Landry, Drake David, and Steven Martin. 2022. “Policy-Level Unreported Frequency Model for Pure IBNR Estimation.” Variance 15 (1).
• Figure 1. Piecewise Uniform-Gamma Distributions
• Figure 2. Diagram of Distribution Fitting and Simulation Process
• Policy-Level Unreported Frequency Model Excel Companion File

## Abstract

The paper develops a policy-level unreported claim frequency distribution for use in individual claim reserving models. Recently, there has been increased interest in using individual claim detail to estimate reserves and to understand variability around reserve estimates. The method we describe can aid in the estimation/simulation of pure incurred but not reported (IBNR) from individual claim and policy data. In addition to a point estimate, the method can provide a full distribution of claim emergence, which can be useful for diagnostic tests (i.e., actual versus expected analyses) and to understand reserve variability.

Accepted: July 05, 2021 EDT

# Appendices

## Appendix A: Reported Frequency Distribution Derivations

### Poisson Case

Let $$n \sim Poisson(\lambda)$$ and $$x|n \sim Binomial(n, q).$$

Recall that the probability mass functions for $$n$$ and $$x|n$$ are

\begin{align} f(n)&=\frac{e^{-\lambda}\lambda^{n}}{n!}, \text{ and} \\ f(x|n)&={n \choose x}q^x(1-q)^{n-x}. \end{align}

Notice that $$f(x),$$ the unconditional probability mass function for $$x,$$ is

\begin{align} f(x)&=\sum_{n=x}^{\infty}f(n)f(x|n) \\ &=\sum_{n=x}^{\infty}\frac{e^{-\lambda}\lambda^{n}}{n!}{n \choose x}q^{x}(1-q)^{n-x} \\ &=\sum_{n=x}^{\infty}\frac{e^{-\lambda}\lambda^{n}}{n!}\frac{n!}{(n-x)!x!}q^{x}(1-q)^{n-x} \\ &=\frac{q^{x}e^{-\lambda}}{x!} \sum_{n=x}^{\infty}\frac{\lambda^{n}(1-q)^{n-x}}{(n-x)!}. \end{align}

Let $$a=n-x.$$ Then

\begin{align} f(x)&=\frac{q^{x}e^{-\lambda}}{x!}\sum_{a=0}^{\infty}\frac{\lambda^{a+x}(1-q)^{a}}{a!} \\ &=\frac{(q\lambda)^{x}e^{-\lambda}}{x!}\sum_{a=0}^{\infty}\frac{(\lambda(1-q))^{a}}{a!} \\ &=e^{\lambda(1-q)}\frac{(q\lambda)^{x}e^{-\lambda}}{x!}\sum_{a=0}^{\infty}\frac{(\lambda(1-q))^{a}}{a!}e^{-\lambda(1-q)} \\ &=\frac{(q\lambda)^{x}e^{-q\lambda}}{x!}\sum_{a=0}^{\infty}\frac{(\lambda(1-q))^{a}}{a!}e^{-\lambda(1-q)}. \end{align}

Notice that $$f(a)=\frac{(\lambda(1-q))^{a}}{a!}e^{-\lambda(1-q)}$$ is the probability mass function for $$a$$ where $$a \sim Poisson(\lambda(1-q))$$ and since $$\sum_{a=0}^{\infty}f(a)=1,$$

$f(x)=\frac{e^{-q\lambda}(q\lambda)^{x}}{x!}.$

Notice that this is the probability mass function for a Poisson distribution. Thus,

$x \sim Poisson(q\lambda).$

### Negative Binomial Case

Let $$n \sim Negative Binomial(k, p)$$ and $$x|n \sim Binomial(n, q).$$

Recall that the probability mass functions for $$n$$ and $$x|n$$ are

\begin{align} f(n)&=\frac{\Gamma(n+k)}{\Gamma(n+1)\Gamma(k)}p^k(1-p)^n, \text{ and} \\ f(x|n)&={n \choose x}q^x(1-q)^{n-x}. \end{align}

Notice that $$f(x),$$ the unconditional probability mass function for $$x,$$ is

\begin{align} f(x)&=\sum_{n=x}^{\infty}f(n)*f(x|n) \\ &=\sum_{n=x}^{\infty}\frac{\Gamma(n+k)}{\Gamma(n+1)\Gamma(k)}p^k(1-p)^n{n \choose x}q^x(1-q)^{n-x} \\ &=\sum_{n=x}^{\infty}\frac{\Gamma(n+k)}{\Gamma(n+1)\Gamma(k)}p^k(1-p)^n \\ &\quad \times \frac{\Gamma(n+1)}{\Gamma(x+1)\Gamma(n-x+1)}q^x(1-q)^{n-x} \\ &=\frac{p^kq^x}{\Gamma(k)\Gamma(x+1)} \\ &\quad \times \sum_{n=x}^{\infty}\frac{\Gamma(n+k)}{\Gamma(n-x+1)}(1-p)^n(1-q)^{n-x}. \end{align}

Let $$a=n-x.$$ Then

\begin{align} f(x)&=\frac{p^kq^x}{\Gamma(k)\Gamma(x+1)} \\ &\quad \times \sum_{a=0}^{\infty}\frac{\Gamma(a+x+k)}{\Gamma(a+1)}(1-p)^{a+x}(1-q)^a \\ &=\frac{p^kq^x(1-p)^x}{\Gamma(k)\Gamma(x+1)} \\ &\quad \times \sum_{a=0}^{\infty}\frac{\Gamma(a+x+k)}{\Gamma(a+1)}((1-p)(1-q))^a \\ &=\frac{p^kq^x(1-p)^x\Gamma(x+k)}{\Gamma(k)\Gamma(x+1)} \\ &\quad \times \sum_{a=0}^{\infty}\frac{\Gamma(a+x+k)}{\Gamma(a+1)\Gamma(x+k)}((1-p)(1-q))^a \\ &=\frac{\Gamma(x+k)}{\Gamma(x+1)\Gamma(k)}p^kq^x(1-p)^x \\ &\quad \times \sum_{a=0}^{\infty}\frac{\Gamma(a+x+k)}{\Gamma(a+1)\Gamma(x+k)}((1-p)(1-q))^a \\ &=\frac{\Gamma(x+k)}{\Gamma(x+1)\Gamma(k)}\frac{p^kq^x(1-p)^x}{(1-(1-p)(1-q))^{x+k}} \\ &\quad \times \sum_{a=0}^{\infty}\frac{\Gamma(a+x+k)}{\Gamma(a+1)\Gamma(x+k)} \\ &\quad \times ((1-p)(1-q))^a(1-(1-p)(1-q))^{x+k}. \end{align}

Notice that $$f(a)$$ $$=$$ $$\frac{\Gamma(a+x+k)}{\Gamma(a+1)\Gamma(x+k)}((1-p)(1-q))^a$$ $$\cdot (1-(1-p)(1-q))^{x+k}$$ is the probability mass function for $$a$$ where $$a$$ $$\sim$$ $$Negative Binomial(x+k,$$ $$(1-p)(1-q))$$ and since $$\sum_{a=0}^{\infty}f(a)=1,$$

\begin{align} f(x)&=\frac{\Gamma(x+k)}{\Gamma(x+1)\Gamma(k)}\frac{p^kq^x(1-p)^x}{(1-(1-p)(1-q))^{x+k}} \\ &=\frac{\Gamma(x+k)}{\Gamma(x+1)\Gamma(k)}(\frac{p}{p+q-pq})^k(\frac{q(1-p)}{p+q-pq})^x. \end{align}

Notice that this is the probability mass function for a negative binomial distribution. Thus, $$x$$ $$\sim$$ $$Negative Binomial(k, \frac{p}{p+q-qp}).$$

## Appendix B: Unreported Frequency Distribution Derivations

### Poisson Case

Let $$n \sim Poisson(\lambda)$$ and $$x|n \sim Binomial(n, q).$$

Recall from Appendix A that this implies $$x \sim Poisson(\lambda)$$ and the probability mass functions for $$n,$$ $$x|n,$$ and $$x$$ are

\begin{align} f(n)&=\frac{e^{-\lambda}\lambda^{n}}{n!}, \\ f(x|n)&={n \choose x}q^x(1-q)^{n-x}, \text{ and} \\ f(x)&=\frac{e^{-q\lambda}(q\lambda)^{x}}{x!}. \end{align}

Recall that Bayes’ theorem states that $$P(A|B)=\frac{P(A \cap B)}{P(B)}.$$ So,

\begin{align} f(n|x)&=\frac{P(n=n \cap x=x)}{f(x)} \\ &=\frac{f(n)f(x|n)}{f(x)} \\ &=\frac{\frac{e^{-\lambda}\lambda^{n}}{n!}{n \choose x}q^{x}(1-q)^{n-x}}{\frac{e^{-q\lambda}(q\lambda)^{x}}{x!}} \\ &=\frac{\frac{e^{-\lambda}\lambda^{n}}{n!}\frac{n!}{(n-x)!x!}q^{x}(1-q)^{n-x}}{\frac{e^{-q\lambda}(q\lambda)^{x}}{x!}} \\ &=e^{-\lambda+q\lambda}\lambda^{n-x}\frac{1}{(n-x)!}(1-q)^{n-x} \\ &=\frac{e^{-(1-q)\lambda}((1-q)\lambda)^{n-x}}{(n-x)!}. \end{align}

Let $$a=n-x.$$ The PMF for $$a|x$$ is

\begin{align} f(a|x)&=P(n-x=a|x) \\ &=P(n=a+x|x) \\ &=\frac{e^{-(1-q)\lambda}((1-q)\lambda)^{a}}{a!}. \end{align}

It follows that $$a|x \sim Poisson((1-q)\lambda).$$

### Negative Binomial Case

Let $$n \sim Negative Binomial(k, p)$$ and $$x|n \sim Binomial(n, q).$$

Recall from Appendix A that this implies $$x \sim Negative Binomial(k, \frac{p}{p+q-qp})$$ and the probability mass functions for $$n,$$ $$x|n,$$ and $$x$$ are

\begin{align} f(n)&=\frac{\Gamma(n+k)}{\Gamma(n+1)\Gamma(k)}p^k(1-p)^n, \\ f(x|n)&={n \choose x}q^x(1-q)^{n-x}, \text{ and} \\ f(x)&=\frac{\Gamma(x+k)}{\Gamma(x+1)\Gamma(k)}(\frac{p}{p+q-pq})^k(\frac{q(1-p)}{p+q-pq})^x. \end{align}

Recall that Bayes’ theorem states that $$P(A|B)=\frac{P(A \cap B)}{P(B)}.$$ So,

\begin{align} f(n|x)&=\frac{P(n=n \cap x=x)}{f(x)} \\ &=\frac{f(n)f(x|n)}{f(x)} \\ &=\frac{\frac{\Gamma(n+k)}{\Gamma(n+1)\Gamma(k)}p^k(1-p)^n{n \choose x}q^x(1-q)^{n-x}}{\frac{\Gamma(x+k)}{\Gamma(x+1)\Gamma(k)}(\frac{p}{p+q-pq})^k(\frac{q-qp}{p+q-pq})^x} \\ &=\frac{\Gamma(n+k)\Gamma(n+1)\Gamma(x+1)\Gamma(k)}{\Gamma(n+1)\Gamma(k)\Gamma(x+1)\Gamma(n-x+1)\Gamma(x+k)} \\ &\quad \times \frac{p^k(1-p)^nq^x(1-q)^{n-x}}{\frac{p^kq^x(1-p)^x}{(p+q-pq)^{k+x}}} \\ &=\frac{\Gamma(n+k)}{\Gamma(n-x+1)\Gamma(x+k)}(p+q-pq)^{k+x} \\ &\quad \times (1-p)^{n-x}(1-q)^{n-x} \\ &=\frac{\Gamma(n+k)}{\Gamma(n-x+1)\Gamma(x+k)}(p+q-pq)^{k+x} \\ &\quad \times ((1-p)(1-q))^{n-x} \\ &=\frac{\Gamma((n-x)+(k+x))}{\Gamma(n-x+1)\Gamma(k+x)}(p+q-pq)^{k+x} \\ &\quad \times (1-p-q+pq)^{n-x}. \end{align}

Let $$a=n-x.$$ The PMF for $$a|x$$ is

\begin{align} f(a|x)&=P(n-x=a|x) \\ &=P(n=a+x|x) \\ &=\frac{\Gamma(a+(k+x))}{\Gamma(a+1)\Gamma(k+x)}(p+q-pq)^{k+x} \\ &\quad \times (1-p-q+pq)^{a}. \end{align}

It follows that $$a|x \sim Negative Binomial(k+x, p+q-pq).$$

## Appendix C: Notation List

• $$A_{i} =$$ policy identifier

• $$B_{i} / O_{i} =$$ ultimate number of claims associated with policy $$A_{i}$$ if it was adjusted so that the frequency exposure equals 1 (used in Section 3)

• $$BP =$$ base-level premium (used in Assumption 2 of Section 3.1)

• $$C_{i} =$$ claim identifier

• $$D_{i} =$$ attachment point on policy $$A_{i}$$

• $$E_{i} =$$ frequency exposure on policy $$A_{i}$$

• $$G_{i} =$$ evaluation date $$-$$ report date for claim $$C_{i}$$

• $$g =$$ policy year frequency trend

• $$j_{i} =$$ probability that an unreported claim on policy $$A_{i}$$ will be reported in the next $$L$$ days

• $$k, p =$$ parameters for a negative binominal distribution

• $$L =$$ length in days of the simulated emergence period in Section 6.2

• $$M =$$ number of policies in the data set

• $$N_{i} =$$ ultimate number of claims on policy $$A_{i}$$

• $$P_{i} =$$ earned premium on policy $$A_{i}$$

• $$py_{i} =$$ policy year for policy $$A_{i}$$

• $$q_{i} =$$ probability that a claim on policy $$A_{i}$$ has been reported by the evaluation date

• $$s =$$ range of uniform portion of the piecewise report lag distribution discussed in Section 4.2

• $$T =$$ evaluation date

• $$U_{i} =$$ number of claims that will be reported in the next $$L$$ days for policy $$A_{i}$$

• $$V_{i} =$$ length of the earned policy period for policy $$A_{i} =$$ min(evaluation date, policy expiration date) $$-$$ policy effective date

• $$W_{i} =$$ the occurrence date for claim $$C_{i} -$$ policy effective date

• $$w =$$ probability that an observation comes from the uniform portion of the piecewise report lag distribution discussed in Section 4.2

• $$X_{i} =$$ number of claims that have been reported on policy $$A_{i}$$ by the evaluation date

• $$Y_{i} =$$ report lag for claim $$C_{i} =$$ report date $$-$$ occurrence date

• $$yr =$$ most recent policy year (used in Assumption 4 of Section 3.1)

• $$Z_{i} =$$ evaluation date $$-$$ policy effective date for policy $$A_{i}$$

• $$\lambda =$$ mean of a Poisson distribution

• $$A(w) =$$ actual percentage of policies that had a reported frequency of $$w$$

• $$b_{x_{i}}(x) =$$ PMF for the reported frequency distribution for policy $$A_{i}$$ implied by the fitted frequency parameters

• $$\mkern 1.5mu\overline{\mkern-1.5mub(w)\mkern-1.5mu}\mkern 1.5mu =$$ average probability of observing a reported frequency $$w$$ implied by the fitted frequency parameters

• $$f(x) =$$ PMF for a Poisson distribution

• $$g(x) =$$ PMF for a negative binomial

• $$h(x) / H(x) =$$ the PDF/CDF for either an exponential, gamma, or Weibull distribution used in the piecewise report lag distribution discussed in Section 4.2

• $$LL(x) =$$ log-likelihood function for parameter(s) $$x$$

• $$r(x) / R(x) =$$ PDF/CDF for the report lag distribution

• $$S(x) =$$ survival function for the severity distribution

1. The materiality of this assumption depends on the selected report lag distribution. Before implementing this simplification, one should evaluate the impact.

2. For more background on why the right truncation adjustment is necessary, please see Section 2.1 of Korn (2016).

3. A full discussion of MCMC techniques is beyond our scope here. If you want to learn more about these methods a good place to start is Appendix B of Meyers (2015).