{2}^{‘}$ defines the variance of $X$,
\ndenoted by $\\sigma^{2}$. The square root of the variance is the standard
\ndeviation $\\sigma$. A further characterization of the shape of the
\ndistribution includes its degree of symmetry as well as its flatness
\ncompared to the standard normal distribution. The ratio of the third
\ncentral moment to the cube of the standard deviation defines the
\ncoefficient of skewness which is a measure of symmetry. A positive
\ncoefficient of skewness indicates that the distribution is skewed to the
\nright (positively skewed). The ratio of the fourth central moment to the
\nfourth power of the standard deviation defines the coefficient of
\nkurtosis which is a measure of the heaviness of the tail distribution
\nrelative to the standard normal curve. High kurtosis is associated with
\na heavy tailed distribution.<\/p>\n

Example 3.1 (SOA)<\/strong><\/p>\n

$$X$$ has a gamma distribution with mean 8 and skewness 1. Find the variance of ( X ).<\/p>\n

Solution<\/strong><\/p>\n

The probability density function of $X$ is given by
\n$f_{X}\\left( x \\right) = \\frac{\\left( \\frac{x}{\\theta} \\right)^{\\alpha}}{x\\Gamma\\left( \\alpha \\right)}exp – \\left( \\frac{x}{\\theta} \\right)$
\nfor $x > 0$*. *<\/p>\n

If $\\alpha$ is an integer, then
\n$\\mu_{k}^{‘} = E\\left( X^{k} \\right) = \\int_{0}^{\\infty}{\\frac{1}{\\left( \\alpha – 1 \\right)!\\theta^{\\alpha}}x^{k + \\alpha – 1}e^{- \\frac{x}{\\theta}}\\text{dx}} = \\frac{\\left( k + \\alpha – 1 \\right)!}{\\left( \\alpha – 1 \\right)!}\\theta^{k}$.<\/p>\n

Thus, $\\mu_{1}^{‘} = E\\left( X \\right) = \\alpha\\theta$,
\n$\\mu_{2}^{‘} = E\\left( X^{2} \\right) = \\left( \\alpha + 1 \\right)\\alpha\\theta^{2}$,
\n$\\mu_{3}^{‘} = E\\left( X^{3} \\right) = \\left( \\alpha + 2 \\right)\\left( \\alpha + 1 \\right)\\alpha\\theta^{3}$<\/p>\n

and $V\\left( X \\right) = \\alpha\\theta^{2}$.<\/p>\n

Skewness$\\ = \\frac{E\\left\\lbrack {(X – \\mu_{1}^{‘})}^{3} \\right\\rbrack}{{V\\left( X \\right)}^{\\frac{3}{2}}} = \\frac{\\mu_{3}^{‘} – 3\\mu_{2}^{‘}\\mu_{1}^{‘} + 2{\\mu_{1}^{‘}}^{3}}{{V\\left( X \\right)}^{\\frac{3}{2}}} = \\frac{\\left( \\alpha + 2 \\right)\\left( \\alpha + 1 \\right)\\alpha\\theta^{3} – 3\\left( \\alpha + 1 \\right)\\alpha^{2}\\theta^{3} + 2\\alpha^{3}\\theta^{3}}{\\left( \\alpha\\theta^{2} \\right)^{\\frac{3}{2}}} = \\frac{2}{\\alpha^{\\frac{1}{2}}} = 1$.<\/p>\n

Hence, $\\alpha = 4$. Since, $E\\left( X \\right) = \\alpha\\theta = 8$, then
\n$\\theta = 2$ and $V\\left( X \\right) = \\alpha\\theta^{2} = 16$.<\/p>\n

Quantiles<\/strong><\/p>\n

Percentiles can also be used to describe the characteristics of the
\ndistribution of $X$. The 100pth<\/em> percentile of the distribution of $X$,
\ndenoted by $\\pi_{p}$, is the value of $X$ which satisfies<\/p>\n

$F_{X}\\left( {\\pi_{p}}^{-} \\right) \\leq p \\leq F\\left( \\pi_{p} \\right)$,
\nfor $0 \\leq p \\leq 1$.<\/p>\n

The 50-th percentile or the middle point of the distribution,
\n$\\pi_{0.5}$, is the median. Unlike discrete random variables,
\npercentiles of continuous variables are unique.<\/p>\n

Example 3.2 (SOA)<\/strong><\/p>\n

Let $X$ be a continuous random variable with density function
\n$f_{X}\\left( x \\right) = \\theta e^{- \\theta x}$, for $x > 0$ and 0
\nelsewhere. If the median of this distribution is $\\frac{1}{3}$, find
\n$\\theta$.<\/p>\n

Solution<\/strong><\/p>\n

$F_{X}\\left( x \\right) = 1 – e^{- \\theta x}$. Then,
\n$F_{X}\\left( \\pi_{0.5} \\right) = 1 – e^{- \\theta\\pi_{0.5}} = 0.5$. Thus,
\n$1 – e^{- \\frac{\\theta}{3}} = 0.5$ and $\\theta = 3ln2$.<\/p>\n

The moment generating function<\/strong><\/p>\n

The moment generating function, denoted by $M_{X}\\left( t \\right)$
\nuniquely characterizes the distribution of $X$. While it is possible for
\ntwo different distributions to have the same moments and yet still
\ndiffer, this is not the case with the moment generating function. That
\nis, if two random variables have the same moment generating function,
\nthen they have the same distribution. The moment generating is a real
\nfunction whose k<\/em>-th derivative at zero is equal to the k<\/em>-th raw
\nmoment\u00a0of $X$.\u00a0The moment generating function is given by<\/p>\n

$M_{X}\\left( t \\right) = E\\left( e^{\\text{tX}} \\right) = \\int_{0}^{\\infty}{e^{\\text{tx}}f_{X}\\left( x \\right)\\text{dx}}$,<\/p>\n

for all $t$ for which the expected value exists.<\/p>\n

Example 3.3 (SOA)<\/strong><\/p>\n

The random variable $X$ has an exponential distribution with mean
\n$\\frac{1}{b}$. It is found that $M_{X}\\left( – b^{2} \\right) = 0.2$.
\nFind $b$.<\/p>\n

Solution<\/strong><\/p>\n

$$M_{X}\\left( t \\right) = E\\left( e^{\\text{tX}} \\right) = \\int_{0}^{\\infty}{e^{\\text{tx}}be^{- bx}\\text{dx}} = \\int_{0}^{\\infty}{be^{- x\\left( b – t \\right)}\\text{dx}} = \\frac{b}{\\left( b – t \\right)}.$$<\/p>\n

Then,
\n$M_{X}\\left( – b^{2} \\right) = \\frac{b}{\\left( b + b^{2} \\right)} = \\frac{1}{\\left( 1 + b \\right)} = 0.2$.
\nThus, $b = 4$.<\/p>\n

**Example 3.4 **<\/p>\n

Let $X_{1}$, $X_{2}$, \u2026, $X_{n}$ be independent
\n$\\text{Ga}\\left( \\alpha_{i},\\theta \\right)$ random variables.<\/p>\n

Find the distribution of $S = \\sum_{i = 1}^{n}X_{i}$.<\/p>\n

Solution<\/strong><\/p>\n

The moment generating function of $S$ is<\/p>\n

$M_{S}\\left( t \\right) = \\text{\\ E}\\left( e^{\\text{tS}} \\right) = E\\left( e^{t\\sum_{i = 1}^{n}X_{i}} \\right) = E\\left( \\prod_{i = 1}^{n}e^{tX_{i}} \\right) = \\prod_{i = 1}^{n}{E\\left( e^{tX_{i}} \\right) = \\prod_{i = 1}^{n}{M_{X_{i}}\\left( t \\right)}}$.<\/p>\n

The moment generating function of $X_{i}$ is
\n$M_{X_{i}}\\left( t \\right) = \\left( 1 – \\theta t \\right)^{- \\alpha_{i}}$.<\/p>\n

Then,
\n$M_{S}\\left( t \\right) = \\prod_{i = 1}^{n}\\left( 1 – \\theta t \\right)^{- \\alpha_{i}} = \\left( 1 – \\theta t \\right)^{- \\sum_{i = 1}^{n}\\alpha_{i}}$,
\nindicating that
\n$S\\sim Ga\\left( \\sum_{i = 1}^{n}\\alpha_{i},\\theta \\right)$.<\/p>\n

By finding the first and second derivatives of $M_{S}\\left( t \\right)$
\nat zero, we can show that
\n$E\\left( S \\right) = \\left. \\ \\frac{\\partial M_{S}\\left( t \\right)}{\\partial t} \\right|{t = 0} = \\alpha\\theta$
\nwhere $\\alpha = \\sum<\/em>{i = 1}^{n}\\alpha_{i}$, and
\n$E\\left( S^{2} \\right) = \\left. \\ \\frac{\\partial^{2}M_{S}\\left( t \\right)}{\\partial t^{2}} \\right|_{t = 0} = \\left( \\alpha + 1 \\right)\\alpha\\theta^{2}$.<\/p>\n

Hence, $V\\left( S \\right) = \\alpha\\theta^{2}$.<\/p>\n

Probability generating function<\/strong><\/p>\n

The probability generating function, denoted by $P_{X}\\left( z \\right)$,
\nalso uniquely characterizes the distribution of $X$. It is defined as<\/p>\n

$P_{X}\\left( z \\right) = E\\left( z^{X} \\right) = \\int_{0}^{\\infty}{z^{x}f_{X}\\left( x \\right)\\text{dx}}$,<\/p>\n

for all $z$ for which the expected value exists.<\/p>\n

We can also use the probability generating function to generate moments
\nof $X$. By taking the k<\/em>-th derivative of $P_{X}\\left( z \\right)$ with
\nrespect to $z$ and evaluate it at $z\\ = \\ 1$, we get<\/p>\n

$E\\left\\lbrack X\\left( X – 1 \\right)\\ldots\\left( X – k + 1 \\right) \\right\\rbrack$.<\/p>\n

3.2.2 Continuous distributions for modeling loss severity<\/strong><\/p>\n

In this section we explain the characteristics of distributions suitable
\nfor modeling severity of losses, including gamma, Pareto, Weibull and
\ngeneralized beta distribution of the second kind. Applications for which
\neach distribution may be used are identified.<\/p>\n

The Gamma distribution<\/strong><\/p>\n

The gamma distribution is commonly used in modeling claim severity. The
\ntraditional approach in modelling losses is to fit separate models for
\nclaim frequency and claim severity. When frequency and severity are
\nmodeled separately it is common for actuaries to use the Poisson
\ndistribution for claim count and the gamma distribution to model
\nseverity. An alternative approach for modelling losses that has recently
\ngained popularity is to create a single model for pure premium (average
\nclaim cost) that will be described in Chapter 4.<\/p>\n

The continuous variable $X$ is said to have the gamma distribution with
\nshape parameter $\\alpha$ and scale parameter $\\theta$ if its probability
\ndensity function is given by<\/p>\n

$f_{X}\\left( x \\right) = \\frac{\\left( \\frac{x}{\\theta} \\right)^{\\alpha}}{x\\Gamma\\left( \\alpha \\right)}exp – \\left( \\frac{x}{\\theta} \\right)$
\n$x > 0,\\ \\alpha > 0,\\ \\theta > 0$.<\/em><\/p>\n

Figures 1 and 2 demonstrate the\u00a0effect\u00a0of the\u00a0shape and\u00a0scale
\nparameters\u00a0on the gamma density function.<\/p>\n

Fig. (1) Gamma density functions with varying scale parameters<\/p>\n

\"\"{width=”3.8915146544181978in”
\nheight=”2.2083333333333335in”}<\/p>\n

Fig. (2) Gamma density functions with varying shape parameters<\/p>\n

\"\"{width=”3.8690748031496063in”
\nheight=”2.138888888888889in”}<\/p>\n

When $\\alpha = 1$ the gamma reduces to an exponential distribution and
\nwhen $\\alpha = \\frac{n}{2}$ and $\\theta = 2$ the gamma reduces to a
\nchi-square distribution with $n$ degrees of freedom. As we will see in
\nSection 3.6.2, the chi-square distribution is used extensively in
\nstatistical hypothesis testing.<\/p>\n

The distribution function of the gamma model is the incomplete gamma
\nfunction, denoted by $\\Gamma\\left( \\frac{\\alpha;x}{\\theta} \\right)$, and
\ndefined as<\/p>\n

$F_{X}\\left( x \\right) = \\Gamma\\left( \\frac{\\alpha;x}{\\theta} \\right) = \\frac{1}{\\Gamma\\left( \\alpha \\right)}\\int_{0}^{\\frac{x}{\\theta}}t^{\\alpha – 1}e^{- t}\\text{dt}$
\n$\\alpha > 0,\\ \\theta > 0$.<\/em><\/p>\n

The $k$-th moment of the gamma distributed random variable for any
\npositive $k$ is given by<\/p>\n

$E\\left( X^{k} \\right) = \\frac{\\theta^{k}\\Gamma\\left( \\alpha + k \\right)}{\\Gamma\\left( \\alpha \\right)}$,
\n$k > 0$.<\/em><\/p>\n

The mean and variance are given by $E\\left( X \\right) = \\alpha\\theta$
\nand $V\\left( X \\right) = \\alpha\\theta^{2}$, respectively.<\/p>\n

Since all moments exist for any positive $k$, the gamma distribution is
\nconsidered a light tailed distribution, which may not be suitable for
\nmodeling risky assets\u00a0as it will not provide a realistic assessment of
\nthe likelihood of severe losses.\u00a0<\/p>\n

The Pareto distribution<\/strong><\/p>\n

The Pareto distribution, named after the Italian economist Vilfredo
\nPareto (1843\u20131923), has many economic and financial applications. It is
\na left skewed and heavy-tailed distribution which makes it suitable for
\nmodeling income, high-risk insurance claims and severity of large
\ncasualty losses. The survival function of the Pareto distribution which
\ndecays slowly to zero was first used to describe the distribution of
\nincome where a small percentage of the population holds a large
\nproportion of the total wealth. For extreme insurance claims, the tail
\nof the severity distribution (losses in excess of a threshold) can be
\nmodelled using a Pareto distribution.<\/p>\n

The continuous variable $X$ is said to have the Pareto distribution with
\nshape parameter $\\alpha$ and scale parameter $\\theta$ if its pdf is
\ngiven by<\/p>\n

$f_{X}\\left( x \\right) = \\frac{\\alpha\\theta^{\\alpha}}{\\left( x + \\theta \\right)^{\\alpha + 1}}$
\n$x > 0,\\ \\alpha > 0,\\ \\theta > 0$.<\/em><\/p>\n

Figures 3 and 4 demonstrate the\u00a0effect\u00a0of the\u00a0shape and\u00a0scale
\nparameters\u00a0on the Pareto density function.<\/p>\n

Fig. (3) Pareto density functions with varying scale parameters<\/p>\n

\"\"{width=”3.8819444444444446in”
\nheight=”2.129678477690289in”}<\/p>\n

Fig. (4) Pareto density functions with varying shape parameters<\/p>\n

\"\"{width=”3.8541666666666665in”
\nheight=”2.1809372265966753in”}<\/p>\n

The distribution function of the Pareto distribution is given by<\/p>\n

$F_{X}\\left( x \\right) = 1 – \\left( \\frac{\\theta}{x + \\theta} \\right)^{\\alpha}$
\n$x > 0,\\ \\alpha > 0,\\ \\theta > 0$.<\/em><\/p>\n

It can be easily seen that the hazard function of the Pareto
\ndistribution is a decreasing function in $x$, another indication that
\nthe distribution is heavy tailed.<\/p>\n

The $k$-th moment of the Pareto distributed random variable exists, if
\nand only if, $\\alpha > k$. If $k$ is a positive integer then<\/p>\n

$E\\left( X^{k} \\right) = \\frac{k!\\theta^{k}}{\\left( \\alpha – 1 \\right)\\cdots\\left( \\alpha – k \\right)}$
\n$\\alpha > k$.<\/em><\/p>\n

The mean and variance are given by
\n$E\\left( X \\right) = \\frac{\\theta}{\\alpha – 1}$ for $\\alpha > 1$ and
\n$V\\left( X \\right) = \\frac{\\alpha\\theta^{2}}{\\left( \\alpha – 1 \\right)^{2}\\left( \\alpha – 2 \\right)}$
\nfor $\\alpha > 2$,<\/em><\/p>\n

respectively.<\/p>\n

Example 3.5<\/strong><\/p>\n

The claim size of an insurance portfolio follows the Pareto distribution
\nwith mean and variance of 40 and 1800 respectively. Find<\/p>\n

i. The shape and scale parameters,<\/p>\n

ii. The 95-th percentile of this distribution.<\/p>\n

Solution<\/strong><\/p>\n

$E\\left( X \\right) = \\frac{\\theta}{\\alpha – 1} = 40$ and
\n$V\\left( X \\right) = \\frac{\\alpha\\theta^{2}}{\\left( \\alpha – 1 \\right)^{2}\\left( \\alpha – 2 \\right)} = 1800$.
\nBy dividing the square of the first equation by the second we get
\n$\\frac{\\alpha – 2}{\\alpha} = \\frac{40^{2}}{1800}$. Thus,
\n$\\alpha = 18.02$ and $\\theta = 680.72$.<\/p>\n

The 95-th percentile, $\\pi_{0.95}$, satisfies the equation
\n$F_{X}\\left( \\pi_{0.95} \\right) = 1 – \\left( \\frac{680.72}{\\pi_{0.95} + 680.72} \\right)^{18.02} = 0.95$.
\nThus, $\\pi_{0.95} = 122.96$.<\/p>\n

The Weibull distribution<\/strong><\/p>\n

The Weibull distribution, named after the Swedish physicist Waloddi
\nWeibull (1887-1979) is widely used in reliability, life data analysis,
\nweather forecasts and general insurance claims.\u00a0Truncated data arise
\nfrequently in insurance studies. The Weibull distribution is
\nparticularly useful in modeling left-truncated claim severity
\ndistributions. Weibull was used to model excess of loss treaty over
\nautomobile insurance as well as earthquake inter-arrival times.<\/p>\n

The continuous variable $X$ is said to have the Weibull distribution
\nwith shape parameter $\\alpha$ and scale parameter $\\theta$ if its
\nprobability density function is given by<\/p>\n

$f_{X}\\left( x \\right) = \\frac{\\alpha}{\\theta}\\left( \\frac{x}{\\theta} \\right)^{\\alpha – 1}exp – \\left( \\frac{x}{\\theta} \\right)^{\\alpha}$
\n$x > 0,\\ \\alpha > 0,\\ \\theta > 0$.<\/em><\/p>\n

Figures 5 and 6 demonstrate the\u00a0effect\u00a0of the\u00a0shape and\u00a0scale
\nparameters\u00a0on the Weibull density function.<\/p>\n

The distribution function of the Weibull distribution is given by<\/p>\n

$F_{X}\\left( x \\right) = 1 – e^{- \\left( \\frac{x}{\\theta} \\right)^{\\alpha}}$
\n$x > 0,\\ \\alpha > 0,\\ \\theta > 0$.<\/em><\/p>\n

It can be easily seen that the shape parameter $\\alpha$ describes the
\nshape of the hazard function of the Weibull distribution. The hazard
\nfunction is a decreasing function when $\\alpha < 1$, constant when
\n$\\alpha = 1$ and increasing when $\\alpha > 1$. This behavior of the
\nhazard function makes the Weibull distribution a suitable model for a
\nwide variety of phenomena such as weather forecasting, electrical and
\nindustrial engineering, insurance modeling and financial risk analysis.<\/p>\n

Fig. (5) Weibull density functions with varying scale parameters<\/p>\n

\"\"{width=”4.082962598425197in”
\nheight=”2.2083333333333335in”}<\/p>\n

Fig. (6) Weibull density functions with varying shape parameters<\/p>\n

\"\"{width=”4.118055555555555in”
\nheight=”2.3484306649168856in”}<\/p>\n

The $k$-th moment of the Weibull distributed random variable is given by<\/p>\n

$E\\left( X^{k} \\right) = \\theta^{k}\\Gamma\\left( 1 + \\frac{k}{\\alpha} \\right)$.<\/em><\/p>\n

The mean and variance are given by<\/p>\n

$E\\left( X \\right) = \\theta\\Gamma\\left( 1 + \\frac{1}{\\alpha} \\right)$,<\/em><\/p>\n

and<\/p>\n

$V\\left( X \\right) = \\theta^{2}\\left{ \\Gamma\\left( 1 + \\frac{2}{\\alpha} \\right) – \\left\\lbrack \\Gamma\\left( 1 + \\frac{1}{\\alpha} \\right) \\right\\rbrack^{2} \\right}$,<\/p>\n

respectively.<\/p>\n

Example 3.6<\/strong><\/p>\n

Suppose that the probability distribution of the lifetime of AIDS
\npatients (in months) from the time of diagnosis is described by the
\nWeibull distribution with shape parameter 1.2 and scale parameter 33.33.<\/p>\n

i. Find the probability that a randomly selected person from this
\n population survives at least 12 months,<\/p>\n

ii. A random sample of 10 patients will be selected from this
\n population. What is the probability that at most two will die within
\n one year of diagnosis.<\/p>\n

iii. Find the 99-th percentile of this distribution.<\/p>\n

Solution<\/strong><\/p>\n

Let $X$ be the lifetime of AIDS patients (in months)<\/p>\n

${P\\left( X \\geq 12 \\right) = S}_{X}\\left( 12 \\right) = e^{- \\left( \\frac{12}{33.33} \\right)^{1.2}} = 0.746$.<\/p>\n

Let $Y$ be the number of patients who die within one year of diagnosis.
\nThen, $Y\\sim Bin\\left( 10,\\ 0.254 \\right)$ and
\n$P\\left( Y \\leq 2 \\right) = 0.514$.<\/p>\n

Let $\\pi_{0.99}$ denote the 99-th percentile of this distribution. Then,<\/p>\n

$S_{X}\\left( \\pi_{0.99} \\right) = e^{- \\left( \\frac{\\pi_{0.99}}{33.33} \\right)^{1.2}} = 0.01$
\nand $\\pi_{0.99} = 118.99$.<\/p>\n

**The Generalized Beta Distribution of the Second Kind **<\/p>\n

The Generalized Beta Distribution of the Second Kind (GB2) was
\nintroduced by Gary Venter (1983) in the context of insurance loss
\nmodeling and by MacDonald (1984) as an income and wealth distribution.
\nIt is a four-parameter very flexible distribution that can model
\npositive as well as negatively skewed distributions.<\/p>\n

The continuous variable $X$ is said to have the GB2 distribution with
\nparameters $a$, $b$, $\\alpha$ and $\\beta$ if its probability density
\nfunction is given by<\/p>\n

$f_{X}\\left( x \\right) = \\frac{ax^{a\\alpha – 1}}{b^{\\text{a\u03b1}}\\left( \\alpha,\\beta \\right)\\left\\lbrack 1 + \\left( \\frac{x}{b} \\right)^{a} \\right\\rbrack^{\\alpha + \\beta}}$
\n$x > 0,\\ a,b,\\alpha,\\beta > 0$,<\/p>\n

where the beta function $\\left( \\alpha,\\beta \\right)$ is defined as<\/p>\n

$\\left( \\alpha,\\beta \\right) = \\int_{0}^{1}{t^{\\alpha – 1}\\left( 1 – t \\right)^{\\beta – 1}}\\text{dt}$.<\/p>\n

The GB2 provides a model for heavy as well as light tailed data. It
\nincludes the exponential, gamma, Weibull, Burr, Lomax, F, chi-square,
\nRayleigh, lognormal and log-logistic as special or limiting cases. For
\nexample, by setting the parameters $a = \\alpha = \\beta = 1$, then the
\nGB2 reduces to the log-logistic distribution. When $a = 1$ and
\n$\\beta \\rightarrow \\infty$, it reduces to the gamma distribution and
\nwhen $\\alpha = 1$ and $\\beta \\rightarrow \\infty$, it reduces to the
\nWeibull distribution.<\/p>\n

The $k$-th moment of the GB2 distributed random variable is given by<\/p>\n

$E\\left( X^{k} \\right) = \\frac{b^{k}\\left( \\alpha + \\frac{k}{a},\\beta – \\frac{k}{a} \\right)}{\\left( \\alpha,\\beta \\right)}$,
\n$k > 0$.<\/p>\n

Earlier applications of the GB2 were on income data and more recently
\nhave been used to model long-tailed claims data. GB2 was used to model
\ndifferent types of automobile insurance claims, severity of fire losses
\nas well as medical insurance claim data.<\/p>\n

3.3 Methods of creating new distributions<\/strong><\/p>\n

In this section we<\/p>\n

i. understand connections among the distributions;<\/p>\n

ii. give insights into when a distribution is preferred when compared to
\n alternatives;<\/p>\n

iii. provide foundations for creating new distributions.<\/p>\n

\u00a0<\/p>\n

3.3.1 Functions of random variables and their distributions<\/strong><\/p>\n

In Section 3.2 we discussed some elementary known distributions. In this
\nsection we discuss means of creating new parametric probability
\ndistributions from existing ones. Let $X$ be a continuous random
\nvariable with a known probability function $f_{X}(x)$ and distribution
\nfunction $F_{X}(x)$. Consider the transformation
\n$Y = g\\left( X \\right)$, where $g(X)$ is a one-to-one transformation
\ndefining a new random variable $Y$. We can use the\u00a0distribution function
\ntechnique, the\u00a0change-of-variable technique\u00a0or the\u00a0moment-generating
\nfunction technique to find the probability density function of the
\nvariable of interest $Y$. In this section we apply the following
\ntechniques for creating new families of distributions: (a)
\nMultiplication by a constant (b) Raising to a power, (c) exponentiation
\nand (d) mixing.<\/p>\n

(a) Multiplication by a constant<\/strong><\/p>\n

If claim data show change over time then such transformation can be
\nuseful to adjust for inflation. If the level of inflation is positive
\nthen claim costs are rising, and if it is negative then costs are
\nfalling. To adjust for inflation we multiply the cost $X$ by 1+
\ninflation rate (negative inflation is deflation). To account for
\ncurrency impact on claim costs we also use a transformation to apply
\ncurrency\u00a0conversion from a base to a counter currency.<\/p>\n

Consider the transformation $Y = cX$, where $c > 0$, then the
\ndistribution function of $Y$ is given by<\/p>\n

$F_{Y}\\left( y \\right) = P\\left( Y \\leq y \\right) = P\\left( cX \\leq y \\right) = P\\left( X \\leq \\frac{y}{c} \\right) = F_{X}\\left( \\frac{y}{c} \\right)$.<\/p>\n

Hence, the probability function of interest $f_{Y}(y)$ can be written as<\/p>\n

$f_{Y}\\left( y \\right) = \\frac{1}{c}f_{X}\\left( \\frac{y}{c} \\right)$.<\/p>\n

Suppose that $X$ has a parametric distribution and define a rescaled
\nversion $Y\\ = \\ cX$, $c\\ > \\ 0$. If $Y$ is in the same parametric
\nfamily then the distribution is said to be a scale distribution. The
\nscale parameter has the following features:<\/p>\n

i. The parameter is changed by multiplying by $c$;<\/p>\n

ii. All other parameter remain unchanged.<\/p>\n

Example 3.7 (SOA)<\/strong><\/p>\n

The aggregate losses of Eiffel Auto Insurance are denoted in euro
\ncurrency and follow a Lognormal distribution with $\\mu = 8$ and
\n$\\sigma = 2$. Given that 1 euro $=$ 1.3 dollars, find the set of
\nlognormal parameters, which describe the distribution of Eiffel’s losses
\nin dollars?<\/p>\n

Solution<\/strong><\/p>\n

Let $X$ and $Y$ denote the aggregate losses of Eiffel Auto Insurance in
\neuro currency and dollars respectively. Then, $Y = 1.3X$.<\/p>\n

$F_{Y}\\left( y \\right) = P\\left( Y \\leq y \\right) = P\\left( 1.3X \\leq y \\right) = P\\left( X \\leq \\frac{y}{1.3} \\right) = F_{X}\\left( \\frac{y}{1.3} \\right)$.<\/p>\n

$X$ follows a lognormal distribution with parameters $\\mu = 8$ and
\n$\\sigma = 2$. The probability density function of $X$ is given by<\/p>\n

$f_{X}\\left( x \\right) = \\frac{1}{\\text{x\u03c3}\\sqrt{2\\pi}}e^{- \\frac{1}{2}\\left( \\frac{lnx – \\mu}{\\sigma} \\right)^{2}}$,
\nfor $x > 0$.<\/p>\n

Then, the probability density function of interest $f_{Y}(y)$ is<\/p>\n

$f_{Y}\\left( y \\right) = \\frac{1}{1.3}f_{X}\\left( \\frac{y}{1.3} \\right) = \\frac{1}{1.3}\\frac{1.3}{\\text{y\u03c3}\\sqrt{2\\pi}}e^{- \\frac{1}{2}\\left( \\frac{\\ln\\left( y\/1.3 \\right) – \\mu}{\\sigma} \\right)^{2}} = \\frac{1}{\\text{y\u03c3}\\sqrt{2\\pi}}e^{- \\frac{1}{2}\\left( \\frac{lny – \\left( ln1.3 + \\mu \\right)}{\\sigma} \\right)^{2}}$.<\/p>\n

Then $Y$ follows a lognormal distribution with parameters
\n$ln1.3 + \\mu = 8.26$ and $\\sigma = 2.00$.<\/p>\n

Example 3.8<\/strong><\/p>\n

Demonstrate that the gamma distribution is a scale distribution.<\/p>\n

Solution<\/strong><\/p>\n

Let $X\\sim Ga(\\alpha,\\theta)$ and $Y = cX$, then
\n$f_{Y}\\left( y \\right) = \\frac{1}{c}f_{X}\\left( \\frac{y}{c} \\right) = \\frac{\\left( \\frac{y}{\\text{c\u03b8}} \\right)^{\\alpha}}{y\\Gamma\\left( \\alpha \\right)}exp – \\left( \\frac{y}{\\text{c\u03b8}} \\right)$.<\/p>\n

We can see that $Y\\sim Ga(\\alpha,c\\theta)$ indicating that gamma is a
\nscale distribution and $\\theta$ is a scale parameter.<\/p>\n

(b) Raising to a power<\/strong><\/p>\n

In the previous section we have talked about the flexibility of the
\nWeibull distribution in fitting reliability data. Looking to the origins
\nof the Weibull distribution, we recognize that the Weibull is a power
\ntransformation of the exponential distribution. This is an application
\nof another type of transformation which involves raising the random
\nvariable to a power.<\/p>\n

Consider the transformation $Y = X^{\\tau}$, where $\\tau > 0$, then the
\ndistribution function of $Y$ is given by<\/p>\n

$F_{Y}\\left( y \\right) = P\\left( Y \\leq y \\right) = P\\left( X^{\\tau} \\leq y \\right) = P\\left( X \\leq y^{\\frac{1}{\\tau}} \\right) = F_{X}\\left( y^{\\frac{1}{\\tau}} \\right)$.<\/p>\n

Hence, the probability function of interest $f_{Y}(y)$ can be written as<\/p>\n

$f_{Y}(y) = \\frac{1}{\\tau}{y^{\\frac{1}{\\tau} – 1}f}_{X}\\left( y^{\\frac{1}{\\tau}} \\right)$.<\/p>\n

On the other hand, if $\\tau < 0$, then the distribution function of $Y$
\nis given by<\/p>\n

$F_{Y}\\left( y \\right) = P\\left( Y \\leq y \\right) = P\\left( X^{\\tau} \\leq y \\right) = P\\left( X \\geq y^{\\frac{1}{\\tau}} \\right) = 1 – F_{X}\\left( y^{\\frac{1}{\\tau}} \\right)$,<\/p>\n

and<\/p>\n

$f_{Y}(y) = \\left| \\frac{1}{\\tau} \\right|{y^{\\frac{1}{\\tau} – 1}f}_{X}\\left( y^{\\frac{1}{\\tau}} \\right)$.<\/p>\n

Example 3.9<\/strong><\/p>\n

We assume that $X$ follows the exponential distribution with mean
\n$\\theta$ and consider the transformed variable $Y = X^{\\tau}$. Show that
\n$Y$ follows the Weibull distribution when $\\tau$ is positive and
\ndetermine the parameters of the Weibull distribution.<\/p>\n

Solution<\/strong><\/p>\n

$f_{X}(x) = \\frac{1}{\\theta}e^{- \\frac{x}{\\theta}}$, $x > 0$.<\/p>\n

$f_{Y}\\left( y \\right) = \\frac{1}{\\tau}{y^{\\frac{1}{\\tau} – 1}f}_{X}\\left( y^{\\frac{1}{\\tau}} \\right) = \\frac{1}{\\text{\u03c4\u03b8}}y^{\\frac{1}{\\tau} – 1}e^{- \\frac{y^{\\frac{1}{\\tau}}}{\\theta}} = \\frac{\\alpha}{\\beta}\\left( \\frac{y}{\\beta} \\right)^{\\alpha – 1}e^{- \\left( \\frac{y}{\\beta} \\right)^{\\alpha}}$.<\/p>\n

where $\\alpha = \\frac{1}{\\tau}$ and $\\beta = \\theta^{\\tau}$. Then, $Y$
\nfollows the Weibull distribution with shape parameter $\\alpha$ and scale
\nparameter $\\beta$.<\/p>\n

(c) Exponentiation<\/strong><\/p>\n

The normal distribution is a very popular model for a wide number of
\napplications and when the sample size is large, it can serve as an
\napproximate distribution for other models. If the random variable $X$
\nhas a normal distribution with mean $\\mu$ and variance $\\sigma^{2}$,
\nthen $Y = e^{X}$ has lognormal distribution with parameters $\\mu$ and
\n$\\sigma^{2}$. The lognormal random variable has a lower bound of zero,
\nis positively skewed and has a long right tail. A lognormal distribution
\nis commonly used to describe distributions of financial assets such as
\nstock prices. It is also used in fitting claim amounts for automobile as
\nwell as health insurance. This is an example of another type of
\ntransformation which involves exponentiation.<\/p>\n

Consider the transformation $Y = e^{X}$, then the distribution function
\nof $Y$ is given by<\/p>\n

$F_{Y}\\left( y \\right) = P\\left( Y \\leq y \\right) = P\\left( e^{X} \\leq y \\right) = P\\left( X \\leq lny \\right) = F_{X}\\left( \\text{lny} \\right)$.<\/p>\n

Hence, the probability function of interest $f_{Y}(y)$ can be written as<\/p>\n

$f_{Y}(y) = \\frac{1}{y}f_{X}\\left( \\text{lny} \\right)$.<\/p>\n

Example 3.10 (SOA)<\/strong><\/p>\n

$X$ has a uniform distribution on the interval $(0,\\ c)$. $Y = e^{X}$.
\nFind the distribution of $Y$.<\/p>\n

Solution<\/strong><\/p>\n

$F_{Y}\\left( y \\right) = P\\left( Y \\leq y \\right) = P\\left( e^{X} \\leq y \\right) = P\\left( X \\leq lny \\right) = F_{X}\\left( \\text{lny} \\right)$.<\/p>\n

Then,
\n$f_{Y}\\left( y \\right) = \\frac{1}{y}f_{X}\\left( \\text{lny} \\right) = \\frac{1}{\\text{cy}}$.
\nSince $0 < x < c$, then $1 < y < e^{c}$.<\/p>\n

3.3.2 Mixture distributions<\/strong><\/p>\n

Mixture distributions represent a useful way of modelling data that are
\ndrawn from a heterogeneous population. This parent population can be
\nthought to be divided into multiple subpopulations with distinct
\ndistributions.<\/p>\n

Finite mixture distributions<\/strong><\/p>\n

Two-point mixture<\/strong><\/p>\n

If the underlying phenomenon is diverse and can actually be described as
\ntwo phenomena representing two subpopulations with different modes, we
\ncan construct the two point mixture random variable $X$. Given random
\nvariables $X_{1}$ and $X_{2}$, with probability density functions
\n$f_{X_{1}}\\left( x \\right)$ and $f_{X_{2}}\\left( x \\right)$
\nrespectively, the probability function of $X$ is the weighted average of
\nthe component probability functions $f_{X_{1}}\\left( x \\right)$ and
\n$f_{X_{2}}\\left( x \\right)$. The probability density function and
\ndistribution function of $X$ are given by<\/p>\n

$f_{X}\\left( x \\right) = af_{X_{1}}\\left( x \\right) + \\left( 1 – a \\right)f_{X_{2}}\\left( x \\right)$,<\/p>\n

and<\/p>\n

$F_{X}\\left( x \\right) = aF_{X_{1}}\\left( x \\right) + \\left( 1 – a \\right)F_{X_{2}}\\left( x \\right)$,<\/p>\n

for $0 < a < 1$, where the mixing parameters $a$ and $(1 – a)$ represent
\nthe proportions of data points that fall under each of the two
\nsubpopulations respectively. This weighted average can be applied to a
\nnumber of other distribution related quantities. The k<\/em>-th moment and
\nmoment generating function of $X$ are given by<\/p>\n

$E\\left( X \\right) = aE\\left( X_{1} \\right) + \\left( 1 – a \\right)E\\left( X_{2} \\right)$,<\/p>\n

and<\/p>\n

$M_{X}\\left( t \\right) = aM_{X_{1}}\\left( t \\right) + \\left( 1 – a \\right)M_{X_{2}}\\left( t \\right)$,<\/p>\n

respectively.<\/p>\n

Example 3.11 (SOA)<\/strong><\/p>\n

The distribution of the random variable $X$ is an equally weighted
\nmixture of two Poisson distributions with parameters $\\lambda_{1}$ and
\n$\\lambda_{2}$ respectively. The mean and variance of $X$ are 4 and 13
\nrespectively. Determine $P\\left( X > 2 \\right)$.<\/p>\n

Solution<\/strong><\/p>\n

$E\\left( X \\right) = 0.5\\lambda_{1} + 0.5\\lambda_{2} = 4$;<\/p>\n

$E\\left( X^{2} \\right) = 0.5\\left( \\lambda_{1} + \\lambda_{1}^{2} \\right) + 0.5\\left( \\lambda_{2} + \\lambda_{2}^{2} \\right) = 13 + 16$;<\/p>\n

Simplifying the two equations we get $\\lambda_{1} + \\lambda_{2} = 8$ and
\n$\\lambda_{1}^{2} + \\lambda_{2}^{2} = 50$. Then, the parameters of the
\ntwo Poisson distributions are 1 and 7.<\/p>\n

$P\\left( X > 2 \\right) = 0.5P\\left( X_{1} > 2 \\right) + 0.5P\\left( X_{2} > 2 \\right) = 0.05$.<\/p>\n

k<\/em>-point mixture<\/strong><\/p>\n

In case of finite mixture distributions, the random variable of interest
\n$X$ has a probability $p_{i}$ of being drawn from homogeneous
\nsubpopulation $i$, where $i = 1,2,\\ldots,k$ and $k$ is the initially
\nspecified number of subpopulations in our mixture. The mixing parameter
\n$p_{i}$ represents the proportion of observations from subpopulation
\n$i$. Consider the random variable $X$ generated from $k$ distinct
\nsubpopulations, where subpopulation $i$ is modeled by the continuous
\ndistribution $f_{X_{i}}\\left( x \\right)$. The probability distribution
\nof $X$ is given by<\/p>\n

$f_{X}\\left( x \\right) = \\sum_{i = 1}^{k}{p_{i}f_{X_{i}}\\left( x \\right)}$,
\n${0 < p}{i} < 1$, $\\sum<\/em>{i = 1}^{k}{p_{i} = 1}$.<\/p>\n

This model is often referred to as a finite mixture or a $k$ point
\nmixture. The distribution function, $k$-th moment and moment generating
\nfunctions of the $k$-th point mixture are given as<\/p>\n

$F_{X}\\left( x \\right) = \\sum_{i = 1}^{k}{p_{i}F_{X_{i}}\\left( x \\right)}$,<\/p>\n

$E\\left( X^{k} \\right) = \\sum_{i = 1}^{k}{p_{i}E\\left( X_{i}^{k} \\right)}$,<\/p>\n

$M_{X}\\left( t \\right) = \\sum_{i = 1}^{k}{p_{i}M_{X_{i}}\\left( t \\right)}$,<\/p>\n

respectively.<\/p>\n

Example 3.12 (SOA)<\/strong><\/p>\n

$Y_{1}$ is a mixture of $X_{1}$ and $X_{2}$ with mixing weights $a$ and
\n$(1 – a)$.<\/p>\n

$Y_{2}$ is a mixture of $X_{3}$ and $X_{4}$ with mixing weights $b$ and
\n$(1 – b)$.<\/p>\n

$Z$ is a mixture of $Y_{1}$ and $Y_{2}$ with mixing weights $c$ and
\n$(1 – c)$.<\/p>\n

Show that $Z$ is a mixture of $X_{1}$, $X_{2}$, $X_{3}$ and $X_{4}$, and
\nfind the mixing weights.<\/p>\n

Solution<\/strong><\/p>\n

$f_{Y_{1}}\\left( x \\right) = af_{X_{1}}\\left( x \\right) + \\left( 1 – a \\right)f_{X_{2}}\\left( x \\right)$,<\/p>\n

$f_{Y_{2}}\\left( x \\right) = bf_{X_{3}}\\left( x \\right) + \\left( 1 – b \\right)f_{X_{4}}\\left( x \\right)$,<\/p>\n

$f_{Z}\\left( x \\right) = cf_{Y_{1}}\\left( x \\right) + \\left( 1 – c \\right)f_{Y_{2}}\\left( x \\right)$,<\/p>\n

$$f_{Z}\\left( x \\right) = c\\left\\lbrack af_{X_{1}}\\left( x \\right) + \\left( 1 – a \\right)f_{X_{2}}\\left( x \\right) \\right\\rbrack + \\left( 1 – c \\right)\\left\\lbrack bf_{X_{3}}\\left( x \\right) + \\left( 1 – b \\right)f_{X_{4}}\\left( x \\right) \\right\\rbrack$$<\/p>\n

$= caf_{X_{1}}\\left( x \\right) + c\\left( 1 – a \\right)f_{X_{2}}\\left( x \\right) + \\left( 1 – c \\right)bf_{X_{3}}\\left( x \\right) + (1 – c)\\left( 1 – b \\right)f_{X_{4}}\\left( x \\right)$.<\/p>\n

Then, $Z$ is a mixture of $X_{1}$, $X_{2}$, $X_{3}$ and $X_{4}$, with
\nmixing weights $\\text{ca}$, $c\\left( 1 – a \\right)$,
\n$\\left( 1 – c \\right)b$ and $(1 – c)\\left( 1 – b \\right)$.<\/p>\n

Continuous mixture<\/strong><\/p>\n

A mixture with a very large number of subpopulations ($k$ goes to
\ninfinity) approaches an infinite mixture, often referred to as a
\ncontinuous mixture. In a continuous mixture, subpopulations are not
\ndistinguished by a discrete mixing parameter but by a continuous
\nvariable $\\theta$, where $\\theta$ plays the role of $p_{i}$ in the
\nfinite mixture. Consider the random variable $X$ with a distribution
\ndepending on a parameter $\\theta$, where $\\theta$ itself is a continuous
\nrandom variable. This description yields the following model for $X$<\/p>\n

$f_{X}\\left( x \\right) = \\int_{0}^{\\infty}{f_{X}\\left( x\\left| \\theta \\right.\\ \\right)g\\left( \\theta \\right)}\\text{d\u03b8}$,<\/p>\n

where $f_{X}\\left( x\\left| \\theta \\right.\\ \\right)$ is the conditional
\ndistribution of $X$ at a particular value of $\\theta$ and
\n$g\\left( \\theta \\right)$ is the probability statement made about the
\nunknown parameter $\\theta$, known as the prior distribution of $\\theta$
\n(the prior information or expert opinion to be used in the analysis).<\/p>\n

This model is often referred to as an infinite mixture or a continuous
\nmixture. The distribution function, $k$-th moment and moment generating
\nfunctions of the continuous mixture are given as<\/p>\n

$F_{X}\\left( x \\right) = \\int_{0}^{\\infty}{F_{X}\\left( x\\left| \\theta \\right.\\ \\right)g\\left( \\theta \\right)}\\text{d\u03b8}$,<\/p>\n

$E\\left( X^{k} \\right) = \\int_{0}^{\\infty}{E\\left( X^{k}\\left| \\theta \\right.\\ \\right)g\\left( \\theta \\right)}\\text{d\u03b8}$,<\/p>\n

$M_{X}\\left( t \\right) = E\\left( e^{\\text{tX}} \\right) = \\int_{0}^{\\infty}{E\\left( e^{\\text{tx}}\\left| \\theta \\right.\\ \\right)g\\left( \\theta \\right)}\\text{d\u03b8}$,<\/p>\n

respectively.<\/p>\n

The $k$-th moments of the mixture distribution can be rewritten as<\/p>\n

$E\\left( X^{k} \\right) = \\int_{0}^{\\infty}{E\\left( X^{k}\\left| \\theta \\right.\\ \\right)g\\left( \\theta \\right)}d\\theta = E\\left\\lbrack E\\left( X^{k}\\left| \\theta \\right.\\ \\right) \\right\\rbrack$.<\/p>\n

In particular the mean and variance of $X$ are given by<\/p>\n

$E\\left( X \\right) = E\\left\\lbrack E\\left( X\\left| \\theta \\right.\\ \\right) \\right\\rbrack$
\nand
\n$V\\left( X \\right) = E\\left\\lbrack V\\left( X\\left| \\theta \\right.\\ \\right) \\right\\rbrack + V\\left\\lbrack E\\left( X\\left| \\theta \\right.\\ \\right) \\right\\rbrack$.<\/p>\n

Example 3.13 (SOA)<\/strong><\/p>\n

$X$ has a binomial distribution with a mean of $100q$ and a variance of
\n$100q\\left( 1 – q \\right)$ and $q$ has a beta distribution with
\nparameters $a = 3$ and $b = 2$. Find the unconditional mean and variance
\nof .<\/p>\n

Solution<\/strong><\/p>\n

$E\\left( q \\right) = \\frac{a}{a + b} = \\frac{3}{5}$ and
\n$E\\left( q^{2} \\right) = \\frac{a\\left( a + 1 \\right)}{\\left( a + b \\right)\\left( a + b + 1 \\right)} = \\frac{2}{5}$.<\/p>\n

$E\\left( X \\right) = E\\left\\lbrack E\\left( X\\left| q \\right.\\ \\right) \\right\\rbrack = E\\left( 100q \\right) = 100E\\left( q \\right) = 60$,<\/p>\n

$$V\\left( X \\right) = E\\left\\lbrack V\\left( X\\left| q \\right.\\ \\right) \\right\\rbrack + V\\left\\lbrack E\\left( X\\left| q \\right.\\ \\right) \\right\\rbrack = E\\left\\lbrack 100q\\left( 1 – q \\right) \\right\\rbrack + V\\left( 100q \\right)$$<\/p>\n

$= 100E\\left( q \\right) – 100E\\left( q^{2} \\right) + 100^{2}V\\left( q \\right) = 420$.<\/p>\n

Exercise 3.14 (SOA)<\/strong><\/p>\n

Claim sizes, , are uniform on for each policyholder. varies by
\npolicyholder according to an exponential distribution with mean 5. Find
\nthe unconditional distribution, mean and variance of .<\/p>\n

Solution<\/strong><\/p>\n

The conditional distribution of $X$ is
\n$f_{X}\\left( \\left. \\ x \\right|\\theta \\right) = \\frac{1}{10}$ for
\n$\\theta < x < \\theta + 10$.<\/p>\n

The prior distribution of $\\theta$ is
\n$g\\left( \\theta \\right) = \\frac{1}{5}e^{- \\frac{\\theta}{5}}$ for
\n$0 < \\theta < \\infty$.<\/p>\n

The conditional mean and variance of $X$ are given by<\/p>\n

$E\\left( \\left. \\ X \\right|\\theta \\right) = \\frac{\\theta + \\theta + 10}{2} = \\theta + 5$
\nand
\n$V\\left( \\left. \\ X \\right|\\theta \\right) = \\frac{\\left\\lbrack \\left( \\theta + 10 \\right) – \\theta \\right\\rbrack^{2}}{12} = \\frac{100}{12}$,
\nrespectively.<\/p>\n

Hence, the unconditional mean and variance of $X$ are given by<\/p>\n

$E\\left( X \\right) = E\\left\\lbrack E\\left( X\\left| \\theta \\right.\\ \\right) \\right\\rbrack = E\\left( \\theta + 5 \\right) = E\\left( \\theta \\right) + 5 = 5 + 5 = 10$,
\nand<\/p>\n

$V\\left( X \\right) = E\\left\\lbrack V\\left( X\\left| \\theta \\right.\\ \\right) \\right\\rbrack + V\\left\\lbrack E\\left( X\\left| \\theta \\right.\\ \\right) \\right\\rbrack = E\\left( \\frac{100}{12} \\right) + V\\left( \\theta + 5 \\right) = 8.33 + V\\left( \\theta \\right) = 33.33$.<\/p>\n

The unconditional distribution of $X$ is
\n$f_{X}\\left( x \\right) = \\int_{}^{}{f_{X}\\left( \\left. \\ x \\right|\\theta \\right)g\\left( \\theta \\right)\\text{d\u03b8}}$.<\/p>\n

\"\"{width=”3.2847222222222223in”
\nheight=”2.0650273403324584in”}<\/p>\n

$$f_{X}\\left( x \\right) = \\left{ \\begin{matrix}
\n\\int_{0}^{x}{\\frac{1}{50}e^{- \\frac{\\theta}{5}}d\\theta = \\frac{1}{10}\\left( 1 – e^{- \\frac{x}{5}} \\right)} & 0 \\leq x \\leq 10, \
\n\\int_{x – 10}^{x}{\\frac{1}{50}e^{- \\frac{\\theta}{5}}\\text{d\u03b8}} = \\frac{1}{10}\\left( e^{- \\frac{\\left( x – 10 \\right)}{5}} – e^{- \\frac{x}{5}} \\right) & 10 < x < \\infty. \
\n\\end{matrix} \\right.\\ $$<\/p>\n

3.4 Coverage Modifications<\/strong><\/p>\n

In this section we evaluate the impacts of coverage modifications: a)
\ndeductibles, b) policy limit, c) coinsurance and inflation on insurer\u2019s
\ncosts.<\/p>\n

3.4.1 Policy deductibles<\/strong><\/p>\n

Under an ordinary deductible policy, the insured (policyholder) agrees
\nto cover a fixed amount of an insurance claim before the insurer starts
\nto pay. This fixed expense paid out of pocket is called the deductible
\nand often denoted by $d$. The insurer is responsible for covering the
\nloss $X$ less the deductible $d$. Depending on the agreement, the
\ndeductible may apply to each covered loss or to a defined benefit period
\n(month, year, etc.)<\/p>\n

Deductibles eliminate a large number of small claims, reduce costs of
\nhandling and processing these claims, reduce premiums for the
\npolicyholders and reduce moral hazard. Moral hazard occurs when the
\ninsured takes more risks, increasing the chances of loss due to perils
\ninsured against, knowing that the insurer will incur the cost (e.g. a
\npolicyholder with collision insurance may be encouraged to drive
\nrecklessly). The larger the deductible, the less the insured pays in
\npremiums for an insurance policy.<\/p>\n

Let $X$ denote the loss incurred to the insured and $Y$ denote the
\namount of paid claim by the insurer. Speaking of the benefit paid to the
\npolicyholder, we differentiate between two variables: The payment per
\nloss and the payment per payment. The payment per loss variable, denoted
\nby $Y^{L}$, includes losses for which a payment is made as well as
\nlosses less than the deductible and hence is defined as<\/p>\n

$$Y^{L} = \\left( X – d \\right)_{+} = \\left{ \\begin{matrix}
\n0 & X \\leq d, \
\nX – d & X > d. \
\n\\end{matrix} \\right.\\ $$<\/p>\n

$Y^{L}$ is often referred to as left censored and shifted variable
\nbecause the values below $d$ are not ignored and all losses are shifted
\nby a value $d$.<\/p>\n

On the other hand, the payment per payment variable, denoted by $Y^{P}$,
\nis not defined when there is no payment and only includes losses for
\nwhich a payment is made. The variable is defined as<\/p>\n

$$Y^{P} = \\left{ \\begin{matrix}
\n\\text{Undefined} & X \\leq d, \
\nX – d & X > d. \
\n\\end{matrix} \\right.\\ $$<\/p>\n

$Y^{P}$ is often referred to as left truncated and shifted variable or
\nexcess loss variable because the claims smaller than $d$ are not
\nreported and values above $d$ are shifted by $d$.<\/p>\n

Even when the distribution of $X$ is continuous, the distribution of
\n$Y^{L}$ is partly discrete and partly continuous. The discrete part of
\nthe distribution is concentrated at $Y = 0$ (when $X \\leq d$) and the
\ncontinuous part is spread over the interval $Y > 0$ (when $X > d$). For
\nthe discrete part, the probability that no payment is made is the
\nprobability that losses fall below the deductible; that is,
\n$P\\left( Y^{L} = 0 \\right) = P\\left( X \\leq d \\right) = F_{X}\\left( d \\right)$.
\nUsing the transformation $Y^{L} = X – d$ for the continuous part of the
\ndistribution, we can find the probability density function of $Y^{L}$
\ngiven by<\/p>\n

$$f_{Y^{L}}\\left( y \\right) = \\left{ \\begin{matrix}
\nF_{X}\\left( d \\right) & y = 0, \
\nf_{X}\\left( y + d \\right) & y > 0. \
\n\\end{matrix} \\right.\\ $$<\/p>\n

We can see that the payment per payment variable is the payment per loss
\nvariable conditioned on the loss exceeding the deductible; that is,
\n$Y^{P} = \\left. \\ Y^{L} \\right|X > d$. Hence, the probability density
\nfunction of $Y^{P}$ is given by<\/p>\n

$f_{Y^{P}}\\left( y \\right) = \\frac{f_{X}\\left( y + d \\right)}{1 – F_{X}\\left( d \\right)}$
\n, for $y > 0$.<\/p>\n

Accordingly, the distribution functions of $Y^{L}$and $Y^{P}$ are given
\nby<\/p>\n

$$F_{Y^{L}}\\left( y \\right) = \\left{ \\begin{matrix}
\nF_{X}\\left( d \\right) & y = 0, \
\nF_{X}\\left( y + d \\right) & y > 0. \
\n\\end{matrix} \\right.\\ $$<\/p>\n

and<\/p>\n

$F_{Y^{P}}\\left( y \\right) = \\frac{F_{X}\\left( y + d \\right) – F_{X}\\left( d \\right)}{1 – F_{X}\\left( d \\right)}$
\n, for $y > 0$.<\/p>\n

respectively.<\/p>\n

The raw moments of $Y^{L}$ and $Y^{P}$ can be found directly using the
\nprobability density function of $X$ as follows<\/p>\n

$E\\left\\lbrack \\left( Y^{L} \\right)^{k} \\right\\rbrack = \\int_{d}^{\\infty}\\left( x – d \\right)^{k}f_{X}\\left( x \\right)\\text{dx}$,<\/p>\n

and<\/p>\n

$E\\left\\lbrack \\left( Y^{P} \\right)^{k} \\right\\rbrack = \\frac{\\int_{d}^{\\infty}\\left( x – d \\right)^{k}f_{X}\\left( x \\right)\\text{dx}}{{1 – F}{X}\\left( d \\right)} = \\frac{E\\left\\lbrack \\left( Y^{L} \\right)^{k} \\right\\rbrack}{{1 – F}<\/em>{X}\\left( d \\right)}$
\n,<\/p>\n

respectively.<\/p>\n

We have seen that the deductible $d$ imposed on an insurance policy is
\nthe amount of loss that has to be paid out of pocket before the insurer
\nmakes any payment. The deductible $d$ imposed on an insurance policy
\nreduces the insurer\u2019s payment. The loss elimination ratio (LER<\/em>) is the
\npercentage decrease in the expected payment of the insurer as a result
\nof imposing the deductible. LER<\/em> is defined as<\/p>\n

$LER = \\frac{E\\left( X \\right) – E\\left( Y^{L} \\right)}{E\\left( X \\right)}$.<\/p>\n

A little less common type of policy deductible is the Franchise
\ndeductible. The Franchise deductible will apply to the policy in the
\nsame way as ordinary deductible except that when the loss exceeds the
\ndeductible $\\text{d\\ }$the full loss is covered by the insurer. The
\npayment per loss and payment per payment variables are defined as<\/p>\n

$$Y^{L} = \\left{ \\begin{matrix}
\n0 & X \\leq d, \
\nX & X > d, \
\n\\end{matrix} \\right.\\ $$<\/p>\n

and<\/p>\n

$$Y^{P} = \\left{ \\begin{matrix}
\n\\text{Undefined} & X \\leq d, \
\nX & X > d, \
\n\\end{matrix} \\right.\\ $$<\/p>\n

respectively.<\/p>\n

Example 3.15 (SOA)<\/strong><\/p>\n

A claim severity distribution is exponential with mean 1000. An
\ninsurance company will pay the amount of each claim in excess of a
\ndeductible of 100. Calculate the variance of the amount paid by the
\ninsurance company for one claim, including the possibility that the
\namount paid is 0.<\/p>\n

Solution<\/strong><\/p>\n

Let $Y^{L}$ denote the amount paid by the insurance company for one
\nclaim.<\/p>\n

$$Y^{L} = \\left( X – 100 \\right)_{+} = \\left{ \\begin{matrix}
\n0 & X \\leq 100, \
\nX – 100 & X > 100. \
\n\\end{matrix} \\right.\\ $$<\/p>\n

The first and second moments of $Y^{L}$ are<\/p>\n

$E\\left( Y^{L} \\right) = \\int_{100}^{\\infty}\\left( x – 100 \\right)f_{X}\\left( x \\right)dx = {\\int_{100}^{\\infty}{S_{X}\\left( x \\right)}dx = 1000e}^{- \\frac{100}{1000}}$,
\nand<\/p>\n

$E\\left\\lbrack \\left( Y^{L} \\right)^{2} \\right\\rbrack = \\int_{100}^{\\infty}\\left( x – 100 \\right)^{2}f_{X}\\left( x \\right)dx = 2 \\times 1000^{2}e^{- \\frac{100}{1000}}$.<\/p>\n

$V\\left( Y^{L} \\right) = \\left( 2 \\times 1000^{2}e^{- \\frac{100}{1000}} \\right) – \\left( {1000e}^{- \\frac{100}{1000}} \\right)^{2} = 990,944$.<\/p>\n

The solution can be simplified if we make use of the relationship
\nbetween $X$ and $Y^{P}$. If $X$ is exponentially distributed with mean
\n1000, then $Y^{P}$ is also exponentially distributed with the same mean.
\nHence, $E\\left( Y^{P} \\right)$=1000 and
\n$E\\left\\lbrack \\left( Y^{P} \\right)^{2} \\right\\rbrack = 2 \\times 1000^{2}$.<\/p>\n

Using the relationship between $Y^{L}$ and $Y^{P}$ we find<\/p>\n

$$E\\left( Y^{L} \\right) = \\ E\\left( Y^{P} \\right)S_{X}\\left( 100 \\right){= 1000e}^{- \\frac{100}{1000}}$$<\/p>\n

$E\\left\\lbrack \\left( Y^{L} \\right)^{2} \\right\\rbrack = E\\left\\lbrack \\left( Y^{P} \\right)^{2} \\right\\rbrack S_{X}\\left( 100 \\right) = 2 \\times 1000^{2}e^{- \\frac{100}{1000}}$.<\/p>\n

Example 3.16 (SOA)<\/strong><\/p>\n

For an insurance:<\/p>\n

i. Losses have a density function
\n $f_{X}\\left( x \\right) = \\left{ \\begin{matrix}
\n 0.02x & 0 < x < 10, \
\n 0 & \\text{elsewhere.} \
\n \\end{matrix} \\right.\\ $<\/p>\n

ii. The insurance has an ordinary deductible of 4 per loss.<\/p>\n

iii. $Y^{P}$ is the claim payment per payment random variable.<\/p>\n

Calculate $E\\left( Y^{P} \\right)$.<\/p>\n

Solution<\/strong><\/p>\n

$$Y^{P} = \\left{ \\begin{matrix}
\n\\text{Undefined} & X \\leq 4, \
\nX – 4 & X > 4. \
\n\\end{matrix} \\right.\\ $$<\/p>\n

$E\\left( Y^{P} \\right) = \\frac{\\int_{4}^{10}\\left( x – 4 \\right)0.02xdx}{{1 – F}_{X}\\left( 4 \\right)} = \\frac{2.88}{0.84} = 3.43$.<\/p>\n

Example 3.17 (SOA)<\/strong><\/p>\n

You are given:<\/p>\n

i. Losses follow an exponential distribution with the same mean in all
\n years.<\/p>\n

ii. The loss elimination ratio this year is 70%.<\/p>\n

iii. The ordinary deductible for the coming year is 4\/3 of the current
\n deductible.<\/p>\n

Compute the loss elimination ratio for the coming year.<\/p>\n

Solution<\/strong><\/p>\n

The LER<\/em> for the current year
\n$= \\frac{E\\left( X \\right) – E\\left( Y^{L} \\right)}{E\\left( X \\right)} = \\frac{\\theta – \\theta e^{- \\frac{d}{\\theta}}}{\\theta} = 1 – e^{- \\frac{d}{\\theta}} = 0.7$.<\/p>\n

Then, $e^{- \\frac{d}{\\theta}} = 0.3$.<\/p>\n

The LER<\/em> for the coming year
\n$= \\frac{\\theta – \\theta e^{- \\frac{\\left( \\frac{4}{3}d \\right)}{\\theta}}}{\\theta} = 1 – e^{- \\frac{\\left( \\frac{4}{3}d \\right)}{\\theta}} = 1 – \\left( e^{- \\frac{d}{\\theta}} \\right)^{\\frac{4}{3}} = 1 – {0.3}^{\\frac{4}{3}} = 0.8$.<\/p>\n

3.4.2 Policy limits<\/strong><\/p>\n

Under a limited policy, the insurer is responsible for covering the
\nactual loss $X$ up to the limit of its coverage. This fixed limit of
\ncoverage is called the policy limit and often denoted by $u$. If the
\nloss exceeds the policy limit, the difference $X – u$ has to be paid by
\nthe policyholder. While a higher policy limit means a higher payout to
\nthe insured, it is associated with a higher premium.<\/p>\n

Let $X$ denote the loss incurred to the insured and $Y$ denote the
\namount of paid claim by the insurer. Then $Y$ is defined as<\/p>\n

$$Y = X \\land u = \\left{ \\begin{matrix}
\nX & X \\leq u, \
\nu & X > u. \
\n\\end{matrix} \\right.\\ $$<\/p>\n

It can be seen that the distinction between $Y^{L}$ and $Y^{P}$ is not
\nneeded under limited policy as the insurer will always make a payment.<\/p>\n

Even when the distribution of $X$ is continuous, the distribution of $Y$
\nis partly discrete and partly continuous. The discrete part of the
\ndistribution is concentrated at $Y = u$ (when $X > u$), while the
\ncontinuous part is spread over the interval $Y < u$ (when $X \\leq u$).
\nFor the discrete part, the probability that the benefit paid is $u$, is
\nthe probability that the loss exceeds the policy limit $u$; that is,
\n$P\\left( Y = u \\right) = P\\left( X > u \\right) = {1 – F}_{X}\\left( u \\right)$.
\nFor the continuous part of the distribution $Y = X$, hence the
\nprobability density function of $Y$ is given by<\/p>\n

$$f_{Y}\\left( y \\right) = \\left{ \\begin{matrix}
\nf_{X}\\left( y \\right) & 0 < y < u, \
\n1 – F_{X}\\left( u \\right) & y = u. \
\n\\end{matrix} \\right.\\ $$<\/p>\n

Accordingly, the distribution function of $Y$ is given by<\/p>\n

$$F_{Y}\\left( y \\right) = \\left{ \\begin{matrix}
\nF_{X}\\left( x \\right) & 0 < y < u, \
\n1 & y \\geq u. \
\n\\end{matrix} \\right.\\ $$<\/p>\n

The raw moments of $Y$ can be found directly using the probability
\ndensity function of $X$ as follows<\/p>\n

$$E\\left( Y^{k} \\right) = E\\left\\lbrack \\left( X \\land u \\right)^{k} \\right\\rbrack = \\int_{0}^{u}x^{k}f_{X}\\left( x \\right)dx + \\int_{u}^{\\infty}{u^{k}f_{X}\\left( x \\right)}\\text{dx}$$<\/p>\n

$\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ = \\int_{0}^{u}x^{k}f_{X}\\left( x \\right)dx + u^{k}\\left\\lbrack {1 – F}_{X}\\left( u \\right) \\right\\rbrack\\text{dx}$.<\/p>\n

Example 3.18 (SOA)<\/strong><\/p>\n

Under a group insurance policy, an insurer agrees to pay 100% of the
\nmedical bills incurred during the year by employees of a small company,
\nup to a maximum total of one million dollars. The total amount of bills
\nincurred, $X$, has probability density function<\/p>\n

$$f_{X}\\left( x \\right) = \\left{ \\begin{matrix}
\n\\frac{x\\left( 4 – x \\right)}{9} & 0 < x < 3, \
\n0 & \\text{elsewhere.} \
\n\\end{matrix} \\right.\\ $$<\/p>\n

where $x$ is measured in millions. Calculate the total amount, in
\nmillions of dollars, the insurer would expect to pay under this policy.<\/p>\n

Solution<\/strong><\/p>\n

$$Y = X \\land 1 = \\left{ \\begin{matrix}
\nX & X \\leq 1, \
\n1 & X > 1. \
\n\\end{matrix} \\right.\\ $$<\/p>\n

$E\\left( Y \\right) = E\\left( X \\land 1 \\right) = \\int_{0}^{1}\\frac{x^{2}(4 – x)}{9}dx + \\int_{1}^{3}\\frac{x\\left( 4 – x \\right)}{9}dx = 0.935$.<\/p>\n

3.4.3 Coinsurance<\/strong><\/p>\n

As we have seen in Section 3.4.1, the amount of loss retained by the
\npolicyholder can be losses up to the deductible $d$. The retained loss
\ncan also be a percentage of the claim. The percentage $\\alpha$, often
\nreferred to as the coinsurance factor, is the percentage of claim the
\ninsurance company is required to cover. If the policy is subject to an
\nordinary deductible and policy limit, coinsurance\u00a0refers to the
\npercentage of claim the insurer is required to cover, after imposing the
\nordinary deductible and policy limit.\u00a0The payment per loss variable,
\n$Y^{L}$, is defined as<\/p>\n

$$Y^{L} = \\left{ \\begin{matrix}
\n0 & X \\leq d, \
\n\\alpha\\left( X – d \\right) & d < X \\leq u, \
\n\\alpha\\left( u – d \\right) & X > u. \
\n\\end{matrix} \\right.\\ $$<\/p>\n

The policy limit (the maximum amount paid by the insurer) in this case
\nis $\\alpha\\left( u – d \\right)$, while $u$ is the maximum covered loss.<\/p>\n

The $k$-th moment of $Y^{L}$ is given by<\/p>\n

$E\\left\\lbrack \\left( Y^{L} \\right)^{k} \\right\\rbrack = \\int_{d}^{u}\\left\\lbrack \\alpha\\left( x – d \\right) \\right\\rbrack^{k}f_{X}\\left( x \\right)dx + \\int_{u}^{\\infty}\\left\\lbrack \\alpha\\left( u – d \\right) \\right\\rbrack^{k}f_{X}\\left( x \\right)\\text{dx}$.<\/p>\n

A growth factor $\\left( 1 + r \\right)$ may be applied to $X$ resulting
\nin an inflated loss random variable $\\left( 1 + r \\right)X$ (the
\nprespecified d<\/em> and u<\/em> remain unchanged). The resulting per loss
\nvariable can be written as<\/p>\n

$$Y^{L} = \\left{ \\begin{matrix}
\n0 & X \\leq \\frac{d}{1 + r}, \
\n\\alpha\\left\\lbrack \\left( 1 + r \\right)X – d \\right\\rbrack & \\frac{d}{1 + r} < X \\leq \\frac{u}{1 + r}, \
\n\\alpha\\left( u – d \\right) & X > \\frac{u}{1 + r}. \
\n\\end{matrix} \\right.\\ $$<\/p>\n

The first and second moments of $Y^{L}$ can be expressed as<\/p>\n

$E\\left( Y^{L} \\right) = \\alpha\\left( 1 + r \\right)\\left\\lbrack E\\left( X \\land \\frac{u}{1 + r} \\right) – E\\left( X \\land \\frac{d}{1 + r} \\right) \\right\\rbrack$,<\/p>\n

and<\/p>\n

$E\\left\\lbrack \\left( Y^{L} \\right)^{2} \\right\\rbrack = \\alpha^{2}\\left( 1 + r \\right)^{2}\\left{ E\\left\\lbrack \\left( X \\land \\frac{u}{1 + r} \\right)^{2} \\right\\rbrack – E\\left\\lbrack \\left( X \\land \\frac{d}{1 + r} \\right)^{2} \\right\\rbrack – 2\\left( \\frac{d}{1 + r} \\right)\\left\\lbrack E\\left( X \\land \\frac{u}{1 + r} \\right) – E\\left( X \\land \\frac{d}{1 + r} \\right) \\right\\rbrack \\right}$,<\/p>\n

respectively.<\/p>\n

The formulae given for the first and second moments of $Y^{L}$ are
\ngeneral. Under full coverage, $\\alpha = 1$, $r = 0$, $u = \\infty$,
\n$d = 0$ and $E\\left( Y^{L} \\right)$ reduces to $E\\left( X \\right)$. If
\nonly an ordinary deductible is imposed, $\\alpha = 1$, $r = 0$,
\n$u = \\infty$ and $E\\left( Y^{L} \\right)$ reduces to
\n$E\\left( X \\right) – E\\left( X \\land d \\right)$. If only a policy limit
\nis imposed $\\alpha = 1$, $r = 0$, $d = 0$ and $E\\left( Y^{L} \\right)$
\nreduces to $E\\left( X \\land u \\right)$.<\/p>\n

Example 3.19 (SOA)<\/strong><\/p>\n

The ground up loss random variable for a health insurance policy in 2006
\nis modeled with X<\/em>, an exponential distribution with mean 1000. An
\ninsurance policy pays the loss above an ordinary deductible of 100, with
\na maximum annual payment of 500. The ground up loss random variable is
\nexpected to be 5% larger in 2007, but the insurance in 2007 has the same
\ndeductible and maximum payment as in 2006. Find the percentage increase
\nin the expected cost per payment from 2006 to 2007.<\/p>\n

Solution<\/strong><\/p>\n

$$Y_{2006}^{L} = \\left{ \\begin{matrix}
\n0 & X \\leq 100, \
\nX – 100 & 100 < X \\leq 600, \
\n500 & X > 600. \
\n\\end{matrix} \\right.\\ $$<\/p>\n

$$Y_{2007}^{L} = \\left{ \\begin{matrix}
\n0 & X \\leq 95.24, \
\n1.05X – 100 & 95.24 < X \\leq 571.43, \
\n500 & X > 571.43. \
\n\\end{matrix} \\right.\\ $$<\/p>\n

$$E\\left( Y_{2006}^{L} \\right) = E\\left( X \\land 600 \\right) – E\\left( X \\land 100 \\right) = 1000\\left( {1 – e}^{- \\frac{600}{1000}} \\right) – 1000\\left( {1 – e}^{- \\frac{100}{1000}} \\right)$$<\/p>\n

$= 356.026$.<\/p>\n

$$E\\left( Y_{2007}^{L} \\right) = 1.05\\left\\lbrack E\\left( X \\land 571.43 \\right) – E\\left( X \\land 95.24 \\right) \\right\\rbrack$$<\/p>\n

$= 1.05\\left\\lbrack 1000\\left( {1 – e}^{- \\frac{571.43}{1000}} \\right) – 1000\\left( {1 – e}^{- \\frac{95.24}{1000}} \\right) \\right\\rbrack$<\/p>\n

$\\mathbf{=}361.659$.<\/p>\n

$E\\left( Y_{2006}^{P} \\right) = \\frac{356.026}{e^{- \\frac{100}{1000}} = 393.469}$.<\/p>\n

$E\\left( Y_{2007}^{P} \\right) = \\frac{361.659}{e^{- \\frac{95.24}{1000}} = 397.797}$.<\/p>\n

There is an increase of 1.1% from 2006 to 2007.<\/p>\n

3.4.4 Reinsurance<\/strong><\/p>\n

In Section 3.4.1 we introduced the policy deductible, which is a
\ncontractual arrangement under which an insured transfers part of the
\nrisk by securing coverage from an insurer in return for an insurance
\npremium. Under that policy, when the loss exceeds the deductible, the
\ninsurer is not required to pay until the insured has paid the fixed
\ndeductible. We now introduce reinsurance, a mechanism of insurance for
\ninsurance companies. Reinsurance is a contractual arrangement under
\nwhich an insurer transfers part of the underlying insured risk by
\nsecuring coverage from another insurer (referred to as a reinsurer) in
\nreturn for a reinsurance premium. Although reinsurance involves a
\nrelationship between three parties: the original insured, the insurer
\n(often referred to as cedent or cedant) and the reinsurer, the parties
\nof the reinsurance agreement are only the primary insurer and the
\nreinsurer. There is no contractual agreement between the original
\ninsured and the reinsurer. The reinsurer is not required to pay under
\nthe reinsurance contract until the insurer has paid a loss to its
\noriginal insured. The amount retained by the primary insurer in the
\nreinsurance agreement (the reinsurance deductible) is called retention.<\/p>\n

Reinsurance arrangements allow insurers with limited financial resources
\nto increase the capacity to write insurance and meet client requests for
\nlarger insurance coverage while reducing the impact of potential losses
\nand protecting the insurance company against catastrophic losses.
\nReinsurance also allows the primary insurer to benefit from underwriting
\nskills, expertize and proficient complex claim file handling of the
\nlarger reinsurance companies.<\/p>\n

Example 3.20 (SOA)<\/strong><\/p>\n

In 2005 a risk has a two-parameter Pareto distribution with $\\alpha = 2$
\nand $\\theta = 3000$. In 2006 losses inflate by 20%. Insurance on the
\nrisk has a deductible of 600 in each year. $P_{i}$, the premium in year
\n$i$, equals 1.2 times expected claims. The risk is reinsured with a
\ndeductible that stays the same in each year. $R_{i}$, the reinsurance
\npremium in year $i$, equals 1.1 times the expected reinsured claims.
\n$\\frac{R_{2005}}{P_{2005} = 0.55}$. Calculate
\n$\\frac{R_{2006}}{P_{2006}}$.<\/p>\n

Solution<\/strong><\/p>\n

$X_{i}:$ The risk in year $i$<\/p>\n

$Y_{i}:$ The insured claim in year $i$<\/p>\n

$P_{i}:$ The insurance premium in year $i$<\/p>\n

$Y_{i}^{R}:$ The reinsured claim in year $i$<\/p>\n

$R_{i}:$ The reinsurance premium in year $i$<\/p>\n

$d:$ The insurance deductible in year $i$ (the insurance deductible is
\nfixed each year, equal to 600)<\/p>\n

$d^{R}:$ The reinsurance deductible or retention in year $i$ (the
\nreinsurance deductible is fixed each year, but unknown)<\/p>\n

where $i = 2005,\\ 2006$<\/p>\n

$Y_{i} = \\left{ \\begin{matrix}
\n0 & X_{i} \\leq 600 \
\nX_{i} – 600 & X_{i} > 600 \
\n\\end{matrix} \\right.\\ $<\/p>\n

where $i = 2005,\\ 2006$<\/p>\n

$$X_{2005}\\sim Pa\\left( 2,3000 \\right)$$<\/p>\n

$$E\\left( Y_{2005} \\right) = E\\left( X_{2005} – 600 \\right){+} = E\\left( X<\/em>{2005} \\right) – E\\left( X_{2005} \\land 600 \\right)$$<\/p>\n

$= 3000 – 3000\\left( 1 – \\frac{3000}{3600} \\right) = 2500$<\/p>\n

$$P_{2005} = 1.2E\\left( Y_{2005} \\right) = 3000$$<\/p>\n

Since $X_{2006} = 1.2X_{2005}$ and Pareto is a scale distribution with
\nscale parameter $\\theta$, then $X_{2006}\\sim Pa\\left( 2,3600 \\right)$<\/p>\n

$$E\\left( Y_{2006} \\right) = E\\left( X_{2006} – 600 \\right){+} = E\\left( X<\/em>{2006} \\right) – E\\left( X_{2006} \\land 600 \\right)$$<\/p>\n

$= 3600 – 3600\\left( 1 – \\frac{3600}{4200} \\right) = 3085.714$<\/p>\n

$$P_{2006} = 1.2E\\left( Y_{2006} \\right) = 3702.857$$<\/p>\n

$$Y_{i}^{R} = \\left{ \\begin{matrix}
\n0 & X_{i} – 600 \\leq d^{R} \
\nX_{i} – 600 – d^{R} & X_{i} – 600 > d^{R} \
\n\\end{matrix} \\right.\\ $$<\/p>\n

Since $\\frac{R_{2005}}{P_{2005}} = 0.55$, then
\n$R_{2005} = 3000 \\times 0.55 = 1650$<\/p>\n

Since $R_{2005} = 1.1E\\left( Y_{2005}^{R} \\right)$, then
\n$E\\left( Y_{2005}^{R} \\right) = \\frac{1650}{1.1} = 1500$<\/p>\n

$$E\\left( Y_{2005}^{R} \\right) = E\\left( X_{2005} – 600 – d^{R} \\right){+} = E\\left( X<\/em>{2005} \\right) – E\\left( X_{2005} \\land \\left( 600 + d^{R} \\right) \\right)$$<\/p>\n

$= 3000 – 3000\\left( 1 – \\frac{3000}{3600 + d^{R}} \\right) = 1500 \\Rightarrow d^{R} = 2400$<\/p>\n

$$E\\left( Y_{2006}^{R} \\right) = E\\left( X_{2006} – 600 – d^{R} \\right){+} = E\\left( X<\/em>{2006} – 3000 \\right){+} = E\\left( X<\/em>{2006} \\right) – E\\left( X_{2006} \\land 3000 \\right)$$<\/p>\n

$= 3600 – 3600\\left( 1 – \\frac{3600}{6600} \\right) = 1963.636$<\/p>\n

$$R_{2006} = 1.1E\\left( Y_{2006}^{R} \\right) = 1.1 \\times 1963.636 = 2160$$<\/p>\n

Therefore $\\frac{R_{2006}}{P_{2006}} = \\frac{2160}{3702.857} = 0.583$<\/p>\n

3.5 Maximum likelihood estimation<\/strong><\/p>\n

In this section we estimate statistical parameters using the method of
\nmaximum likelihood. Maximum likelihood estimates in the presence of
\ngrouping, truncation or censoring are calculated.<\/p>\n

3.5.1 Maximum likelihood estimators for complete data<\/strong><\/p>\n

Pricing of insurance premiums and estimation of claim reserving are
\namong many actuarial problems that involve modeling the severity of loss
\n(claim size). The principles for using maximum likelihood to estimate
\nmodel parameters were introduced in Chapter 2. In
\nthis\u00a0section,\u00a0we\u00a0present a few examples to illustrate how actuaries fit
\na parametric distribution model to a set of claim data using maximum
\nlikelihood. In these examples we derive the asymptotic
\nvariance\u00a0of\u00a0maximum-likelihood estimators of the model parameters. We
\nuse the delta method to derive the asymptotic variances of functions of
\nthese parameters.<\/p>\n

Example 3.21<\/strong><\/p>\n

You are given the following:<\/p>\n

A random sample of claim amounts: 8,000 10,000 12,000 15,000.<\/p>\n

Claim amounts follow an inverse exponential distribution, with parameter
\n$\\theta$.<\/p>\n

i. Calculate the maximum likelihood estimator for $\\theta$.<\/p>\n

ii. Approximate the variance of the maximum likelihood estimator.<\/p>\n

iii. Determine an approximate 95% confidence interval for $\\theta$.<\/p>\n

iv. Determine an approximate 95% confidence interval for
\n $P\\left( X \\leq 9,000 \\right).$<\/p>\n

Solution<\/strong><\/p>\n

$f_{X}\\left( x \\right) = \\frac{\\theta e^{- \\frac{\\theta}{x}}}{x^{2}}$,
\n$x > 0$.<\/p>\n

The likelihood function, $L\\left( \\theta \\right)$, can be viewed as the
\nprobability of the observed data, written as a function of the model\u2019s
\nparameter $\\theta$<\/p>\n

$L\\left( \\theta \\right) = \\prod_{i = 1}^{4}{f_{X_{i}}\\left( x_{i} \\right)} = \\frac{\\theta^{4}e^{- \\theta\\sum_{i = 1}^{4}\\frac{1}{x_{i}}}}{\\prod_{i = 1}^{4}x_{i}^{2}}$.<\/em><\/p>\n

The loglikelihood function, $\\text{lnL}\\left( \\theta \\right)$, is the
\nsum of the individual logarithms.<\/p>\n

$\\text{lnL}\\left( \\theta \\right) = 4ln\\theta – \\theta\\sum_{i = 1}^{4}\\frac{1}{x_{i}} – 2\\sum_{i = 1}^{4}\\ln x_{i}$.<\/p>\n

$\\frac{\\text{dlnL}\\left( \\theta \\right)}{\\text{d\u03b8}} = \\frac{4}{\\theta} – \\sum_{i = 1}^{4}\\frac{1}{x_{i}}$.<\/p>\n

The maximum likelihood estimator of $\\theta$, denoted by $\\hat{\\theta}$,
\nis the solution to the equation<\/p>\n

$\\frac{4}{\\hat{\\theta}} – \\sum_{i = 1}^{4}{\\frac{1}{x_{i}} = 0}$. Thus,
\n$\\hat{\\theta} = \\frac{4}{\\sum_{i = 1}^{4}\\frac{1}{x_{i}}} = 10,667$<\/p>\n

The second derivative of $\\text{lnL}\\left( \\theta \\right)$ is given by<\/p>\n

$\\frac{d^{2}\\text{lnL}\\left( \\theta \\right)}{d\\theta^{2}} = \\frac{- 4}{\\theta^{2}}$.<\/p>\n

Evaluating the second derivative of the loglikelihood function at
\n$\\hat{\\theta} = 10,667$ gives a negative value, indicating
\n$\\hat{\\theta}$ as the value that maximizes the loglikelihood function.<\/p>\n

Taking reciprocal of negative expectation of the second derivative of
\n$\\text{lnL}\\left( \\theta \\right)$, we obtain an estimate of the variance
\nof $\\hat{\\theta}$<\/p>\n

$\\hat{V}\\left( \\hat{\\theta} \\right) = \\left. \\ \\left\\lbrack E\\left( \\frac{d^{2}\\text{lnL}\\left( \\theta \\right)}{d\\theta^{2}} \\right) \\right\\rbrack^{- 1} \\right|_{\\theta = \\hat{\\theta}} = \\frac{{\\hat{\\theta}}^{2}}{4} = 28,446,222$.<\/p>\n

It should be noted that as the sample size $n \\rightarrow \\infty$, the
\ndistribution of the maximum likelihood estimator $\\hat{\\theta}$
\nconverges to a normal distribution with mean $\\theta$ and variance
\n$\\hat{V}\\left( \\hat{\\theta} \\right)$. The approximate confidence
\ninterval in this example is based on the assumption of normality,
\ndespite the small sample size, only for the purpose of illustration.<\/p>\n

The 95% confidence interval for $\\theta$ is given by<\/p>\n

$10,667 \\pm 1.96\\sqrt{28,446,222} = \\left( 213.34,\\ 21,120.66 \\right)$.<\/p>\n

The distribution function of X<\/em> is
\n$F\\left( x \\right) = 1 – e^{- \\frac{x}{\\theta}}$. Then, the maximum
\nlikelihood estimate of $g\\left( \\theta \\right) = F\\left( 9,000 \\right)$
\nis
\n$g\\left( \\hat{\\theta} \\right) = 1 – e^{- \\frac{9,000}{10,667}} = 0.57$.<\/p>\n

We use the delta method to approximate the variance of
\n$g\\left( \\hat{\\theta} \\right)$.<\/p>\n

$\\frac{\\text{dg}\\left( \\theta \\right)}{\\text{d\u03b8}} = {- \\frac{9,000}{\\theta^{2}}e}^{- \\frac{9,000}{\\theta}}$.<\/p>\n

$\\hat{V}\\left\\lbrack g\\left( \\hat{\\theta} \\right) \\right\\rbrack = \\left( – {\\frac{9,000}{{\\hat{\\theta}}^{2}}e}^{- \\frac{9,000}{\\hat{\\theta}}} \\right)^{2}\\hat{V}\\left( \\hat{\\theta} \\right) = 0.0329$.<\/p>\n

The 95% confidence interval for $F\\left( 9,000 \\right)$ is given by<\/p>\n

$0.57 \\pm 1.96\\sqrt{0.0329} = \\left( 0.214,\\ 0.926 \\right)$.<\/p>\n

Example 3.22<\/strong><\/p>\n

A random sample of size 6 is from a lognormal distribution with
\nparameters $\\mu$ and $\\sigma$. The sample values are 200, 3,000, 8,000,
\n60,000, 60,000, 160,000.<\/p>\n

i. Calculate the maximum likelihood estimator for $\\mu$ and $\\sigma$.<\/p>\n

ii. Estimate the covariance matrix of the maximum likelihood estimator.<\/p>\n

iii. Determine approximate 95% confidence intervals for $\\mu$ and
\n $\\sigma$.<\/p>\n

iv. Determine an approximate 95% confidence interval for the mean of the
\n lognormal distribution.<\/p>\n

Solution<\/strong><\/p>\n

$f_{X}\\left( x \\right) = \\frac{1}{\\text{x\u03c3}\\sqrt{2\\pi}}exp – \\frac{1}{2}\\left( \\frac{lnx – \\mu}{\\sigma} \\right)^{2}$,
\n$x > 0$.<\/p>\n

The likelihood function, $L\\left( \\mu,\\sigma \\right)$, is the product of
\nthe pdf for each data point.<\/p>\n

$L\\left( \\mu,\\sigma \\right) = \\prod_{i = 1}^{6}{f_{X_{i}}\\left( x_{i} \\right)} = \\frac{1}{\\sigma^{6}\\left( 2\\pi \\right)^{3}\\prod_{i = 1}^{6}x_{i}}exp – \\frac{1}{2}\\sum_{i = 1}^{6}\\left( \\frac{\\ln x_{i} – \\mu}{\\sigma} \\right)^{2}$.<\/p>\n

The loglikelihood function, $\\text{lnL}\\left( \\mu,\\sigma \\right)$, is
\nthe sum of the individual logarithms.<\/p>\n

$\\text{lnL}\\left( \\mu,\\sigma \\right) = – 6ln\\sigma – 3ln\\left( 2\\pi \\right) – \\sum_{i = 1}^{6}\\ln x_{i} – \\frac{1}{2}\\sum_{i = 1}^{6}\\left( \\frac{\\ln x_{i} – \\mu}{\\sigma} \\right)^{2}$.<\/p>\n

The first partial derivatives are<\/p>\n

$\\frac{\\partial lnL\\left( \\mu,\\sigma \\right)}{\\partial\\mu} = \\frac{1}{\\sigma^{2}}\\sum_{i = 1}^{6}\\left( \\ln x_{i} – \\mu \\right)$.<\/p>\n

$\\frac{\\partial lnL\\left( \\mu,\\sigma \\right)}{\\partial\\sigma} = \\frac{- 6}{\\sigma} + \\frac{1}{\\sigma^{3}}\\sum_{i = 1}^{6}\\left( \\ln x_{i} – \\mu \\right)^{2}$.<\/p>\n

The maximum likelihood estimators of $\\mu$ and $\\sigma$, denoted by
\n$\\hat{\\mu}$ and $\\hat{\\sigma}$, are the solutions to the equations<\/p>\n

$\\frac{1}{{\\hat{\\sigma}}^{2}}\\sum_{i = 1}^{6}\\left( lnx_{i} – \\hat{\\mu} \\right) = 0$.<\/p>\n

$\\frac{- 6}{\\hat{\\sigma}} + \\frac{1}{{\\hat{\\sigma}}^{3}}\\sum_{i = 1}^{6}\\left( \\ln x_{i} – \\hat{\\mu} \\right)^{2} = 0$.<\/p>\n

These yield the estimates<\/p>\n

$\\hat{\\mu} = \\frac{\\sum_{i = 1}^{6}{\\ln x_{i}}}{6} = 9.38$ and
\n${\\hat{\\sigma}}^{2} = \\frac{\\sum_{i = 1}^{6}\\left( \\ln x_{i} – \\hat{\\mu} \\right)^{2}}{6} = 5.12$.<\/p>\n

The second partial derivatives are<\/p>\n

$\\frac{\\partial^{2}\\text{lnL}\\left( \\mu,\\sigma \\right)}{\\partial\\mu^{2}} = \\frac{- 6}{\\sigma^{2}}$,
\n$\\frac{\\partial^{2}\\text{lnL}\\left( \\mu,\\sigma \\right)}{\\partial\\mu\\partial\\sigma} = \\frac{- 2}{\\sigma^{3}}\\sum_{i = 1}^{6}\\left( \\ln x_{i} – \\mu \\right)$
\nand
\n$\\frac{\\partial^{2}\\text{lnL}\\left( \\mu,\\sigma \\right)}{\\partial\\sigma^{2}} = \\frac{6}{\\sigma^{2}} – \\frac{3}{\\sigma^{4}}\\sum_{i = 1}^{6}\\left( \\ln x_{i} – \\mu \\right)^{2}$.<\/p>\n

To derive the covariance matrix of the mle we need to find the
\nexpectations of the second derivatives. Since the random variable $X$ is
\nfrom a lognormal distribution with parameters $\\mu$ and $\\sigma$, then
\n$\\text{lnX}$ is normally distributed with mean $\\mu$ and variance
\n$\\sigma^{2}$.<\/p>\n

$E\\left( \\frac{\\partial^{2}\\text{lnL}\\left( \\mu,\\sigma \\right)}{\\partial\\mu^{2}} \\right) = E\\left( \\frac{- 6}{\\sigma^{2}} \\right) = \\frac{- 6}{\\sigma^{2}}$,<\/p>\n

$E\\left( \\frac{\\partial^{2}\\text{lnL}\\left( \\mu,\\sigma \\right)}{\\partial\\mu\\partial\\sigma} \\right) = \\frac{- 2}{\\sigma^{3}}\\sum_{i = 1}^{6}{E\\left( \\ln x_{i} – \\mu \\right)} = \\frac{- 2}{\\sigma^{3}}\\sum_{i = 1}^{6}\\left\\lbrack E\\left( \\ln x_{i} \\right) – \\mu \\right\\rbrack$=$\\frac{- 2}{\\sigma^{3}}\\sum_{i = 1}^{6}\\left( \\mu – \\mu \\right) = 0$,<\/p>\n

and<\/p>\n

$E\\left( \\frac{\\partial^{2}\\text{lnL}\\left( \\mu,\\sigma \\right)}{\\partial\\sigma^{2}} \\right) = \\frac{6}{\\sigma^{2}} – \\frac{3}{\\sigma^{4}}\\sum_{i = 1}^{6}{E\\left( \\ln x_{i} – \\mu \\right)}^{2} = \\frac{6}{\\sigma^{2}} – \\frac{3}{\\sigma^{4}}\\sum_{i = 1}^{6}{V\\left( \\ln x_{i} \\right) = \\frac{6}{\\sigma^{2}} – \\frac{3}{\\sigma^{4}}\\sum_{i = 1}^{6}{\\sigma^{2} = \\frac{- 12}{\\sigma^{2}}}}$.<\/p>\n

Using the negatives of these expectations we obtain the Fisher
\ninformation matrix $\\begin{bmatrix}
\n\\frac{6}{\\sigma^{2}} & 0 \
\n0 & \\frac{12}{\\sigma^{2}} \
\n\\end{bmatrix}$.<\/p>\n

The covariance matrix, $\\Sigma$, is the inverse of the Fisher
\ninformation matrix $\\Sigma = \\begin{bmatrix}
\n\\frac{\\sigma^{2}}{6} & 0 \
\n0 & \\frac{\\sigma^{2}}{12} \
\n\\end{bmatrix}$.<\/p>\n

The estimated matrix is given by $\\hat{\\Sigma} = \\begin{bmatrix}
\n0.8533 & 0 \
\n0 & 0.4267 \
\n\\end{bmatrix}$.<\/p>\n

The 95% confidence interval for $\\mu$ is given by
\n$9.38 \\pm 1.96\\sqrt{0.8533} = \\left( 7.57,\\ 11.19 \\right)$.<\/p>\n

The 95% confidence interval for $\\sigma^{2}$ is given by
\n$5.12 \\pm 1.96\\sqrt{0.4267} = \\left( 3.84,\\ 6.40 \\right)$.<\/p>\n

The mean of X<\/em> is $\\exp\\left( \\mu + \\frac{\\sigma^{2}}{2} \\right)$.
\nThen, the maximum likelihood estimate of
\n$g\\left( \\mu,\\sigma \\right) = \\exp\\left( \\mu + \\frac{\\sigma^{2}}{2} \\right)$
\nis
\n$g\\left( \\hat{\\mu},\\hat{\\sigma} \\right) = \\exp\\left( \\hat{\\mu} + \\frac{{\\hat{\\sigma}}^{2}}{2} \\right) = 153,277$.<\/p>\n

We use the delta method to approximate the variance of the mle
\n$g\\left( \\hat{\\mu},\\hat{\\sigma} \\right)$.<\/p>\n

$\\frac{\\partial g\\left( \\mu,\\sigma \\right)}{\\partial\\mu} = exp\\left( \\mu + \\frac{\\sigma^{2}}{2} \\right)$
\nand
\n$\\frac{\\partial g\\left( \\mu,\\sigma \\right)}{\\partial\\sigma} = \\sigma exp\\left( \\mu + \\frac{\\sigma^{2}}{2} \\right)$.<\/p>\n

Using the delta method, the approximate variance of
\n$g\\left( \\hat{\\mu},\\hat{\\sigma} \\right)$ is given by<\/p>\n

$$\\left. \\ \\hat{V}\\left( g\\left( \\hat{\\mu},\\hat{\\sigma} \\right) \\right) = \\begin{bmatrix}
\n\\frac{\\partial g\\left( \\mu,\\sigma \\right)}{\\partial\\mu} & \\frac{\\partial g\\left( \\mu,\\sigma \\right)}{\\partial\\sigma} \
\n\\end{bmatrix}\\Sigma\\begin{bmatrix}
\n\\frac{\\partial g\\left( \\mu,\\sigma \\right)}{\\partial\\mu} \
\n\\frac{\\partial g\\left( \\mu,\\sigma \\right)}{\\partial\\sigma} \
\n\\end{bmatrix} \\right|_{\\mu = \\hat{\\mu},\\sigma = \\hat{\\sigma}}$$<\/p>\n

$= \\begin{bmatrix}
\n153,277 & 346,826 \
\n\\end{bmatrix}\\begin{bmatrix}
\n0.8533 & 0 \
\n0 & 0.4267 \
\n\\end{bmatrix}\\begin{bmatrix}
\n153,277 \
\n346,826 \
\n\\end{bmatrix} =$71,374,380,000<\/p>\n

The 95% confidence interval for
\n$\\exp\\left( \\mu + \\frac{\\sigma^{2}}{2} \\right)$ is given by<\/p>\n

$153,277 \\pm 1.96\\sqrt{71,374,380,000} = \\left( – 370,356,\\ 676,910 \\right)$.<\/p>\n

Since the mean of the lognormal distribution cannot be negative, we
\nshould replace the negative lower limit in the previous interval by a
\nzero.<\/p>\n

3.5.2 Maximum likelihood estimators for grouped data<\/strong><\/p>\n

In the previous section we considered the maximum likelihood estimation
\nof continuous models from complete (individual) data. Each individual
\nobservation is recorded, and its contribution to the likelihood function
\nis the density at that value. In this section we consider the problem of
\nobtaining maximum likelihood estimates of parameters from grouped data.
\nThe observations are only available in grouped form, and the
\ncontribution of each observation to the likelihood function is the
\nprobability of falling in a specific group (interval). Let $n_{j}$
\nrepresent the number of observations in the interval
\n$\\left( \\left. \\ c_{j – 1},c_{j} \\right\\rbrack \\right.\\ \\text{.\\ }$The
\ngrouped data likelihood function is thus given by<\/p>\n

$L\\left( \\theta \\right) = \\prod_{j = 1}^{k}\\left\\lbrack F\\left( \\left. \\ c_{j} \\right|\\theta \\right) – F\\left( \\left. \\ c_{j – 1} \\right|\\theta \\right) \\right\\rbrack^{n_{j}}$,<\/p>\n

where $c_{0}$ is the smallest possible observation (often set to zero)
\nand $c_{k}$ is the largest possible observation (often set to infinity).<\/p>\n

Example 3.23 (SOA)<\/strong><\/p>\n

For a group of policies, you are given:<\/p>\n

i. Losses follow the distribution function
\n $F\\left( x \\right) = 1 – \\frac{\\theta}{x}$, $\\theta < x < \\infty$.<\/p>\n

ii. A sample of 20 losses resulted in the following:<\/p>\n

Interval Number of Losses<\/p>\n

\n

$$x \\leq 10$$ 9
\n $$10 < x \\leq 25$$ 6
\n $$x > 25$$ 5<\/p>\n

Calculate the maximum likelihood estimate of $\\theta$.<\/p>\n

Solution<\/strong><\/p>\n

The contribution of each of the 9 observations in the first interval to
\nthe likelihood function is the probability of $X \\leq 10$; that is,
\n$P\\left( X \\leq 10 \\right) = F\\left( 10 \\right)$. Similarly, the
\ncontributions of each of 6 and 5 observations in the second and third
\nintervals are
\n$P\\left( 10 < X \\leq 25 \\right) = F\\left( 25 \\right) – F(10)$ and
\n$P\\left( X > 25 \\right) = 1 – F(25)$, respectively. The likelihood
\nfunction is thus given by<\/p>\n

$$L\\left( \\theta \\right) = \\left\\lbrack F\\left( 10 \\right) \\right\\rbrack^{9}\\left\\lbrack F\\left( 25 \\right) – F(10) \\right\\rbrack^{6}\\left\\lbrack 1 – F(25) \\right\\rbrack^{5}$$<\/p>\n

${= \\left( 1 – \\frac{\\theta}{10} \\right)}^{9}\\left( \\frac{\\theta}{10} – \\frac{\\theta}{25} \\right)^{6}\\left( \\frac{\\theta}{25} \\right)^{5}$<\/p>\n

${= \\left( \\frac{10 – \\theta}{10} \\right)}^{9}\\left( \\frac{15\\theta}{250} \\right)^{6}\\left( \\frac{\\theta}{25} \\right)^{5}$.<\/p>\n

Then,
\n$\\text{lnL}\\left( \\theta \\right) = 9ln\\left( 10 – \\theta \\right) + 6ln\\theta + 5ln\\theta – 9ln10 + 6ln15 – 6ln250 – 5ln25$.<\/p>\n

$\\frac{\\text{dlnL}\\left( \\theta \\right)}{\\text{d\u03b8}} = \\frac{- 9}{\\left( 10 – \\theta \\right)} + \\frac{6}{\\theta} + \\frac{5}{\\theta}$.<\/p>\n

The maximum likelihood estimator, $\\hat{\\theta}$, is the solution to the
\nequation
\n$\\frac{- 9}{\\left( 10 – \\hat{\\theta} \\right)} + \\frac{11}{\\hat{\\theta}} = 0$;
\nand $\\hat{\\theta} = 5.5$.<\/p>\n

3.5.3 Maximum likelihood estimators for censored data<\/strong><\/p>\n

Another distinguishing feature of data gathering mechanism is censoring.
\nWhile for some event of interest (losses, claims, lifetimes, etc.) the
\ncomplete data maybe available, for others only partial information is
\navailable; information that the observation exceeds a specific value.
\nThe limited policy introduced in Section 3.4.2 is an example of right
\ncensoring. Any loss greater than or equal to the policy limit is
\nrecorded at the limit. The contribution of the censored observation to
\nthe likelihood function is the probability of the random variable
\nexceeding this specific limit. Note that contributions of both complete
\nand censored data share the survivor function, for a complete point this
\nsurvivor function is multiplied by the hazard function, but for a
\ncensored observation it is not.<\/p>\n

**Example 3.24 (SOA) **<\/p>\n

The random variable has survival function:
\n$S_{X}\\left( x \\right) = \\frac{\\theta^{4}}{\\left( \\theta^{2} + x^{2} \\right)^{2}}$.<\/p>\n

Two values of $X$ are observed to be 2 and 4. One other value exceeds 4.<\/p>\n

Calculate the maximum likelihood estimate of $\\theta$.<\/p>\n

Solution<\/strong><\/p>\n

The contributions of the two observations 2 and 4 are
\n$f_{X}\\left( 2 \\right)$ and $f_{X}\\left( 4 \\right)$ respectively. The
\ncontribution of the third observation, which is only known to exceed 4
\nis $S_{X}\\left( 4 \\right)$. The likelihood function is thus given by<\/p>\n

$L\\left( \\theta \\right) = f_{X}\\left( 2 \\right)f_{X}\\left( 4 \\right)S_{X}\\left( 4 \\right)$.<\/p>\n

The probability density function of $X$ is given by
\n$f_{X}\\left( x \\right) = \\frac{4x\\theta^{4}}{\\left( \\theta^{2} + x^{2} \\right)^{3}}$.
\nThus,<\/p>\n

$L\\left( \\theta \\right) = \\frac{8\\theta^{4}}{\\left( \\theta^{2} + 4 \\right)^{3}}\\frac{16\\theta^{4}}{\\left( \\theta^{2} + 16 \\right)^{3}}\\frac{\\theta^{4}}{\\left( \\theta^{2} + 16 \\right)^{2}} = \\frac{128\\theta^{12}}{\\left( \\theta^{2} + 4 \\right)^{3}\\left( \\theta^{2} + 16 \\right)^{5}}$,<\/p>\n

$\\text{lnL}\\left( \\theta \\right) = ln128 + 12ln\\theta – 3ln\\left( \\theta^{2} + 4 \\right) – 5ln\\left( \\theta^{2} + 16 \\right)$,<\/p>\n

and<\/p>\n

$\\frac{\\text{dlnL}\\left( \\theta \\right)}{\\text{d\u03b8}} = \\frac{12}{\\theta} – \\frac{6\\theta}{\\left( \\theta^{2} + 4 \\right)} – \\frac{10\\theta}{\\left( \\theta^{2} + 16 \\right)}$.<\/p>\n

The maximum likelihood estimator, $\\hat{\\theta}$, is the solution to the
\nequation
\n$\\frac{12}{\\hat{\\theta}} – \\frac{6\\hat{\\theta}}{\\left( {\\hat{\\theta}}^{2} + 4 \\right)} – \\frac{10\\hat{\\theta}}{\\left( {\\hat{\\theta}}^{2} + 16 \\right)} = 0$
\nor
\n$12\\left( {\\hat{\\theta}}^{2} + 4 \\right)\\left( {\\hat{\\theta}}^{2} + 16 \\right) – 6{\\hat{\\theta}}^{2}\\left( {\\hat{\\theta}}^{2} + 16 \\right) – 10{\\hat{\\theta}}^{2}\\left( {\\hat{\\theta}}^{2} + 4 \\right) = – 4{\\hat{\\theta}}^{4} + 104{\\hat{\\theta}}^{2} + 768 = 0$,
\nwhich yields ${\\hat{\\theta}}^{2} = 32$ and $\\hat{\\theta} = 5.7$.<\/p>\n

3.5.4 Maximum likelihood estimators for truncated data<\/strong><\/p>\n

This section is concerned with the maximum likelihood estimation of the
\ncontinuous distribution of the random variable $X$ when the data is
\nincomplete due to truncation. If the values of $X$ are truncated at $d$,
\nthen it should be noted that we would not have been aware of the
\nexistence of these values had they not exceeded $d$. The policy
\ndeductible introduced in Section 3.4.1 is an example of left truncation.
\nAny loss less than or equal to the deductible is not recorded. The
\ncontribution to the likelihood function of an observation $x$ truncated
\nat $d$ will be a conditional probability and the $f_{X}\\left( x \\right)$
\nwill be replaced by
\n$\\frac{f_{X}\\left( x \\right)}{S_{X}\\left( d \\right)}$.<\/p>\n

Example 3.25 (SOA)<\/strong><\/p>\n

For the single parameter Pareto distribution with $\\theta = 2$, maximum
\nlikelihood estimation is applied to estimate the parameter $\\alpha$.
\nFind the estimated mean of the ground up loss distribution based on the
\nmaximum likelihood estimate of $\\alpha$ for the following data set:<\/p>\n

Ordinary policy deductible of 5, maximum covered loss of 25 (policy
\nlimit 20).<\/p>\n

8 insurance payment amounts: 2, 4, 5, 5, 8, 10, 12, 15<\/p>\n

2 limit payments: 20, 20.<\/p>\n

Solution<\/strong><\/p>\n

The contributions of the different observations can be summarized as
\nfollows:<\/p>\n

For the exact loss: $f_{X}\\left( x \\right)$<\/p>\n

For censored observations: $S_{X}\\left( 25 \\right)$.<\/p>\n

For truncated observations:
\n$\\frac{f_{X}\\left( x \\right)}{S_{X}\\left( 5 \\right)}$.<\/p>\n

Given that ground up losses smaller than 5 are omitted from the data
\nset, the contribution of all observations should be conditional on
\nexceeding 5. The likelihood function becomes<\/p>\n

$L\\left( \\alpha \\right) = \\frac{\\prod_{i = 1}^{8}{f_{X}\\left( x_{i} \\right)}}{\\left\\lbrack S_{X}\\left( 5 \\right) \\right\\rbrack^{8}}\\left\\lbrack \\frac{S_{X}\\left( 25 \\right)}{S_{X}\\left( 5 \\right)} \\right\\rbrack^{2}$.<\/p>\n

For the single parameter Pareto the probability density and distribution
\nfunctions are given by<\/p>\n

$f_{X}\\left( x \\right) = \\frac{\\alpha\\theta^{\\alpha}}{x^{\\alpha + 1}}$
\nand
\n$F_{X}\\left( x \\right) = 1 – \\left( \\frac{\\theta}{x} \\right)^{\\alpha}$
\nfor $x > \\theta$, respectively.<\/p>\n

Then, the likelihood and loglikelihood functions are given by<\/p>\n

$L\\left( \\alpha \\right) = \\frac{\\alpha^{8}}{\\prod_{i = 1}^{8}x_{i}^{\\alpha + 1}}\\frac{5^{10\\alpha}}{25^{2\\alpha}}$,<\/p>\n

$\\text{lnL}\\left( \\alpha \\right) = 8ln\\alpha – \\left( \\alpha + 1 \\right)\\sum_{i = 1}^{8}{\\ln x_{i}} + 10\\alpha ln5 – 2\\alpha ln25$.<\/p>\n

$\\frac{\\text{dlnL}\\left( \\alpha \\right)}{\\text{d\u03b1}} = \\frac{8}{\\alpha} – \\sum_{i = 1}^{8}{\\ln x_{i}} + 10ln5 – 2ln25$.<\/p>\n

The maximum likelihood estimator, $\\hat{\\alpha}$, is the solution to the
\nequation<\/p>\n

$\\frac{8}{\\hat{\\alpha}} – \\sum_{i = 1}^{8}{\\ln x_{i}} + 10ln5 – 2ln25 = 0$,<\/p>\n

which yields
\n$\\hat{\\alpha} = \\frac{8}{\\sum_{i = 1}^{8}{\\ln x_{i}} – 10ln5 + 2ln25} = \\frac{8}{(ln7 + ln9 + \\ldots + ln20) – 10ln5 + 2ln25} = 0.785$.<\/p>\n

The mean of the Pareto only exists for $\\alpha > 1$. Since
\n$\\hat{\\alpha} = 0.785 < 1$. Then, the mean does not exist.<\/p>\n

3.5.5 Concluding remarks<\/strong><\/p>\n

In describing losses, actuaries fit appropriate parametric distribution
\nmodels for the frequency and severity of loss. This involves finding
\nappropriate statistical distributions that could efficiently model the
\ndata in hand. After fitting a distribution model to a data set, the
\nmodel should be validated. Model validation is a crucial step in the
\nmodel building sequence. It assesses how well these statistical
\ndistributions fit the data in hand and how well can we expect this model
\nto perform in the future. If the selected\u00a0model\u00a0does not\u00a0fit\u00a0the data,
\nanother distribution\u00a0is to be chosen. If more than one model seems to be
\na good fit for the data, we then have to make the choice on which model
\nto use. It should be noted though that the same data should not serve
\nfor both purposes (fitting and validating the model). Additional data
\nshould be used to assess the performance of the model. There are many
\nstatistical tools for model validation. Alternative goodness of fit
\ntests used to determine whether sample data are consistent with the
\ncandidate model, will be presented in a separate chapter.<\/p>\n

Further readings and references<\/strong><\/p>\n

Cummins, J. D. and Derrig, R. A. 1991. Managing the Insolvency Risk of
\nInsurance Companies<\/em>, Springer Science+ Business Media, LLC.<\/p>\n

Frees, E. W. and Valdez, E. A. 2008. Hierarchical Insurance Claims
\nModeling, Journal of the American Statistical Association<\/em>, 103,
\n1457-1469.<\/p>\n

Klugman, S. A., Panjer, H. H. and Willmot, G. E. 2008. Loss Models from
\nData to Decisions<\/em>, Wiley.<\/p>\n

Kreer, M., K\u0131z\u0131lers\u00fc, A., Thomas, A. W. and Eg\u00eddio dos Reis, A. D. 2015.
\nGoodness-of-fit tests and applications for left-truncated Weibull
\ndistributions to non-life insurance, European Actuarial Journal<\/em>, 5,
\n139\u2013163.<\/p>\n

McDonald, J. B. 1984. Some Generalized Functions for the Size
\nDistribution of Income, Econometrica<\/em>, 52, 647\u2013663.<\/p>\n

McDonald, J. B. and Xu, Y. J. 1995. A Generalization of the Beta
\nDistribution with Applications, Journal of Econometrics<\/em>, 66, 133\u201352.<\/p>\n

Tevet, D. 2016. Applying Generalized Linear Models to Insurance Daata:
\nFrequency\/Severity versus Premium Modeling in: Frees, E. W., Derrig, A.
\nR. and Meyers G. (Eds.) Predictive Modeling Applications in Actuarial
\nScience Vol. II Case Studies in Insurance<\/em>. Cambridge University Press.<\/p>\n

Venter, G. 1983. Transformed Beta and Gamma Distributions and Aggregate
\nLosses. Proceedings of the Casualty Actuarial Society<\/em>, 70: 156\u2013193.<\/p>\n","protected":false},"excerpt":{"rendered":"

Chapter 3: Modeling loss severity Zeinab Amin September 15, 2016 3.1 Chapter preview The traditional loss distribution approach to modeling aggregate losses starts by separately fitting a frequency distribution to the number of losses and …<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1372,"menu_order":10,"comment_status":"closed","ping_status":"closed","template":"","meta":{"jetpack_post_was_ever_published":false},"jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/P8cLPd-1xe","acf":[],"_links":{"self":[{"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/5904"}],"collection":[{"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/comments?post=5904"}],"version-history":[{"count":3,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/5904\/revisions"}],"predecessor-version":[{"id":5906,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/5904\/revisions\/5906"}],"up":[{"embeddable":true,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/1372"}],"wp:attachment":[{"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/media?parent=5904"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}