Recognizing an Affine Probability Ratio

Why an Affine Probability Ratio Is Useful

A frequency distribution can be recognized from the way successive probabilities change. Instead of comparing $p_n$ directly with $p_{n-1}$, the Panjer $(a,b,0)$ class scales their ratio by $n$. If the resulting values follow a straight line in $n$, the probabilities satisfy a compact recurrence. The crucial qualification is that the recurrence must hold at every positive integer index, not just where a ratio can be calculated.

This tutorial focuses on identifying that affine pattern and checking its boundaries. A ratio is undefined when its denominator is zero, so ratio calculations alone cannot establish membership. The recurrence itself supplies the necessary test at those indices.

Definition: Let $N$ be a random variable taking values in the nonnegative integers, with $p_n=P(N=n)$. The distribution is in the Panjer $(a,b,0)$ class if there are constants $a$ and $b$ such that $$ p_n=\left(a+\frac{b}{n}\right)p_{n-1}, \qquad \text{for every integer } n\geq 1. $$ The condition applies even when $p_{n-1}=0$. The probabilities must also be a valid probability distribution: they are nonnegative and sum to 1.

From the Recurrence to a Straight Line

Whenever $p_{n-1}>0$, divide the recurrence by $p_{n-1}$ and multiply by $n$. Define the scaled ratio $$ R_n=\frac{n p_n}{p_{n-1}}. $$ Then the recurrence is equivalent, at that index, to $$ R_n=an+b. $$ Thus $R_n$, plotted against $n$, must lie on a straight line with slope $a$ and intercept $b$. Two distinct indices with defined ratios determine a candidate line; every other defined ratio must agree with it.

This ratio test is only part of the check. If $p_{n-1}=0$, the defining recurrence requires $$ p_n=\left(a+\frac{b}{n}\right)0=0. $$ A zero probability therefore cannot be followed by a positive probability. In particular, if a distribution has a gap in its support, an affine fit to the ratios before the gap does not make it a class member.

List the probabilities and support.
Identify which $p_n$ are positive and which are zero. The support is the set of indices with positive probability.

Calculate defined scaled ratios.
For each $n\geq1$ with $p_{n-1}>0$, calculate $R_n=n p_n/p_{n-1}$, including an index just beyond a finite endpoint when applicable.

Check the affine pattern.
Determine whether all the calculated values satisfy $R_n=an+b$ for the same constants $a$ and $b$.

Check every zero-denominator step.
At each $n$ with $p_{n-1}=0$, verify that $p_n=0$. This completes the recurrence check where ratios cannot be formed.

Why the Endpoint Matters

Suppose a distribution has finite support ending at $m$: $p_m>0$ and $p_n=0$ for every $n>m$. The recurrence at $n=m+1$ still has a positive preceding probability, so the ratio is defined there. Since $p_{m+1}=0$, it requires $$ R_{m+1}=\frac{(m+1)p_{m+1}}{p_m}=0 \quad\text{and therefore}\quad a(m+1)+b=0. $$ The fitted line must reach zero at the first index beyond the support. This endpoint condition is not optional: if the line predicts a nonzero ratio there, the recurrence predicts a nonzero $p_{m+1}$, contrary to the stated support.

After that first zero, the recurrence is automatic if all later probabilities are zero: both $p_n$ and $p_{n-1}$ are zero. More generally, once a zero occurs, every subsequent probability must remain zero. The ratio is undefined at a zero denominator, but the recurrence remains meaningful.

Worked Example: Checking a Finite Endpoint

A hypothetical risk portfolio has claim count $N$ with $p_0=1/8,\ p_1=3/8,\ p_2=3/8,\ p_3=1/8$, and $p_n=0$ for $n\geq4$. The probabilities are nonnegative and sum to $1/8+3/8+3/8+1/8=1$. Determine whether the distribution is in the Panjer $(a,b,0)$ class.

Method. Calculate the scaled ratios while the preceding probability is positive, fit a line, and then test the first index beyond the support.

At $n=1$, $$ R_1=\frac{1(3/8)}{1/8}=3. $$ At $n=2$, $$ R_2=\frac{2(3/8)}{3/8}=2. $$ At $n=3$, $$ R_3=\frac{3(1/8)}{3/8}=1. $$ The values decrease by $1$ each time $n$ increases by $1$, so the candidate slope is $a=-1$. Using $R_1=a+b=3$ gives $b=4$. The candidate line is $R_n=-n+4$.

At the endpoint index $n=4$, the line predicts $R_4=-4+4=0$. The actual scaled ratio is also $$ R_4=\frac{4p_4}{p_3}=\frac{4(0)}{1/8}=0. $$ Equivalently, the recurrence predicts $p_4=(-1+4/4)p_3=0$, as required. For $n\geq5$, the preceding and current probabilities are both zero, so the recurrence holds automatically. Therefore the distribution is in the class with $a=-1,\ b=4$.

Check and trap avoided. The line agrees at all indices with positive denominators, including $n=4$. Stopping at $n=3$ would miss the endpoint test, which is essential for finite support.

Zeros and Gaps in the Support

A positive-probability sequence that begins at zero and continues without gaps is straightforward to test by ratios. A gap requires extra care. For example, if $p_1=0$ but $p_2>0$, then $R_2$ cannot be calculated, but the recurrence at $n=2$ says $p_2=(a+b/2)p_1=0$. No choice of $a$ and $b$ can make that equation equal a positive $p_2$.

Worked Example: A Gap That Ratios Cannot Reveal

A hypothetical event count has $p_0=1/2,\ p_1=0,\ p_2=1/2$, and $p_n=0$ for $n\geq3$. The probabilities sum to $1$. Determine whether any constants $a,b$ put this distribution in the class.

At $n=1$, the recurrence would require $$ 0=p_1=(a+b)p_0=\frac{a+b}{2}, $$ so $a+b=0$. This is a possible constraint on the coefficients, but it is not a complete membership test.

At $n=2$, the recurrence requires $$ p_2=\left(a+\frac{b}{2}\right)p_1 =\left(a+\frac{b}{2}\right)0=0. $$ The stated distribution has $p_2=1/2$, a contradiction. Hence there are no suitable $a,b$, and the distribution is not in the class.

Check and trap avoided. $R_2=2p_2/p_1$ is undefined, not an affine ratio that can be ignored. The recurrence at $n=2$ rules out membership directly. Checking only indices for which $p_{n-1}>0$ would wrongly accept this distribution.

Complete Recognition Requires Verification

A useful way to organize the test is to distinguish a candidate fit from a proof of membership. A line through two ratio points only identifies candidate coefficients. Those coefficients must work for every other defined ratio, at the finite-support endpoint if one exists, and at every step with a zero preceding probability. If only one ratio is available, it generally does not determine both $a$ and $b$; further probability information or recurrence equations are needed.

Worked Example: A Candidate Line Fails at the Endpoint

A hypothetical count has $p_0=p_1=p_2=1/3$, with $p_n=0$ for $n\geq3$. Its probabilities sum to $1$. Test whether it is in the class.

For $n=1$, $$ R_1=\frac{1(1/3)}{1/3}=1. $$ For $n=2$, $$ R_2=\frac{2(1/3)}{1/3}=2. $$ The line through $(1,1)$ and $(2,2)$ is $R_n=n$, which gives candidate coefficients $a=1,\ b=0$.

At $n=3$, the actual ratio is defined because $p_2>0$, and it equals $$ R_3=\frac{3p_3}{p_2}=\frac{3(0)}{1/3}=0. $$ The candidate line instead predicts $R_3=3$. In recurrence form, it predicts $p_3=(1+0/3)p_2=1/3$, not zero. Thus the candidate fails at the endpoint, and the distribution is not in the class.

Check and trap avoided. The first two ratios determine a line, but do not establish that it fits the entire distribution. The endpoint is an observed ratio too, because its denominator $p_2$ is positive.

Worked Example: Verifying an Unbounded Sequence

A hypothetical count has probabilities $p_n=(1/3)(2/3)^n$ for every $n\geq0$. These are nonnegative and sum to $$ \sum_{n=0}^{\infty}\frac{1}{3}\left(\frac{2}{3}\right)^n =\frac{1/3}{1-2/3}=1. $$ Determine its affine ratio and verify the recurrence.

The first probabilities are $p_0=1/3,\ p_1=2/9,\ p_2=4/27,\ p_3=8/81$. Thus $$ R_1=\frac{2/9}{1/3}=\frac{2}{3},\qquad R_2=\frac{2(4/27)}{2/9}=\frac{4}{3},\qquad R_3=\frac{3(8/81)}{4/27}=2. $$ These values increase by $2/3$ per unit increase in $n$, so they lie on the line $R_n=(2/3)n$. The candidate coefficients are $a=2/3,\ b=0$.

To verify the pattern for every $n\geq1$, use the stated probability formula: $$ \frac{n p_n}{p_{n-1}} =n\frac{(1/3)(2/3)^n}{(1/3)(2/3)^{n-1}} =\frac{2n}{3}. $$ Therefore $p_n=(2/3)p_{n-1}$ at every positive integer index. All probabilities are positive, so there are no zero-denominator steps to check. The distribution is in the class with $a=2/3,\ b=0$.

Check and trap avoided. The first three ratios suggest the line, while the general calculation verifies all indices. A few matching values alone would not establish an infinite sequence's recurrence.

Common Exam Traps

Dropping the factor $n$. The quantity required to be affine is $n p_n/p_{n-1}$, not $p_n/p_{n-1}$ itself.
Dividing by zero. If $p_{n-1}=0$, do not form a ratio. Apply the recurrence, which requires $p_n=0$.
Stopping at the last positive probability. At a finite endpoint $m$, test $n=m+1$. Its preceding probability is positive, and the fitted line must give zero there.
Assuming a line fit proves membership. Verify every defined ratio and every recurrence step with zero preceding probability.
Using non-normalized values as a distribution. A proposed probability sequence must be nonnegative and sum to $1$, in addition to satisfying the recurrence.

Key takeaway: The affine-ratio test is $R_n=n p_n/p_{n-1}=an+b$ wherever $p_{n-1}>0$, together with the full recurrence at every $n\geq1$. At a finite endpoint the line must reach zero; after any zero probability, no later probability may be positive.

Check Your Understanding

For each distribution, test the affine ratios and all zero-probability steps before deciding whether it is in the Panjer $(a,b,0)$ class.

A count has $p_0=1/4,\ p_1=1/2,\ p_2=1/4$, and $p_n=0$ for $n\geq3$. Find candidate $a,b$ and check the endpoint. (Answer: Yes; $a=-1,\ b=3$, and $R_3=0$.)
A count has $p_0=1/5,\ p_1=2/5,\ p_2=2/5$, and $p_n=0$ for $n\geq3$. Do the first two ratios give a valid class fit? (Answer: No; $R_1=2,\ R_2=2$ imply $a=0,\ b=2$, but the endpoint has $R_3=0$, not $2$.)
A count has $p_0=1/3,\ p_1=0,\ p_2=2/3$, and $p_n=0$ for $n\geq3$. Can any affine ratio coefficients make it a class member? (Answer: No; the recurrence at $n=2$ forces $p_2=0$ because $p_1=0$.)
A count has $p_0=1/4,\ p_1=1/4,\ p_2=1/8,\ p_3=1/8,\ p_4=1/4$. Test whether it is in the class. (Answer: No; $R_1=1,\ R_2=1$ imply $a=0,\ b=1$, which predicts $R_3=1$, but the actual $R_3=3$.)
A finite-support count has positive probabilities at $n=0,1,2,3$, zero probabilities for $n\geq4$, and scaled ratios $R_1=3,\ R_2=2,\ R_3=1$. Does the affine pattern satisfy the endpoint condition? (Answer: Yes; the ratios give $a=-1,\ b=4$, and $R_4=0$, as required at the first index beyond the support.)

What You'll Learn