### Practice Problem Set 2 – finding median losses

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This post has two practice problems to find the median of a distribution that models insurance losses.

 Practice Problem 2a Losses have a distribution with the following density function: $\displaystyle f(x)=\frac{1}{6} e^{-\frac{x}{12}}-\frac{1}{6} e^{-\frac{x}{6}} \ \ \ x>0$. Calculate the median loss amount.
 Practice Problem 2b Losses are modeled by a distribution that is a mixture of two exponential distributions, one with mean 6 and the other with mean 12. The weight of each distribution is 50%. Calculate the median loss amount.

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Solutions

Problem 2a

Median $\displaystyle =-12 \times \log \biggl(\frac{2-\sqrt{2}}{2} \biggr)=14.7354$

log is logarithm to base $e$ = 2.718281828…

Problem 2b

Median $\displaystyle =-12 \times \log \biggl(\frac{-1+\sqrt{5}}{2} \biggr)=5.7745$

log is logarithm to base $e$ = 2.718281828…

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$\copyright$ 2017 – Dan Ma

### Practice Problem Set 1 – working with moments

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This post has two practice problems to complement the previous post Working with moments.

 Practice Problem 1a Losses are modeled by a distribution that is the independent sum of two exponential distributions, one with mean 6 and the other with mean 12. Calculate the skewness of the loss distribution. Calculate the kurtosis of the loss distribution.
 Practice Problem 1b Losses are modeled by a distribution that is a mixture of two exponential distributions, one with mean 6 and the other with mean 12. The weight of each distribution is 50%. Calculate the skewness of the loss distribution. Calculate the kurtosis of the loss distribution.

Comment
As the previous post Working with moments shows, this is primarily an exercise in finding moments. There is no need to first find the probability density function of the loss distribution. For more information on exponential distribution, see here.

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Solutions

Problem 1a

$E(X)=18$

$E(X^2)=504$

$E(X^3)=19440$

$E(X^4)=964224$

$\displaystyle \gamma_1=\frac{3888}{180^{1.5}}=1.61$

$\displaystyle \text{Kurt}[X]=\frac{229392}{180^{2}}=7.08$

Problem 1b

$E(X)=9$

$E(X^2)=180$

$E(X^3)=5832$

$E(X^4)=264384$

$\displaystyle \gamma_1=\frac{2430}{99^{1.5}}=2.4669$

$\displaystyle \text{Kurt}[X]=\frac{122229}{99^{2}}=12.47$

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$\copyright$ 2017 – Dan Ma

### Working with moments

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This post gives some background information on calculation involving moments.

Let $X$ be a random variable. Let $\mu=E(X)$ be its mean and $\sigma^2=Var(X)$ be its variance. Thus $\sigma$ is the standard deviation of $X$.

The expectation $\displaystyle E[X^k]$ is the $k$th raw moment. It is also called the $k$th moment about zero. The expectation $\displaystyle E[(X-\mu)^k]$ is the $k$th central moment. It is also called the $k$th moment about the mean. Given $X$, its standardized random variable is $\displaystyle \frac{X-\mu}{\sigma}$. Then the $k$th standardized moment is $\displaystyle E \biggl[\biggl(\frac{X-\mu}{\sigma} \biggr)^k \biggr]$.

The mean of $X$ is the first raw moment $\mu=E(X)$. The variance of $X$ is the second central moment $\displaystyle Var(X)=E[(X-\mu)^2]$. It is equivalent to $Var(X)=E(X^2)-\mu^2$. In words, the variance is the second raw moment subtracting the square of the mean.

The skewness of $X$ is the third standardized moment of $X$. The kurtosis is the fourth standardized moment of $X$. The excess kurtosis is the kurtosis subtracting 3. They are:

$\displaystyle \gamma_1 =E \biggl[\biggl(\frac{X-\mu}{\sigma} \biggr)^3 \biggr]$

$\displaystyle \text{kurt}[X] =E \biggl[\biggl(\frac{X-\mu}{\sigma} \biggr)^4 \biggr]$

$\displaystyle \text{Ex Kurt}[X] = \text{kurt}[X]-3$

The calculation is usually done by expanding the expression inside the expectation. As a result, the calculation would then be a function of the individual raw moments up to the third or fourth order.

\displaystyle \begin{aligned} \gamma_1&=E \biggl[\biggl(\frac{X-\mu}{\sigma} \biggr)^3 \biggr] \\&=\frac{E(X^3)-3 \mu E(X^2)+3 \mu^2 E(X)-\mu^3}{(\sigma^2)^{1.5}} \\&=\frac{E(X^3)-3 \mu \sigma^2 - \mu^3}{(\sigma^2)^{1.5}} \end{aligned}

\displaystyle \begin{aligned} \text{kurt}[X]&=E \biggl[\biggl(\frac{X-\mu}{\sigma} \biggr)^4 \biggr] \\&=\frac{E(X^4)-4 \mu E(X^3)+6 \mu^2 E(X^2)-4 \mu^3 E(X)+\mu^4}{\sigma^4} \\&=\frac{E(X^4)-4 \mu E(X^3)+6 \mu^2 E(X^2)-3 \mu^4}{\sigma^4} \end{aligned}

Note that the last line in skewness is a version that depends on the mean, the variance and the third raw moment. The calculation is illustrated with some examples. Practice problems are given in subsequent posts.

Example 1
Losses are modeled by a distribution that has the following density function. Calculate the mean, variance, skewness and kurtosis of the loss distribution.

$\displaystyle f(x)=\left\{ \begin{array}{ll} \displaystyle 0.15 &\ 0

The following shows the calculation of the first four raw moments.

$\displaystyle E(X)=\int_0^5 0.15 x \ dx+\int_5^{10} 0.05 x \ dx=3.75$

$\displaystyle E(X^2)=\int_0^5 0.15 x^2 \ dx+\int_5^{10} 0.05 x^2 \ dx=\frac{62.5}{3}$

$\displaystyle E(X^3)=\int_0^5 0.15 x^3 \ dx+\int_5^{10} 0.05 x^3 \ dx=140.625$

$\displaystyle E(X^4)=\int_0^5 0.15 x^4 \ dx+\int_5^{10} 0.05 x^4 \ dx=1062.5$

The following shows the results.

$\displaystyle Var(X)=\frac{62.5}{3}-3.75^2=\frac{20.3125}{3}=6.77083$

$\displaystyle \gamma_1=\frac{140.625-3 \times 3.75 \times \displaystyle \frac{20.3125}{3} - 3.75^3}{\biggl(\displaystyle \frac{20.3125}{3} \biggr)^{1.5}}=0.66515$

$\displaystyle \text{kurt}[X]=\frac{1062.5-4 \times 3.75 \times 140.625+ 6 \times 3.75^2 \times \displaystyle \frac{62.5}{3}-3 \times 3.75^4}{\displaystyle \biggl(\frac{20.3125}{3} \biggr)^2}=2.56686$

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$\copyright$ 2017 – Dan Ma