Boundary Probabilities in the (a,b,1) Class

Why the First Two Probabilities Matter

A Panjer recursion specifies how probabilities relate after the sequence has started, but a recursion that begins at count 2 does not itself set the probabilities of counts 0 and 1. Those two values are the boundary probabilities $p_0=P(N=0)$ and $p_1=P(N=1)$. They determine the scale of the remaining probabilities, while the requirement that all probabilities sum to one constrains their possible values.

The key distinction is between a finite and an infinite total weight for the positive-count terms. When that weight is finite and the terms are nonnegative, normalization allows a range of boundary pairs unless one boundary value or another condition is supplied. When the weight is infinite, any positive $p_1$ would give infinite total probability, so the only normalized choice is $p_1=0,\ p_0=1$.

Definition: A count $N$ is in the Panjer $(a,b,1)$ class if its probabilities $p_n=P(N=n)$ satisfy $$ p_n=\left(a+\frac{b}{n}\right)p_{n-1},\qquad n\ge 2. $$ The recursion starts at $n=2$, so it does not specify $p_0$ or $p_1$. Those are boundary probabilities, and a valid probability distribution must have $p_n\ge 0$ for every $n$ and $\sum_{n=0}^{\infty}p_n=1$.

Weights Generated From the Recursion

For fixed $a$ and $b$, define the weight $w_1=1$ and generate further weights using the same factors as the probability recursion. This separates the shape of the positive-count sequence, determined by $a,b$, from its scale, determined by $p_1$.

w_n=\prod_{j=2}^{n}\left(a+\frac{b}{j}\right)\quad(n\ge 2), \qquad w_1=1, \qquad p_n=p_1w_n\quad(n\ge 1).

If all the weights are nonnegative, define their total by $W=\sum_{n=1}^{\infty}w_n$, allowing $W=\infty$. Because $w_1=1$, the weights are not all zero. The sum $W$ is the total positive-count mass per unit of $p_1$; it is not itself a probability.

Start at one.
Set $w_1=1$. This makes the formula $p_1=p_1w_1$ hold at the boundary.

Apply each recursion factor.
For each $n\ge 2$, multiply the preceding weight by $a+b/n$. Then $w_n=(a+b/n)w_{n-1}$.

Scale by the boundary probability.
Induction gives $p_n=p_1w_n$ for all $n\ge1$, including cases where a weight becomes zero.

Normalize.
Add the zero-count probability to the positive-count probabilities: $p_0+p_1W=1$ when $W$ is finite.

Normalization and the Boundary Pair

The recursion does not involve $p_0$. Consequently, normalization is applied to the full sequence, not just to its first two terms: the probabilities for counts 2 and above must also be included. The following result gives the exact constraint when the weights are nonnegative.

Theorem: Suppose the $(a,b,1)$ weights $w_n$ are nonnegative.

If $W=\sum_{n=1}^{\infty}w_n<\infty$, a normalized sequence has $p_1=t$ and $p_0=1-Wt$, where $0\le t\le 1/W$. Conversely, every such $t$ gives a normalized nonnegative sequence.
If $W=\infty$, normalization requires $p_1=0$ and $p_0=1$. This gives the distribution concentrated at zero.

For finite $W$, the result follows by summing the positive-count probabilities: $\sum_{n=1}^{\infty}p_n=\sum_{n=1}^{\infty}p_1w_n=p_1W$. Thus total probability is one exactly when $p_0+p_1W=1$. Nonnegativity requires $p_1\ge0$ and $p_0=1-p_1W\ge0$, giving $0\le p_1\le1/W$. Since $w_1=1$, $W\ge1$, so this interval is well-defined. The recursion then verifies that all the terms have the required form.

If $W=\infty$ and $p_1>0$, the positive-count sum is $p_1W=\infty$, which cannot be part of a probability distribution. Therefore $p_1=0$. The recursion then gives $p_n=0$ for every $n\ge2$, and normalization forces $p_0=1$. This infinite-weight case is not an exception to normalization: it makes the normalized boundary pair unique.

If any weight is negative, a positive $p_1$ would make the corresponding $p_n=p_1w_n$ negative. Thus a valid distribution in that case must have $p_1=0,\ p_0=1$. In applications, check nonnegativity as well as the weight sum; a finite sum alone does not make a sequence a valid distribution.

Worked Example: A Finite Weight Sum and a Given Zero Probability

Let $N$ have a Poisson distribution with parameter 1, so $p_n=e^{-1}/n!$ for $n\ge0$. Its probabilities satisfy the $(a,b,1)$ recursion with $a=0,\ b=1$. Suppose only the recursion coefficients and $p_0=e^{-1}$ are used to recover $p_1$. Find $p_1$, $p_2$, and $p_3$. Method: construct the weights and normalize. Mistake avoided: setting $p_0+p_1=1$ while omitting the probabilities for counts 2 and above.

With $a=0,\ b=1$, the recursion factor at count $n$ is $1/n$. Starting with $w_1=1$, the next weights are $$ w_2=\frac{1}{2},\qquad w_3=\frac{1}{3}\cdot\frac{1}{2}=\frac{1}{6},\qquad w_4=\frac{1}{4}\cdot\frac{1}{6}=\frac{1}{24}. $$ In general $w_n=1/n!$, so $$ W=\sum_{n=1}^{\infty}\frac{1}{n!}=e-1. $$

Since $p_0=e^{-1}$, normalization gives $$ p_1=\frac{1-p_0}{W} =\frac{1-e^{-1}}{e-1} =\frac{(e-1)/e}{e-1} =\frac{1}{e}. $$ Then $$ p_2=p_1w_2=\frac{1}{2e},\qquad p_3=p_1w_3=\frac{1}{6e}. $$ As a check, the full positive-count mass is $p_1W=(1/e)(e-1)=1-e^{-1}$; adding $p_0=e^{-1}$ gives 1. The example has a finite weight sum, and the supplied $p_0$ selects one member of the allowable boundary family.

Worked Example: A Finite-Support Sequence and Its Endpoint

Let $N\sim\operatorname{Binomial}(3,1/3)$. Show how its recursion determines the positive-count weights, recover $p_1$ from $p_0$, and check the endpoint. Method: use the binomial probability ratio to identify $a,b$, then apply the finite-weight normalization equation. Mistake avoided: continuing a finite-support sequence with nonzero probabilities after its endpoint.

For $n=2,3$, the ratio of consecutive binomial probabilities is $$ \frac{p_n}{p_{n-1}} =\frac{4-n}{n}\cdot\frac{1/3}{2/3} =-\frac{1}{2}+\frac{2}{n}. $$ Thus $a=-1/2$ and $b=2$. The weights are $$ w_1=1,\qquad w_2=-\frac12+\frac22=\frac12,\qquad w_3=\left(-\frac12+\frac23\right)w_2 =\frac16\cdot\frac12=\frac{1}{12}, $$ and $$ w_4=\left(-\frac12+\frac24\right)w_3=0. $$ Every subsequent weight is also zero, because its recurrence multiplies the preceding zero. Therefore $$ W=1+\frac12+\frac{1}{12}=\frac{19}{12}. $$

The zero probability is $p_0=(2/3)^3=8/27$. Hence $$ p_1=\frac{1-p_0}{W} =\frac{1-8/27}{19/12} =\frac{19}{27}\cdot\frac{12}{19} =\frac{12}{27}=\frac49. $$ The next probabilities are $$ p_2=p_1w_2=\frac49\cdot\frac12=\frac29,\qquad p_3=p_1w_3=\frac49\cdot\frac{1}{12}=\frac{1}{27},\qquad p_4=p_1w_4=0. $$ Check the total: $8/27+12/27+6/27+1/27=27/27=1$. The zero weight at count 4 is the endpoint; the class recursion does not imply positive probabilities beyond it.

Worked Example: Infinite Total Weight Forces Degeneracy

Suppose $a=1,\ b=0$, so $p_n=p_{n-1}$ for every $n\ge2$. Determine the only normalized distribution in this class. Method: examine the weights before applying normalization. Mistake avoided: treating the normalization equation as if it always left a free choice of $p_0,p_1$.

Here each recursion factor equals 1. Therefore $$ w_1=1,\quad w_2=1,\quad w_3=1,\quad\ldots,\qquad W=\sum_{n=1}^{\infty}1=\infty. $$ If $p_1>0$, the recursion gives $p_n=p_1$ for every $n\ge1$, and the sum of positive-count probabilities diverges. A probability distribution cannot have that total, so $p_1=0$. The recursion then gives $p_n=0$ for all $n\ge2$, and normalization gives $p_0=1$. The sequence $(1,0,0,\ldots)$ satisfies the recursion and sums to one, confirming the result.

How to Use the Boundary Constraint

When the weights are nonnegative and $W$ is finite, knowing either boundary probability determines the other: $$ p_1=\frac{1-p_0}{W}\quad\text{if }p_0\text{ is given},\qquad p_0=1-p_1W\quad\text{if }p_1\text{ is given}. $$ The result must satisfy $0\le p_1\le1/W$. If it does not, the proposed boundary value is inconsistent with a normalized nonnegative sequence.

If only $a,b$ are given and $W$ is finite with nonnegative weights, normalization constrains the boundary pair but does not generally select one pair: every $p_1$ in $[0,1/W]$, with $p_0=1-Wp_1$, is possible under the recursion. An additional condition, such as a specified $p_0$, is needed to select a particular pair. In contrast, when the nonnegative weights have infinite sum, normalization itself forces $p_1=0,\ p_0=1$. When the weights include a negative term, nonnegativity also rules out a positive $p_1$.

A useful exam check is to calculate the first few weights, including any zero or negative factor, before manipulating a sum. Then decide whether $W$ is finite or infinite and only after that apply normalization. This order prevents both overlooking a finite-support endpoint and assuming a positive-count distribution exists when the weights do not permit one.

Key takeaway: For a valid nondegenerate $(a,b,1)$ sequence, $p_n=p_1w_n$ for $n\ge1$, with nonnegative weights and finite total $W$. Normalization then gives $p_0+p_1W=1$. If the nonnegative weights sum to infinity, the only normalized sequence is $p_0=1,\ p_1=0$.

Check Your Understanding

For each problem, first check the weights, then use normalization where applicable.

For $a=0,\ b=2$, the weights are $w_n=2^{n-1}/n!$. If $p_0=e^{-2}$, find $p_1$. (Answer: $2e^{-2}$)
For $a=0,\ b=1$, the weights are $w_n=1/n!$. If $p_1=1/4$, find $p_0$. (Answer: $1-(e-1)/4$)
For $a=-1/2,\ b=2$, the weights are $1,1/2,1/12,0,\ldots$. If $p_1=1/3$, find $p_0$. (Answer: $17/36$)
For $a=1,\ b=0$, can a normalized sequence have $p_0=1/5$? (Answer: No; the weights sum to infinity, so the only normalized sequence has $p_0=1,\ p_1=0$.)
For $a=-1,\ b=1$, determine whether a valid normalized sequence can have $p_1>0$. The first generated weight is $w_2=-1/2$. (Answer: No; $p_2=p_1w_2<0$ if $p_1>0$, so the only valid sequence has $p_1=0,\ p_0=1$.)

What You'll Learn