Poisson Tail Probabilities

Why Poisson Tail Probabilities Need Careful Cutoffs

A claim-frequency question often asks for the chance of several counts at once: no more than three claims, at least five claims, or more than a specified number. These are tail probabilities. The arithmetic is a sum of Poisson point probabilities, but a small change in wording can change which integer counts belong in that sum.

The first task is therefore to translate the wording into an event for the count. Only then should you calculate. For a Poisson count, a lower tail contains counts from zero up to a cutoff; an upper tail contains counts from a cutoff onward. The cumulative distribution function (CDF) packages the lower-tail sum and lets us calculate upper tails without adding infinitely many terms.

Definition: If $N\sim\operatorname{Poisson}(\lambda)$, where $\lambda>0$ is the expected number of events in the specified exposure period, then $N$ takes nonnegative integer values and $$ P(N=j)=e^{-\lambda}\frac{\lambda^j}{j!},\qquad j=0,1,2,\ldots. $$ Its cumulative distribution function at an integer $k\geq 0$ is $$ F_N(k)=P(N\leq k)=\sum_{j=0}^{k}e^{-\lambda}\frac{\lambda^j}{j!}. $$

The Poisson model is appropriate when the count is modeled over a specified exposure under a constant event rate, with events occurring independently. The parameter $\lambda$ must refer to that same exposure period. If the rate is stated per year but the question concerns several months, first convert the rate to the period in question; do not use the annual mean unchanged.

Translate the Wording Before Calculating

For an integer-valued count, “at most” includes its stated boundary, while “fewer than” excludes it. Similarly, “at least” includes its boundary, but “more than” excludes it. The following identities make the distinction explicit.

Wording	Event	Expression using the Poisson CDF
At most $k$	$N\leq k$	$F_N(k)$
Fewer than $k$	$N<k$, or $N\leq k-1$	$F_N(k-1)$
At least $r$	$N\geq r$	$1-F_N(r-1)$
More than $r$	$N>r$, or $N\geq r+1$	$1-F_N(r)$

The boundary shifts in the upper-tail expressions follow from the counts that are excluded. The event $N\geq r$ excludes $0,1,\ldots,r-1$, whose total probability is $F_N(r-1)$. The event $N>r$ excludes $0,1,\ldots,r$, whose total probability is $F_N(r)$. In each case, the desired probability is one minus the probability of the excluded counts.

P(N\geq r)=\sum_{j=r}^{\infty}P(N=j)=1-F_N(r-1),\qquad r\geq 1.

This also justifies why the CDF is useful for an upper tail: its complement turns an infinite sum into a finite sum over counts from zero through the correct cutoff. The formula applies to integer cutoffs; if a question gives a non-integer threshold, first identify the integer counts that satisfy the event.

Worked Example: An Upper Tail Starting at Three Claims

A hypothetical portfolio has a Poisson claim count $N$ for a one-year period, with $\lambda=2.5$. Calculate the probabilities of at most two claims and at least three claims. Give each probability to four decimal places.

Method. Calculate the lower tail by summing the masses for counts $0,1,2$, then use the CDF identity for the upper tail. The model is $N\sim\operatorname{Poisson}(2.5)$ for one year.

The required point probabilities are

\begin{aligned} P(N=0)&=e^{-2.5}=0.0820850,\\ P(N=1)&=e^{-2.5}\frac{2.5^1}{1!}=0.2052125,\\ P(N=2)&=e^{-2.5}\frac{2.5^2}{2!} =e^{-2.5}(3.125)=0.2565156. \end{aligned}

Thus, the probability of at most two claims is

P(N\leq 2)=0.0820850+0.2052125+0.2565156=0.5438131\approx 0.5438.

“At least three” means $N\geq 3$, so the excluded counts are $0,1,2$. Therefore,

P(N\geq 3)=1-F_N(2)=1-0.5438131=0.4561869\approx 0.4562.

Check. The lower and upper events partition all possible counts: $N\leq2$ or $N\geq3$. Their probabilities add to $0.5438131+0.4561869=1.0000000$. A common mistake is to treat “at least three” as $1-F_N(3)$, which excludes count three as well and instead calculates $P(N>3)$.

Reading the CDF Cutoff Correctly

An upper-tail calculation is only as accurate as its cutoff. To calculate $P(N\geq r)$, stop the lower-tail sum at $r-1$, not at $r$. To calculate $P(N>r)$, stop at $r$. The mass at the boundary is the difference between the two answers:

P(N\geq r)-P(N>r)=P(N=r).

This identity is a useful check. If two answers for “at least $r$” and “more than $r$” are equal, the probability of exactly $r$ has likely been omitted or the events have been translated incorrectly. The next example calculates both tails from the same Poisson model.

Worked Example: Distinguishing “At Least” From “More Than”

For a hypothetical monthly count $N\sim\operatorname{Poisson}(3)$, calculate $P(N\geq4)$ and $P(N>4)$. Give each answer to four decimal places.

Method. Use the CDF complement for each event, with separate cutoffs. For $N\geq4$, the excluded counts are $0$ through $3$. For $N>4$, they are $0$ through $4$.

The masses for counts zero through three are

\begin{aligned} P(N=0)&=e^{-3}=0.0497871,\\ P(N=1)&=3e^{-3}=0.1493612,\\ P(N=2)&=\frac{3^2}{2!}e^{-3}=4.5e^{-3}=0.2240418,\\ P(N=3)&=\frac{3^3}{3!}e^{-3}=4.5e^{-3}=0.2240418. \end{aligned}

Therefore,

$$ F_N(3)=0.0497871+0.1493612+0.2240418+0.2240418=0.6472319, $$

and

P(N\geq4)=1-F_N(3)=1-0.6472319=0.3527681\approx0.3528.

For the strict upper tail, first calculate the mass at four:

P(N=4)=\frac{3^4}{4!}e^{-3}=3.375e^{-3}=0.1680314.

Then $F_N(4)=0.6472319+0.1680314=0.8152633$, giving

P(N>4)=1-F_N(4)=1-0.8152633=0.1847367\approx0.1847.

Check. The difference is $0.3527681-0.1847367=0.1680314$, which equals $P(N=4)$. The common error is using $1-F_N(4)$ for $P(N\geq4)$; that expression gives $P(N>4)$ and wrongly removes the probability of exactly four claims.

Match the Parameter to the Exposure Period

When a Poisson process has a constant rate, its mean count is proportional to the length of the exposure period. If the expected count is $\lambda_0$ over a period of length $t_0$, then over a period of length $t$ the mean is $\lambda_0(t/t_0)$. This scaling relies on the constant-rate assumption; it should not be applied if the event rate varies over time without an appropriate model.

Once the mean is expressed for the requested period, calculate the tail just as before. The following example combines rate conversion with an inclusive upper-tail event.

Worked Example: Converting an Annual Rate to Six Months

A hypothetical book of policies has a constant expected claim rate of $2.4$ claims per year. Assume the claim count follows a Poisson model with independent events and constant rate. Find the probability of at least two claims in six months. Give the result to four decimal places.

Method. First scale the annual mean to six months, then calculate the complement of the counts below two. Six months is $6/12=0.5$ of a year, so the six-month Poisson mean is

\lambda=2.4\left(\frac{6}{12}\right)=1.2.

Thus, for the six-month period, $N\sim\operatorname{Poisson}(1.2)$. “At least two” means $N\geq2$, so the excluded counts are zero and one. Their probabilities are

\begin{aligned} P(N=0)&=e^{-1.2}=0.3011942,\\ P(N=1)&=1.2e^{-1.2}=0.3614331. \end{aligned}

The lower-tail probability is $F_N(1)=0.3011942+0.3614331=0.6626273$. Therefore,

P(N\geq2)=1-F_N(1)=1-0.6626273=0.3373727\approx0.3374.

Check. The complementary event is $N\leq1$, and $0.3373727+0.6626273=1$. A common mistake is to use $\lambda=2.4$ for six months, which incorrectly retains the full-year mean. Another is to use $1-F_N(2)$, which calculates $P(N>2)$, not $P(N\geq2)$.

A Reliable Tail-Calculation Process

A short sequence of decisions prevents most errors. The CDF is a sum of point probabilities, so the calculation remains transparent even when a calculator or software package supplies the final numerical value.

Set the model and exposure.
Identify the count $N$, state $N\sim\operatorname{Poisson}(\lambda)$, and ensure $\lambda$ refers to the period in the question.

Translate the wording to an event.
Write an inequality such as $N\leq k$, $N\geq r$, or $N>r$. This makes the boundary count explicit.

Choose the finite sum.
For a lower tail, add masses through the stated upper cutoff. For an upper tail, subtract the lower-tail sum ending one count below the inclusive threshold.

Check the result.
Confirm that the included and excluded counts match the wording, that the probability lies between zero and one, and, where useful, that a tail difference equals the boundary mass.

For a very small upper tail, it may also be possible to sum the relevant masses directly, but this requires handling every count from the threshold onward. The CDF complement is usually convenient because it uses a finite lower-tail sum. Whichever method is used, retain sufficient precision during intermediate steps and round only the final probability. In exam work, showing the event and the cutoff often matters as much as showing the arithmetic.

Key takeaway: For $N\sim\operatorname{Poisson}(\lambda)$, $P(N\leq k)=F_N(k)$, $P(N\geq r)=1-F_N(r-1)$, and $P(N>r)=1-F_N(r)$. Translate the words to an integer event before selecting the CDF cutoff, and match $\lambda$ to the stated exposure.

Check Your Understanding

For each problem, identify the event and its cutoff before calculating. Round probabilities to four decimal places.

A hypothetical monthly claim count has a Poisson distribution with mean $1.5$. Find the probability of at most one claim. (Answer: $0.5578$)
A hypothetical quarterly claim count is Poisson with mean $2$. Find the probability of at least two claims. (Answer: $0.5940$)
A hypothetical count $N$ has distribution $\operatorname{Poisson}(4)$. Find the probability of more than three events. (Answer: $0.5665$)
A hypothetical portfolio has a constant expected claim rate of $3.6$ claims per year. Under a Poisson model, find the probability of fewer than two claims in five months. (Answer: $0.5578$)
A hypothetical one-year count has distribution $\operatorname{Poisson}(6)$. Find the probability of at least five claims. (Answer: $0.7149$)

These methods are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. The central discipline is to translate the tail wording correctly, use the matching Poisson mean, and include or exclude the boundary count exactly as specified.

Wording	Event	Expression using the Poisson CDF
At most \(k\)	\(N\leq k\)	\(F_N(k)\)
Fewer than \(k\)	\(N<k\), or \(N\leq k-1\)	\(F_N(k-1)\)
At least \(r\)	\(N\geq r\)	\(1-F_N(r-1)\)
More than \(r\)	\(N>r\), or \(N\geq r+1\)	\(1-F_N(r)\)

What You'll Learn