Poisson Counts as an (a,b,0) Distribution

Why the Poisson Distribution Belongs to the (a,b,0) Class

A Panjer recursion describes a sequence of count probabilities by relating each probability to the one immediately before it. For a Poisson count, the probability ratio between consecutive counts has a particularly simple form: it is the Poisson parameter divided by the new count. This makes the Poisson distribution a direct example of the $(a,b,0)$ class, with $a=0$.

The connection matters because a distribution’s class parameters let us recognize its probability structure and apply the corresponding recursion. Here the key is to keep the count index in the right place: the coefficient $b$ is constant, while the factor $b/n$ changes with $n$.

Definition: A probability sequence $\{p_n\}_{n\geq 0}$ is in the $(a,b,0)$ class if, for $n\geq 1$, its probabilities satisfy $$ p_n=\left(a+\frac{b}{n}\right)p_{n-1}, $$ with an initial probability $p_0$ and parameters for which the sequence is a valid probability distribution. For a Poisson random variable $N\sim\operatorname{Poisson}(\lambda)$, where $\lambda>0$, the class parameters are $a=0$ and $b=\lambda$, and $p_0=e^{-\lambda}$.

Deriving the Poisson Recursion

Assume the count $N$ has a Poisson distribution with parameter $\lambda>0$. Its probability mass function is

p_n=P(N=n)=e^{-\lambda}\frac{\lambda^n}{n!}, \qquad n=0,1,2,\ldots.

For $n\geq 1$, divide the probability at count $n$ by the probability at count $n-1$. Both probabilities are positive when $\lambda>0$, so this ratio is defined:

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

Since $p_n/p_{n-1}=\lambda/n$, multiplying by $p_{n-1}$ gives the recursion. Matching its coefficient with the defining $(a,b,0)$ form identifies the parameters.

p_n=\frac{\lambda}{n}p_{n-1} =\left(0+\frac{\lambda}{n}\right)p_{n-1}, \qquad a=0,\quad b=\lambda.

The initial probability comes from setting $n=0$ in the Poisson mass function: $p_0=e^{-\lambda}$. It is not determined by the recurrence alone; it must be specified to start the calculation. For the Poisson model, this initial probability also ensures that the sequence sums to one, because

\sum_{n=0}^{\infty}p_n = p_0\sum_{n=0}^{\infty}\frac{\lambda^n}{n!} = p_0e^\lambda = e^{-\lambda}e^\lambda = 1.

Key result: If $N\sim\operatorname{Poisson}(\lambda)$ with $\lambda>0$, then its $(a,b,0)$ parameters are $a=0$, $b=\lambda$; its initial probability is $p_0=e^{-\lambda}$; its mean is $E[N]=\lambda$; and its support is $\{0,1,2,\ldots\}$.

Mean and Support

The Poisson mean can be checked directly from the mass function. The term for $n=0$ contributes zero to the expectation. For $n\geq1$, use $n/n!=1/(n-1)!$, and then re-index the sum with $j=n-1$:

E[N] = \sum_{n=1}^{\infty}n e^{-\lambda}\frac{\lambda^n}{n!} = \lambda e^{-\lambda}\sum_{n=1}^{\infty}\frac{\lambda^{n-1}}{(n-1)!} = \lambda e^{-\lambda}\sum_{j=0}^{\infty}\frac{\lambda^j}{j!} = \lambda.

For $\lambda>0$, every mass $e^{-\lambda}\lambda^n/n!$ is positive for each nonnegative integer $n$. Thus every such count is in the support, and no non-integer value is possible. If $\lambda=0$, the distribution is degenerate: $P(N=0)=1$, its mean is zero, and its support is just $\{0\}$. The recurrence with $a=b=0$ gives $p_n=0$ for $n\geq1$, consistent with that degenerate case.

The recurrence also provides a practical check on calculations: once $p_0$ is known, each successive probability follows by multiplying the previous one by $\lambda/n$. The factor changes at every step. The recurrence is not $p_n=\lambda p_{n-1}$; that would omit the count index and produce incorrect probabilities.

Applying the Recursion

Worked Example: Building Probabilities and a Tail

A hypothetical portfolio has a Poisson annual claim count with parameter $\lambda=2$. Use the recursion to find $p_0,p_1,p_2$, then calculate $P(N\geq3)$. Give probabilities to four decimal places.

Method. Start from the Poisson initial probability and apply $p_n=(2/n)p_{n-1}$ for each successive count. Then use the complement of the first three probabilities. This avoids summing the entire infinite tail.

The initial probability is $p_0=e^{-2}=0.1353352832\ldots$. The next probabilities are

p_1=\frac{2}{1}p_0 =2(0.1353352832\ldots) =0.2706705665\ldots,

p_2=\frac{2}{2}p_1 =1(0.2706705665\ldots) =0.2706705665\ldots.

Therefore

P(N\geq3) =1-(p_0+p_1+p_2) =1-(0.1353352832\ldots+0.2706705665\ldots+0.2706705665\ldots) =1-5e^{-2} =0.3233235838\ldots.

Rounded to four decimal places, $p_0=0.1353$, $p_1=0.2707$, $p_2=0.2707$, and $P(N\geq3)=0.3233$. As a check, the probabilities for counts 0, 1, and 2 sum to $0.6766764162\ldots$, and that value plus the tail is 1.

Worked Example: Interpreting Class Parameters

A hypothetical count model is specified by $p_n=(3.5/n)p_{n-1}$ for $n\geq1$, with $p_0=e^{-3.5}$. Identify the distribution and its class parameters, calculate $P(N=1\text{ or }N=2)$, and state its mean and support.

Method. Compare the given recursion with $p_n=(a+b/n)p_{n-1}$, and match it to the Poisson recursion. Then calculate the two requested probabilities from $p_0$.

The constant part of the coefficient is zero, and the coefficient of $1/n$ is 3.5. Thus $a=0$, $b=3.5$, and $\lambda=3.5$. Since the initial probability is $e^{-3.5}$, the model is $N\sim\operatorname{Poisson}(3.5)$.

p_0=e^{-3.5}=0.0301973834\ldots,

p_1=\frac{3.5}{1}p_0 =3.5(0.0301973834\ldots) =0.1056908419\ldots,

p_2=\frac{3.5}{2}p_1 =1.75(0.1056908419\ldots) =0.1849589733\ldots.

The events $N=1$ and $N=2$ are disjoint, so

P(N=1\text{ or }N=2)=p_1+p_2 =0.1056908419\ldots+0.1849589733\ldots =0.2906498152\ldots \approx 0.2906.

The mean is $E[N]=\lambda=3.5$, and the support is all nonnegative integers. As a check on the recursion, $p_2/p_1=1.75=3.5/2$, as required for count 2.

Worked Example: Recovering the Parameter from a Probability Ratio

For a hypothetical Poisson count, the ratio $p_4/p_3$ is observed to be $0.65$. Determine $\lambda$, then calculate $P(N\leq2)$, to four decimal places. Also state the expected count.

Method. Use the consecutive-probability ratio at the specified count, not the ratio at count 3. For a Poisson count, $p_n/p_{n-1}=\lambda/n$.

At $n=4$, the ratio is $\lambda/4$. Therefore

0.65=\frac{\lambda}{4} \quad\Longrightarrow\quad \lambda=4(0.65)=2.6.

The initial probability and the next two probabilities are

p_0=e^{-2.6}=0.0742735782\ldots,

p_1=\frac{2.6}{1}p_0 =2.6(0.0742735782\ldots) =0.1931113033\ldots,

p_2=\frac{2.6}{2}p_1 =1.3(0.1931113033\ldots) =0.2510446944\ldots.

Thus

P(N\leq2)=p_0+p_1+p_2 =0.0742735782\ldots+0.1931113033\ldots+0.2510446944\ldots =0.5184295759\ldots \approx0.5184.

The expected count is $E[N]=2.6$. The ratio check is $p_4/p_3=2.6/4=0.65$, agreeing with the given information.

Common Exam Trap: Losing the Count Index

The Poisson recursion has $b=\lambda$, but the multiplier at step $n$ is $b/n$, not $b$ alone. For example, with $\lambda=2$, the correct first probability after $p_0$ is

p_1=\frac{2}{1}e^{-2}=0.2707\ldots.

A wrong calculation that uses $p_1=2p_0$ happens to give the same result at $n=1$, so it may not reveal the error immediately. At the next step, the correct value is $p_2=(2/2)p_1=p_1$. Continuing to multiply by 2 would instead give $2p_1$, which is wrong. Always substitute the current count $n$ into the denominator.

A second trap is to identify $b$ as $\lambda/n$. That expression is the full count-specific multiplier, not a class parameter: $b$ must stay fixed as $n$ changes. For a Poisson model, $b=\lambda$, while the recurrence uses $b/n$.

Key takeaway: A Poisson count with parameter $\lambda>0$ is in the $(a,b,0)$ class because its consecutive-probability ratio is $\lambda/n$. Identify $a=0$, $b=\lambda$, and $p_0=e^{-\lambda}$; then use the count index at each step. Its mean is $\lambda$, and its support is the nonnegative integers.

Check Your Understanding

For each problem, use the Poisson recursion where appropriate. Round probabilities to four decimal places.

A hypothetical Poisson count has $\lambda=1.6$. State its $(a,b,0)$ parameters and mean, and calculate $P(N\leq1)$. (Answer: $a=0,\ b=1.6;\ E[N]=1.6;\ P(N\leq1)=0.5249$)
A count sequence has $a=0$, $b=2.25$, and $p_0=e^{-2.25}$. Identify the Poisson parameter and calculate $p_1$. (Answer: $\lambda=2.25;\ p_1=2.25e^{-2.25}=0.2371$)
For a Poisson count, $p_2/p_1=1.4$. Find $\lambda$ and calculate $P(N=0\text{ or }N=1)$. (Answer: $\lambda=2.8;\ P(N=0\text{ or }N=1)=3.8e^{-2.8}=0.2311$)
A hypothetical event count over a fixed exposure is Poisson with parameter $\lambda=1.5$. Use the recursion to find $p_0,p_1,p_2,p_3$, then calculate $P(N=3)$. (Answer: $p_0=0.2231,\ p_1=0.3347,\ p_2=0.2510,\ p_3=0.1255;\ P(N=3)=0.1255$)
A proposed count model has $p_0=e^{-3}$ and $p_n=(3/n)p_{n-1}$ for $n\geq1$. Identify its distribution, class parameters, mean, and support. (Answer: $N\sim\operatorname{Poisson}(3);\ a=0,\ b=3;\ E[N]=3;\ \{0,1,2,\ldots\}$)

What You'll Learn