Classifying a Distribution From Its Probability Ratios

Why Probability Ratios Help Classify Counts

A count distribution is often easier to recognize from how its probabilities change from one count to the next than from its full probability mass function. For a random count $N$, the relevant pattern is not usually the probability ratio by itself: it is that ratio multiplied by the count index. Testing this scaled ratio gives a direct way to decide whether a distribution has the $(a,b,0)$ recursion. The method is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Let $p_n=P(N=n)$ for $n=0,1,2,\ldots$. At any index $n\geq 1$ for which $p_{n-1}>0$, define the scaled consecutive-probability ratio $$ R_n=\frac{n p_n}{p_{n-1}}. $$ A distribution with $p_0>0$ belongs to the $(a,b,0)$ class if constants $a$ and $b$ exist such that $$ p_n=\left(a+\frac{b}{n}\right)p_{n-1},\qquad n\geq 1. $$ Equivalently, $R_n=an+b$ at every relevant index.

The classification test is therefore an affine-pattern test: as $n$ changes, the values of $R_n$ must lie on one straight line. For three or more consecutive values, this means their first differences must be constant. Two values alone never establish a pattern, because any two points determine a line. The test must also respect the distribution’s entire support, not just a convenient selection of probabilities.

What the Test Requires

For an infinite-support distribution with positive probabilities at every nonnegative integer, calculate $R_n$ for all $n\geq 1$. The distribution is in the $(a,b,0)$ class exactly when those values follow one affine rule. For a finite-support distribution with positive probabilities from 0 through $M$, also check the terminating index $M+1$: since $p_{M+1}=0$ and $p_M>0$, its scaled ratio is $R_{M+1}=0$. The affine rule must give zero there as well.

A support gap is another warning. If $p_{n-1}=0$ but $p_n>0$, the defining recursion would give $p_n=(a+b/n)0=0$, a contradiction. Thus a distribution with $p_0>0$ and a positive probability appearing after a zero cannot belong to the $(a,b,0)$ class. If $p_0=0$, the ratio test beginning at zero is not available, and the distribution is not an $(a,b,0)$ distribution under this definition.

Check the starting probability and support.
Confirm $p_0>0$, and check that positive probabilities do not resume after a zero.

Calculate scaled ratios.
For each applicable index, compute $R_n=n p_n/p_{n-1}$, not only $p_n/p_{n-1}$.

Test the full pattern.
Check whether the scaled ratios are affine in $n$, including the terminating ratio for finite support.

State the conclusion and its scope.
Report whether the distribution is in the class, and identify any values or endpoint that fail the test.

Why an Affine Ratio Is the Right Criterion

The recursion in the definition can be rearranged by multiplying both sides by $n/p_{n-1}$, wherever $p_{n-1}>0$. This gives the scaled ratio directly:

p_n=\left(a+\frac{b}{n}\right)p_{n-1} \quad\Longleftrightarrow\quad \frac{n p_n}{p_{n-1}}=an+b.

The right side is an affine function of the index $n$, with slope $a$ and intercept $b$. This explains why the test uses $n p_n/p_{n-1}$: multiplying by $n$ removes the $1/n$ term in the recursion. If the scaled ratios follow the affine rule across the support, dividing by $n$ recovers the recursion. For finite support, the rule must also be consistent with the zero probability at the first index beyond the support.

This equivalence provides a recognition test, not permission to check only a few probabilities. A line that fits the first two ratios but misses a later one does not describe the distribution’s recursion. Nor is an affine sequence of ratios sufficient if it implies negative probabilities or fails at an endpoint; the supplied probabilities must still form a valid distribution.

Worked Examples

Worked Example: Testing a Finite Binomial Count

A hypothetical inspection team finds defective components in a batch. Suppose the defect count $N$ has a binomial distribution with $m=4$ trials and defect probability $q=1/2$. Determine whether it is in the $(a,b,0)$ class, including the finite-support endpoint.

Method. Write the probabilities, calculate the scaled ratios, and test whether a single affine rule holds through the first zero probability. This endpoint check avoids treating the last positive probability as if it were the end of the recursion.

The binomial probabilities are $p_n=\binom{4}{n}(1/2)^4$ for $n=0,1,2,3,4$. They are $1/16,4/16,6/16,4/16,1/16$, respectively, and $p_5=0$. Their sum is $(1+4+6+4+1)/16=1$.

Calculate each ratio in order:

R_1=\frac{1(4/16)}{1/16}=4,\quad R_2=\frac{2(6/16)}{4/16}=3,\quad R_3=\frac{3(4/16)}{6/16}=2,\quad R_4=\frac{4(1/16)}{4/16}=1,\quad R_5=\frac{5(0)}{1/16}=0.

The values are $4,3,2,1,0$, so they follow $R_n=5-n$, including at $n=5$. Therefore $a=-1$ and $b=5$, and the distribution is in the $(a,b,0)$ class. The check succeeds: $(a+b/n)=(5-n)/n$, which is nonnegative for $n=1,2,3,4$ and zero at $n=5$, as required for termination.

Check. The recurrence gives $p_1=(4/1)p_0=4/16$, and at $n=5$ it gives $p_5=0p_4=0$. The common mistake is to use $4-n$, which misses the ratios and fails at the endpoint; the correct intercept is $5$.

Worked Example: Testing a Negative Binomial Count

A hypothetical maintenance log records the number $N$ of unsuccessful attempts before two successful tests. Suppose attempts are independent, each succeeds with probability $1/3$, and $N$ counts failures. Determine whether the resulting count distribution is in the $(a,b,0)$ class.

Method. Use the negative binomial probability mass function, compute several consecutive scaled ratios, and check that the formula holds for general $n$. This avoids deciding from only two observed ratios.

Write $r=2$, success probability $s=1/3$, and failure probability $1-s=2/3$. For $n=0,1,2,\ldots$,

p_n=\binom{n+r-1}{n}s^r(1-s)^n =\frac{n+1}{9}\left(\frac{2}{3}\right)^n.

For example, $p_0=1/9$, $p_1=4/27$, $p_2=4/27$, and $p_3=32/243$. The first three scaled ratios are

R_1=\frac{1(4/27)}{1/9}=\frac{4}{3},\qquad R_2=\frac{2(4/27)}{4/27}=2,\qquad R_3=\frac{3(32/243)}{4/27}=\frac{8}{3}.

To verify the pattern generally, divide adjacent probabilities:

\frac{p_n}{p_{n-1}} =\frac{(n+1)(2/3)^n}{n(2/3)^{n-1}} =\frac{2(n+1)}{3n}, \qquad R_n=n\frac{p_n}{p_{n-1}}=\frac{2}{3}n+\frac{2}{3}.

The scaled ratios are affine for every $n\geq 1$, with $a=2/3$ and $b=2/3$. Since this distribution has positive probabilities for all nonnegative counts, there is no finite endpoint to check. It belongs to the $(a,b,0)$ class.

Check. Substituting the parameters into the recursion gives $p_n=(2/3+2/(3n))p_{n-1}$, agreeing with the adjacent-probability ratio derived above. A common error is to report $p_n/p_{n-1}$ as the affine quantity; it is $n p_n/p_{n-1}$ that follows the straight line.

Worked Example: Rejecting a Distribution That Fits Two Ratios

A hypothetical count of alerts has probabilities $p_0=1/4$, $p_1=1/4$, $p_2=1/8$, $p_3=1/8$, and $p_4=1/4$, with all other probabilities zero. The probabilities sum to $1$. Determine whether this distribution is in the $(a,b,0)$ class.

Method. Calculate the scaled ratios at every positive denominator and test whether the first differences are constant. This guards against inferring a line from only the first two ratios.

The scaled ratios are

R_1=\frac{1(1/4)}{1/4}=1,\quad R_2=\frac{2(1/8)}{1/4}=1,\quad R_3=\frac{3(1/8)}{1/8}=3,\quad R_4=\frac{4(1/4)}{1/8}=8,\quad R_5=\frac{5(0)}{1/4}=0.

The first differences are $0,2,5,-8$, which are not constant. In particular, the first two ratios would suggest the constant line $R_n=1$, but that line already fails at $n=3$. The distribution is not in the $(a,b,0)$ class.

Check. The support is contiguous through $4$, so a support gap does not explain the failure; the ratios themselves fail the affine test. The endpoint ratio is also $0$, which does not restore the pattern. The common mistake is to stop after the first two ratios, which cannot establish that all subsequent values lie on the same line.

Common Classification Traps

Testing the unscaled ratio. The defining recursion makes $p_n/p_{n-1}=a+b/n$; it is generally not affine. Multiply by $n$ before checking for a straight-line pattern.
Using too few indices. Two ratio values always determine a line. Check every available ratio, or use a formula that establishes the pattern for every index.
Ignoring a finite endpoint. If the maximum possible count is $M$, include $R_{M+1}=0$. A proposed line that does not reach zero there cannot describe the terminating recursion.
Overlooking support gaps. A positive probability after a zero probability is incompatible with the $(a,b,0)$ recursion from $p_0>0$.
Confusing class membership with a familiar label. A distribution’s name does not replace the ratio test. State the affine ratio rule and check its range of validity.

Key takeaway: For a distribution with $p_0>0$, calculate $R_n=n p_n/p_{n-1}$. Membership in the $(a,b,0)$ class requires one affine rule across the full support, with no support gaps and with the terminating zero included when support is finite.

Check Your Understanding

For each distribution, test the scaled ratios across the stated support before deciding whether it belongs to the $(a,b,0)$ class.

A count is Poisson with mean $3$. Find $R_n$ for $n\geq1$ and state the class parameters. (Answer: $R_n=3$, so it is in the class with $a=0,\ b=3$.)
A hypothetical batch has a binomial defect count with $m=5$ trials and defect probability $q=2/5$. Find the affine rule, including the first index beyond the support. (Answer: $R_n=(2/3)(6-n)$ for $1\leq n\leq6$, so $a=-2/3,\ b=4$, and the count is in the class.)
A count has probabilities $p_0=1/4,\ p_1=1/2,\ p_2=1/4,\ p_3=0$, with all later probabilities zero. Is it in the class? (Answer: Yes. $R_1=2,\ R_2=1,\ R_3=0$, so $R_n=3-n$, giving $a=-1,\ b=3$.)
A count has $p_0=1/2,\ p_1=1/4,\ p_2=0,\ p_3=1/4$, with all other probabilities zero. Is it in the class? (Answer: No. The positive probability at $3$ after $p_2=0$ violates the recursion.)

What You'll Learn