Normalizing an (a,b,1) Probability Sequence

Why Normalization Requires Two Checks

A recurrence can generate a sequence of nonnegative numbers without making those numbers probabilities. To define a count distribution, the probabilities must also sum to one. For an $(a,b,1)$ sequence, these checks involve both the starting probabilities and the terms generated from them. In particular, the recurrence begins at $n=2$, so $p_0$ and $p_1$ are boundary values rather than values determined by that recurrence.

This distinction is useful when checking a proposed frequency model or finding a missing boundary probability. A few terms may look plausible, but normalization concerns the entire sequence, including its infinite tail when the support is unbounded.

Definition: An $(a,b,1)$ probability sequence is specified by starting values $p_0$ and $p_1$, and by the recurrence $$ p_n=\left(a+\frac{b}{n}\right)p_{n-1},\qquad n\ge 2. $$ For the sequence to define a probability distribution on the nonnegative integers, every $p_n$ must be nonnegative and $\sum_{n=0}^{\infty}p_n=1$.

Write $c_n=a+b/n$ for the multiplier at step $n$. Repeated application of the recurrence gives $$ p_n=p_1\prod_{j=2}^{n}c_j,\qquad n\ge 2, $$ where the corresponding product for $n=1$ is empty and equals one. Thus all probabilities from $p_1$ onward are determined by $p_1$, $a$, and $b$; $p_0$ does not enter those terms.

If $p_1=0$, then the recurrence gives $p_n=0$ for every $n\ge2$. In that special case the only normalized sequence has $p_0=1$. When $p_1>0$, each term must be checked: a negative multiplier before any zero term would produce a negative probability. If a term becomes zero, every subsequent term is zero under the recurrence. For a positive, nonterminating tail, all its multipliers must be nonnegative.

Writing the Normalization Condition

Define the tail sum $H(a,b)$ by adding the relative weights generated from $p_1$:

H(a,b)=\sum_{n=1}^{\infty}\prod_{j=2}^{n}\left(a+\frac{b}{j}\right).

The $n=1$ term in this sum is one, so it accounts for $p_1$ itself. If the series is finite, the total probability is

\sum_{n=0}^{\infty}p_n=p_0+p_1H(a,b).

Consequently, when $H(a,b)$ is finite, the normalization equation is $p_0+p_1H(a,b)=1$. If $p_1$ is given, the required value is $p_0=1-p_1H(a,b)$, which must be nonnegative. If $p_0$ is given, the equation instead determines $p_1$, provided the resulting value is nonnegative. If $H(a,b)$ is infinite, no positive $p_1$ can give a normalized infinite tail.

Key result: For $p_1>0$, check validity in two stages: first confirm that the recursively generated terms are nonnegative; then check whether their sum is finite and whether $p_0+p_1H(a,b)=1$. Nonnegative terms alone do not imply a probability distribution.

Why the Tail-Sum Test Works

The normalization formula follows directly by factoring $p_1$ out of each term. For every $n\ge1$, the recurrence gives $p_n=p_1\prod_{j=2}^{n}c_j$, with the empty-product convention covering $n=1$. Summing these expressions and adding $p_0$ yields $p_0+p_1H(a,b)$. This also explains why choosing $p_0$ cannot repair an infinite positive tail: a finite boundary probability cannot offset an infinite sum.

The ratio of consecutive terms in a positive, nonterminating tail is $p_n/p_{n-1}=a+b/n$. When $0\le a<1$, this ratio tends to a value below one, so the ratio test implies that the tail sum converges. When $a>1$, the ratio eventually exceeds one, and positive terms cannot tend to zero; the tail sum diverges.

At the boundary $a=1$, the ratio tends to one, so the ratio test does not decide convergence. For a positive nonterminating tail, the product has the approximate order $n^b$. To see the reason, take logarithms:

\log\!\left(\prod_{j=2}^{n}\left(1+\frac{b}{j}\right)\right) =\sum_{j=2}^{n}\log\left(1+\frac{b}{j}\right) =b\sum_{j=2}^{n}\frac{1}{j}+O(1) =b\log n+O(1).

Here the bounded remainder follows because the terms beyond $b/j$ in the logarithm are of order $1/j^2$, whose series converges. Thus the product is of order $n^b$, and the tail behaves like a power series with terms of that order. For a positive nonterminating tail, its sum converges when $b<-1$ and diverges when $b\ge-1$. This boundary conclusion is useful only after checking that the terms remain nonnegative. In all cases, the actual starting probabilities must still satisfy the normalization equation.

Worked Examples

Worked Example: Checking a Geometric-Type Sequence

A hypothetical claim count $N$ has $p_0=0.40$, $p_1=0.30$, and recurrence parameters $a=0.50$, $b=0$. Determine whether these values define a probability distribution.

Method. Generate the multipliers, then use the entire tail sum rather than checking only the first few terms. Here $c_n=0.50$ at every step $n\ge2$, so all generated terms are nonnegative.

The first terms are $p_2=0.30(0.50)=0.15$, $p_3=0.15(0.50)=0.075$, and $p_4=0.075(0.50)=0.0375$. In general, $p_n=0.30(0.50)^{n-1}$ for $n\ge1$. Therefore

H(0.50,0)=\sum_{n=1}^{\infty}(0.50)^{n-1} =\frac{1}{1-0.50}=2.

The total probability is $p_0+p_1H=0.40+0.30(2)=1.00$. The sequence is nonnegative and normalized, so it is a valid distribution. As a check, the mass at zero is $0.40$ and the positive-count probabilities sum to $0.60$. The common mistake avoided is to add just $p_0,p_1,p_2,\ldots$ for a few terms and treat that partial sum as the total.

Worked Example: Finding the Missing Probability at Zero

A hypothetical frequency model has $a=0.50$, $b=0.50$, and $p_1=0.20$. Find the value of $p_0$ that normalizes the sequence, and check that all terms are nonnegative.

Method. First simplify the recurrence product, then sum the resulting series. For $j\ge2$, $c_j=0.50+0.50/j=(j+1)/(2j)$, which is positive. For $n\ge1$, multiplying the factors gives

\prod_{j=2}^{n}\frac{j+1}{2j} =\frac{n+1}{2^n}.

For $n=1$, this formula gives $1$, as required. Hence $p_n=0.20(n+1)/2^n$ for $n\ge1$. The tail sum is

p_1H(0.50,0.50) =0.20\sum_{n=1}^{\infty}\frac{n+1}{2^n} =0.20\left(\sum_{n=1}^{\infty}\frac{n}{2^n} +\sum_{n=1}^{\infty}\frac{1}{2^n}\right) =0.20(2+1)=0.60.

The two series in parentheses sum to $2$ and $1$, respectively. Thus $p_0=1-0.60=0.40$. The terms are nonnegative because $p_1>0$ and every multiplier is positive; the total is $0.40+0.60=1.00$. The common mistake avoided is using $p_0=1-p_1$, which would ignore the probability mass at counts $n\ge2$.

Worked Example: Positive Terms That Cannot Be Normalized

A proposed count model has $p_0=0.20$, $p_1=0.40$, $a=1$, and $b=-0.50$. Determine whether it is a probability distribution.

Method. Check nonnegativity, then test convergence of the full tail. The multiplier is $c_n=1-0.50/n$. For every $n\ge2$, $c_n\ge0.75$, so the terms are positive. The first generated term is $p_2=0.40(0.75)=0.30$.

However, the tail does not have a finite sum. The product from the recurrence satisfies

\log\left(\frac{p_n}{p_1}\right) =\sum_{j=2}^{n}\log\left(1-\frac{0.50}{j}\right) =-0.50\log n+O(1).

It follows that $p_n$ is of order $n^{-0.50}$. The series $\sum_{n=1}^{\infty}n^{-0.50}$ diverges, so the positive tail probabilities have an infinite sum. No choice of the finite value $p_0=0.20$ can make the total equal one. The sequence is not a probability distribution. The common mistake avoided is to infer normalization from a few terms: here $p_0+p_1+p_2=0.90$, but that partial sum gives no assurance that the infinite tail is finite.

Common Normalization Traps

Starting the recurrence at the wrong index. For this class, the stated recurrence applies for $n\ge2$. It does not determine $p_1$ from $p_0$.
Checking only the first few probabilities. A finite partial sum below one does not establish that the infinite series converges or has total one.
Ignoring signs. If $p_1>0$ and a multiplier is negative before the sequence reaches zero, the resulting term is negative, so the candidate fails immediately.
Assuming that nonnegative terms are enough. A positive sequence with a divergent sum is not a probability distribution.
Forgetting to check $p_0$. Even if the tail is finite, the value $1-p_1H(a,b)$ must be nonnegative.

Key takeaway: For an $(a,b,1)$ sequence, preserve the boundary values $p_0,p_1$, recursively check signs from $n=2$, and normalize using $p_0+p_1H(a,b)=1$. A valid sequence needs both nonnegative terms and a finite total of exactly one.

Check Your Understanding

For each candidate, check the recurrence and the total probability. Answers are given at the end of each question.

A hypothetical model has $a=0.25$, $b=0$, $p_1=0.30$, and $p_0=0.60$. Is it normalized? (Answer: Yes; $H=1/(1-0.25)=4/3$, so $p_0+p_1H=0.60+0.40=1$.)
For $a=0.50$, $b=0.50$, and $p_1=0.25$, find the value of $p_0$ that normalizes the sequence. (Answer: $p_0=0.25$, since $H=3$ and $1-0.25(3)=0.25$.)
Let $a=0$, $b=1$, $p_1=1/(e-1)$, and $p_0=0$. Check whether this is a normalized sequence. (Answer: Yes; $p_n=p_1/n!$ for $n\ge1$, so the positive-count probabilities sum to $p_1(e-1)=1$.)
A candidate has $a=1$, $b=0$, $p_0=0.90$, and $p_1=0.10$. Are these starting probabilities sufficient for normalization? (Answer: No; $p_n=0.10$ for every $n\ge1$, so the tail sum diverges.)
A proposed model has $a=0.40$, $b=-1$, and $p_1=0.20$. Check its first generated term and decide whether it can be a probability sequence. (Answer: No; $c_2=0.40-1/2=-0.10$, so $p_2=0.20(-0.10)=-0.02$ is negative.)

What You'll Learn