Checking Poisson Standards with Exact Probabilities

Why Check a Poisson Credibility Standard Exactly?

A limited-fluctuation standard uses a normal approximation to choose an expected claim count large enough that random variation falls within a specified tolerance with a desired probability. That calculation is useful for setting a standard, but its probability statement is approximate: a Poisson claim count is discrete, while the normal distribution is continuous. When the expected count is small, the discreteness and skewness can make the actual probability noticeably different from the target.

An exact check evaluates the probability under the Poisson model at the proposed count standard. It does not replace the normal-based standard; it tells us how well that standard performs for the particular expected count being considered. The calculation is especially useful near a threshold and when a count is small enough that the normal approximation deserves scrutiny.

Definition: If the count $N$ has a Poisson distribution with expected count $\lambda>0$, then $$ P(N=k)=e^{-\lambda}\frac{\lambda^k}{k!},\qquad k=0,1,2,\ldots. $$ For a standardized deviation threshold $c>0$, the acceptance event is $|N-\lambda|\le c\sqrt{\lambda}$. Since $N$ must be a nonnegative integer, its exact lower and upper bounds are $$ L=\max\!\left(0,\left\lceil\lambda-c\sqrt{\lambda}\right\rceil\right), \qquad U=\left\lfloor\lambda+c\sqrt{\lambda}\right\rfloor. $$ The exact probability of meeting the criterion is $P(L\le N\le U)$.

Here $\lceil x\rceil$ is the smallest integer at least $x$, and $\lfloor x\rfloor$ is the largest integer at most $x$. Both are essential: the endpoints of the real-valued interval are not generally possible claim counts. For example, if the interval is $[1.2,7.8]$, the qualifying counts are 2 through 7, not 1 through 8.

For the standardized criterion, the normal approximation treats $(N-\lambda)/\sqrt{\lambda}$ as approximately standard normal. Thus the approximate probability of meeting the criterion is $P(|Z|\le c)$, where $Z\sim N(0,1)$. If $c=z_{(1+p)/2}$, this target is $p$. The exact Poisson probability need not equal $p$, particularly when $\lambda$ is small.

Computing the Exact Probability

Once the integer bounds are determined, add the Poisson probabilities for every qualifying count:

P(L\le N\le U) =e^{-\lambda}\sum_{k=L}^{U}\frac{\lambda^k}{k!}.

This finite sum is the exact model probability. A convenient way to calculate its terms is to start with $P(N=0)=e^{-\lambda}$ and use the recurrence

P(N=k+1)=P(N=k)\frac{\lambda}{k+1}.

The recurrence follows directly by dividing consecutive Poisson probabilities:

\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}.

Thus each next probability is obtained by multiplying the previous one by $\lambda/(k+1)$. This is often less error-prone than separately evaluating factorials, especially when several adjacent probabilities are needed. If the qualifying interval excludes zero, do not include $P(N=0)$ in the sum merely because it is the starting value used to generate the recurrence.

Worked Example: A Two-Sided Threshold with Mean Four

A hypothetical portfolio has a Poisson claim count with expected count $\lambda=4$. A proposed standard accepts the count when it is within $c=1.5$ standard deviations of its mean. Find the exact probability of acceptance and compare it with the normal approximation.

Method: integer bounds first, then sum the exact Poisson probabilities. Since $\sqrt{4}=2$, the deviation allowance is $1.5(2)=3$. The real-valued acceptance interval is

4-3\le N\le4+3,

so the qualifying integer counts are $1\le N\le7$. The normal approximation gives $P(|Z|\le1.5)=0.8664$, to four decimal places.

For the exact probability, factor out $e^{-4}$. The sum of the terms from $k=1$ through $k=7$ is

\begin{aligned} \sum_{k=1}^{7}\frac{4^k}{k!} &=4+8+\frac{64}{6}+\frac{256}{24} +\frac{1024}{120}+\frac{4096}{720}+\frac{16384}{5040}\\ &=4+8+10.6666667+10.6666667+8.5333333 +5.6888889+3.2507937\\ &=50.8063492. \end{aligned}

Therefore,

P(1\le N\le7)=e^{-4}(50.8063492)=0.9305507\ldots\approx0.9306.

The exact probability is about 0.9306, compared with the normal approximation of 0.8664. For this expected count, the normal value is not a close estimate of the probability of this particular acceptance event.

Check. The event excludes $N=0$, so the sum begins with $k=1$. The displayed sum is $50.8063492$, not $51.8063492$; the latter would incorrectly include the $k=0$ term, which equals 1. The calculated probability is also less than 1, as required.

Small Expected Counts: Discreteness Matters

When $\lambda$ is small, the lower part of a symmetric real-valued interval may extend below zero, but a Poisson count can never be negative. Also, the integer interval may contain relatively few possible values. Both facts can cause the exact probability to differ from the nominal normal probability. The following examples illustrate why the bounds and the sum should be checked separately.

Worked Example: A 1.96-Standard-Deviation Criterion at Mean One

For a hypothetical class of policies, suppose the claim count satisfies $N\sim\operatorname{Poisson}(1)$. Evaluate the exact probability of meeting $|N-1|\le1.96$, then compare it with the nominal 95% normal probability.

Method: apply the integer bounds, including the nonnegative-count restriction. The real interval is

1-1.96\le N\le1+1.96, \qquad\text{or}\qquad -0.96\le N\le2.96.

The possible count values are therefore $N=0,1,2$. The exact probability is

\begin{aligned} P(0\le N\le2) &=e^{-1}\left(\frac{1^0}{0!}+\frac{1^1}{1!}+\frac{1^2}{2!}\right)\\ &=e^{-1}(1+1+0.5)\\ &=\frac{2.5}{e}=0.9196986\ldots\approx0.9197. \end{aligned}

The normal approximation assigns $P(|Z|\le1.96)\approx0.9500$. The exact probability is about 0.9197, lower by approximately 0.0303. The count cannot take negative values, and the event consists of just three integer outcomes; a continuous symmetric approximation does not capture those features exactly.

Check. The lower real endpoint is negative, but it does not create negative count outcomes. The accepted values begin at zero, and the sum contains exactly three probabilities.

Worked Example: A One-Standard-Deviation Criterion at Mean Two

A hypothetical exposure group has $N\sim\operatorname{Poisson}(2)$. Find the exact probability that the count is within one standard deviation of its expected value, and compare it with the normal approximation.

Method: convert the inclusive deviation condition to integer endpoints. Here $\sqrt{2}$ is the standard deviation, so the criterion is

|N-2|\le\sqrt{2}, \qquad 2-\sqrt{2}\le N\le2+\sqrt{2}.

Since $2-\sqrt{2}\approx0.5858$ and $2+\sqrt{2}\approx3.4142$, the qualifying counts are $1,2,3$. Hence

\begin{aligned} P(1\le N\le3) &=e^{-2}\left(\frac{2^1}{1!}+\frac{2^2}{2!}+\frac{2^3}{3!}\right)\\ &=e^{-2}\left(2+2+\frac{4}{3}\right)\\ &=e^{-2}\left(\frac{16}{3}\right) =0.7217882\ldots\approx0.7218. \end{aligned}

The normal approximation gives $P(|Z|\le1)=0.6827$. The exact value is about 0.7218, higher by approximately 0.0391. The direction of the difference is not fixed: discreteness and skewness can make the exact probability either above or below a normal target.

Check. The count $N=0$ is not accepted because its deviation from the mean is 2, which is greater than $\sqrt{2}$. The exact sum therefore begins at $k=1$, not at $k=0$.

Interpreting the Comparison

An exact check answers a specific question: under the stated Poisson model, what is the probability that the count meets the stated criterion? It does not establish that a normal approximation is always poor for small counts, nor that an exact probability should replace every normal-based credibility standard. Instead, it quantifies the approximation’s performance at the selected expected count and threshold.

For a relative-error criterion, the same logic applies. If the acceptable relative deviation is $\epsilon$, the event is $|N-\lambda|\le\epsilon\lambda$. Its integer bounds are

L=\max\!\left(0,\left\lceil(1-\epsilon)\lambda\right\rceil\right), \qquad U=\left\lfloor(1+\epsilon)\lambda\right\rfloor,

and the exact Poisson probability is summed from $L$ through $U$. A normal-based full-credibility threshold may be used to select a candidate $\lambda$; this exact calculation then checks the probability at that candidate. The threshold and the exact probability calculation are related, but they are not the same step.

State the model and criterion.
Write $N\sim\operatorname{Poisson}(\lambda)$ and specify the deviation or relative-error condition being checked.

Find the integer bounds.
Apply the ceiling to the lower endpoint, the floor to the upper endpoint, and restrict the lower count to at least zero.

Sum the probabilities.
Evaluate $e^{-\lambda}\sum_{k=L}^{U}\lambda^k/k!$, using a recurrence for consecutive terms if helpful.

Compare and interpret.
Compare the exact result with the normal target, report the difference, and state whether the approximation is adequate for the intended use.

Common Exam Traps

Rounding a bound in the wrong direction. The lower endpoint requires a ceiling and the upper endpoint requires a floor. Rounding both to the nearest integer can include a count outside the event or exclude one inside it.
Including an excluded endpoint. If the event begins at $N=1$, adding the $k=0$ term changes the probability. The initial probability $P(N=0)$ may help generate later terms, but it is not part of every sum.
Assuming the normal target is exact. A normal quantile defines an approximate threshold. The exact Poisson probability must be calculated to assess its performance.
Assuming the approximation error has one direction. The exact probability can be greater or less than the normal target. Determine the direction from the calculation rather than from a rule of thumb.
Forgetting the count support. Negative values are impossible for a claim count. A real lower endpoint below zero does not justify adding probabilities for negative integers.

Key takeaway: A normal-based Poisson standard gives an approximate probability target. To check it exactly, convert the event to nonnegative integer bounds and sum every Poisson probability within those bounds—no more and no fewer.

Check Your Understanding

For each question, find the integer bounds before calculating or interpreting the probability. Probabilities are rounded to four decimal places.

Let $N\sim\operatorname{Poisson}(3)$. Find the exact probability that $|N-3|\le2$. (Answer: bounds 1 through 5; probability 0.8663.)
Let $N\sim\operatorname{Poisson}(0.8)$. Find the exact probability that $|N-0.8|\le1.6$. (Answer: bounds 0 through 2; probability 0.9526.)
For $N\sim\operatorname{Poisson}(5)$, calculate the probability that $2\le N\le8$ by summing the Poisson probabilities. (Answer: 0.8915.)
For $N\sim\operatorname{Poisson}(4)$, calculate $P(1\le N\le7)$, taking care not to include the probability of zero. (Answer: $e^{-4}(50.8063492\ldots)=0.9306$.)
Suppose $N\sim\operatorname{Poisson}(2)$. Find the exact one-sided probability $P(N\ge4)$ using the complement $P(N\le3)$. (Answer: $1-e^{-2}(19/3)=0.1429$.)

What You'll Learn