Creating New Distributions

Study guide: Exam ASTAM

Creating New Distributions means building a new severity model from a known one by five operations that Loss Models describes: multiplication by a constant, raising to a power, exponentiation, mixing and splicing. Each of the first three applies a function to a random variable $X$ with a known distribution, and the CDF of the new variable $Y$ follows by a change of variable. Most of the named distributions in the exam tables arise in one of these ways.

$Y = \theta X: \quad F_Y(y) = F_X\!\left(\frac{y}{\theta}\right)$

$Y = X^{1/\tau}: \quad F_Y(y) = \begin{cases} F_X(y^{\tau}), & \tau > 0 \\ 1 - F_X(y^{\tau}), & \tau < 0 \end{cases}$

$Y = e^{X}: \quad F_Y(y) = F_X(\ln y)$

$f_Y(y) = \int f_{Y \mid \Lambda}(y \mid \lambda)\,f_\Lambda(\lambda)\,d\lambda$

Multiplication by a constant: Multiplying by $\theta > 0$ changes the units or applies uniform inflation, and the density becomes $f_Y(y) = f_X(y/\theta)/\theta$ . A family that is closed under this operation is a scale family, and $\theta$ is its scale parameter.
Raising to a power: Loss Models calls $Y = X^{1/\tau}$ transformed when $\tau > 0$ , inverse when $\tau = -1$ , and inverse transformed for any other $\tau < 0$ . The Weibull is a transformed exponential, and the inverse exponential has $F(y) = e^{-\theta/y}$ . A negative power reverses the order of the values, so the small values of $X$ become the large values of $Y$ , and an inverse distribution usually has a heavy right tail with only some of its moments finite.
Exponentiation: If $X$ is normal with parameters $\mu$ and $\sigma$ , then $Y = e^X$ is lognormal with the same parameters. Exponentiation stretches the right tail a great deal, because each unit step in $X$ multiplies $Y$ by $e$ . The moments of $Y$ are values of the moment generating function of $X$ , since $E[Y^k] = E[e^{kX}]$ .
Mixing and splicing: The last block mixes a conditional density over a random parameter $\Lambda$ , as when a gamma-distributed rate turns an exponential into a Pareto (see Mixing Distribution). Splicing joins different densities on different intervals, such as a light-tailed body for small losses and a heavy tail for large ones.

Weibull and Inverse Exponential From an Exponential $X$ is exponential with mean 1. Let $Y = 1{,}000\,X^{1/2}$ and $Z = 1{,}000/X$ . Find the CDF of each, the probability that each exceeds 1,500, and the mean of each.
Answer Both CDFs follow from $F_X(x) = 1 - e^{-x}$ . The power $1/2$ is positive, so the inequality keeps its direction for $Y$ , while dividing into 1,000 reverses it for $Z$ .
 $\begin{align*} F_Y(y) &= \Pr\big[X \le (y/1{,}000)^2\big] \\ &= 1 - e^{-(y/1{,}000)^2} \\ F_Z(z) &= \Pr\big[X \ge 1{,}000/z\big] \\ &= e^{-1{,}000/z} \end{align*}$ 
$Y$ is Weibull with $\tau = 2$ and $\theta = 1{,}000$ , and $Z$ is inverse exponential with $\theta = 1{,}000$ . Then $\Pr(Y > 1{,}500) = e^{-2.25} = 0.1054$ and $\Pr(Z > 1{,}500) = 1 - e^{-2/3} = 0.4866$ . The mean of $Y$ is $1{,}000\,\Gamma(1.5) = 886.23$ . The mean of $Z$ is $1{,}000\,E[1/X] = \int_0^\infty (1{,}000/x)\,e^{-x}\,dx$ , which diverges near $x = 0$ , so $Z$ has no mean. The same exponential gives a light-tailed Weibull or a severity too heavy-tailed to have a mean, depending on the sign of the power.

A Lognormal Severity and Its Inflation The logarithm of a claim amount is normal with $\mu = 7$ and $\sigma = 1.2$ . Find the probability that a claim exceeds 5,000 and the mean claim. Then repeat both after 8% inflation.
Answer The claim $Y = e^X$ is lognormal, so $\Pr(Y > 5{,}000) = \Pr(X > \ln 5{,}000)$ .
 $\begin{align*} z &= \frac{\ln 5{,}000 - 7}{1.2} \\ &= \frac{8.5172 - 7}{1.2} \\ &= 1.2643 \\ \Pr(Y > 5{,}000) &= 1 - \Phi(1.2643) \\ &= 0.1031 \\ E[Y] &= e^{7 + 1.2^2/2} \\ &= 2{,}252.96 \end{align*}$ 
Inflation multiplies $Y$ by 1.08, which adds $\ln 1.08 = 0.0770$ to $X$ , so $\mu$ becomes 7.0770 and $\sigma$ stays at 1.2. Then $z = (8.5172 - 7.0770)/1.2 = 1.2002$ , so $\Pr(1.08\,Y > 5{,}000) = 0.1150$ , and the mean becomes $1.08 \times 2{,}252.96 = 2{,}433.20$ . The mean rises by exactly 8%, but the chance of a claim above the fixed threshold of 5,000 rises by about 12%.

How it appears on the exam

ASTAM asks candidates to create new distributions by multiplication by a constant, raising to a power, exponentiation, mixing and splicing. A question names an operation and a starting distribution and asks for the new CDF or density, a probability, a moment or a percentile.

FAM asks candidates to recognize classes of distributions and their relationships. Knowing that the Weibull is a transformed exponential, or that the lognormal is an exponentiated normal, places a distribution in its family, which tells which of its moments exist and how heavy its tail is.

Moments without integrating: For a power or an exponential of $X$ , the moments of the new variable come straight from the moments of $X$ , through $E[Y^k] = E[X^{k/\tau}]$ or $E[Y^k] = M_X(k)$ .

Exam traps

Negative powers: For $\tau < 0$ , $F_Y(y) = 1 - F_X(y^{\tau})$ , because the transformation is decreasing. Using $F_X(y^{\tau})$ gives a "CDF" that falls as $y$ rises.

Missing moments: An inverse distribution has finite moments only below some order. The inverse exponential has no mean at all, so a premium or comparison based on it has to use a limited expected value or a percentile.

Inflating a lognormal: Multiplying a lognormal claim by $c$ adds $\ln c$ to $\mu$ and leaves $\sigma$ alone. Multiplying $\mu$ by $c$ is wrong.

Mixing is not adding: A mixture averages densities. The distribution of the sum of two claims is a convolution, which is a different object with a different variance.