Deriving the Poisson Full-Credibility Standard

Why a Claim-Count Standard Needs a Probability Requirement

A claim count varies from period to period, even when the underlying expected frequency is unchanged. A full-credibility standard makes that variability explicit: it specifies how close the observed count should be to its expected value and how often that degree of closeness should occur. For a Poisson count, the relationship between the mean and variance makes it possible to derive a simple approximate threshold for the expected count.

The result is a normal-approximation standard, not an exact Poisson guarantee. Its main use is to translate a chosen relative-error tolerance and probability level into a required expected claim count. The derivation also explains why the threshold increases with the confidence level and decreases as the permitted relative error widens.

Definition: Let $N\sim\operatorname{Poisson}(\lambda)$, where $\lambda>0$ is the expected claim count. For a two-sided relative-error tolerance $k>0$ and target probability $q$, with $0<q<1$, the full-credibility requirement is $$ P\left(\frac{|N-\lambda|}{\lambda}\leq k\right)\geq q. $$ The requirement concerns the observed count $N$'s deviation from its expected count $\lambda$. The condition $\lambda>0$ is necessary because the relative error divides by $\lambda$.

The standard assumes a Poisson count: claims occur according to a process for which the count over the period has mean $\lambda$ and variance $\lambda$. It also uses a normal approximation to that count. In particular, the derivation treats the standardized count as approximately standard normal; it does not assert that a finite Poisson distribution is exactly normal. The usual approximation is most defensible when the expected count is sufficiently large for the Poisson distribution to be reasonably represented by a normal curve.

Here $k$ is a proportion, not a percentage number. For example, a 4% relative error means $k=0.04$. The parameter $\lambda$ is an expected count and can be noninteger; the observed count $N$, by contrast, is an integer.

For a Poisson random variable, $E[N]=\lambda$ and $\operatorname{Var}(N)=\lambda$. Its standard deviation is therefore $\sqrt{\lambda}$. Standardizing $N$ gives a random variable that is approximately standard normal when the normal approximation is appropriate.

Standardize the Poisson count.
Use the Poisson mean and variance to write $(N-\lambda)/\sqrt{\lambda}\approx Z$, where $Z\sim N(0,1)$.

Rewrite the relative-error event.
Because $\lambda>0$, multiplying the inequality by $\lambda$ gives $|N-\lambda|\leq k\lambda$. Dividing by $\sqrt{\lambda}>0$ gives $|(N-\lambda)/\sqrt{\lambda}|\leq k\sqrt{\lambda}$.

Apply the normal approximation.
For a symmetric interval around zero, the approximate probability is $P(|Z|\leq k\sqrt{\lambda})=2\Phi(k\sqrt{\lambda})-1$, where $\Phi$ is the standard normal cumulative distribution function.

Impose the target probability.
Set the approximate central probability at least equal to $q$, then solve for $\lambda$ using the corresponding standard normal quantile.

To make the last step precise, let $z_p$ satisfy $P(Z\leq z_p)=p$ for $Z\sim N(0,1)$. Symmetry of the standard normal distribution means that a central probability $q$ has upper endpoint $z_{(1+q)/2}$: the probability below that endpoint is $(1+q)/2$. Thus the approximate requirement becomes $k\sqrt{\lambda}\geq z_{(1+q)/2}$. Since both sides are nonnegative, squaring preserves the inequality.

Theorem: Under the normal approximation to a Poisson count, the two-sided relative-error requirement is met approximately when $$ \lambda\geq \left(\frac{z_{(1+q)/2}}{k}\right)^2. $$ This is the Poisson full-credibility standard for expected claim count. If a problem requires the reported threshold to be a whole number of claims, round the threshold upward; do not round down, because that could fail the inequality.

Why the Formula Has This Form

The square in the formula follows directly from the Poisson variance. A count's typical random deviation is measured in units of $\sqrt{\lambda}$, while the allowed absolute deviation under a relative tolerance $k$ is $k\lambda$. Comparing those two scales gives $k\lambda/\sqrt{\lambda}=k\sqrt{\lambda}$. To make the allowed deviation sufficiently large in standard-deviation units, this quantity must reach the normal quantile required by the probability level. Solving that condition requires squaring the quantile-to-tolerance ratio.

This reasoning shows the direction of the effects without memorizing them. Increasing $q$ requires a larger normal quantile, so the required $\lambda$ increases. Increasing $k$ permits a larger relative deviation, so the required $\lambda$ decreases. In particular, doubling the permitted relative error divides the required expected count by four, all else equal.

Worked Applications

Worked Example: Four-Percent Tolerance at 90% Probability

A hypothetical benefit program wants its Poisson claim count to be within 4% of its expected count with approximately 90% probability. Using $z_{0.95}=1.645$, find the full-credibility threshold. If the threshold is to be stated as a whole-number expected-count benchmark, identify the minimum whole number.

Method. This is a two-sided relative-error requirement, so $k=0.04$ and $q=0.90$. The central 90% interval uses $z_{(1+q)/2}=z_{0.95}=1.645$.

Substitute into the derived standard: $$ \lambda\geq\left(\frac{1.645}{0.04}\right)^2 =\left(41.125\right)^2 =1691.265625. $$

The continuous threshold is 1691.265625 expected claims. If a whole-number benchmark is required, round upward to obtain 1692. This upward rounding is justified because 1691 is below the calculated threshold, whereas 1692 is above it. As a check, $0.04\sqrt{1692}\approx1.6454$, which exceeds 1.645.

Common mistake avoided. The 90% requirement is central, not one-sided. It uses $z_{0.95}$, not $z_{0.90}$. Using $z_{0.90}\approx1.282$ would give $(1.282/0.04)^2\approx1027.2$, substantially understating the required count.

Worked Example: A Five-Percent Standard at 95% Probability

A hypothetical portfolio has a Poisson claim count. Determine the approximate expected-count threshold for a two-sided relative-error tolerance of 5% at 95% probability, using $z_{0.975}=1.96$. Also check whether a whole-number benchmark of 1537 would meet the approximate standard.

Method. Here $k=0.05$, $q=0.95$, and the central probability requires $z_{(1+0.95)/2}=z_{0.975}=1.96$. The threshold is $$ \lambda\geq\left(\frac{1.96}{0.05}\right)^2 =(39.2)^2 =1536.64. $$

Because 1537 is greater than 1536.64, it meets the approximate standard. A direct check gives $0.05\sqrt{1537}\approx0.05(39.2046)=1.9602$, slightly above the required 1.96. A benchmark of 1536 would be below the threshold, so if a whole-number benchmark is required, the minimum is 1537.

Common mistake avoided. Do not round 1536.64 to the nearest integer, which would produce 1537 in this instance by coincidence but is not a reliable rule. The criterion is an inequality: when an integer threshold is required, take the ceiling of the unrounded value.

Worked Example: Approximate Probability at a Given Expected Count

A hypothetical service plan has a Poisson claim count with expected value $\lambda=1200$. Estimate the probability that the count is within 6% of its expected value. Use the normal approximation and report the probability to four decimal places.

Method. The relative-error event is $|N-1200|/1200\leq0.06$. Under the normal approximation, its standardized half-width is $$ k\sqrt{\lambda}=0.06\sqrt{1200} =0.06(34.6410) \approx2.07846. $$

The approximate central probability is therefore $$ 2\Phi(2.07846)-1 \approx2(0.98117)-1 =0.96234. $$

Rounded to four decimal places, the estimated probability is 0.9623. This is the estimated chance of meeting the 6% standard, not a 95% probability simply because 1.96 appears in some common normal-approximation standards.

Common mistake avoided. Do not substitute a quantile for a probability calculation. Here $\lambda$ and $k$ are given, so first calculate $k\sqrt{\lambda}$, then evaluate the two-sided probability $2\Phi(k\sqrt{\lambda})-1$.

Assumptions, Interpretation, and Exam Traps

The formula is useful because it gives a direct planning standard from three inputs: the error tolerance $k$, the probability level $q$, and the Poisson model. It can help compare how much exposure or expected claim volume is needed to make a frequency estimate sufficiently stable under a specified criterion. It is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

The derivation has limits. It assumes a Poisson count with variance equal to its mean and approximates the count distribution by a normal distribution. If the count is not reasonably approximated by a normal distribution, or if the variance differs materially from the Poisson variance, this standard may not describe the desired probability well. The displayed threshold is approximate, not a claim that the exact Poisson probability equals $q$ at the threshold. Also, this standard concerns a count's deviation from its own expected value; it does not by itself establish a standard for aggregate dollars or for an estimated frequency rate.

A reliable calculation sequence is: express a percentage as a proportion, identify whether the target probability is central, choose the corresponding normal quantile, calculate the threshold without intermediate rounding, and round upward only if a whole-number benchmark is required. For a probability calculation with a given $\lambda$, use the same normal approximation in the reverse direction: calculate $k\sqrt{\lambda}$ and evaluate the central area.

Key takeaway: For a Poisson count, the normal approximation turns a two-sided relative-error requirement into $\lambda\geq(z_{(1+q)/2}/k)^2$. The square arises because the Poisson standard deviation is $\sqrt{\lambda}$, and an integer threshold, when requested, must be rounded up.

Check Your Understanding

For each problem, use the stated normal quantile and the Poisson normal approximation. Treat a requested whole-number threshold as a ceiling.

A hypothetical count must be within 10% of its expected value with 90% probability. Use $z_{0.95}=1.645$. Find the minimum whole-number expected-count benchmark. (Answer: 271)
For a 4% relative-error tolerance at 95% probability, use $z_{0.975}=1.96$. Find the expected-count threshold and its minimum whole-number benchmark. (Answer: 2401)
A hypothetical process requires a 3% tolerance at 90% probability. Use $z_{0.95}=1.645$. Find the minimum whole-number expected-count benchmark. (Answer: 3007)
A Poisson count has expected value 900. Estimate the probability that it is within 7% of its mean, using the normal approximation and rounding the probability to four decimals. (Answer: 0.9643)
A hypothetical count standard requires a 5% tolerance at 99% probability. Use $z_{0.995}=2.576$. Find the minimum whole-number expected-count benchmark. (Answer: 2655)

Deriving the Poisson Full-Credibility Standard

What You'll Learn

Why a Claim-Count Standard Needs a Probability Requirement