Checking Recursion Indices and Probability Ratios

Why Recursion Indices Matter

A probability recursion can make a long sequence of count probabilities quick to calculate. But the formula is valid only at specified indices. Starting it one step too early, replacing its denominator, or treating a boundary probability as part of the recurrence can produce plausible-looking numbers that do not describe the stated distribution.

This tutorial focuses on auditing those details for the Panjer frequency classes $(a,b,0)$ and $(a,b,1)$. Write $p_n=P(N=n)$ for the probability that a claim count $N$ equals $n$. The central checks are: identify the first index at which the recurrence applies, preserve the $n$ in its denominator, and handle $p_0$ and $p_1$ according to the class being used.

Definition: An $(a,b,0)$ probability sequence satisfies $$ p_n=\left(a+\frac{b}{n}\right)p_{n-1},\qquad n\geq 1, $$ where $p_0$ is supplied separately. An $(a,b,1)$ probability sequence satisfies $$ p_n=\left(a+\frac{b}{n}\right)p_{n-1},\qquad n\geq 2, $$ where $p_0$ and $p_1$ are supplied separately. The recurrence is not a formula at indices outside its stated range.

For either class, a probability ratio can be written $p_n/p_{n-1}=a+b/n$ whenever the recurrence applies and $p_{n-1}>0$. This ratio form is useful for checking a calculation, but it does not remove the index restriction. In particular, the expression $b/n$ is undefined at $n=0$, and the $(a,b,1)$ recurrence does not define $p_1$.

What the Ratio Tells You

The denominator is the current index $n$, the index of the probability on the left side of the recurrence. It is not the preceding index $n-1$. One way to audit the notation is to write the ratio explicitly as $p_n/p_{n-1}$ before substituting a value of $n$.

For example, for a Poisson count with parameter $\lambda>0$, $p_n=e^{-\lambda}\lambda^n/n!$. Dividing the probability at $n$ by the one at $n-1$ gives

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

Thus the Poisson parameters are $a=0$ and $b=\lambda$. The factorial ratio $(n-1)!/n!=1/n$ explains why the denominator is $n$. Replacing it by $n-1$ is not a harmless reindexing; it changes the ratio and therefore the probability sequence.

Worked Example: Starting a Poisson Recursion at the Right Index

A hypothetical portfolio has a Poisson claim count with parameter $\lambda=2$. Calculate $p_0$ through $p_4$ recursively, and explain why the recurrence must not be started at $n=0$.

Method and assumptions. The count is Poisson with parameter $2$, so $p_n=(2/n)p_{n-1}$ for $n\geq1$, and $p_0=e^{-2}$. Use $e^{-2}\approx0.1353353$ for the displayed checks.

The boundary probability is $p_0=e^{-2}\approx0.1353353$. Apply the recurrence at $n=1$: $p_1=(2/1)p_0=2e^{-2}\approx0.2706706$.

At $n=2$, $p_2=(2/2)p_1=p_1=2e^{-2}\approx0.2706706$. At $n=3$, $p_3=(2/3)p_2=(4/3)e^{-2}\approx0.1804471$.

At $n=4$, keep the denominator equal to $4$: $p_4=(2/4)p_3=(1/2)(4/3)e^{-2}=(2/3)e^{-2}\approx0.0902235$. The four-decimal probabilities are $p_0=0.1353$, $p_1=0.2707$, $p_2=0.2707$, $p_3=0.1804$, and $p_4=0.0902$.

Check and mistake avoided. Each ratio $p_n/p_{n-1}$ is $2/n$ at its stated index. At $n=0$, the expression $2/0$ is undefined, so the recurrence cannot calculate $p_0$; $p_0$ comes from the Poisson model. Also, at $n=4$, $2/4=1/2$, not $2/3$. The value $0.0902$ follows from $(1/2)(0.1804471)$, not from multiplying by $2/3$.

Check the Denominator and the Support

The binomial model provides a second useful audit because its ratio contains both $a$ and $b/n$, and the count has a finite endpoint. If $N$ has a binomial distribution with parameters $m$ and $q$, where $m$ is a positive integer and $0<q<1$, then $p_n=\binom{m}{n}q^n(1-q)^{m-n}$ for $0\leq n\leq m$. For $1\leq n\leq m$, division by $p_{n-1}$ gives

\frac{p_n}{p_{n-1}} = \frac{m-n+1}{n}\frac{q}{1-q} = -\frac{q}{1-q} + \frac{(m+1)q}{(1-q)n}.

Therefore $a=-q/(1-q)$ and $b=(m+1)q/(1-q)$, with the recursion used only for $1\leq n\leq m$. The upper endpoint matters: $p_n=0$ for $n>m$, and one should not keep applying this ratio beyond the binomial support.

Worked Example: Auditing a Binomial Ratio and Endpoint

A hypothetical inspection process records the number of defective units in a sample of four. Assume independent unit outcomes, each defective with probability $q=0.2$. Use the Panjer recursion to calculate $p_0$ through $p_4$, then check the endpoint.

Method and assumptions. The count is binomial with $m=4$ and $q=0.2$. Thus $1-q=0.8$, $p_0=(0.8)^4=0.4096$, $a=-0.2/0.8=-0.25$, and $b=5(0.2)/0.8=1.25$. For $1\leq n\leq4$, use $p_n=(-0.25+1.25/n)p_{n-1}$.

At $n=1$, the multiplier is $-0.25+1.25=1$, so $p_1=1(0.4096)=0.4096$. At $n=2$, it is $-0.25+1.25/2=0.375$, so $p_2=0.375(0.4096)=0.1536$.

At $n=3$, the multiplier is $-0.25+1.25/3=1/6$, so $p_3=(1/6)(0.1536)=0.0256$. At $n=4$, the multiplier is $-0.25+1.25/4=0.0625$, so $p_4=0.0625(0.0256)=0.0016$.

The probabilities sum to $0.4096+0.4096+0.1536+0.0256+0.0016=1.0000$. The four-unit sample cannot have more than four defectives, so $p_5=0$. The recursion is used through $n=4$, not extended as though the binomial count had infinite support.

Check and mistake avoided. The $n=2$ denominator is $2$, giving a multiplier of $0.375$. Using $n-1=1$ instead would give a multiplier of $1$ and the incorrect value $p_2=0.4096$. The correct ratio is obtained directly from $p_n/p_{n-1}$, with $n$ on the left-hand probability.

The Separate Boundary in an (a,b,1) Recursion

For an $(a,b,1)$ sequence, the recurrence begins at $n=2$, not $n=1$. Both $p_0$ and $p_1$ are boundary probabilities; the recurrence uses $p_1$ to calculate $p_2$, and then proceeds upward. This distinction is especially important for zero-modified or hurdle-type models, where the probability at zero and the positive-count probabilities are set separately.

The following example uses a simple geometric tail to make the boundary role visible. The distribution is explicitly specified, so no assumption that $p_1$ must follow from $p_0$ is needed.

Worked Example: Keeping the (a,b,1) Boundary at n=1

A hypothetical count distribution is specified by $p_0=0.20$, $p_1=0.40$, and the recursion $p_n=(0.5+0/n)p_{n-1}$ for $n\geq2$. Calculate $p_2,p_3,p_4$, verify the total probability, and identify the error made by applying the recursion at $n=1$.

Method and assumptions. This is an $(a,b,1)$ sequence with $a=0.5$ and $b=0$, and the given boundary probabilities are $p_0=0.20$ and $p_1=0.40$. Since $b=0$, the multiplier is $0.5$ for every permitted index $n\geq2$.

At $n=2$, $p_2=0.5p_1=0.5(0.40)=0.20$. At $n=3$, $p_3=0.5p_2=0.5(0.20)=0.10$. At $n=4$, $p_4=0.5p_3=0.5(0.10)=0.05$.

The remaining positive tail is geometric: $p_5=0.025$, $p_6=0.0125$, and so on. Its sum from $n=2$ onward is $0.20+0.10+0.05+\cdots=0.20/(1-0.5)=0.40$. Therefore the total is $p_0+p_1+\sum_{n=2}^{\infty}p_n=0.20+0.40+0.40=1.00$.

Check and mistake avoided. Applying the recurrence at $n=1$ would assert $p_1=0.5p_0=0.10$, contradicting the specified boundary $p_1=0.40$. The stated $(a,b,1)$ recurrence starts at $n=2$; it does not replace or determine $p_1$.

A Reliable Audit Procedure

When a problem gives or asks for a recursive probability sequence, separate the model information from the arithmetic. The following procedure catches the most common index mistakes before they propagate through later probabilities.

Identify the class and its starting index.
For $(a,b,0)$, start at $n=1$ using $p_0$. For $(a,b,1)$, start at $n=2$ using $p_1$.

Write the ratio before calculating.
Set $p_n/p_{n-1}=a+b/n$. Keep $n$ as the denominator and substitute the current index only after writing the ratio.

Respect the support.
For a finite-support count, stop at the largest possible count. Do not interpret an algebraic ratio as permission to extend the distribution past that endpoint.

Check boundary probabilities and totals.
Confirm that the supplied $p_0$ and $p_1$ have been used in their proper roles, and check that all probabilities are nonnegative and sum to one when the full sequence is available.

These checks are related but not interchangeable. A sequence may follow the right ratio at every applicable index and still fail to be a probability distribution if the boundary values are wrong or the probabilities do not normalize. Conversely, a total of one does not establish that an incorrectly indexed recurrence is valid. Verify the formula, its index range, and the probability conditions separately.

A common exam trap is to see a familiar ratio and apply it automatically at the first visible probability. Instead, read the subscript condition as part of the formula. The statement $p_n=(a+b/n)p_{n-1}$ is incomplete unless its valid range of $n$ is also known. For $(a,b,0)$, $p_0$ is the starting value; for $(a,b,1)$, both $p_0$ and $p_1$ are boundary values. These distinctions are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Key takeaway: Before using a Panjer probability recursion, write down its class, starting index, denominator, boundary probabilities, and support. Then calculate one step at a time and check the resulting probabilities against the model.

Check Your Understanding

For each problem, state or use the valid index carefully. Probabilities in final answers are rounded to four decimal places where needed.

A hypothetical claim count is Poisson with parameter $\lambda=3$. Find $p_1$ from $p_0=e^{-3}$, and state whether the recursion can be evaluated at $n=0$. (Answer: $p_1=3e^{-3}\approx0.1494$; no, the term $3/0$ is undefined.)
A count is binomial with $m=5$ and $q=0.25$. Find $a,b$, then calculate $p_2$ recursively from $p_0$. (Answer: $a=-1/3,\ b=2;\ p_0=0.2373,\ p_1=0.3955,\ p_2=0.2637$.)
An $(a,b,1)$ sequence has $p_0=0.15,\ p_1=0.25,\ a=0.2,\ b=0.6$. Calculate $p_2$ and $p_3$, and give the first index at which the recurrence applies. (Answer: it applies at $n=2$; $p_2=0.1250,\ p_3=0.0500$.)
A hypothetical binomial count has $m=3$ and $q=0.5$. Starting with $p_0=0.125$, calculate $p_1,p_2,p_3$ recursively and state the probability beyond the support. (Answer: $p_1=0.3750,\ p_2=0.3750,\ p_3=0.1250;\ p_n=0$ for $n>3$.)
A sequence is specified as $(a,b,1)$ with $p_0=0.30,\ p_1=0.20,\ a=0.4,\ b=0.8$. Calculate $p_2$, and explain why $p_1$ is not recalculated from $p_0$. (Answer: $p_2=(0.4+0.8/2)(0.20)=0.1600$; the recurrence applies from $n=2$, while $p_1$ is a supplied boundary probability.)

What You'll Learn