Recurrence Relations for Poisson Probabilities

Why Use a Recurrence?

A Poisson probability can be calculated directly from its probability mass function, but evaluating a factorial and a power separately for every count can be inefficient. When several consecutive counts are needed, the probabilities are linked: each one can be calculated from the preceding one with a single multiplication and division. This is especially useful for range probabilities, where the event includes a sequence of neighboring integer counts.

The method depends on the model being Poisson with a fixed parameter for the count interval under study. For example, if the count is recorded over a different exposure period, the parameter must first be adjusted to that period. The recurrence does not determine the parameter; it only generates probabilities once the correct parameter is known.

Definition: If $X\sim\operatorname{Poisson}(\lambda)$, where $\lambda>0$, then $X$ takes nonnegative integer values and $$ P(X=k)=e^{-\lambda}\frac{\lambda^k}{k!},\qquad k=0,1,2,\ldots. $$ Writing $p_k=P(X=k)$, the forward recurrence is $$ p_{k+1}=p_k\frac{\lambda}{k+1},\qquad k=0,1,2,\ldots. $$

The denominator is the new count, $k+1$, not the old count $k$. To start a sequence from zero, use $p_0=e^{-\lambda}$. To start from another known point probability, apply the recurrence at that count and proceed forward. If $\lambda>0$, it can also be rearranged to work backward: $p_k=p_{k+1}(k+1)/\lambda$.

Why the Recurrence Holds

The recurrence follows directly by taking the ratio of two adjacent Poisson probabilities. This short derivation also makes clear why the denominator increases by one at each step.

Write adjacent probabilities.
For $k\geq 0$, the Poisson mass function gives $p_k=e^{-\lambda}\lambda^k/k!$ and $p_{k+1}=e^{-\lambda}\lambda^{k+1}/(k+1)!$.

Divide to form the ratio.
Since $\lambda>0$, $p_k>0$, so the ratio is defined: $p_{k+1}/p_k=[e^{-\lambda}\lambda^{k+1}/(k+1)!]/[e^{-\lambda}\lambda^k/k!]$.

Cancel common factors.
The exponential terms cancel, $\lambda^{k+1}/\lambda^k=\lambda$, and $k!/(k+1)!=1/(k+1)$. Thus $p_{k+1}/p_k=\lambda/(k+1)$, or $p_{k+1}=p_k\lambda/(k+1)$.

This result is for a Poisson distribution with one fixed parameter $\lambda$. It should not be used by substituting a changing parameter at each count. The ratio also helps check arithmetic: the next probability is larger if $\lambda>k+1$, equal if $\lambda=k+1$, and smaller if $\lambda<k+1$.

Building a Sequence from Zero

When an event involves consecutive counts, a reliable calculation is to find $p_0$, generate each probability up to the largest count required, and then add only the probabilities in the event. Do not round intermediate probabilities to the final reporting precision. Retain extra digits through the recurrence and round the final event probability once.

Worked Example: A Consecutive Range of Claim Counts

Hypothetical scenario. A portfolio is modeled as having a Poisson number $X$ of reported claims in a specified month, with parameter $\lambda=4$. Find $P(2\leq X\leq5)$. The distribution and parameter are Poisson with $\lambda=4$ for this month; no interest rate or mortality assumption is involved.

Method. Start from $p_0=e^{-4}$, use the forward recurrence to generate $p_1$ through $p_5$, and add $p_2,p_3,p_4,p_5$. This avoids recomputing factorials for each count.

The starting probability is $p_0=e^{-4}\approx0.0183156389$. Applying $p_{k+1}=p_k(4)/(k+1)$ gives:

\begin{aligned} p_1&=0.0183156389\left(\frac{4}{1}\right)=0.0732625556,\\ p_2&=0.0732625556\left(\frac{4}{2}\right)=0.1465251112,\\ p_3&=0.1465251112\left(\frac{4}{3}\right)=0.1953668149,\\ p_4&=0.1953668149\left(\frac{4}{4}\right)=0.1953668149,\\ p_5&=0.1953668149\left(\frac{4}{5}\right)=0.1562934519. \end{aligned}

Therefore, $$ P(2\leq X\leq5)=p_2+p_3+p_4+p_5 =0.1465251112+0.1953668149+0.1953668149+0.1562934519 =0.6935521929. $$ Rounded to four decimal places, the answer is 0.6936. The sequence is also plausible: the probabilities rise from $p_2$ to $p_3$, remain equal from $p_3$ to $p_4$, then decrease to $p_5$, in agreement with the adjacent ratios.

Common mistake avoided: The range begins at 2, so $p_0$ and $p_1$ are used to build the sequence but are not added to the requested event.

Starting at a Nonzero Count

Sometimes a problem supplies a point probability or asks for a range whose lower endpoint is not zero. In that case, use the probability at the starting count as the recurrence seed. Each successive value must still be calculated using the denominator corresponding to the new count. The following calculation illustrates why retaining precision matters when several probabilities are added.

Worked Example: Probabilities from the Fifth Count Onward

Hypothetical scenario. The number $X$ of service interruptions in a fixed monitoring period follows a Poisson distribution with parameter $\lambda=6$. Find $P(5\leq X\leq8)$. The model is Poisson with $\lambda=6$ for that period; interest and mortality assumptions do not apply.

Method. Calculate $p_5$ from the mass function, then use the recurrence for $p_6,p_7,p_8$. This is useful when a short consecutive range is required and checks that the recurrence is applied with the correct count in the denominator.

First, $$ p_5=e^{-6}\frac{6^5}{5!} =0.0024787521767\left(\frac{7776}{120}\right) =0.0024787521767(64.8) \approx0.1606231410. $$ Now apply the recurrence, retaining precision in the calculation:

\begin{aligned} p_6&=p_5\frac{6}{6}\approx0.1606231410,\\ p_7&=p_6\frac{6}{7}\approx0.1376769780,\\ p_8&=p_7\frac{6}{8}\approx0.1032577335. \end{aligned}

Thus, $$ P(5\leq X\leq8)=p_5+p_6+p_7+p_8 \approx0.1606231410+0.1606231410+0.1376769780+0.1032577335 =0.5621809935. $$ Rounded to four decimal places, the probability is 0.5622. As a check, $p_6=p_5$ because the recurrence ratio at $k=5$ is $6/6=1$; the following probabilities decrease because their recurrence ratios are $6/7$ and $6/8$, both less than one.

Common mistake avoided: At the step from $p_5$ to $p_6$, the denominator is 6, not 5. Also, $0.0024787521767\times64.8\approx0.1606231410$; rounding this product inaccurately can affect the displayed later probabilities, even if the final four-decimal answer happens to remain unchanged.

Adjusting the Parameter for Exposure

The recurrence requires the parameter for the interval in the question. If a count model specifies a constant rate per unit of exposure, the Poisson parameter for an exposure of $t$ units is the rate multiplied by $t$. Use this conversion before generating probabilities. A rate stated per year, for example, cannot be used unchanged for a six-month count.

Worked Example: A Rate Given per Year

Hypothetical scenario. A monitoring system has a constant event rate of 3.2 events per year. Assume the count during six months follows a Poisson model with parameter equal to rate times exposure. Find the probability of exactly three or exactly four events in six months. The distribution is Poisson; the annual rate is 3.2 events per year and the exposure is $6/12=0.5$ year, so the six-month parameter is $\lambda=3.2(0.5)=1.6$. Interest and mortality assumptions do not apply.

Method. Convert the exposure first, then start from $p_0=e^{-1.6}$ and recur through $p_4$. This prevents mixing a yearly parameter with a half-year event.

\begin{aligned} p_0&=e^{-1.6}\approx0.2018965180,\\ p_1&=p_0\frac{1.6}{1}\approx0.3230344288,\\ p_2&=p_1\frac{1.6}{2}\approx0.2584275430,\\ p_3&=p_2\frac{1.6}{3}\approx0.1378280229,\\ p_4&=p_3\frac{1.6}{4}\approx0.0551312092. \end{aligned}

The requested outcomes are disjoint, so their probabilities add: $$ P(X=3\text{ or }X=4)=p_3+p_4 \approx0.1378280229+0.0551312092 =0.1929592321. $$ Rounded to four decimal places, the answer is 0.1930. As a basic check, the requested probability is positive and less than one, and $p_4<p_3$ because the recurrence ratio $1.6/4$ is less than one.

Common mistake avoided: Using $\lambda=3.2$ would model a full year, not six months. The exposure conversion must precede the recurrence.

Common Errors and Checks

The recurrence is simple, but a small indexing error propagates through every subsequent probability. The following checks help catch mistakes before a range is summed.

Use $k+1$ in the denominator. From $p_k$ to $p_{k+1}$, multiply by $\lambda/(k+1)$. For example, from $p_5$ to $p_6$, the multiplier is $\lambda/6$, not $\lambda/5$.
Use one fixed parameter. Every step in a sequence uses the same $\lambda$, appropriate to the stated observation interval.
Respect the event endpoints. For $a\leq X\leq b$, add $p_a,p_{a+1},\ldots,p_b$, including both endpoints. A recurrence may generate values outside that range, but those values are not automatically part of the event.
Keep guard digits. Carry several decimal places through each step and round only the final probability, usually to four decimal places here.
Check the direction of change. The ratio $p_{k+1}/p_k=\lambda/(k+1)$ tells whether the next probability should increase, remain equal, or decrease. This is a local arithmetic check, not a substitute for verifying the model.

If a needed starting probability is supplied rather than computed from $e^{-\lambda}$, verify that it is consistent with the stated Poisson model before extending the sequence. The recurrence propagates the supplied value; it cannot correct an inconsistent starting value or an incorrectly chosen parameter.

Key takeaway: For a Poisson count with fixed parameter $\lambda$, generate each successive point probability by multiplying the preceding probability by $\lambda$ and dividing by the new count. Use the correct exposure-specific parameter, sum exactly the requested counts, and delay rounding until the end. This recurrence technique is aligned with the SOA VEE Mathematical Statistics syllabus.

Check Your Understanding

For each problem, use the recurrence and round the final probability to four decimal places unless the question asks otherwise.

A Poisson count has parameter $\lambda=3$. Starting from $p_0=e^{-3}$, calculate $P(X=2)$ by recurrence. (Answer: 0.2240)
A Poisson count has parameter $\lambda=4$, and $p_2\approx0.1465251$. Use one recurrence step to find $P(X=3)$. (Answer: 0.1954)
A Poisson count has parameter $\lambda=2$. Use $p_0=e^{-2}$ and the recurrence to find $P(2\leq X\leq3)$. (Answer: 0.4511)
A Poisson count has parameter $\lambda=5$. First calculate $p_5=e^{-5}5^5/5!$, then use the recurrence to find $P(X=6)$. (Answer: 0.1462)
A hypothetical system has a constant event rate of 3 events per year. Assume its six-month count is Poisson with parameter equal to rate times exposure. Find the probability of at least two events in six months, using a complement and the recurrence for $p_0$ and $p_1$. (Answer: 0.4422)

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