Poisson Probability And Tail Calculations

Introduction

A claim-frequency model is useful only if its probabilities are interpreted correctly. An insurer may need the chance of no claims, exactly a specified number of claims, or a count that exceeds a threshold. For a Poisson count, each calculation follows from one probability function, but the wording of the event determines which terms to include.

This tutorial focuses on calculating those probabilities and checking the boundaries. Assume $N$ is the number of claims during a specified observation period and has a Poisson distribution with mean $\lambda$ claims in that same period. The examples are hypothetical.

Definition: A count $N$ has a Poisson distribution with mean $\lambda>0$, written $N\sim\operatorname{Poisson}(\lambda)$, if its possible values are the nonnegative integers and $$ P(N=k)=e^{-\lambda}\frac{\lambda^k}{k!},\qquad k=0,1,2,\ldots. $$ Here $\lambda=E[N]$. The mean must refer to the observation period represented by $N$. The distribution is appropriate only when the Poisson frequency model is assumed for that count; the formula does not itself establish that the model fits a particular portfolio.

Exact Counts and the Recurrence

For an exact count $k$, substitute $\lambda$ and $k$ into the probability function. The probability of zero claims is especially simple: $P(N=0)=e^{-\lambda}$. For larger counts, calculating a factorial and a power separately is possible, but a recurrence is often faster and reduces arithmetic errors.

Result: If $p_k=P(N=k)$, then $$ p_0=e^{-\lambda},\qquad p_k=p_{k-1}\frac{\lambda}{k}\quad(k\geq 1). $$ This recurrence gives probabilities successively. It does not change the Poisson model or its parameters.

To justify the recurrence, divide the probability for $k$ by the probability for $k-1$:

\frac{p_k}{p_{k-1}} = \frac{e^{-\lambda}\lambda^k/k!}{e^{-\lambda}\lambda^{k-1}/(k-1)!} = \frac{\lambda}{k}.

Thus $p_k=p_{k-1}\lambda/k$. The exponential factor cancels, and the factorial ratio is $(k-1)!/k!=1/k$. Starting with $p_0$, the recurrence generates the exact-count probabilities in order.

Worked Example: No Claims, Exactly Two, and More Than Two

Suppose a hypothetical warranty portfolio has a claim count $N\sim\operatorname{Poisson}(1.5)$ over a specified six-month period. Calculate the probabilities of no claims, exactly two claims, at most two claims, and more than two claims.

Method. Use the Poisson probability function and recurrence for $\lambda=1.5$. Retain full precision in the calculations, then report probabilities to four decimal places.

First, $p_0=e^{-1.5}=0.2231301601$. Then the recurrence gives $p_1=p_0(1.5/1)=0.3346952402$ and $p_2=p_1(1.5/2)=0.2510214302$. These values also agree with direct substitution into the probability function.

Therefore, the probability of no claims is $p_0$, or 0.2231, and the probability of exactly two claims is $p_2$, or 0.2510.

“At most two” includes 0, 1, and 2 claims: $P(N\leq2)=p_0+p_1+p_2 =0.2231301601+0.3346952402+0.2510214302 =0.8088468305$. To four decimal places, this is 0.8088.

“More than two” means 3 or more, so it is the complement of “at most two”: $P(N>2)=1-P(N\leq2)=1-0.8088468305=0.1911531695$. To four decimal places, this is 0.1912.

Check. The events $N\leq2$ and $N>2$ are disjoint and together cover every possible count. Their probabilities sum to $0.8088468305+0.1911531695=1$, as required.

Intervals and Upper Tails

A finite interval requires adding every exact-count probability in the interval, including both endpoints when the wording is inclusive. For example, $P(a\leq N\leq b)$ contains the terms for $a,a+1,\ldots,b$. When the range has many terms, a cumulative probability or a complement may be more efficient than a long direct sum.

Result: For nonnegative integers $a\leq b$, $$ P(a\leq N\leq b)=\sum_{k=a}^{b}p_k. $$ For an integer $r\geq0$, $$ P(N>r)=1-\sum_{k=0}^{r}p_k, \qquad P(N\geq r)=1-\sum_{k=0}^{r-1}p_k\quad(r\geq1). $$ The distinction between $>$ and $\geq$ changes the boundary term.

A cumulative probability is $P(N\leq r)=\sum_{k=0}^{r}p_k$. A software function may calculate a cumulative probability directly, but the event must still be specified correctly. When calculating by hand, the recurrence can supply the terms, and a complement can reduce the amount of addition. Keep enough digits in intermediate values: summing rounded terms can shift the final result.

Worked Example: An Inclusive Claim-Count Interval

For a hypothetical equipment-insurance segment, suppose $N\sim\operatorname{Poisson}(3.2)$ during a stated reporting period. Find the probability that the segment has between two and five claims, inclusive.

Method. Sum $p_2,p_3,p_4,p_5$. The endpoints are included because the event is $2\leq N\leq5$. Begin with $p_0=e^{-3.2}=0.04076220398$, then apply the recurrence:

\begin{aligned} p_1&=p_0\frac{3.2}{1}=0.13043905273,\\ p_2&=p_1\frac{3.2}{2}=0.20870248437,\\ p_3&=p_2\frac{3.2}{3}=0.22261598333,\\ p_4&=p_3\frac{3.2}{4}=0.17809278666,\\ p_5&=p_4\frac{3.2}{5}=0.11397938346. \end{aligned}

Using the unrounded probabilities, $P(2\leq N\leq5)=p_2+p_3+p_4+p_5 =0.20870248437+0.22261598333+0.17809278666+0.11397938346 =0.72339063782$. Rounded to four decimal places, the answer is 0.7234.

Check. The interval consists of four disjoint exact-count events, so its probability is their sum. It is less than 1, as it must be; the excluded counts are 0, 1, and 6 or more.

Worked Example: At Least Five Claims and the Boundary Complement

A hypothetical service plan has a claim count $N\sim\operatorname{Poisson}(4)$ for its specified observation period. Calculate the probability of at least five claims.

Method. “At least five” means $N\geq5$. Its complement is $N\leq4$, so calculate the probabilities from zero through four and subtract their sum from 1.

Starting from $p_0=e^{-4}=0.01831563889$, the recurrence produces $p_1=0.01831563889(4/1)=0.07326255555$, $p_2=0.07326255555(4/2)=0.14652511111$, $p_3=0.14652511111(4/3)=0.19536681481$, and $p_4=0.19536681481(4/4)=0.19536681481$.

Thus $P(N\leq4)=p_0+p_1+p_2+p_3+p_4=0.62883693518$, and $P(N\geq5)=1-0.62883693518=0.37116306482$. Rounded to four decimal places, the probability is 0.3712.

Check. The complement contains counts 0 through 4, so the remaining event begins at 5. The result is between 0 and 1 and is larger than $P(N=5)$, since it includes every count above 5 as well.

Choosing a Calculation and Avoiding Common Traps

For a short interval, direct addition is transparent. For an upper tail beginning at a large count, a complement is often more efficient if the cumulative terms below the threshold are manageable. For a lower tail, direct cumulative addition is natural. In all cases, write the event in symbols before calculating: this makes the included counts explicit.

Identify the count and its period.
Confirm that the stated $\lambda$ is the mean for the exact observation period represented by $N$.

Translate the wording.
Write an exact count as $N=k$, a range as $a\leq N\leq b$, and a tail using either $>$ or $\geq$ exactly as stated.

Choose the calculation.
Use the probability function or recurrence for exact counts, sum the included terms for an interval, and consider a complement for a tail.

Check and round.
Check that probabilities lie between 0 and 1, verify complementary events sum to 1, and round only after retaining full precision through the calculation.

A frequent exam trap is confusing “at least” with “more than.” The wrong calculation for “at least five” is $1-P(N\leq5)$; that expression equals $P(N\geq6)$, because it excludes the outcome of exactly five. The correction is $P(N\geq5)=1-P(N\leq4)$. Conversely, “more than five” is $P(N>5)=1-P(N\leq5)$.

Another trap is rounding each recurrence value before using it in the next step. Small rounding differences can accumulate across a sum, especially when several probabilities are involved. Maintain full calculator precision (or enough guard digits), sum unrounded values, and round the final answer to the requested number of decimal places. A recurrence calculated from an already rounded $p_{k-1}$ should not be presented as though it were the exact $p_k$.

The model’s mean also needs attention. If $\lambda$ is specified for one period but the question asks about a different period, do not silently use the given mean without checking that it corresponds to the count being analyzed. Use the period and parameter explicitly given or established by the problem. Probability calculations cannot repair a mismatch between the modeled count and its mean.

Key takeaway: Poisson exact-count probabilities come from $e^{-\lambda}\lambda^k/k!$; intervals are sums of the included terms, and tails can be calculated by complements. The most reliable routine is to identify the count boundary first, preserve precision, and check the result against its complement or included events.

Check Your Understanding

For each problem, first express the event using the count variable, then calculate. Answers are rounded to four decimal places.

A hypothetical roadside-assistance program has $N\sim\operatorname{Poisson}(0.8)$ calls during its stated observation period. Find the probability of no calls. (Answer: 0.4493)
A hypothetical repair pool has $N\sim\operatorname{Poisson}(2.5)$ claims for a specified period. Find the probability of exactly three claims. (Answer: 0.2138)
A hypothetical benefits plan has $N\sim\operatorname{Poisson}(1.7)$ claims during a stated period. Find the probability of at most one claim. (Answer: 0.4932)
A hypothetical glass-protection program has $N\sim\operatorname{Poisson}(3)$ claims. Find the probability of between two and four claims, inclusive. (Answer: 0.6161)
A hypothetical equipment portfolio has $N\sim\operatorname{Poisson}(2.4)$ claims during its specified period. Find the probability of at least five claims, using a complement. (Answer: 0.0959)

What You'll Learn