Generating Independent Frequency-Severity Pairs

Introduction

A simulation of a claim-generating portfolio needs two kinds of random outcomes: the number of claims in a period and the severity of each claim. The count controls how many severity values are generated, but it must not determine their distribution. In particular, a simulation should preserve the specified frequency model while generating claim severities independently of the count and of one another.

This tutorial develops a repeatable procedure for generating those inputs. It uses inverse-CDF sampling, Bernoulli trials, and mixture selection. The central discipline is to match each random input to the event it is intended to generate and to use independent random inputs wherever the model requires independence.

Definition: Let $N$ be the claim count for one period, and let $X_1,X_2,\ldots$ denote successive claim severities. A frequency-severity model specifies the distribution of $N$, the severity distribution $F_X$, and the independence assumptions. In the usual independent model, $N$ is independent of all severities, and the $X_j$ are independent and identically distributed with CDF $F_X$. In one trial, generate $N$ first, then generate exactly $N$ severity values.

Turning Independent Uniforms Into Frequency and Severity

Inverse-CDF sampling is a general way to generate a variable from a specified distribution. If $U$ is uniform on $(0,1)$, define the generalized inverse $F^{-1}(u)=\inf\{x:F(x)\ge u\}$. Then $F^{-1}(U)$ has CDF $F$. For a discrete distribution, this means selecting the first value whose cumulative probability is at least $U$.

For independent frequency and severity, use separate independent uniform inputs. Use one to generate $N$, then use additional independent uniforms for the severities. If the frequency is discrete, the count is selected from its cumulative probabilities. If severity is continuous, apply the severity inverse CDF to each assigned uniform.

P(N=n,\ X_1\le x_1,\ldots,X_n\le x_n) =P(N=n)\prod_{j=1}^{n}F_X(x_j)

This factorization holds under the stated independence assumptions. It also shows why the simulation procedure must preserve more than the marginal distributions: the frequency draw and the severity draws must have the required joint independence.

Procedure: For each trial, generate one frequency input and use it to obtain $N$. Then generate $N$ severity inputs, one for each claim. These inputs must be independent of the frequency input and of one another when the model specifies independent, identically distributed severities. If severity follows a mixture, each claim also needs an independent component-selection draw.

Worked Examples

Worked Example: Poisson Frequency With Exponential Severity

A hypothetical portfolio has a Poisson claim count with parameter $\lambda=1.5$ per year. Claim severity is exponential with mean $\theta=\$200$. For one trial, the independent uniform input for the count is $U_N=0.50$, and the severity input is $U_1=0.75$. Generate the count and all severities required in this trial.

Method. Use Poisson cumulative probabilities to select the count, and the exponential inverse CDF to generate the severity. The inputs are stipulated to be independent.

For the Poisson count, the first probabilities are

P(N=0)=e^{-1.5}=0.2231,\qquad P(N=1)=1.5e^{-1.5}=0.3347.

The cumulative probability through one claim is $0.2231+0.3347=0.5578$. Since $0.2231<0.50\le0.5578$, the selected count is $N=1$.

For an exponential severity with mean $\theta$, $F_X(x)=1-e^{-x/\theta}$ for $x\ge0$. Solving $u=1-e^{-x/\theta}$ gives $x=-\theta\ln(1-u)$. Thus

X_1=-200\ln(1-0.75) =-200\ln(0.25) =277.2589\ldots

The trial therefore has one claim with severity $\$277.26$, rounded to the nearest cent. As a check, $F_X(277.2589\ldots)=1-e^{-277.2589\ldots/200}=0.75$, the supplied severity uniform input.

Common mistake avoided: Do not generate a fixed number of severities regardless of $N$. This trial requires exactly one severity because the simulated count is one.

Worked Example: Binomial Frequency With Tabulated Severity

A hypothetical group of four exposures can each produce at most one claim in a period. Each exposure independently has claim probability $0.25$, so $N\sim\operatorname{Binomial}(4,0.25)$. Severity is $\$100$ with probability $0.50$, $\$300$ with probability $0.30$, and $\$900$ with probability $0.20$. Independent frequency uniforms are $0.12,0.61,0.21,0.88$; the severity uniforms, if needed, are $0.42,0.73,\ldots$. Generate this trial.

Method. Generate each exposure's claim indicator by comparing its uniform with $0.25$. Then sample the severity for each resulting claim from the tabulated cumulative distribution.

An exposure produces a claim if its uniform is less than $0.25$. The four comparisons are $0.12<0.25$, $0.61>0.25$, $0.21<0.25$, and $0.88>0.25$. The indicators are therefore $1,0,1,0$, giving $N=1+0+1+0=2$.

The severity cumulative probabilities are $0.50$ at $\$100$, $0.50+0.30=0.80$ at $\$300$, and $1.00$ at $\$900$. Thus $0.42\le0.50$ selects $\$100$, while $0.50<0.73\le0.80$ selects $\$300$. The two severities, in the order of the two claims, are $\$100$ and $\$300$.

The check is that exactly two of the four Bernoulli inputs fall below $0.25$, and both severity uniforms fall in the correct tabulated intervals. The severity inputs must also be independent of the four frequency inputs.

Common mistake avoided: The binomial count is not generated by treating one uniform as though it were a single claim indicator. Here the model has four independent exposure-level indicators.

Worked Example: Poisson Frequency With Mixture Severity

A hypothetical portfolio has $N\sim\operatorname{Poisson}(0.8)$. Each claim independently belongs to severity component A with probability $0.70$, where severity is exponential with mean $\$100$, or component B with probability $0.30$, where severity is exponential with mean $\$400$. For a trial, the count uniform is $0.50$. If a claim occurs, its component-selection uniform is $0.80$, and its conditional exponential uniform is $0.50$. Generate the count and severity.

Method. First invert the Poisson cumulative distribution. For each claim, select a mixture component using its own uniform, then invert the exponential distribution conditional on that component.

The Poisson probabilities for zero and one claim are

P(N=0)=e^{-0.8}=0.4493,\qquad P(N=1)=0.8e^{-0.8}=0.3595.

The cumulative probability through one claim is $0.4493+0.3595=0.8088$. Since $0.4493<0.50\le0.8088$, $N=1$.

The component-selection uniform $0.80$ exceeds $0.70$, so component B is selected. Applying the exponential inverse with mean $\$400$ and uniform $0.50$ gives

X_1=-400\ln(1-0.50) =-400\ln(0.50) =277.2589\ldots

The generated severity is $\$277.26$, rounded to the nearest cent. It is a component B severity. As a check, the component-selection probability for B is $1-0.70=0.30$, and the conditional CDF of this severity under B is $1-e^{-277.2589\ldots/400}=0.50$.

Common mistake avoided: Do not use the component-selection uniform as the exponential uniform. They serve different purposes and must be independent under this model.

Why the Independence Requirement Matters

A correct marginal frequency distribution and a correct marginal severity distribution do not by themselves guarantee a correct simulation. The relationship between the generated count and severities also matters. For example, using the same uniform $U$ to generate both a Poisson count and an exponential severity can make them dependent: a large value of $U$ tends to select a larger count and, under the exponential inverse, a larger severity as well. Each marginal transformation could still have its intended distribution, but their joint behavior would not represent the independent model.

The justification for the recommended procedure follows directly from inverse-transform sampling. If the count input $U_N$ is uniform, inverse transformation gives $N$ its specified distribution. Each independent severity input $U_j$ transforms to a variable with CDF $F_X$. Since the inputs are independent, applying a deterministic transformation separately to each one preserves independence. Therefore the generated count and severities have the required distributions and factorization.

P(N=n,\ X_1\le x_1,\ldots,X_n\le x_n) =P(N=n)\prod_{j=1}^{n}P(X_j\le x_j) =P(N=n)\prod_{j=1}^{n}F_X(x_j)

In practice, a computer may produce a deterministic sequence of pseudo-random numbers rather than physical independent uniforms. A simulation design should nevertheless assign distinct draws to frequency, component selection, and severity, using a sound random-number generator. Reusing a draw for convenience can introduce dependence that the model did not specify.

Key takeaway: Specify the frequency and severity models first. Generate the count from its own random input, then generate exactly that many severities using independent inputs. For a mixture severity, make a separate component choice for each claim before generating that claim's conditional severity.

Check Your Understanding

For each problem, assume the stated uniform inputs are independent unless the problem says otherwise. Round dollar amounts to the nearest cent.

A count is Poisson with $\lambda=2$, and its uniform input is $0.80$. Using cumulative probabilities $P(N\le2)=0.6767$ and $P(N\le3)=0.8571$, find the simulated count. (Answer: $N=3$)
A claim severity is exponential with mean $\$250$. Its uniform input is $0.40$. Apply the inverse CDF and give the severity. (Answer: $\$127.71$)
A count is $\operatorname{Binomial}(3,0.30)$, generated from uniforms $0.05,0.42,0.28$. Severity is $\$80$ with probability $0.60$ and $\$200$ otherwise. For severity uniforms $0.55$ and $0.75$, give the count and severities. (Answer: $N=2$; severities $\$80$ and $\$200$)
A count has $P(N=n)=0.70(0.30)^n$, for $n=0,1,\ldots$, and its uniform input is $0.75$. If the count is positive, a mixture selects component A with probability $0.60$, otherwise component B; the selection uniform is $0.65$. Component B is exponential with mean $\$300$, and its severity uniform is $0.50$. Give the count, selected component, and severity. (Answer: $N=1$, component B, severity $\$207.94$)
A simulation uses one uniform input both to select a Poisson frequency and to generate an exponential severity. Explain why this may fail to represent an independent frequency-severity model, and state the correction. (Answer: Reusing the input can create dependence between count and severity; use separate independent uniform inputs for frequency and severity.)

What You'll Learn