Frequency Distribution

Study guide: Exam FAM

A Frequency Distribution is the probability distribution of the number of claims $N$ in a period, for one policy or for a portfolio, and paired with a severity distribution it builds the aggregate loss model. Loss Models' main claim-count models are the Poisson, binomial and negative binomial, which make up the (a,b,0) class, together with their zero-truncated and zero-modified versions in the (a,b,1) class. It differs from Frequency, Exam 5's observed ratio of claims to exposure, which is an average rather than a distribution.

$p_k = \left(a + \frac{b}{k}\right) p_{k-1}, \quad k = 1, 2, 3, \ldots$

$p_k^M = \frac{1 - p_0^M}{1 - p_0}\,p_k, \quad k = 1, 2, 3, \ldots$

Distribution	$p_0$	$a$	$b$
Poisson $(\lambda)$	$e^{-\lambda}$	$0$	$\lambda$
Binomial $(m, q)$	$(1-q)^m$	$-\frac{q}{1-q}$	$\frac{(m+1)q}{1-q}$
Neg. binomial $(r, \beta)$	$(1+\beta)^{-r}$	$\frac{\beta}{1+\beta}$	$\frac{(r-1)\beta}{1+\beta}$
Geometric $(\beta)$	$(1+\beta)^{-1}$	$\frac{\beta}{1+\beta}$	$0$

Symbols: $p_k = \Pr(N = k)$ , and the table, from Loss Models' Appendix B as printed in the exam tables, gives each member's $a$ , $b$ and starting value. Every member has $E[N] = (a+b)/(1-a)$ and $\text{Var}(N) = (a+b)/(1-a)^2$ , which give $\lambda$ and $\lambda$ for the Poisson, $mq$ and $mq(1-q)$ for the binomial, and $r\beta$ and $r\beta(1+\beta)$ for the negative binomial.
Choosing a member: The variance is below the mean for the binomial, equal to it for the Poisson and above it for the negative binomial, because $\text{Var}(N)/E[N] = 1/(1-a)$ and $a$ is negative, zero and positive in turn. Equivalently, $k\,p_k/p_{k-1} = ak + b$ is linear in $k$ , and its slope and intercept on data suggest which member fits.
The (a,b,1) class: Its members follow the same relation only from $k = 2$ , so $p_0$ is free. A zero-truncated distribution has $p_0^T = 0$ and $p_k^T = p_k/(1 - p_0)$ . A zero-modified one puts an arbitrary $p_0^M$ at zero and scales the rest as in the second block, so its mean is $1 - p_0^M$ times the zero-truncated mean. The class also holds the logarithmic distribution and the zero-truncated negative binomial with $-1 < r < 0$ , which have no (a,b,0) counterpart.
Exposure, deductibles and estimation: When exposure grows by a factor, the Poisson $\lambda$ , the negative binomial $r$ and the binomial $m$ grow by it. When a deductible lets only a fraction $v$ of losses through, $\lambda$ , $\beta$ and $q$ are multiplied by $v$ instead (Coverage Modifications). By maximum likelihood, $\hat\lambda = \bar{x}$ for the Poisson, $\hat q = \bar{x}/m$ for the binomial with $m$ known, and $\hat\beta = \bar{x}/r$ for the negative binomial with $r$ known.

Identifying the Member From Two Ratios For a claim-count distribution in the (a,b,0) class, $p_1/p_0 = 1$ and $p_2/p_1 = 2/3$ . Identify the distribution and its parameters, and find $p_3$ .
Answer The ratios give $a + b = 1$ and $a + b/2 = 2/3$ , so $b/2 = 1/3$ .
 $\begin{align*} b &= \tfrac{2}{3} \\ a &= \tfrac{1}{3} \end{align*}$ 
Because $a > 0$ , the distribution is negative binomial. Then $\beta/(1+\beta) = 1/3$ gives $\beta = 0.5$ , and $(r-1)a = b$ gives $r = 3$ . So $p_0 = 1.5^{-3} = 0.296296$ , $p_1 = p_0 = 0.296296$ and $p_2 = \tfrac{2}{3}p_1 = 0.197531$ .
 $\begin{align*} p_3 &= \left(\tfrac{1}{3} + \tfrac{2/3}{3}\right)(0.197531) \\ &= \tfrac{5}{9}(0.197531) \\ &= 0.109739 \end{align*}$ 
The mean is $r\beta = 1.5$ and the variance is $r\beta(1+\beta) = 2.25$ , so the claim counts are overdispersed relative to a Poisson with the same mean.

Zero-Modified Poisson Claim Counts A policy has no claims in a year with probability 0.4. Given at least one claim, the number of claims follows a zero-truncated Poisson with $\lambda = 1.5$ . Find the probabilities of one and two claims, and the mean.
Answer This is a zero-modified Poisson with $p_0^M = 0.4$ , and the corresponding Poisson has $p_0 = e^{-1.5} = 0.223130$ . Each positive probability is the Poisson one times $(1 - 0.4)/(1 - 0.223130) = 0.772330$ .
 $\begin{align*} p_1^M &= 0.772330 \times 1.5\,e^{-1.5} \\ &= 0.772330 \times 0.334695 \\ &= 0.258495 \\ p_2^M &= \tfrac{1.5}{2}\,p_1^M \\ &= 0.193871 \\ E[N] &= \frac{0.6 \times 1.5}{0.776870} \\ &= 1.1585 \end{align*}$ 
The second probability comes from the recursion with $a = 0$ and $b = 1.5$ , which still holds from $k = 2$ . The mean is 0.6 times the zero-truncated mean $1.5/0.776870 = 1.9308$ , and it is below the Poisson mean of 1.5 because the policy is more likely than a Poisson to have no claims.

Maximum Likelihood for Claim Counts Over a year, 500 policies had the following numbers of claims: 300 had none, 140 had one, 50 had two and 10 had three. Fit a Poisson by maximum likelihood, and say whether the data suggest a negative binomial instead.
Answer The loglikelihood is $\ell(\lambda) = -500\lambda + \left(\sum x_i\right)\ln\lambda + \text{constant}$ , and setting its derivative to zero gives $\hat\lambda = \bar{x}$ .
 $\begin{align*} \hat\lambda &= \frac{140(1) + 50(2) + 10(3)}{500} \\ &= 0.54 \\ \hat\sigma^2 &= \frac{140 + 50(4) + 10(9)}{500} - 0.54^2 \\ &= 0.5684 \end{align*}$ 
The variance of the data is only 5% above the mean, so a Poisson is a reasonable choice. A negative binomial would be preferred if the variance were well above the mean, as it is when policyholders differ in their claim rates.

How it appears on the exam

FAM asks candidates to recognize the (a,b,0) and (a,b,1) classes and their relationships, identify the role of their parameters, perform calculations with them, and choose a distribution for a given application. A question gives ratios of successive probabilities, a zero-modified distribution's $p_0^M$ , or a change in exposure or deductible, and asks for a probability, a mean or the new parameters.

FAM parametric estimation: A question gives claim-count data, sometimes grouped, and asks for the maximum likelihood estimate of a Poisson, binomial or negative binomial parameter.

ASTAM uses the same classes as the frequency in aggregate models, where the values of $a$ and $b$ drive the recursive formula.

Exam traps

Exposure versus deductible: Doubling the exposure doubles the negative binomial's $r$ and leaves $\beta$ alone, while a deductible multiplies $\beta$ by the probability of a payment and leaves $r$ alone. Swapping the two gives the right mean but the wrong variance.

Where the recursion starts: For an (a,b,1) distribution, $p_1^M \neq (a+b)\,p_0^M$ , so $p_1$ must come from the scaling formula before the recursion can continue from $k = 2$ .

The negative binomial's parameterization: The exam tables count failures, with $p_0 = (1+\beta)^{-r}$ and mean $r\beta$ . The trials form from probability courses has a different mean, and mixing the two shifts every probability.

The binomial's $a$ : For the binomial, $a = -q/(1-q)$ is negative. Dropping the sign makes the distribution look like a negative binomial.