Loss distributions for modeling claim severity

In this part, we will focus on the following aspects to discuss the distributions:

• Characteristics of loss distributions: we will explore the Moment Generating Function (MGF), probability density function (PDF), cumulative distribution function (CDF), survival function and etc.
• Propositions of loss distributions: we will discuss the properties of the distributions, such as the mean, variance, skewness, and kurtosis.
• Estimation of parameters: we will introduce the methods to estimate the parameters of the distributions, such as the method of moments (MM), maximum likelihood estimation (MLE), and etc.

• Applications in insurance: we will discuss the applications of the distributions in insurance contexts, such as pricing insurance policies, setting reserves, and etc.

Note

More details will be added into Probability and Statistics section.

Preliminaries

Before we delve into the specific distributions, let's first introduce some basic concepts that are essential for understanding the subsequent discussions.

When you study the loss distributions, you should have already known that the modeling is based on the random variables and the probability. We use $P(X)$ to denote the probability of the random variable $X$.

Then we have the following definitions:

• Probability density function (PDF): The probability density function (PDF) of a continuous random variable $X$ is a function $f(x)$ that describes the likelihood of the variable taking on a particular value $x$. The PDF is non-negative and integrates to 1 over the entire range of possible values of $X$.

  $$f(x) = P(X = x)$$

• Cumulative distribution function (CDF): The cumulative distribution function (CDF) of a random variable $X$ is a function $F(x)$ that describes the probability that the variable takes on a value less than or equal to $x$.

  $$F(x) = P(X \leq x)$$

• Survival function: The survival function of a random variable $X$ is a function $S(x)$ that describes the probability that the variable takes on a value greater than $x$.

  $$\bar{F}(x) = P(X > x) = 1 - F(x)$$

Exponential distribution

The exponential distribution is a continuous probability distribution that describes the time between events in a Poisson process, where events occur continuously and independently at a constant average rate. The exponential distribution is characterised by a single parameter $\lambda$, which represents the rate at which events occur.

Characteristics of the exponential distribution

If a random variable $X$ is exponential distributed with rate parameter $\lambda$, we write $X \sim \text{Exp}(\lambda)$.

• PDF:

  $$f(x) = \lambda e^{-\lambda x}, \quad x \geq 0$$

• CDF:

  $$F(x) = 1 - e^{-\lambda x}, \quad x \geq 0$$

• Survival function:

  $$\bar{F}(x) = e^{-\lambda x}, \quad x \geq 0$$

Propositions of the exponential distribution

The exponential distribution has the following properties:

• MGF(Moment Generating Function):

  $$M_X(t) = E(e^{tX}) = \frac{\lambda}{\lambda - t}, \quad t < \lambda$$

• CGF (Cumulant Generating Function):

  $$C_X(t) = \log M_X(t) = \log \left( \frac{\lambda}{\lambda - t} \right)$$

• Mean: The mean can be generated from the MGF:

  $$E(X) = m_1 = M_X'(0) = \frac{d}{dt} \left( \frac{\lambda}{\lambda - t} \right) \bigg|_{t=0} = \frac{1}{\lambda}$$

• Variance:

  $$\text{Var}(X) = m_2 = M_X''(0) - m_1^2 = \frac{2}{\lambda^2} - \frac{1}{\lambda^2} = \frac{1}{\lambda^2}$$

  or we can use CGF to calculate the variance:

  $$\text{Var}(X) = C_X''(0) = \frac{d^2}{dt^2} \log \left( \frac{\lambda}{\lambda - t} \right) \bigg|_{t=0} = \frac{2}{\lambda^2}$$

• Skewness: According to the definition of CDF, we can calculate the skewness:

  $$\text{Skew}(X) = \frac{m_3}{m_2^{3/2}} = \frac{C'''_X(0)}{Var^{3/2}(X)} = \frac{2/\lambda^3}{(1/\lambda^2)^{3/2}} = 2$$

Estimation of parameters

Method of moments (MM)

From the above propositions, we can see that the first moment of the exponential distribution is $\frac{1}{\lambda}$. We only have one parameter to estimate, so we can use the method of moments to estimate the parameter $\lambda$. It can be solved by the following criteria, the first origin moment is equal to the first sample moment:

$$E(X) = \frac{1}{\lambda} = \frac{1}{n} \sum_{i=1}^n X_i = \bar{X}$$

Then we can get the estimator of $\lambda$:

$$\hat{\lambda} = \frac{1}{\bar{X}}$$

Maximum likelihood estimation (MLE)

The likelihood function of the exponential distribution is:

$$L(\lambda) = \prod_{i=1}^n \lambda e^{-\lambda x_i} = \lambda^n e^{-\lambda \sum_{i=1}^n x_i}$$

Then the log-likelihood function is:

$$\log L(\lambda) = n \log \lambda - \lambda \sum_{i=1}^n x_i$$

To find the MLE of $\lambda$, we need to solve the following equation:

$$\frac{d}{d\lambda} \log L(\lambda) = \frac{n}{\lambda} - \sum_{i=1}^n x_i = 0$$

Then we can get the MLE of $\lambda$:

$$\hat{\lambda} = \frac{n}{\sum_{i=1}^n x_i} = \frac{1}{\bar{X}}$$

Gamma distribution

The gamma distribution is a continuous probability distribution that generalises the exponential distribution. It is characterized by two parameters: a shape parameter $\alpha$ and a rate parameter $\beta$ (or $\lambda$). The exponential distribution is a special case of the gamma distribution when $\alpha = 1$.

Characteristics of the gamma distribution

If a random variable $X$ is gamma distributed with shape parameter $\alpha$ and rate parameter $\lambda$, we write $X \sim \text{Gamma}(\alpha, \lambda)$ or $X \sim \Gamma(\alpha, \lambda)$.

• PDF:

  $$f(x) = \frac{\lambda^\alpha}{\Gamma(\alpha)} x^{\alpha - 1} e^{-\lambda x}, \quad x \geq 0$$

  where $\Gamma(\alpha)$ is the gamma function defined as $\Gamma(\alpha) = \int_0^\infty x^{\alpha - 1} e^{-x} dx$.

• CDF:

  $$F(x) = \frac{1}{\Gamma(\alpha)} \gamma(\alpha, \lambda x), \quad x \geq 0$$

• Survival function:

  $$\bar{F}(x) = 1 - F(x) = 1 - \frac{1}{\Gamma(\alpha)} \gamma(\alpha, \lambda x), \quad x \geq 0$$

  where $\gamma(\alpha, \lambda x)$ is the lower incomplete gamma function defined as $\gamma(\alpha, \lambda x) = \int_0^{x} t^{\alpha - 1} e^{-\lambda t} dt$.

Propositions of the gamma distribution

The gamma distribution has the following properties:

• MGF:

  $$M_X(t) = E(e^{tX}) = \left( \frac{\lambda}{\lambda - t} \right)^\alpha, \quad t < \lambda$$

• CGF:

  $$C_X(t) = \log M_X(t) = \alpha \log \left( \frac{\lambda}{\lambda - t} \right)$$

• Mean: The mean can be generated from the MGF:

  $$E(X) = m_1 = M_X'(0) = \frac{d}{dt} \left( \left( \frac{\lambda}{\lambda - t} \right)^\alpha \right) \bigg|_{t=0} = \frac{\alpha}{\lambda}$$

• Variance:

  $$\text{Var}(X) = m_2 = M_X''(0) - m_1^2 = \frac{\alpha}{\lambda^2}$$

• Skewness:

  $$\text{Skew}(X) = \frac{m_3}{m_2^{3/2}} = \frac{C'''_X(0)}{Var^{3/2}(X)} = \frac{2}{\sqrt{\alpha}}$$

Estimation of parameters

Method of moments (MM)

In the gamma distribution, we have two parameters to estimate, so we need to use the method of moments to estimate the parameters $\alpha$ and $\lambda$. The first moment of the gamma distribution is $\frac{\alpha}{\lambda}$, and the second moment is $\frac{\alpha}{\lambda^2}$. Both of them should be equal to the sample moments.

We can solve the following criteria:

$$E(X) = \frac{\alpha}{\lambda} = \frac{1}{n} \sum_{i=1}^n X_i = \bar{X}$$

$$Var(X) = \frac{\alpha}{\lambda^2} = \frac{1}{n} \sum_{i=1}^n (X_i - \bar{X})^2 = S^2$$

Solve the first moment,

$$\hat{\lambda} = \frac{\alpha}{\bar{X}}$$

Then, replace $\lambda$ with $\frac{\alpha}{\bar{X}}$ in the second moment, we can get the estimator of $\alpha$:

$$\hat{\alpha} = \frac{\bar{X}^2}{S^2}$$

and the estimator of $\lambda$:

$$\hat{\lambda} = \frac{\hat{\alpha}}{\bar{X}} = \frac{\bar{X}^2 / S^2}{\bar{X}} = \frac{\bar{X}}{S^2}$$

The method of moments estimators for $\alpha$ and $\lambda$ are:

$$\hat{\alpha} = \frac{\bar{X}^2}{S^2}, \quad \hat{\lambda} = \frac{\bar{X}}{S^2}$$

Maximum likelihood estimation (MLE)

The likelihood function of the gamma distribution is:

$$L(\alpha, \lambda) = \prod_{i=1}^n \frac{\lambda^\alpha}{\Gamma(\alpha)} x_i^{\alpha - 1} e^{-\lambda x_i}$$

Then the log-likelihood function is given by:

$$\begin{array}{rl} l & = \sum_{i=1}^n \left( \alpha \log(\lambda) - \log(\Gamma(\alpha)) + (\alpha - 1) \log(x_i) - \lambda x_i \right) \\ \\ & = n \alpha \log(\lambda) - n \log(\Gamma(\alpha)) + (\alpha - 1) \sum_{i=1}^n \log(x_i) - \lambda \sum_{i=1}^n x_i \end{array}$$

Taking the derivative of the log-likelihood function with respect to $\alpha$ and $\lambda$,

we have the following equations:

$$\frac{dl}{d\lambda} = \frac{n\alpha}{\lambda} - \sum_{i=1}^n x_i = 0$$

Solve the equation, we have the MLE of $\lambda$:

$$\hat{\lambda} = \frac{n\alpha}{\sum_{i=1}^n x_i} = \frac{\alpha}{\bar{X}}$$

Then, taking the derivative of the log-likelihood function with respect to $\alpha$,

$$\begin{array}{rl} \frac{dl}{d\alpha} & = n \log(\lambda) - n \frac{\Gamma'(\alpha)}{\Gamma(\alpha)} + \sum_{i=1}^n \log(x_i) = 0 \\ \\ & = n(\log(\lambda) - \psi(\alpha) + \frac{\sum_{i=1}^n \log(x_i)}{n}) = 0 \end{array}$$

where $\psi(\alpha) = \frac{\Gamma'(\alpha)}{\Gamma(\alpha)}$ is the digamma function.

Replace $\lambda$ with $\frac{\alpha}{\bar{X}}$, we can get the MLE of $\alpha$:

$$log(\hat{\alpha}) - log(\bar{X}) - \psi(\hat{\alpha}) + \frac{\sum_{i=1}^n \log(x_i)}{n} = 0$$

This equatio can be solved by R or Python.

x <- rgamma(1000, shape = 5, rate = 0.5) # nolint R
aux <- log(mean(x)) - mean(log(x)) # nolint: assignment_linter.
f <- function(z) {
log(z) - digamma(z) - aux
}
alpha <- uniroot(f, c(1e-8, 1e8))$root
lambda <- alpha / mean(x)

Lognormal distribution

The lognormal distribution is a continuous probability distribution of a random variable whose logarithm is normally distributed. It is characterised by two parameters: the mean $\mu$ and the standard deviation $\sigma$ of the logarithm of the variable.

Characteristics of the lognormal distribution

If a random variable $X$ is lognormally distributed with parameters $\mu$ and $\sigma$, we write $X \sim \text{Lognormal}(\mu, \sigma)$ or $Y = logX \sim \text{N}(\mu, \sigma^2)$.

• PDF: Letting $X = e^Y$, we can derive the PDF of the lognormal distribution:

  $$f(x) = \frac{1}{x} \left[\frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(\log(x) - \mu)^2}{2\sigma^2}}\right], \quad x > 0$$

• CDF:

  $$F(x) = \frac{1}{2} + \frac{1}{2} \text{erf}\left(\frac{\log(x) - \mu}{\sqrt{2}\sigma}\right)$$

Note

CDF of the lognormal distribution will not be used normally in the actuarial practice.

• Survival function:

  $$\bar{F}(x) = 1 - F(x) = \frac{1}{2} - \frac{1}{2} \text{erf}\left(\frac{\log(x) - \mu}{\sqrt{2}\sigma}\right)$$

Propositions of the lognormal distribution

The lognormal distribution has the following properties:

• MGF: The lognormal distribution does not have a closed-form MGF.

• Mean:

  According to the relationship between $X$ and $e^Y$, we can derive the mean for lognormal distribution by:

  $$E(X) = E(e^Y) = M_Y(1) = e^{\mu + \frac{\sigma^2}{2}}$$

• Variance: Similar to the mean, we can derive the variance for lognormal distribution by:

  $$\begin{array}{rl} Var(X) & = E(X^2) - E^2(X) = E(e^{2Y}) - E^2(e^Y) \\ \\ & = M_Y(2) - M_Y^2(1) = e^{2\mu + 2\sigma^2} - e^{2\mu + \sigma^2} \\ \\ & = e^{2\mu + \sigma^2} (e^{\sigma^2} - 1) \end{array}$$

  where $M_Y(t)$ is the MGF of the normal distribution $Y \sim N(\mu, \sigma^2)$.

• Skewness: The skewness of lognormal distribution should be derived by the standard formula:

Estimation of parameters

The lognormal distributions are similar to the normal distribution that they have two parameters $\mu$ and $\sigma^2$.

Method of moments (MM)

like the previous estimation, we can have the following first tww origin moments from the above propositions:

$$\begin{array}{l} \mu_1 = E(X) = exp(\mu + \frac{\sigma^2}{2}) \\ \\ \mu_2 = E(X^2) = E(e^{2Y}) = exp(2\mu + 2\sigma^2) \end{array}$$

Solve the first equation with taking logarithem:

$$log(\mu_1) = \mu + \frac{\sigma^2}{2} \rightarrow \mu = log(\mu_1) - \frac{\sigma^2}{2}$$

Then, replace $\mu$ with $log(\mu_1) - \frac{\sigma^2}{2}$ in the second equation, we can get the estimator of $\sigma^2$:

$$log(\mu_2) = 2\mu + 2\sigma^2 \rightarrow log(\mu_2) = 2\left(log(\mu_1) - \frac{\sigma^2}{2}\right) + 2\sigma^2 = 2log(\mu_1) + \sigma^2$$

$$\sigma = log(\mu_2) - 2log(\mu_1)$$

Then we use the sample moments to estimate the parameters $\mu$ and $\sigma^2$.

$$\begin{array}{l} \hat{\mu}_1 = \frac{\sum_{i=1}^n X_i}{n} \\ \\ \hat{\mu}_2 = \frac{\sum_{i=1}^n X_i^2}{n} \end{array}$$

Replace the above estimators into the equations, we can get the estimators of $\mu$ and $\sigma^2$:

$$\hat{\sigma}^2 = log(\hat{\mu}_2) - 2log(\hat{\mu}_1)$$

$$\begin{array}{rl} \hat{\mu} & = log(\hat{\mu}_1) - \frac{\hat{\sigma}^2}{2} \\ \\ & = log\left(\frac{\sum_{i=1}^n X_i}{n}\right) - \frac{log\left(\frac{\sum_{i=1}^n X_i^2}{n}\right) - 2log\left(\frac{\sum_{i=1}^n X_i}{n}\right)}{2} \\ \\ & = log\left(\frac{\sum_{i=1}^n X_i}{n}\right) - \frac{log\left(\sum_{i=1}^n X_i^2\right)}{2} + log\left(\frac{\sum_{i=1}^n X_i}{n}\right) \\ \\ & = 2log\left(\frac{\sum_{i=1}^n X_i}{n}\right) - \frac{log\left(\sum_{i=1}^n X_i^2\right)}{2} \end{array}$$

Maximum likelihood estimation (MLE)

The likelihood function of the lognormal distribution is:

$$\begin{array}{rl} L(\mu, \sigma) & = \prod_{i=1}^n \frac{1}{x_i} \left[\frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(\log(x_i) - \mu)^2}{2\sigma^2}}\right] \\ \\ & = \prod_{i=1}^n \left((2\pi \sigma^2)^{-\frac{1}{2}} x_i^{-1} \text{exp}(-\frac{(log(x_i) - \mu)^2}{2\sigma^2}) \right) \end{array}$$

Then the log-likelihood function is:

$$\begin{array}{rl} l &= \sum_{i=1}^{n} \left(-\frac{1}{2} \log(2 \pi) - \frac{1}{2} \log(\sigma^2) - \log(x_i) - \frac{(\log(x_i) - \mu)^2}{2 \sigma^2}\right) \\ \\ &= -\frac{n}{2} \log(2 \pi) - \frac{n}{2} \log(\sigma^2) - \sum_{i=1}^n \log(x_i) - \sum_{i=1}^n \frac{(\log(x_i) - \mu)^2}{2 \sigma^2} \\ \\ &= -\frac{n}{2} \log(2 \pi) - \frac{n}{2} \log(\sigma^2) - \sum_{i=1}^n \log(x_i) - \sum_{i=1}^n \frac{(\log(x_i)^2 - 2 \mu \log(x_i) + \mu^2)}{2 \sigma^2} \\ \\ &= -\frac{n}{2} \log(2 \pi) - \frac{n}{2} \log(\sigma^2) - \sum_{i=1}^n \log(x_i) - \frac{\sum_{i=1}^n \log(x_i)^2}{2 \sigma^2} + \frac{\mu \sum_{i=1}^n \log(x_i)}{\sigma^2} - \frac{n \mu^2}{2 \sigma^2} \end{array}$$

Taking the derivative of the log-likelihood function with respect to $\mu$ and $\sigma^2$, we have:

$$\begin{align*} \frac{dl}{d\mu} &= \frac{\sum_{i=1}^n \log(x_i)}{\sigma^2} - \frac{2n\mu}{2\sigma^2} = 0 \\ \\ &\Rightarrow \frac{\sum_{i=1}^n \log(x_i)}{\sigma^2} = \frac{n\mu}{\sigma^2} \\ \\ &\Rightarrow n\mu = \sum_{i=1}^n \log(x_i) \\ \\ &\Rightarrow \hat{\mu} = \frac{\sum_{i=1}^n \log(x_i)}{n} = \overline{\log(x_i)} \end{align*}$$

Then, solve for $\sigma^2$, we have

$$\begin{align*} \frac{dl}{d\sigma^2} &= -\frac{n}{2\sigma^2} - \sum_{i=1}^n \frac{(\log(x_i) - \mu)^2}{2} (-\sigma^2)^{-2} = 0 \\ &\Rightarrow \frac{n}{2\sigma^2} = \frac{\sum_{i=1}^n (\log(x_i) - \mu)^2}{2\sigma^4} \\ &\Rightarrow n\sigma^2 = \sum_{i=1}^n (\log(x_i) - \mu)^2 \\ &\Rightarrow \hat{\sigma}^2 = \frac{\sum_{i=1}^n (\log(x_i) - \hat{\mu})^2}{n} \\ &\Rightarrow \hat{\sigma} = \sqrt{\frac{\sum_{i=1}^n (\log(x_i) - \overline{\log(x_i)})^2}{n}} \end{align*}$$

The MLE of $\mu$ and $\sigma$ are:

$$\hat{\mu} = \overline{\log(x_i)}, \quad \hat{\sigma} = \sqrt{\frac{\sum_{i=1}^n (\log(x_i) - \overline{\log(x_i)})^2}{n}}$$

Conclusion

In this part, we have discussed the loss distributions that are commonly used in actuarial practice, including the exponential distribution, gamma distribution and lognormal distribution. We have explored the characteristics of these distributions, their key properties, and the methods for estimating their parameters. These distributions are essential for modeling claim severity in insurance contexts and are widely used for pricing insurance policies, setting reserves, and assessing risk.