Common Errors In Frequency Parameters And Supports

Why Parameters and Supports Must Agree

A frequency model can be named correctly and still be used incorrectly. The most common causes are treating a rate as though it were already a mean count, allowing a binomial count to exceed its number of trials, or confusing negative binomial failures with the total number of trials. A quick check of the parameter’s meaning and the random variable’s possible values can catch these errors before they affect a calculation.

The models below use different kinds of parameters. A Poisson parameter is an expected count over a stated exposure; a binomial parameter includes a fixed number of trials; and, under the negative binomial failure-count convention used here, the random variable counts failures before a specified number of successes. Keep the units and the count convention attached to each parameter.

Definition: The support of a discrete random variable is the set of values it can take with positive probability. For a Poisson count with mean \(\lambda>0\), the support is \(0,1,2,\ldots\). For a binomial count with \(n\) trials, the support is \(0,1,\ldots,n\). For a negative binomial count \(K\) of failures before the \(r\)th success, the support is \(0,1,2,\ldots\). The model’s parameters determine the probabilities on its support; they do not make values outside that support possible.

Keep the Model Parameters and Supports Straight

For a Poisson process with a constant rate \(\lambda\) per exposure unit, the count over exposure \(e\) has a Poisson distribution with mean \(\lambda e\). The rate and exposure must use compatible units. For example, a rate per policy-year must be multiplied by policy-years, not by the number of policies alone if the coverage periods differ.

A binomial count \(N\) counts successes in \(n\) fixed trials, each with success probability \(p\), where trials are independent and have the same success probability. Thus \(N\sim\operatorname{Binomial}(n,p)\), and only \(0\) through \(n\) successes are possible. The parameter \(n\) is the number of opportunities, not the expected number of successes; the mean is \(np\).

For the negative binomial model in this tutorial, \(K\) counts failures before the \(r\)th success in independent trials with success probability \(p\). Its probability function is

P(K=k)=\binom{k+r-1}{k}p^r(1-p)^k,\qquad k=0,1,2,\ldots

Here \(r\) is the required number of successes, not the number of failures. If a question instead asks for the total number of trials \(T\) needed to obtain \(r\) successes, then \(T=K+r\), so its support is \(r,r+1,r+2,\ldots\). The two random variables describe related counts, but their values differ by \(r\).

Why the Poisson Mean Scales With Exposure

Suppose a homogeneous Poisson process has rate \(\lambda\) per unit of exposure. Counts over disjoint exposure intervals are independent, and the expected count in an interval of length \(e\) is \(\lambda e\). Dividing a fixed exposure into pieces therefore does not change the mean total: the expected counts for the pieces add.

N(e)\sim\operatorname{Poisson}(\lambda e), \qquad P(N(e)=k)=e^{-\lambda e}\frac{(\lambda e)^k}{k!}, \quad k=0,1,2,\ldots

The product \(\lambda e\) is dimensionless as a count mean: for example, claims per policy-year multiplied by policy-years. Using \(\lambda\) alone as the mean is correct only when the exposure is one unit. This scaling result assumes a constant rate over the exposure being modeled; if the rate changes over time, the appropriate mean must account for that changing rate rather than mechanically applying one constant rate.

Worked Example: Scaling a Poisson Rate to Partial-Year Exposure

A hypothetical portfolio has a claim frequency rate of \(0.4\) claims per policy-year. Fifteen policies are each covered for six months. Under a constant-rate Poisson model, find the probability of exactly two claims in the combined coverage.

Method and exposure. Convert the coverage to policy-years first. Each policy contributes \(0.5\) policy-year, so the combined exposure is

e=15(0.5)=7.5\text{ policy-years}.

The rate is \(0.4\) claims per policy-year, so the Poisson mean is

\lambda e=0.4(7.5)=3\text{ claims}.

Therefore the total count \(N\) is Poisson with mean \(3\), with support \(0,1,2,\ldots\). The probability of exactly two claims is

P(N=2)=e^{-3}\frac{3^2}{2!} =e^{-3}\frac{9}{2} =4.5e^{-3} =0.2240418\ldots

Rounded to four decimal places, the probability is \(0.2240\). Check: the mean is three claims over \(7.5\) policy-years, consistent with the stated rate of \(0.4\) per policy-year. A common mistake is to use \(0.4\) as the mean, as though the portfolio had exactly one policy-year of exposure. That would give \(e^{-0.4}(0.4)^2/2=0.0536\), which uses the wrong exposure.

Respect the Binomial Trial Limit

The binomial support follows directly from what the count represents. In \(n\) trials, the number of successes cannot be negative and cannot exceed \(n\). Thus a calculation assigning positive probability to \(n+1\) successes, for example, cannot be a binomial calculation with \(n\) trials.

A second check is to distinguish the number of trials from the number expected to succeed. If \(n=9\) and \(p=0.20\), then the expected number of successes is \(np=1.8\), but the support is still \(0,1,\ldots,9\). The mean is not a replacement for \(n\).

Worked Example: Finding a Binomial Probability Without Extending Its Support

A hypothetical insurer reviews nine separate policies for a fixed period. Each policy has probability \(0.20\) of producing at least one claim during that period, independently of the others, and each policy can contribute at most one success to this count. Let \(N\) be the number of policies with a claim. Find \(P(N\leq 2)\), and state the support.

Method. There are \(n=9\) independent trials, each with success probability \(p=0.20\). Hence \(N\sim\operatorname{Binomial}(9,0.20)\), with support \(0,1,\ldots,9\). The event \(N\leq2\) includes \(N=0,1,2\).

First,

P(N=0)=\binom{9}{0}(0.20)^0(0.80)^9 =0.134217728.

Next,

P(N=1)=\binom{9}{1}(0.20)(0.80)^8 =9(0.20)(0.16777216) =0.301989888.

Finally,

P(N=2)=\binom{9}{2}(0.20)^2(0.80)^7 =36(0.04)(0.2097152) =0.301989888.

Add the three disjoint probabilities:

P(N\leq2)=0.134217728+0.301989888+0.301989888 =0.738197504.

Rounded to four decimal places, \(P(N\leq2)=0.7382\). Check: the result is between zero and one, and it sums only probabilities for values in the requested event. The common error of treating the count as unbounded would allow more than nine policies to have a claim, contradicting the setup. In particular, the correct value for \(P(N=2)\) here is \(0.301989888\), or \(0.3020\) to four decimal places.

Distinguish Negative Binomial Failures From Total Trials

The negative binomial support depends on which count the question defines. Under the failure-count convention, zero failures are possible: the first \(r\) trials could all be successes. The failure count has no finite upper limit because any number of failures could occur before the \(r\)th success. By contrast, the total number of trials must be at least \(r\), since \(r\) successes require at least \(r\) trials.

The conversion \(T=K+r\) is useful when a problem describes a stopping time in terms of attempts but the model is parameterized by failures. Translate the event into the model’s variable before applying its probability formula. For example, “the \(r\)th success occurs on trial five” means \(T=5\), or \(K=5-r\), not \(K=5\).

Worked Example: Converting a Negative Binomial Count to Attempts

In a hypothetical repeated-trial model, each trial succeeds independently with probability \(p=0.60\). Let \(K\) be the number of failures before the third success, so \(r=3\). Find the probability that the third success occurs by the fifth trial, and give the supports of \(K\) and the total trial count \(T\).

Method and count conversion. The failure-count support is \(K=0,1,2,\ldots\). Since \(T=K+3\), the total-trial support is \(T=3,4,5,\ldots\). The event \(T\leq5\) is therefore \(K\leq2\).

For zero failures,

P(K=0)=\binom{2}{0}(0.60)^3(0.40)^0=0.216.

For one failure,

P(K=1)=\binom{3}{1}(0.60)^3(0.40) =3(0.216)(0.40)=0.2592.

For two failures,

P(K=2)=\binom{4}{2}(0.60)^3(0.40)^2 =6(0.216)(0.16)=0.20736.

Thus,

P(T\leq5)=P(K\leq2) =0.216+0.2592+0.20736 =0.68256.

Rounded to four decimal places, the probability is \(0.6826\). Check: the event includes total trial counts \(3,4,5\), equivalently failure counts \(0,1,2\). A common mistake is to set \(K=5\) when the third success occurs on trial five. The correction is \(K=5-3=2\); the probability of exactly five trials is therefore \(P(K=2)=0.20736\), not \(P(K=5)\).

A Short Parameter-and-Support Checklist

Before using a frequency probability formula, state what the random variable counts and check the parameter units. This takes little time and prevents errors that a correct-looking formula cannot repair. In the Poisson case, explicitly multiply rate by exposure. In the binomial case, identify the fixed trial count and verify that the requested value is no greater than it. In the negative binomial case, write down whether the variable counts failures or total trials.

Name the count.
Write one sentence defining exactly what the random variable counts: claims, successes, failures, or total attempts.

Check parameter meaning and units.
For a Poisson rate, multiply by exposure in matching units. For a binomial variable, identify \(n\) opportunities and the per-trial probability \(p\). For a negative binomial variable, identify the required number \(r\) of successes and the success probability \(p\).

Write the support before calculating.
Compare the requested event with the possible values. Convert total trials to failures, or failures to total trials, before substituting into a negative binomial probability.

Check the result.
Confirm that a probability is between zero and one and that the count values included in a sum are exactly those in the event and in the model’s support.

These checks are useful because the same word, such as “count,” can refer to different quantities in different models. A rate is not a count, an expected count is not a maximum count, and failures are not total attempts. Keeping those distinctions explicit is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Key takeaway: Attach the units and definition to every parameter. A Poisson mean is rate times exposure; a binomial count is bounded by its fixed trial count; and a negative binomial failure count starts at zero, while the corresponding total-trial count starts at \(r\).

Check Your Understanding

For each problem, state the relevant support as well as the requested probability or parameter. Treat all scenarios as hypothetical.

A Poisson claim rate is \(0.3\) claims per policy-month. A group contributes ten policy-months of exposure. Find the mean count and state the support. (Answer: mean \(=3\); support \(0,1,2,\ldots\).)
A count \(N\) is binomial with \(n=12\) and \(p=0.10\). Find \(P(N=0)\) to four decimal places and state the support. (Answer: \(0.9^{12}=0.282429536481\), or \(0.2824\); support \(0,1,\ldots,12\).)
A hypothetical Poisson frequency rate is \(2.4\) events per 1,000 device-years. For 500 device-years, find the probability of exactly one event to four decimal places. (Answer: mean \(=2.4(500/1000)=1.2\); \(P(N=1)=1.2e^{-1.2}=0.361433\ldots\), or \(0.3614\); support \(0,1,2,\ldots\).)
Let \(N\sim\operatorname{Binomial}(9,0.20)\). Find \(P(N=2)\) to four decimal places, and give the largest possible value of \(N\). (Answer: \(P(N=2)=\binom{9}{2}(0.20)^2(0.80)^7=0.301989888\), or \(0.3020\); largest value \(=9\).)
Let \(K\) count failures before the fourth success, with success probability \(p=0.50\) on each independent trial. Find the probability of at most two failures, to four decimal places, and state the support of the total number of trials. (Answer: \(P(K\leq2)=0.0625+0.125+0.15625=0.34375\), or \(0.3438\); total-trial support \(4,5,6,\ldots\).)

What You'll Learn