Tutorials › Exam FAM - Fundamentals of Actuarial Mathematics › Exact Frequency Calculation Workshop

Home  /  Tutorials  /  Exam FAM / Frequency Probabilities and Approximations

Exam FAM · Frequency Probabilities and Approximations · Tutorial 556 of 988

Exact Frequency Calculation Workshop

Build reliable exact frequency calculations by combining probability mass recursions, interval sums, complements, and distribution moments.

Intermediate 10 min read

What You'll Learn

  • Calculate point probabilities recursively for binomial, Poisson, and negative binomial counts
  • Find inclusive interval probabilities by summing the required probability masses
  • Use complements to calculate upper-tail probabilities efficiently
  • Apply each model’s assumptions and parameterization correctly
  • Check exact probability calculations against the corresponding mean and variance

Exact Frequency Calculations

A frequency problem may ask for the probability of one claim count, a range of counts, or an event such as at least a specified number of claims. It may also ask for the mean or variance. The main challenge is to keep the model, event boundaries, and calculation method consistent. When neighboring probability masses are related by a simple ratio, a recursion can be faster and less error-prone than evaluating each mass from its formula.

This workshop uses the binomial, Poisson, and negative binomial models. In each case, identify the random variable and its parameterization first. Then write the event precisely, calculate the required masses or complement, and use the model’s moments as a check. All scenarios below are hypothetical.

Definitions: A binomial count \(N\sim\operatorname{Bin}(n,p)\) counts successes in \(n\) independent trials, each with success probability \(p\). A Poisson count \(N\sim\operatorname{Poisson}(\lambda)\) has parameter \(\lambda>0\), which is both its mean and variance. A negative binomial count \(N\) here means the number of failures before the \(r\)th success in independent trials with success probability \(p\), where \(r\) is a positive integer and \(0<p<1\). This negative binomial parameterization has support \(0,1,2,\ldots\).

The model assumptions matter. A binomial model has a fixed number of independent trials and the same success probability on every trial. A Poisson model is specified by its rate parameter. Under the negative binomial model defined above, the count stops when the \(r\)th success occurs; it does not count the total number of trials. Their probability masses and moments are:

$$ \begin{aligned} \text{Binomial:}\quad &P(N=k)=\binom{n}{k}p^k(1-p)^{n-k}, & E[N]&=np,\quad \operatorname{Var}(N)=np(1-p),\\ \text{Poisson:}\quad &P(N=k)=e^{-\lambda}\frac{\lambda^k}{k!}, & E[N]&=\lambda,\quad \operatorname{Var}(N)=\lambda,\\ \text{Negative binomial:}\quad &P(N=k)=\binom{k+r-1}{k}p^r(1-p)^k, & E[N]&=\frac{r(1-p)}{p},\quad \operatorname{Var}(N)=\frac{r(1-p)}{p^2}. \end{aligned} $$

For the binomial model, \(k\) ranges from 0 to \(n\); for the other two models, \(k\) ranges over the nonnegative integers. In the negative binomial formula, the binomial coefficient counts the possible placements of the \(r-1\) earlier successes among the first \(k+r-1\) trials, with the last trial being the \(r\)th success.

Probability Recursions and Their Justification

A recursion calculates a probability mass from the preceding one. It is particularly useful when the problem asks for several consecutive counts. The starting mass must be correct, and the range of the recursion must respect the model’s support.

Probability recursions: For \(N\sim\operatorname{Bin}(n,p)\) with \(0<p<1\), for \(0\le k<n\), \(P(N=k+1)=P(N=k)\dfrac{n-k}{k+1}\dfrac{p}{1-p}\). For \(p=0\), \(P(N=0)=1\); for \(p=1\), \(P(N=n)=1\). For \(N\sim\operatorname{Poisson}(\lambda)\), \(P(N=k+1)=P(N=k)\dfrac{\lambda}{k+1}\) for \(k\ge0\). For the negative binomial convention above, \(P(N=k+1)=P(N=k)\dfrac{k+r}{k+1}(1-p)\) for \(k\ge0\).

For example, the Poisson recursion follows directly by dividing adjacent probability masses. This derivation also shows why the starting probability and parameter are essential:

$$ \frac{P(N=k+1)}{P(N=k)} = \frac{e^{-\lambda}\lambda^{k+1}/(k+1)!} {e^{-\lambda}\lambda^k/k!} = \frac{\lambda}{k+1}. $$

The binomial and negative binomial recursions follow in the same way by taking the ratio of their adjacent probability mass formulas. The binomial ratio contains \(1-p\) in the denominator, so that form is not used at \(p=0\) or \(p=1\); in those cases the count is deterministic. For all three models, recursion is an arithmetic shortcut, not a change to the distribution.

Worked Example: A Poisson Interval and Upper Tail

Worked Example: A Poisson Interval and Upper Tail

A hypothetical service center models its number of claims in one month by \(N\sim\operatorname{Poisson}(2.5)\). Find \(P(1\le N\le4)\) and \(P(N\ge4)\). The model assumes the monthly count follows a Poisson distribution with parameter \(\lambda=2.5\).

Method. Start from \(P(N=0)=e^{-2.5}=0.0820849986\), then use the Poisson recursion. Keep the calculation at full precision until the final answers.

$$ \begin{aligned} P(N=1)&=0.0820849986\left(\frac{2.5}{1}\right)=0.2052124966,\\ P(N=2)&=0.2052124966\left(\frac{2.5}{2}\right)=0.2565156207,\\ P(N=3)&=0.2565156207\left(\frac{2.5}{3}\right)=0.2137630172,\\ P(N=4)&=0.2137630172\left(\frac{2.5}{4}\right)=0.1336018858. \end{aligned} $$

The requested interval includes counts 1, 2, 3, and 4, so add exactly those four masses:

$$ P(1\le N\le4) =0.2052124966+0.2565156207+0.2137630172+0.1336018858 =0.8090930203\approx0.8091. $$

For the upper tail starting at 4, use the complement of counts 0 through 3:

$$ P(N\ge4) =1-(0.0820849986+0.2052124966+0.2565156207+0.2137630172) =0.2424238669\approx0.2424. $$

The mean and variance are both \(\lambda=2.5\), as a model check. The common mistake is to treat “at least 4” as “greater than 4,” or to include \(P(N=4)\) in the complement. The complement of \(N\ge4\) is \(N\le3\), which does not include count 4.

Intervals, Complements, and Model-Specific Calculations

An inclusive interval \(a\le N\le b\) means the sum of the masses at \(a,a+1,\ldots,b\). An event such as \(N\ge c\) can instead be calculated as \(1-P(N\le c-1)\). The complement is often more efficient when it contains fewer masses. For a finite-support binomial count, the same principle applies at either end; for example, \(P(N\ge c)=1-\sum_{k=0}^{c-1}P(N=k)\).

Write down the event before calculating. In particular, “more than \(c\)” means \(N\ge c+1\), whereas “at least \(c\)” includes \(c\). If an event has both limits, such as \(a<N\le b\), its integer counts run from \(a+1\) through \(b\), not from \(a\).

Worked Example: A Binomial Interval and Complement

A hypothetical insurer models claims from eight independent, identical short-term policies. Each policy has probability \(p=0.20\) of producing a claim during the period, so \(N\sim\operatorname{Bin}(8,0.20)\). Find \(P(1\le N\le3)\), \(P(N\ge4)\), and the mean and variance.

Method. Use the binomial recursion, starting with \(P(N=0)=(0.8)^8=0.16777216\). Since \(0<p<1\), the binomial recursion applies.

$$ \begin{aligned} P(N=1)&=0.16777216\left(\frac{8}{1}\frac{0.2}{0.8}\right)=0.33554432,\\ P(N=2)&=0.33554432\left(\frac{7}{2}\frac{0.2}{0.8}\right)=0.29360128,\\ P(N=3)&=0.29360128\left(\frac{6}{3}\frac{0.2}{0.8}\right)=0.14680064. \end{aligned} $$

Add the masses for counts 1 through 3:

$$ P(1\le N\le3) =0.33554432+0.29360128+0.14680064 =0.77594624\approx0.7759. $$

For \(N\ge4\), the complement is \(N\le3\), including zero claims:

$$ P(N\ge4) =1-(0.16777216+0.33554432+0.29360128+0.14680064) =0.05628160\approx0.0563. $$

The moments are \(E[N]=np=8(0.20)=1.6\) and \(\operatorname{Var}(N)=np(1-p)=8(0.20)(0.80)=1.28\). The common mistake is to omit \(P(N=0)\) from the complement: the event complementary to \(N\ge4\) includes 0, 1, 2, and 3 claims.

Worked Example: Negative Binomial Counts Before a Target Number of Successes

A hypothetical monitoring process records failures before its third successful inspection. Each inspection independently succeeds with probability \(p=0.60\). Let \(N\) be the number of failures before the third success, so \(r=3\). Find \(P(1\le N\le3)\), \(P(N\ge4)\), and the mean and variance.

Method. Under this parameterization \(P(N=0)=p^r=(0.60)^3=0.216\). Apply the negative binomial recursion, with failure probability \(1-p=0.40\):

$$ \begin{aligned} P(N=1)&=0.216\left(\frac{3}{1}\right)(0.40)=0.2592,\\ P(N=2)&=0.2592\left(\frac{4}{2}\right)(0.40)=0.20736,\\ P(N=3)&=0.20736\left(\frac{5}{3}\right)(0.40)=0.13824. \end{aligned} $$

Thus the interval probability is

$$ P(1\le N\le3)=0.2592+0.20736+0.13824=0.6048. $$

To obtain the tail, first sum the probabilities for \(N=0,1,2,3\): \(0.216+0.2592+0.20736+0.13824=0.8208\). Therefore

$$ P(N\ge4)=1-0.8208=0.1792. $$

The mean is \(r(1-p)/p=3(0.40)/0.60=2\). The variance is \(r(1-p)/p^2=3(0.40)/(0.60)^2=1.2/0.36=3.3333\) to four decimal places. In particular, the variance is not \(2.2222\): the numerator is \(1.2\), and dividing by \(0.36\) gives \(3.3333\). A common modeling error is to confuse failures before the third success with the total number of inspections through the third success. The latter count is \(N+3\), not \(N\).

A Reliable Calculation Sequence

A short sequence of checks helps prevent boundary, parameter, and arithmetic errors. Use it whether the problem asks for a point mass, an interval, a tail, or moments.

1
Name the model and count.
State what \(N\) counts and identify all parameters. For a negative binomial model, specify whether the count is failures before a target number of successes.
2
Translate the event into integer counts.
For example, \(2<N\le5\) means \(N=3,4,5\), while \(N\ge5\) has complement \(N\le4\).
3
Choose masses or a complement.
Use a recursion for consecutive masses and a complement when it replaces a longer tail by a shorter sum.
4
Check the result.
Confirm the probability lies between 0 and 1, the summed counts match the event, and the mean and variance use the same model parameterization.

Common Exam Traps and Why Exact Work Matters

Exact calculations are not always the shortest calculations, but they provide a dependable benchmark. A recursion avoids repeatedly evaluating factorials or powers, while a complement avoids summing a long upper tail. Exact work also makes it easier to locate an error: if a result is too large, inspect whether an endpoint was counted twice or whether a complement omitted a mass.

  • Dropping an endpoint. \(a\le N\le b\) includes both \(a\) and \(b\). List the integer counts before summing.
  • Using the wrong complement. \(N\ge c\) complements to \(N\le c-1\), not \(N\le c\).
  • Mixing parameterizations. In the negative binomial model used here, the mean counts failures, so it is \(r(1-p)/p\). A total-trials count includes the \(r\) successes and has a different mean.
  • Applying an interior recursion at a boundary. The binomial recursion with division by \(1-p\) requires \(0<p<1\); at \(p=0\) or \(p=1\), use the degenerate distribution directly.
  • Using moments to replace a requested exact probability. The mean and variance summarize a distribution but generally do not determine a particular point or interval probability. Calculate the probability from the specified model.
Key takeaway: Exact frequency work is a sequence of model identification, event translation, probability calculation, and verification. Recursions generate adjacent masses efficiently; complements and moments provide separate checks, but each must match the stated distribution and event.

Check Your Understanding

For each problem, identify the event boundaries before calculating. Probabilities are rounded to four decimal places where appropriate.

  1. A hypothetical portfolio has \(N\sim\operatorname{Bin}(12,0.05)\). Find \(P(N\ge1)\), \(E[N]\), and \(\operatorname{Var}(N)\). (Answer: \(P(N\ge1)=0.4596\), mean \(=0.6\), variance \(=0.57\).)
  2. A hypothetical count follows \(\operatorname{Poisson}(1.8)\). Find \(P(N=2)\) and \(P(N\le2)\). (Answer: \(P(N=2)=0.2678\), \(P(N\le2)=0.7306\).)
  3. A hypothetical group of six independent contracts each has claim probability \(0.30\). Find the probability of at least two claims, along with the mean and variance. (Answer: \(P(N\ge2)=0.5798\), mean \(=1.8\), variance \(=1.26\).)
  4. Let \(N\) be failures before the third success, with independent success probability \(p=0.75\). Find \(P(N\ge2)\), the mean, and the variance. (Answer: \(P(N\ge2)=0.2617\), mean \(=1\), variance \(=1.3333\).)
  5. Let \(N\) be failures before the second success, with success probability \(p=0.50\). Find \(P(N\le2)\), \(E[N]\), and \(\operatorname{Var}(N)\). (Answer: \(P(N\le2)=0.6875\), mean \(=2\), variance \(=4\).)

This material is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. Exact frequency practice is most effective when every calculation is accompanied by a precise event statement and a check against the chosen model.

Previous

Next

Back to Tutorials
Clinical Biostats