# A catalog of parametric severity models

Various parametric continuous probability models have been presented and discussed in this blog. The number of parameters in these models ranges from one to two, and in a small number of cases three. They are all potential candidates for models of severity in insurance applications and in other actuarial applications. This post highlights these models. The list presented here is not exhaustive; it is only a brief catalog. There are other models that are also suitable for actuarial applications but not accounted for here. However, the list is a good place to begin. This post also serves a navigation device (the table shown below contains links to the blog posts).

A Catalog

Many of the models highlighted here are related to gamma distribution either directly or indirectly. So the catalog starts with the gamma distribution at the top and then branches out to the other related models. Mathematically, the gamma distribution is a two-parameter continuous distribution defined using the gamma function. The gamma sub family includes the exponential distribution, Erlang distribution and chi-squared distribution. These are distributions that are gamma distributions with certain restrictions on the one or both of the gamma parameters. Other distributions are obtained by raising a distribution to a power. Others are obtained by mixing distributions.

Here’s a listing of the models. Click on the links to find out more about the distributions.

……Derived From ………………….Model
Gamma function
Gamma sub families
Independent sum of gamma
Exponentiation
Raising to a power Raising exponential to a positive power

Raising exponential to a power

Raising gamma to a power

Raising Pareto to a power

Burr sub families
Mixture
Others

The above table categorizes the distributions according to how they are mathematically derived. For example, the gamma distribution is derived from the gamma function. The Pareto distribution is mathematically an exponential-gamma mixture. The Burr distribution is a transformed Pareto distribution, i.e. obtained by raising a Pareto distribution to a positive power. Even though these distributions can be defined simply by giving the PDF and CDF, knowing how their mathematical origins informs us of the specific mathematical properties of the distributions. Organizing according to the mathematical origin gives us a concise summary of the models.

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From a mathematical standpoint, the gamma distribution is defined using the gamma function.

$\displaystyle \Gamma(\alpha)=\int_0^\infty t^{\alpha-1} \ e^{-t} \ dt$

In this above integral, the argument $\alpha$ is a positive number. The expression $t^{\alpha-1} \ e^{-t}$ in the integrand is always positive. The area in between the curve $t^{\alpha-1} \ e^{-t}$ and the x-axis is $\Gamma(\alpha)$. When this expression is normalized, i.e. divided by $\Gamma(\alpha)$, it becomes a density function.

$\displaystyle f(t)=\frac{1}{\Gamma(\alpha)} \ t^{\alpha-1} \ e^{-t}$

The above function $f(t)$ is defined over all positive $t$. The integral of $f(t)$ over all positive $t$ is 1. Thus $f(t)$ is a density function. It only has one parameter, the $\alpha$, which is the shape parameter. Adding the scale parameter $\theta$ making it a two-parameter distribution. The result is called the gamma distribution. The following is the density function.

$\displaystyle f(x)=\frac{1}{\Gamma(\alpha)} \ \biggl(\frac{1}{\theta}\biggr)^\alpha \ x^{\alpha-1} \ e^{-\frac{x}{\theta}} \ \ \ \ \ \ \ x>0$

Both parameters $\alpha$ and $\theta$ are positive real numbers. The first parameter $\alpha$ is the shape parameter and $\theta$ is the scale parameter.

As mentioned above, many of the distributions listed in the above table is related to the gamma distribution. Some of the distributions are sub families of gamma. For example, when $\alpha$ are positive integers, the resulting distributions are called Erlang distribution (important in queuing theory). When $\alpha=1$, the results are the exponential distributions. When $\alpha=\frac{k}{2}$ and $\theta=2$ where $k$ is a positive integer, the results are the chi-squared distributions (the parameter $k$ is referred to the degrees of freedom). The chi-squared distribution plays an important role in statistics.

Taking independent sum of $n$ independent and identically distributed exponential random variables produces the Erlang distribution, a sub gamma family of distribution. Taking independent sum of $n$ exponential random variables, with pairwise distinct means, produces the hypoexponential distributions. On the other hand, the mixture of $n$ independent exponential random variables produces the hyperexponential distribution.

The Pareto distribution (Pareto Type II Lomax) is the mixture of exponential distributions with gamma mixing weights. Despite the connection with the gamma distribution, the Pareto distribution is a heavy tailed distribution. Thus the Pareto distribution is suitable for modeling extreme losses, e.g. in modeling rare but potentially catastrophic losses.

As mentioned earlier, raising a Pareto distribution to a positive power generates the Burr distribution. Restricting the parameters in a Burr distribution in a certain way will produces the paralogistic distribution. The table indicates the relationships in a concise way. For details, go into the blog posts to get more information.

Tail Weight

Another informative way to categorize the distributions listed in the table is through looking at the tail weight. At first glance, all the distributions may look similar. For example, the distributions in the table are right skewed distributions. Upon closer look, some of the distributions put more weights (probabilities) on the larger values. Hence some of the models are more suitable for models of phenomena with significantly higher probabilities of large or extreme values.

When a distribution significantly puts more probabilities on larger values, the distribution is said to be a heavy tailed distribution (or said to have a larger tail weight). In general tail weight is a relative concept. For example, we say model A has a larger tail weight than model B (or model A has a heavier tail than model B). However, there are several ways to check for tail weight of a given distribution. Here are the four criteria.

Tail Weight Measure What to Look for
1 Existence of moments The existence of more positive moments indicates a lighter tailed distribution.
2 Hazard rate function An increasing hazard rate function indicates a lighter tailed distribution.
3 Mean excess loss function An increasing mean excess loss function indicates a heavier tailed distribution.
4 Speed of decay of survival function A survival function that decays rapidly to zero (as compared to another distribution) indicates a lighter tailed distribution.

Existence of moments
For a positive real number $k$, the moment $E(X^k)$ is defined by the integral $\int_0^\infty x^k \ f(x) \ dx$ where $f(x)$ is the density function of the distribution in question. If the distribution puts significantly more probabilities in the larger values in the right tail, this integral may not exist (may not converge) for some $k$. Thus the existence of moments $E(X^k)$ for all positive $k$ is an indication that the distribution is a light tailed distribution.

In the above table, the only distributions for which all positive moments exist are gamma (including all gamma sub families such as exponential), Weibull, lognormal, hyperexponential, hypoexponential and beta. Such distributions are considered light tailed distributions.

The existence of positive moments exists only up to a certain value of a positive integer $k$ is an indication that the distribution has a heavy right tail. All the other distributions in the table are considered heavy tailed distribution as compared to gamma, Weibull and lognormal. Consider a Pareto distribution with shape parameter $\alpha$ and scale parameter $\theta$. Note that the existence of the Pareto higher moments $E(X^k)$ is capped by the shape parameter $\alpha$. If the Pareto distribution is to model a random loss, and if the mean is infinite (when $\alpha=1$), the risk is uninsurable! On the other hand, when $\alpha \le 2$, the Pareto variance does not exist. This shows that for a heavy tailed distribution, the variance may not be a good measure of risk.

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

$\displaystyle h(x)=\frac{f(x)}{S(x)}$

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

Another indication of heavy tail weight is that the distribution has a decreasing hazard rate function. On the other hand, a distribution with an increasing hazard rate function has a light tailed distribution. If the hazard rate function is decreasing (over time if the random variable is a time variable), then the population die off at a decreasing rate, hence a heavier tail for the distribution in question.

The Pareto distribution is a heavy tailed distribution since the hazard rate is $h(x)=\alpha/x$ (Pareto Type I) and $h(x)=\alpha/(x+\theta)$ (Pareto Type II Lomax). Both hazard rates are decreasing function.

The Weibull distribution is a flexible model in that when its shape parameter is $0<\tau<1$, the Weibull hazard rate is decreasing and when $\tau>1$, the hazard rate is increasing. When $\tau=1$, Weibull is the exponential distribution, which has a constant hazard rate.

The point about decreasing hazard rate as an indication of a heavy tailed distribution has a connection with the fourth criterion. The idea is that a decreasing hazard rate means that the survival function decays to zero slowly. This point is due to the fact that the hazard rate function generates the survival function through the following.

$\displaystyle S(x)=e^{\displaystyle -\int_0^x h(t) \ dt}$

Thus if the hazard rate function is decreasing in $x$, then the survival function will decay more slowly to zero. To see this, let $H(x)=\int_0^x h(t) \ dt$, which is called the cumulative hazard rate function. As indicated above, $S(x)=e^{-H(x)}$. If $h(x)$ is decreasing in $x$, $H(x)$ has a lower rate of increase and consequently $S(x)=e^{-H(x)}$ has a slower rate of decrease to zero.

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

The mean excess loss function
The mean excess loss is the conditional expectation $e_X(d)=E(X-d \lvert X>d)$. If the random variable $X$ represents insurance losses, mean excess loss is the expected loss in excess of a threshold conditional on the event that the threshold has been exceeded. Suppose that the threshold $d$ is an ordinary deductible that is part of an insurance coverage. Then $e_X(d)$ is the expected payment made by the insurer in the event that the loss exceeds the deductible.

Whenever $e_X(d)$ is an increasing function of the deductible $d$, the loss $X$ is a heavy tailed distribution. If the mean excess loss function is a decreasing function of $d$, then the loss $X$ is a lighter tailed distribution.

The Pareto distribution can also be classified as a heavy tailed distribution based on an increasing mean excess loss function. For a Pareto distribution (Type I) with shape parameter $\alpha$ and scale parameter $\theta$, the mean excess loss is $e(X)=d/(\alpha-1)$, which is increasing. The mean excess loss for Pareto Type II Lomax is $e(X)=(d+\theta)/(\alpha-1)$, which is also decreasing. They are both increasing functions of the deductible $d$! This means that the larger the deductible, the larger the expected claim if such a large loss occurs! If the underlying distribution for a random loss is Pareto, it is a catastrophic risk situation.

In general, an increasing mean excess loss function is an indication of a heavy tailed distribution. On the other hand, a decreasing mean excess loss function indicates a light tailed distribution. The exponential distribution has a constant mean excess loss function and is considered a medium tailed distribution.

Speed of decay of the survival function to zero
The survival function $S(x)=P(X>x)$ captures the probability of the tail of a distribution. If a distribution whose survival function decays slowly to zero (equivalently the cdf goes slowly to one), it is another indication that the distribution is heavy tailed. This point is touched on when discussing hazard rate function.

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

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

Note that at the large values, the Pareto right tails retain much more probabilities. This is also confirmed by the ratio of the two survival functions, with the ratio approaching infinity. Using an exponential distribution to model a Pareto random phenomenon would be a severe modeling error even though the exponential distribution may be a good model for describing the loss up to the 75th percentile (in the above comparison). It is the large right tail that is problematic (and catastrophic)!

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

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

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

In general, whenever the ratio of two survival functions diverges to infinity, it is an indication that the distribution in the numerator of the ratio has a heavier tail. When the ratio goes to infinity, the survival function in the numerator is said to decay slowly to zero as compared to the denominator.

It is important to examine the tail behavior of a distribution when considering it as a candidate for a model. The four criteria discussed here provide a crucial way to classify parametric models according to the tail weight.

severity models
math

Daniel Ma
mathematics

$\copyright$ 2017 – Dan Ma

# Lognormal distribution

This post discusses the basic properties of the lognormal distribution. The lognormal distribution is a transformation of the normal distribution through exponentiation. As a result, some of the mathematical properties of the lognormal distribution can be derived from the normal distribution. The normal distribution is applicable in many situations but not in all situations. The normal density curve is a bell-shaped curve and is thus not appropriate in phenomena that are skewed to the right. In such situations, the lognormal distribution can be a good alternative to the normal distribution. In an actuarial setting, the lognormal distribution is an excellent candidate for a model of insurance claim sizes.

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Defining the distribution

In this post, the notation log refers to the natural log function, i.e., logarithm to the base $e=2.718281828459 \cdots$. Thus log(2) = 0.693147181.

A random variable $Y$ is said to follow a lognormal distribution if $\log(Y)$ follows a normal distribution. A lognormal distribution has two parameters $\mu$ and $\sigma$, which are the mean and standard deviation of the normal random variable $\log(Y)$. To be more precise, the definition is restated as follows:

A random variable $Y$ is said to follow a lognormal distribution with parameters $\mu$ and $\sigma$ if $\log(Y)$ follows a normal distribution with mean $\mu$ and standard deviation $\sigma$.

Many useful probability distributions are transformations of other known distributions. The above definition shows that a normal distribution is the transformation of a lognormal distribution under the natural logarithm. Start with a lognormal distribution, taking the natural log of it gives you a normal distribution. The other direction is actually more informative, i.e., a lognormal distribution is the transformation of a normal distribution by the exponential function. Start with a normal random variable $X$, the exponentiation of it is a lognormal distribution, i.e., $Y=e^{X}$ is a lognormal distribution. The following summarizes these two transformations.

If the random variable $Y$ is lognormal, then the random variable $X=\log(Y)$ is normal.
If the random variable $X$ is normal, then the random variable $Y=e^{X}$ is lognormal.

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Comparing Normal and Lognormal

Normally one of the first things to focus on is the probability density function when studying a continuous probability model. In the case of the lognormal distribution, a natural way to start is to focus on the relationship between lognormal distribution and normal distribution. In this section, we compare the following:

• The lognormal distribution with parameters $\mu$ = 0 and $\sigma$ = 1 (standard lognormal distribution).
• The normal distribution with mean 0 and standard deviation 1 (standard normal distribution).

Suppose that the random variable $Y$ follows the standard lognormal distribution. This means that $X=\log(Y)$ is the normal distribution with mean 0 and standard deviation 1, i.e. the standard normal distribution. Going the other direction, $Y=e^X$ is the lognormal distribution we start with.

The standard normal $X$ takes on values from $-\infty$ to $\infty$. But most likely, about 99.7% of the times, $X$ takes on values between -3 and 3 (in general, a normal random variable takes on values between -3 standard deviations and +3 standard deviation about 99.7% of the time).

Since exponentiation always gives positive values, the standard lognormal random variable $Y$ always takes on positive values. Since $X$ is rarely outside of -3 and +3, the standard lognormal random variable $Y$ will takes on values between $e^{-3}$ = 0.049787068 to $e^{3}$ = 20.08553692 about 99.7% of the time. Thus observing a standard lognormal value over 20 would be an extremely rare event. The following figure shows the graphs of the standard normal density function and the standard lognormal density function.

Figure 1 – normal and lognormal density curves

In Figure 1, the standard normal density curve is symmetric bell shaped curve, with mean and median located at x = 0. The standard lognormal density is located entirely over the positive x-axis. This is because the exponential function always gives positive values regardless of the sign of the argument. The lognormal density curve in Figure 1 is not symmetric and is a uni-modal curve skewed toward the right. All the standard normal values on the negative x-axis, when exponentiated, are in the interval (0, 1). Thus in Figure 1, the lower half of the lognormal probabilities lie in the interval x = 0 to x = 1 (i.e. the median of this lognormal distribution is x = 1). The other half of the lognormal probabilities lie in the interval $(1, \infty)$. Such lopsided assignment of probabilities shows that lognormal distribution is a positively skewed distribution (skewed to the right).

In the above paragraph, the lower half of the normal distribution on $(-\infty,0)$ is matched with the lognormal distribution on the interval $(0, 1)$. Such interval matching can tell us a great deal about the lognormal distribution. Another example: about 75% of the standard normal distributional values lie below x = 0.67. Thus in Figure 1, about 75% of the lognormal probabilities lie in the interval $(0, 1.95)$ where $e^{0.67} \approx 1.95$. Another example: what is the probability that the lognormal distribution in Figure 1 lie between 1 and 3.5? Then the normal matching interval is $(0, 1.25)$ where $\log(3.5) \approx 1.25$. The normal probability in this interval is 0.8944 – 0.5 = 0.3944. Thus randomly generated a value in the standard lognormal distribution, there is a 39.44% percent chance that it is between 1 and 3.5.

The interval matching idea is very useful for computing lognormal probabilities (e.g. cumulative distribution function) and for finding lognormal percentiles. This idea is discussed further below to make it work for any lognormal distribution, not just the standard lognormal distribution. The following compares the two cumulative distribution functions (CDFs).

Figure 2 – normal and lognormal CDFs

Note that the standard normal CDF basically reaches the level of y = 1 when the x-values get close to 3.0. The lognormal CDF approaches 1.0 too, but at a much slower rate. The lognormal CDF is close to 1 when x = 10 and is rapidly approaching 1 after that point.

Though lognormal distribution is a skewed distribution, some are less skewed than others. The lognormal distributions with larger parameter value of $\sigma$ tend to be more skewed. The following is a diagram of three lognormal density curves that demonstrates this point. Note that the small $\sigma$ of 0.25 relatively resembles a bell curve.

Figure 3 – three lognormal density curves

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How to compute lognormal probabilities and percentiles

Let $Y$ be a random variable that follows a lognormal distribution with parameters $\mu$ and $\sigma$. Then the related normal random variable is $X=\log(Y)$, which has mean $\mu$ and standard deviation $\sigma$. If we raise $e$ to $X$, we get back the lognormal $Y$.

Continuing the interval matching idea, the lognormal interval $Y \le y$ will match with the normal interval $X \le \log(y)$. Both intervals receive the same probability in their respective distributions. The following states this more clearly.

$\displaystyle P\biggl(Y \le y\biggr)=P\biggl(X=\log(Y) \le \log(y)\biggr) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

On the other hand, the normal interval $X \le x$ will match with the lognormal interval $Y \le e^x$. The same probability is assigned to both intervals in their respective distributions. This idea is stated as follows:

$\displaystyle P\biggl(X \le x\biggr)=P\biggl(Y=e^X \le e^x\biggr) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

The idea of $(1)$ gives the cumulative distribution of the lognormal distribution (argument $y$), which is evaluated as the CDF of the corresponding normal distribution at $\log(y)$. One obvious application of $(2)$ is to have an easy way to find percentiles for the lognormal distribution. It is relatively easy to find the corresponding percentile of the normal distribution. Then the lognormal percentile is $e$ raised to the corresponding percentile of the normal distribution. For example, the median of the normal distribution is at the mean $\mu$. Then the median of the lognormal distribution is $e^{\mu}$.

The calculation in both $(1)$ and $(2)$ involve finding normal probabilities, which can be obtained using software or using a table of probability values of the standard normal distribution. To do the table approach, each normal CDF is converted to the standard normal CDF as follows:

$\displaystyle P\biggl(X \le x \biggr)=P\biggl(\frac{X-\mu}{\sigma} \le \frac{x-\mu}{\sigma} \biggr)=\Phi(z) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$

where $z$ is the z-score which is the ratio $(x-\mu)/\sigma$ and $\Phi(z)$ is the cumulative distribution function of the standard normal distribution, which can be looked up from a table based on the z-score. In light of this, $(1)$ can be expressed as follows:

$\displaystyle P\biggl(Y \le y\biggr)=\Phi\biggl(\frac{\log(y)-\mu}{\sigma} \biggr) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$

A quick example to demonstrate how this works.

Example 1
If $Y$ is lognormally distributed with parameters $\mu=2.5$ and $\sigma=1.5$,

• what is the probability $P(1.9 \le Y \le 31.34)$?
• what is the 95th percentile of this lognormal distribution?

The first answer is $P(1.9 \le Y \le 31.34)=P(Y \le 31.34)-P(Y \le 1.9)$, which is calculated as follows:

$\displaystyle P(Y \le 31.34)=\Phi \biggl( \frac{\log(31.34)-2.5}{1.5}\biggr)=\Phi(0.63)=0.7357$

$\displaystyle P(Y \le 1.9)=\Phi \biggl( \frac{\log(1.9)-2.5}{1.5}\biggr)=\Phi(-1.24)=1-0.8925=0.1075$

\displaystyle \begin{aligned} P(1.9 \le Y \le 31.34)&=P(Y \le 31.34)-P(Y \le 1.9) \\&=0.7357-0.1075 \\&=0.6282 \end{aligned}

The z-score for the 95th percentile for the standard normal distribution is z = 1.645. Then the 95th percentile for the normal distribution with mean 2.5 and standard deviation 1.5 is x = 2.5 + 1.645 (1.5) = 4.9675. Then apply the exponential function to obtain $e^{4.9675} \approx 143.67$, which is the desired lognormal 95th percentile. $\square$

As $(2)$ and Example 1 suggest, to find a lognormal percentile, first find the percentile for the corresponding normal distribution. If $x_p$ is the $100p$th percentile of the normal distribution, then $e^{x_p}$ is the $100p$th percentile of the lognormal distribution. Usually, we can first find the $100p$th percentile for the standard normal distribution $z_p$. Then the normal percentile we need is $x=\mu + z_p \cdot \sigma$. The lognormal percentile is then:

$\displaystyle e^{\displaystyle \mu + z_p \cdot \sigma}=\text{lognormal } 100p \text{th percentile} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5)$

The above discussion shows that the explicit form of the lognormal density curve is not needed in computing lognormal probabilities and percentiles. For the sake of completeness, the following shows the probability density functions of both the normal distribution and the lognormal distribution.

Normal PDF
$\displaystyle f_X(x)=\frac{1}{\sigma \sqrt{2 \pi}} \ \ \text{\Large e}^{\displaystyle -\frac{(x-\mu)^2}{2 \sigma^2}} \ \ \ \ \ \ \ \ \ \ \ -\infty

Lognormal PDF
$\displaystyle f_Y(y)=\frac{1}{y \ \sigma \sqrt{2 \pi}} \ \ \text{\Large e}^{\displaystyle -\frac{(\log(y)-\mu)^2}{2 \sigma^2}} \ \ \ \ \ 0

The cumulative distribution function for the lognormal distribution is then

$\displaystyle F_Y(y)=\int_0^y \frac{1}{t \ \sigma \sqrt{2 \pi}} \ \ \text{\Large e}^{\displaystyle -\frac{(\log(t)-\mu)^2}{2 \sigma^2}} \ dt \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (8)$

Of course, we do not have to use $(8)$ since the lognormal CDF can be obtained based on the corresponding normal CDF.

One application of the lognormal PDF in $(7)$ is to use it to find the mode (by taking its derivative and finding the critical value). The mode of the lognormal distribution with parameters $\mu$ and $\sigma$ is $\displaystyle e^{\mu - \sigma^2}$.

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How to find lognormal moments

To find the mean and higher moments of the lognormal distribution, we once again rely on basic information about normal distribution. For any random variable $T$ (normal or otherwise), its moment generating function, if exists, is defined by $M_T(t)=E(e^{t \ T})$. The following is the moment generating function of the normal distribution with mean $\mu$ and standard deviation $\sigma$.

$\displaystyle M_X(t)=\text{\Large e}^{\displaystyle \biggl[ \mu t + (1/2) \sigma^2 t^2 \biggr]} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (9)$

As before, let $Y$ be a random variable that follows a lognormal distribution with parameters $\mu$ and $\sigma$. Then $Y=e^X$ where $X$ is normal with mean $\mu$ and standard deviation $\sigma$. Then $E(Y)=E(e^X)$ is simply the normal moment generating function evaluated at 1. In fact, the kth moment of $Y$, $E(Y^k)=E(e^{k X})$, is simply the normal mgf evaluated at $k$. Because the mgf of the normal distribution is defined at any real number, all moments for the lognormal distribution exist. The following gives the moments explicitly.

$E(Y)=\text{\Large e}^{\displaystyle \biggl[ \mu+(1/2) \sigma^2 \biggr]}$

$E(Y^k)=\text{\Large e}^{\displaystyle \biggl[ k \mu+(k^2/2) \sigma^2 \biggr]} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (10)$

In particular, the variance and standard deviation are:

\displaystyle \begin{aligned}Var(Y)&=\text{\Large e}^{\displaystyle \biggl[ 2 \mu+2 \sigma^2 \biggr]}-\text{\Large e}^{\displaystyle \biggl[ 2 \mu+\sigma^2 \biggr]} \\&=\biggl(\text{\Large e}^{\displaystyle \sigma^2}-1\biggr) \text{\Large e}^{\displaystyle \biggl[ 2 \mu+\sigma^2 \biggr]} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (11) \end{aligned}

$\displaystyle \sigma_Y=\sqrt{\text{\Large e}^{\displaystyle \sigma^2}-1} \ \ \text{\Large e}^{\displaystyle \biggl[ \mu+\frac{1}{2} \sigma^2 \biggr]}=\sqrt{\text{\Large e}^{\displaystyle \sigma^2}-1} \ E(Y) \ \ \ \ \ \ \ \ \ \ \ \ (12)$

The formulas $(11)$ and $(12)$ give the variance and standard deviation if the parameters $\mu$ and $\sigma$ are known. They do not need to be committed to memory, since they can always be generated from knowing the moments in $(10)$. As indicated before, the lognormal median is $e^{\mu}$, which is always less than the mean, which is $e$ raised to $\mu+(1/2) \sigma^2$. So the mean is greater than the median by a factor of $e$ raised to $(1/2) \sigma^2$. The mean being greater than the median is another sign that the lognormal distribution is skewed right.

Example 2
Suppose $Y$ follows a lognormal distribution with mean 12.18 and variance 255.02. Determine the probability that $Y$ is greater than its mean.

With the given information, we have:

$E(Y)=\text{\Large e}^{\displaystyle \biggl[ \mu+(1/2) \sigma^2 \biggr]}=12.18$

$\biggl(\text{\Large e}^{\displaystyle \sigma^2}-1\biggr) \text{\Large e}^{\displaystyle \biggl[ 2 \mu+\sigma^2 \biggr]}=\biggl(\text{\Large e}^{\displaystyle \sigma^2}-1\biggr) \biggl(\text{\Large e}^{\displaystyle \biggl[ \mu+(1/2) \sigma^2 \biggr]}\biggr)^2=255.02$

\displaystyle \begin{aligned}Var(Y)&=\biggl(\text{\Large e}^{\displaystyle \sigma^2}-1\biggr) \text{\Large e}^{\displaystyle \biggl[ 2 \mu+\sigma^2 \biggr]} \\&=\biggl(\text{\Large e}^{\displaystyle \sigma^2}-1\biggr) \biggl(\text{\Large e}^{\displaystyle \biggl[ \mu+(1/2) \sigma^2 \biggr]}\biggr)^2 \\&= \biggl(\text{\Large e}^{\displaystyle \sigma^2}-1\biggr) 12.18^2=255.02 \end{aligned}

From the last equation, we can solve for $\sigma$. The following shows the derivation.

\displaystyle \begin{aligned} &\biggl(\text{\Large e}^{\displaystyle \sigma^2}-1\biggr)=\frac{255.02}{12.18^2} \\&\text{\Large e}^{\displaystyle \sigma^2}=2.719014994 \\&\sigma^2=\log(2.719014994)=1.00026968 \end{aligned}

Thus we can take $\sigma=1$. Then plug $\sigma=1$ into $E(Y)$ to get $\mu=2$. The desired probability is:

\displaystyle \begin{aligned} P(Y>12.18)&=1-\Phi \biggl(\frac{\log(12.18)-2}{1} \biggr) \\&=1-\Phi(0.499795262) \\&=1-\Phi(0.5)=1-0.6915=0.3085 \ \square \end{aligned}

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Linear transformations

For any random variable $W$, a linear transformation of $W$ is the random variable $aW+b$ where $a$ and $b$ are real constants. It is well known that if $X$ follows a normal distribution, any linear transformation of $X$ also follows a normal distribution. Does this apply to lognormal distribution? A linear transformation of a lognormal distribution may not have distributional values over the entire positive x-axis. For example, if $Y$ is lognormal, then $Y+1$ is technically not lognormal since the values of $Y+1$ lie in $(1, \infty)$ and not $(0, \infty)$. Instead, we focus on the transformations $cY$ where $c>0$ is a constant. We have the following fact.

If $Y$ has a lognormal distribution with parameters $\mu$ and $\sigma$, then $cY$ has a lognormal distribution with parameters $\mu+\log(c)$ and $\sigma$.

The effect of the constant adjustment of the lognormal distribution is on the $\mu$ parameter, which is adjusted by adding the natural log of the constant $c$. Note that the adjustment on $\mu$ is addition and not multiplication. The $\sigma$ parameter is unchanged.

One application of the transformation $cY$ is that of inflation. For example, suppose $Y$ represents claim amounts in a given calendar year arising from a group of insurance policies. If the insurance company expects that the claims amounts in the next year will increase by 10%, then $1.1Y$ is the random variable that models next year’s claim amounts. If $Y$ is assumed to be lognormal, then the effect of the 10% inflation is on the $\mu$ parameter as indicated above.

To see why the inflation on $Y$ works as described, let’s look at the cumulative distribution function of $T=cY$.

\displaystyle \begin{aligned} P(T \le y)&=P(Y \le \frac{y}{c})=F_Y(y/c) \\&=\int_0^{y/c} \frac{1}{t \ \sigma \sqrt{2 \pi}} \ \ \text{\Large e}^{\displaystyle -\frac{(\log(t)-\mu)^2}{2 \sigma^2}} \ dt \end{aligned}

Taking derivative of the last item above, we obtain the probability density function $f_T(y)$.

\displaystyle \begin{aligned} f_T(y)&=\frac{1}{y/c \ \sigma \sqrt{2 \pi}} \ \ \text{\Large e}^{\displaystyle -\frac{(\log(y/c)-\mu)^2}{2 \sigma^2}} (1/c) \\&=\frac{1}{y \ \sigma \sqrt{2 \pi}} \ \ \text{\Large e}^{\displaystyle -\frac{\biggl(\log(y)-(\mu+\log(c))\biggr)^2}{2 \sigma^2}} \end{aligned}

Comparing with the density function $(8)$, the last line is the lognormal density function with parameters $\mu + \log(c)$ and $\sigma$.

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Distributional quantities involving higher moments

As the formula $(10)$ shows, all moments exist for the lognormal distribution. As a result, any distributional quantity that is defined using moments can be described explicitly in terms of the parameters $\mu$ and $\sigma$. We highlight three such distributional quantities: coefficient of variation, coefficient of skewness and kurtosis. The following shows their definitions. The calculation is done by plugging in the moments obtained from $(10)$.

$\displaystyle CV=\frac{\sigma_Y}{\mu_Y} \ \ \ \ \ \text{(Coefficient of variation)}$

\displaystyle \begin{aligned} \gamma_1&=\frac{E[ (Y-\mu_Y)^3 ]}{\sigma_Y^3} \\&=\frac{E(Y^3)-3 \mu_Y \sigma_Y^2-\mu_Y^3}{(\sigma_Y^2)^{\frac{3}{2}}} \ \ \ \ \ \text{(Coefficient of skewness)} \end{aligned}

\displaystyle \begin{aligned} \beta_2&=\frac{E[ (Y-\mu_Y)^4 ]}{\biggl(E[ (Y-\mu_Y)^2]\biggr)^2} \\&=\frac{E(Y^4)-4 \ \mu_Y \ E(Y^3) + 6 \ \mu_Y^2 \ E(Y^2) - 3 \ \mu_Y^4}{(\sigma_Y^2)^{2}} \ \ \ \ \ \text{(Kurtosis)} \end{aligned}

The above definitions are made for any random variable $Y$. The notations $\mu_Y$ and $\sigma_Y$ are the mean and standard deviation of $Y$, respectively. Coefficient of variation is the ratio the standard deviation to the mean. It is a standardized measure of dispersion of a probability distribution or frequency distribution. The coefficient of skewness is defined as the ratio of the third central moment about the mean to the cube of the standard deviation. The second line in the above definition is an equivalent form that is in terms of the mean, variance and the third raw moment, which may be easier to calculate in some circumstances. Kurtosis is defined to be the ratio of the fourth central moment about the mean to the square of the second central moment about the mean. The second line in the definition gives an equivalent form that is in terms of the mean, variance and the third and fourth raw moments.

The above general definitions of CV, $\gamma_1$ and $\beta_2$ can be obtained for the lognormal distribution. The mean and variance and higher raw moments can be obtained by using $(10)$. Then it is a matter of plugging in the relevant items into the above definitions. The following example shows how this is done.

Example 3
Determine the CV, $\gamma_1$ and $\beta_2$ of the lognormal distribution in Example 2.

The calculation in Example 2 shows that the lognormal parameters are $\mu=2$ and $\sigma=1$. Now use formula $(10)$ to get the ingredients.

$\mu_Y=e^{2.5}$

$Var(Y)=(e-1) e^5 \ \ \ \ \ \ \ \ \sigma_Y=\sqrt{e-1} \ e^{2.5}$

$E(Y^2)=e^6$

$E(Y^3)=e^{10.5}$

$E(Y^4)=e^{16}$

Right away, CV = $\sqrt{e-1}=1.31$. The following shows the calculation for skewness and kurtosis.

\displaystyle \begin{aligned} \gamma_1&=\frac{E(Y^3)-3 \mu_Y \sigma_Y^2-\mu_Y^3}{(\sigma_Y^2)^{\frac{3}{2}}} \\&=\frac{e^{10.5}-3 \ e^{2.5} \ (e-1) \ e^5-(e^{2.5})^3}{[(e-1) \ e^5]^{\frac{3}{2}}}\\&=6.1849 \end{aligned}

\displaystyle \begin{aligned} \beta_2&=\frac{E(Y^4)-4 \ \mu_Y \ E(Y^3) + 6 \ \mu_Y^2 \ E(Y^2) - 3 \ \mu_Y^4}{(\sigma_Y^2)^{2}} \\&=\frac{e^{16}-4 \ e^{2.5} \ e^{10.5} + 6 \ (e^{2.5})^2 \ e^{6} - 3 \ (e^{2.5})^4}{[(e-1) \ e^5]^{2}}\\&=113.9364 \end{aligned}

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Is there a moment generating function for the lognormal distribution?

Because the normal distribution has a moment generating function, all moments exist for the lognormal distribution (see formula $(10)$ above). Does the moment generating function exist for the lognormal distribution? Whenever the mgf exists for a distribution, its moments can be derived from the mgf. What about the converse? That is, when all moments exist for a given distribution, does it mean that its moment generating function would always exist? The answer is no. It turns out that the lognormal distribution is a counterexample. We conclude this post by showing this fact.

Let $Y$ be the standard lognormal distribution, i.e., $X=\log(Y)$ has the standard normal distribution. We show that the expectation $E(e^{tY})$ converges to infinity when $t>0$.

\displaystyle \begin{aligned} E(e^{t \ Y})&=E(e^{t \ e^X}) \\&=\frac{1}{\sqrt{2 \pi}} \ \int_{-\infty}^\infty e^{t \ e^x} \ e^{-0.5 x^2} \ dx \\&=\frac{1}{\sqrt{2 \pi}} \ \int_{-\infty}^\infty e^{t \ e^x - 0.5x^2} \ dx \\&> \frac{1}{\sqrt{2 \pi}} \ \int_{0}^\infty e^{t \ e^x - 0.5x^2} \ dx \\&> \frac{1}{\sqrt{2 \pi}} \ \int_{0}^\infty e^{\biggl(t \ ( \displaystyle 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}) - 0.5x^2 \biggr)} \ dx=\infty \end{aligned}

The last integral in the above derivation converges to infinity. Note that the Taylor’s series expansion of $e^x$ is $\displaystyle 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots$. In the last step, $e^x$ is replaced by $\displaystyle 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}$. Then the exponent in the last integral is a third degree polynomial with a positive coefficient for the $x^3$ term. Thus this third degree polynomial converges to infinity as x goes to infinity. With the last integral goes to infinity, the mgf $E(e^{t \ Y})$ goes to infinity as well.

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Practice problems

Practice problems to reinforce the calculation are found in here.

More practice problems are available: A summary of the lognormal basic properties, more lognormal practice problems.

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$\copyright$ 2017 – Dan Ma