Actuarial Frequency Approximation Workshop

From Policy Exposure to an Aggregate Claim Count

Actuarial frequency problems often describe a portfolio in pieces: one group has a full year of coverage, another has a shorter exposure, and each policy has a small chance of a claim. The calculation becomes manageable when each piece is translated into a claim probability and then into an expected count. A Poisson model can approximate the aggregate count when the individual risks are independent and their probabilities are sufficiently small.

The approximation is a model, not an automatic consequence of having many policies. First identify the independent opportunities for a claim, their probabilities over the stated exposure period, and the expected total count. Then calculate the event probability under the Poisson model and, when useful, compare it with the exact binomial calculation or a stated error bound.

Definition: If $N$ is a Poisson count with parameter $\lambda>0$, then $P(N=k)=e^{-\lambda}\lambda^k/k!$ for $k=0,1,2,\ldots$. In a rare-event approximation to independent Bernoulli risks with probabilities $p_1,\ldots,p_m$, use $\lambda=\sum_{j=1}^m p_j$. For $n$ identical risks with probability $p$, this is $\lambda=np$.

The probabilities must correspond to the exposure period being modeled. For example, if the problem gives a probability per policy-year but counts claims over six months, that annual probability cannot simply be used as the six-month probability. A conversion requires an explicit assumption about how risk accumulates over time.

Why Independent Poisson Counts Combine

A portfolio may consist of separate groups with different exposures or claim probabilities. If the groups are independent and their counts are modeled as Poisson, their total is also Poisson, with parameter equal to the sum of their parameters. This gives a precise reason for adding the expected counts across groups.

Derivation: Suppose independent counts $N_1$ and $N_2$ are Poisson with parameters $\lambda_1$ and $\lambda_2$. For an integer $k\geq 0$, independence gives $$ \begin{aligned} P(N_1+N_2=k) &=\sum_{j=0}^{k}P(N_1=j)P(N_2=k-j)\\ &=e^{-(\lambda_1+\lambda_2)} \sum_{j=0}^{k}\frac{\lambda_1^j\lambda_2^{k-j}}{j!(k-j)!}\\ &=\frac{e^{-(\lambda_1+\lambda_2)}}{k!} \sum_{j=0}^{k}\binom{k}{j}\lambda_1^j\lambda_2^{k-j}\\ &=\frac{e^{-(\lambda_1+\lambda_2)} (\lambda_1+\lambda_2)^k}{k!}. \end{aligned} $$ The last equality is the binomial theorem. This is the Poisson probability formula with parameter $\lambda_1+\lambda_2$. Applying the result repeatedly gives the sum of the parameters for any finite number of independent groups.

For independent policy-level Bernoulli risks, the expected total count is also $\sum_j p_j$, whether or not the total count is exactly Poisson. That expected count supplies the Poisson parameter. The quality of the approximation depends on the risk probabilities and the event being calculated; a matching mean alone does not guarantee a close approximation.

Worked Examples

Worked Example: Comparing a Portfolio Approximation With the Exact Result

A hypothetical insurer covers 100 independent policies for one year. Each policy has probability $0.01$ of producing one claim during the year, and no policy can produce more than one claim in that period. Approximate the probability of at least two claims using a Poisson model, then calculate the exact binomial probability and the absolute error.

Method and parameter. The exact count $X$ is binomial with $n=100$ and $p=0.01$. Its mean is $np=100(0.01)=1$, so the Poisson approximation uses $N\sim\operatorname{Poisson}(1)$.

Poisson approximation. The event is the complement of zero or one claim:

\begin{aligned} P(N\geq 2) &=1-P(N=0)-P(N=1)\\ &=1-e^{-1}-e^{-1}\\ &=1-2e^{-1}\\ &=1-2(0.3678794412)\\ &\approx 0.2642411. \end{aligned}

Exact binomial calculation. Using the binomial probabilities for zero and one claim,

\begin{aligned} P(X\geq2) &=1-(0.99)^{100} -\binom{100}{1}(0.01)(0.99)^{99}\\ &=1-(0.99)^{100}-(0.99)^{99}\\ &\approx 1-0.3660323413-0.3697296376\\ &\approx 0.2642380. \end{aligned}

The absolute error is approximately $ |0.2642411-0.2642380|=0.0000031$. To four decimal places, both probabilities are $0.2642$; the more detailed comparison shows that the Poisson value is slightly higher. The method avoids treating an approximation as an exact equality. In particular, the exact result is about $0.2642380$, not $0.2642381$: retain enough digits in the intermediate terms before subtracting.

Worked Example: Combining Two Policy Groups With Different Exposures

A hypothetical program has 400 independent policies each exposed for one year, with claim probability $0.005$ per policy. A separate group has 300 independent policies exposed for six months, with claim probability $0.0025$ per policy over that six-month period. Assume independence across all policies. Approximate the probability of at least two claims in the combined period and give a standard upper bound on the absolute approximation error.

Method and parameter. The probabilities are already stated for each group’s coverage period, so no time conversion is needed. The first group contributes an expected count of $400(0.005)=2$. The second contributes $300(0.0025)=0.75$. Thus

\lambda=400(0.005)+300(0.0025)=2+0.75=2.75.

For $N\sim\operatorname{Poisson}(2.75)$, the event of at least two claims is the complement of zero or one:

\begin{aligned} P(N\geq2) &=1-e^{-2.75}(1+2.75)\\ &=1-(0.0639278612)(3.75)\\ &\approx 0.7602705\\ &\approx 0.7603. \end{aligned}

For a sum of independent Bernoulli indicators with probabilities $p_j$, a commonly used Le Cam bound is $|P(X\in A)-P(N\in A)|\leq 2\sum_j p_j^2$ for any event $A$, where $N$ is Poisson with parameter $\sum_jp_j$. Here,

\begin{aligned} 2\sum_jp_j^2 &=2\left[400(0.005)^2+300(0.0025)^2\right]\\ &=2(400(0.000025)+300(0.00000625))\\ &=2(0.010000+0.001875)\\ &=0.02375. \end{aligned}

Therefore the absolute error for the at-least-two event is bounded by $0.02375$ using this bound. This is a conservative bound, not an estimate that the actual error equals $0.02375$. The method avoids assigning the six-month group the one-year probability: doing so would incorrectly use $0.005$ for all 700 policies.

Worked Example: Converting an Annual Probability to Four Months

A hypothetical portfolio has 600 independent policies. For each policy, the probability of at least one claim over a full year is $0.012$. Assume a constant hazard over time, so the probability of no claim over an exposure fraction $t$ is $(1-0.012)^t$. Approximate the probability of at least one claim across the portfolio during four months, and compare it with the exact probability under these assumptions.

Method and exposure conversion. Four months is $4/12=1/3$ of a year. The probability of a claim on one policy over that period is

\begin{aligned} q_{4\text{ mo}} &=1-(1-0.012)^{1/3}\\ &=1-(0.988)^{1/3}\\ &\approx 0.0040161. \end{aligned}

The Poisson parameter for 600 policies is $\lambda=600(0.0040161)\approx2.4097$, using the unrounded converted probability for the calculation. Therefore,

\begin{aligned} P(N\geq1) &=1-P(N=0)\\ &=1-e^{-2.4096646}\\ &\approx 1-0.0898455\\ &\approx 0.9102. \end{aligned}

The exact probability of no claims is the product of the 600 independent no-claim probabilities:

\begin{aligned} P(X=0) &=(1-q_{4\text{ mo}})^{600}\\ &=\left((0.988)^{1/3}\right)^{600}\\ &=(0.988)^{200}\\ &\approx 0.0894106. \end{aligned}

Thus the exact probability of at least one claim is $1-0.0894106\approx0.9105894$, or $0.9106$ to four decimal places. The absolute difference between the Poisson and exact probabilities is about $0.000435$. The method avoids using the linear shortcut $0.012/3=0.004$ as though it were the exact four-month probability. Under the stated constant-hazard assumption, the correct conversion is $1-(1-0.012)^{1/3}$; it gives about $0.0040161$, and the corresponding Poisson mean is about $2.4097$.

A Reliable Workflow for Frequency Approximation

The same sequence of decisions works for a single homogeneous group or a portfolio divided into exposure classes. In particular, keep probabilities, exposures, and the event definition tied to the same time period.

Specify the count and period.
State what constitutes a claim, the policies or risks being counted, and the time window. Identify any assumption that limits each policy to at most one claim in the period.

Find each group’s probability.
Use the probability given for that exposure period. Convert from an annual probability only when a time model is supplied or justified, such as the constant-hazard assumption used above.

Add expected counts.
For group $g$ with $n_g$ independent policies and probability $p_g$, calculate $n_gp_g$. Then set $\lambda=\sum_g n_gp_g$.

Translate the event before calculating.
For example, “at least two” is $1-P(N=0)-P(N=1)$, while “at most two” is $P(N=0)+P(N=1)+P(N=2)$. Use the Poisson formula for the required terms.

Check the approximation when requested.
Compare with an exact binomial calculation when the risks are homogeneous, or apply a suitable bound for independent, non-identically distributed risks. Report the approximation and the comparison separately.

Common Exam Traps

Adding policy counts instead of expected counts. If groups have different probabilities, the parameter is $\sum_g n_gp_g$, not the total number of policies and not an unweighted average probability multiplied by a group count.
Mixing time periods. A one-year probability and a six-month count cannot be combined until probabilities are expressed for the periods being counted. A probability does not generally scale linearly with time.
Using a Poisson formula without defining $\lambda$. Write the expected count calculation first. This makes unit and exposure mistakes easier to catch.
Confusing a bound with actual error. An upper bound says the error cannot exceed a stated value under its assumptions; it does not say the error is that value.
Rounding too early. Keep extra digits in powers, exponentials, and intermediate probabilities. Round the reported probability at the final step.

A Poisson approximation is especially useful when the count is built from many independent opportunities, each with a small probability of success. It is less compelling when individual probabilities are large, independence is not plausible, or an exact binomial calculation is straightforward and required. These checks make the result more than a mechanical substitution into a formula.

Key takeaway: Align every probability with its exposure period, calculate $\lambda$ by adding expected claims across groups, and translate the event carefully into Poisson probabilities. When accuracy matters, compare with the exact binomial result or state an appropriate error bound.

Check Your Understanding

For each problem, state the Poisson parameter before calculating the requested probability. Round probabilities to four decimal places.

A hypothetical block has 500 independent policies, each with claim probability $0.004$ over the stated period. Use a Poisson approximation to find the probability of at least one claim. (Answer: $\lambda=2$; probability $0.8647$.)
A portfolio has 250 independent policies with claim probability $0.006$ and 150 independent policies with probability $0.002$, all for the same period. Approximate the probability of exactly two claims. (Answer: $\lambda=1.8$; probability $e^{-1.8}(1.8)^2/2\approx0.2678$.)
For a Poisson claim count with mean $3.2$, calculate the probability of at most one claim. (Answer: $e^{-3.2}(1+3.2)\approx0.1712$.)
A hypothetical group has 1,200 independent policies, each with probability $0.0015$ of a claim in the period. Approximate the probability of at least two claims. (Answer: $\lambda=1.8$; probability $1-e^{-1.8}(1+1.8)\approx0.5372$.)
For 200 independent risks with probability $0.01$ each, use the Poisson model to approximate the probability of at least three successes, and calculate the bound $2\sum_jp_j^2$ on absolute event-probability error. (Answer: $\lambda=2$; probability $1-5e^{-2}\approx0.3233$; bound $2(200)(0.01)^2=0.0400$.)

What You'll Learn