Poisson Probability Recursion

Why Recurse Between Poisson Probabilities?

A Poisson probability calculation can involve factorials and powers of the parameter. When several probabilities are needed, calculating each one separately repeats work and creates opportunities for arithmetic errors. The ratio of consecutive Poisson probabilities avoids that repetition: one known probability can be carried forward or backward through the count values.

This method is useful for point probabilities, finite intervals, and tail probabilities. The main discipline is to keep the count index attached to each probability. In particular, the factor used to move from count $k$ to count $k+1$ is not the same as the factor used to move from count $k$ to count $k-1$.

Definition: Let $N$ be a Poisson random variable with parameter $\lambda>0$, so that $$ P(N=k)=p_k=e^{-\lambda}\frac{\lambda^k}{k!},\qquad k=0,1,2,\ldots. $$ The values $p_k$ are the Poisson point probabilities. For $\lambda=0$, $P(N=0)=1$ and $P(N=k)=0$ for $k\geq 1$; the reverse recursion below is not used in that degenerate case.

The Consecutive-Probability Recursion

Start with the Poisson formula for $p_k$. For $k\geq 0$, divide the expression for $p_{k+1}$ by the expression for $p_k$. The exponential terms cancel, and the factorial in the denominator contributes one additional factor of $k+1$:

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

Thus, if $p_k$ is known, multiply it by $\lambda/(k+1)$ to obtain $p_{k+1}$. Solving the same relation for the preceding probability gives the reverse recursion, valid for $k\geq 1$:

p_{k+1}=\frac{\lambda}{k+1}p_k, \qquad p_{k-1}=\frac{k}{\lambda}p_k.

The forward multiplier decreases as $k$ increases. Probabilities rise while $\lambda/(k+1)>1$, remain level when the multiplier is 1, and fall when it is less than 1. This is a useful reasonableness check, not a substitute for using the correct index.

Key result: If one Poisson point probability $p_j$ is known, neighboring probabilities can be generated in either direction using the appropriate recursion. For a finite interval, add the generated point probabilities. For an upper tail, either subtract the probabilities below the threshold from 1 or add the probabilities in the tail.

A Practical Calculation Procedure

First identify the parameter $\lambda$, the starting count and probability, and the event being requested. Then decide which direction the recursion must move. A finite interval that straddles the starting count can be handled by moving downward to its lower endpoint and upward to its upper endpoint. An upper tail can be calculated from a complement when only a few probabilities lie below its threshold; otherwise, recursively add terms in the tail.

Set the count index.
Write the starting probability as $p_j=P(N=j)$, and mark the requested count or range.

Choose the recursion direction.
Use $p_{k+1}=\lambda p_k/(k+1)$ to move upward, and $p_{k-1}=kp_k/\lambda$ to move downward.

Combine the required terms.
Add the point probabilities in a finite interval. For $P(N\geq r)$, either use $1-\sum_{k=0}^{r-1}p_k$ or add $p_r,p_{r+1},\ldots$.

Check the result.
Confirm the recursion used the correct index, the event includes the intended endpoint, and the probability lies between 0 and 1.

Worked Calculations

Worked Example: A Finite Interval Around the Starting Count

A hypothetical roadside-assistance portfolio has a Poisson number $N$ of covered callouts in a specified month, with $\lambda=4$. Assume the count follows the Poisson model and the parameter is constant for this month. Given $p_0=e^{-4}=0.0183156$, calculate $P(2\leq N\leq 5)$. Give the answer to four decimal places.

Method. Recurse upward from the given starting probability. The event includes counts 2, 3, 4, and 5, so all four probabilities must be added.

From $p_0$, the first probability is

p_1=\frac{4}{1}p_0=4(0.0183156)=0.0732624.

The next probabilities follow one at a time:

\begin{aligned} p_2&=\frac{4}{2}p_1=2(0.0732624)=0.1465248,\\ p_3&=\frac{4}{3}p_2=\frac{4}{3}(0.1465248)=0.1953664,\\ p_4&=\frac{4}{4}p_3=1(0.1953664)=0.1953664,\\ p_5&=\frac{4}{5}p_4=0.8(0.1953664)=0.1562931. \end{aligned}

Therefore,

P(2\leq N\leq 5) =p_2+p_3+p_4+p_5 =0.1465248+0.1953664+0.1953664+0.1562931 =0.6935507 \approx 0.6936.

The calculation is consistent with the probabilities increasing through $p_3$, tying at $p_4$, and then decreasing. The common mistake avoided here is using the same multiplier for every step: the denominator is the new count, so it changes from 2 to 3 to 4 to 5.

Worked Example: An Upper Tail by Complement

For a hypothetical small group of insured devices, suppose the number $N$ of breakdown claims during a month follows a Poisson distribution with $\lambda=2$. Calculate the probability of at least three claims. Use $e^{-2}=0.1353353$, and give the answer to four decimal places.

Method. The complement of $N\geq 3$ is $N=0,1,2$. Calculate those three probabilities by recursion, then subtract their sum from 1.

Starting at $p_0=e^{-2}$, the upward recursion gives

\begin{aligned} p_1&=\frac{2}{1}p_0=2(0.1353353)=0.2706706,\\ p_2&=\frac{2}{2}p_1=1(0.2706706)=0.2706706. \end{aligned}

Hence,

\begin{aligned} P(N\geq 3) &=1-(p_0+p_1+p_2)\\ &=1-(0.1353353+0.2706706+0.2706706)\\ &=1-0.6766765\\ &=0.3233235\\ &\approx 0.3233. \end{aligned}

As a check, the tail begins with $p_3=(2/3)p_2=0.1804471$, followed by $p_4=(2/4)p_3=0.0902236$, and continues with smaller positive terms. The tail is therefore greater than $p_3$, as the answer is. A common mistake is to report $0.6767$, which is the probability of the complement, $N\leq 2$, not the requested tail.

Worked Example: Moving Both Downward and Upward

A hypothetical monitoring system records a Poisson number $N$ of alerts per shift with $\lambda=5$. Suppose the starting probability $p_6=P(N=6)$ is $0.1462228$. Calculate $P(4\leq N\leq 8)$, rounding to four decimal places.

Method. The interval contains the starting count. Use the reverse recursion for counts 5 and 4, and the forward recursion for counts 7 and 8.

Moving downward from $p_6$:

\begin{aligned} p_5&=\frac{6}{5}p_6=1.2(0.1462228)=0.1754674,\\ p_4&=\frac{5}{5}p_5=1(0.1754674)=0.1754674. \end{aligned}

Moving upward from $p_6$:

\begin{aligned} p_7&=\frac{5}{7}p_6=\frac{5}{7}(0.1462228)=0.1044449,\\ p_8&=\frac{5}{8}p_7=\frac{5}{8}(0.1044449)=0.0652781. \end{aligned}

The requested interval includes every count from 4 through 8:

\begin{aligned} P(4\leq N\leq 8) &=p_4+p_5+p_6+p_7+p_8\\ &=0.1754674+0.1754674+0.1462228+0.1044449+0.0652781\\ &=0.6668806\\ &\approx 0.6669. \end{aligned}

The two probabilities $p_4$ and $p_5$ are equal because the reverse multiplier from $p_5$ to $p_4$ is $5/5=1$. This is also consistent with the forward multiplier from $p_4$ to $p_5$ being $5/5=1$. The common mistake avoided is using the forward factor to move downward; moving from $p_6$ to $p_5$ requires the factor $6/5$, not $5/6$.

Common Indexing Traps

Recursion is short, but a one-count indexing error can affect every probability that follows. Keep the following distinctions explicit:

Forward step: from $p_k$ to $p_{k+1}$, multiply by $\lambda/(k+1)$, using the destination count in the denominator.
Reverse step: from $p_k$ to $p_{k-1}$, multiply by $k/\lambda$, using the current count in the numerator.
Tail endpoint: $P(N\geq r)$ includes $p_r$, while $P(N>r)$ begins at $p_{r+1}$. The complement for $P(N\geq r)$ ends at $r-1$.
Range endpoints: for integer counts, $P(a\leq N\leq b)$ includes both $p_a$ and $p_b$. Count the terms before summing.

For an upper tail, the complement is often efficient when $r$ is small because only $r$ probabilities, from $p_0$ through $p_{r-1}$, are required. If the starting value is near the tail and the lower probabilities are numerous, direct recursive addition from $p_r$ may be more convenient. Whichever route is used, the event definition determines the terms; convenience does not change the endpoint.

Key takeaway: Poisson recursion replaces repeated factorial calculations with a sequence of simple ratios. Write down the current count before each step, choose the direction carefully, and verify that the summed terms match the event exactly.

Check Your Understanding

Use the Poisson recursion where appropriate. Round each probability to four decimal places.

A hypothetical service desk has a Poisson count $N$ of daily requests with $\lambda=3$. Starting from $p_0=e^{-3}$, calculate $P(N=4)$. (Answer: 0.1680)
A hypothetical sensor has a Poisson count $N$ of alarms per day with $\lambda=1.5$. Given $p_0=e^{-1.5}$, calculate $P(N\geq 2)$ by a complement. (Answer: 0.4422)
For a hypothetical repair portfolio, $N\sim\operatorname{Poisson}(4)$. Starting from $p_0=e^{-4}$, calculate $P(2\leq N\leq 4)$. (Answer: 0.5373)
A hypothetical fleet has $N\sim\operatorname{Poisson}(6)$, and $p_5=0.1606231$. Use recursion in both directions to calculate $P(4\leq N\leq 6)$. (Answer: 0.4551)
A hypothetical claims unit has $N\sim\operatorname{Poisson}(2.5)$. Starting from $p_0=e^{-2.5}$, calculate $P(N\leq 3)$. (Answer: 0.7576)

What You'll Learn