Binomial Counts as an (a,b,0) Distribution

Why the Binomial Recursion Matters

A binomial count models the number of successes in a fixed number of independent trials when each trial has the same success probability. Its probability mass function is familiar, but computing each probability separately can be inefficient. The Panjer recursion instead builds each probability from the preceding one. To use it correctly, you need the right coefficients, a starting probability, and a clear understanding of where the finite support ends.

This tutorial derives the recursion, shows how its parameters depend on the trial count and success probability, and checks the recursion at the upper endpoint. The method is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Let \(N\) be the number of successes in \(n\) independent trials, each with success probability \(p\), where \(n\) is a positive integer and \(0<p<1\). Then \(N\sim\operatorname{Bin}(n,p)\), and \(P(N=k)=p_k=\binom{n}{k}p^k(1-p)^{n-k}\) for \(k=0,1,\ldots,n\). Write \(q=1-p\). The count has finite support: \(p_k=0\) for \(k>n\).

The notation \(p_k\) means the probability of exactly \(k\) successes; it is distinct from \(p\), the success probability on one trial. A probability sequence is in the \((a,b,0)\) class when its probabilities satisfy \(p_k=(a+b/k)p_{k-1}\) for positive integers \(k\), together with its probability at zero. For the binomial model, the recursion will hold through the endpoint and give zero for all probabilities beyond it.

Deriving the Binomial Recursion

For \(1\leq k\leq n\), divide the binomial probability at \(k\) by the probability at \(k-1\). The ratio is well-defined because \(0<p<1\) and \(k\) is within the support.

\begin{aligned} \frac{p_k}{p_{k-1}} &= \frac{\binom{n}{k}p^kq^{n-k}} {\binom{n}{k-1}p^{k-1}q^{n-k+1}}\\ &= \frac{n-k+1}{k}\frac{p}{q}\\ &= -\frac{p}{q}+\frac{(n+1)p}{kq}. \end{aligned}

The first equality compares successive binomial probabilities. In the second, the ratio of the binomial coefficients is \((n-k+1)/k\), while the powers of \(p\) and \(q\) leave the factor \(p/q\). Finally, \(n-k+1=(n+1)-k\); splitting this expression over \(k\) gives the affine form in \(1/k\).

Theorem: If \(N\sim\operatorname{Bin}(n,p)\) with \(n\geq1\) and \(0<p<1\), then its Panjer parameters are \(a=-p/q\) and \(b=(n+1)p/q\), where \(q=1-p\). Thus \(p_k=(a+b/k)p_{k-1}\) for \(1\leq k\leq n+1\), with \(p_0=q^n\). At \(k=n+1\), the multiplier is zero, so \(p_{n+1}=0\); all probabilities for \(k>n\) are zero.

The starting probability follows directly from the event of no successes: all \(n\) trials must fail, so \(p_0=q^n\). The ratio derivation applies through \(k=n\). At the next value, \(k=n+1\), the binomial formula gives probability zero, and the recursion also gives zero because \(a+b/(n+1)=-p/q+p/q=0\). For \(k>n+1\), the preceding probability is already zero, so the recursion continues to give zero.

The endpoint matters: the model cannot produce more than \(n\) successes. The zero multiplier at \(k=n+1\) is a useful algebraic check that the parameters and the trial count have been handled correctly. The recursion does not replace the need for a starting value; without \(p_0\), it determines only ratios between successive probabilities.

Worked Applications

Worked Example: Building the Probabilities and Checking the Endpoint

A hypothetical quality-control process examines four components. Treat each component as independently meeting a specified standard with probability \(p=0.2\). Let \(N\) be the number that meet the standard. Use the recursion to calculate all probabilities, then verify that it terminates at the maximum count and sums to one.

Method and parameters. There are \(n=4\) independent trials, \(p=0.2\), and \(q=1-p=0.8\). Therefore \(a=-p/q=-0.2/0.8=-0.25\) and \(b=(n+1)p/q=5(0.2)/0.8=1.25\). The initial probability is \(p_0=q^n=0.8^4=0.4096\).

Apply the recursion. For each \(k\), multiply the preceding probability by \(a+b/k\):

\begin{aligned} p_1&=(-0.25+1.25/1)(0.4096)=1(0.4096)=0.4096,\\ p_2&=(-0.25+1.25/2)(0.4096)=0.375(0.4096)=0.1536,\\ p_3&=(-0.25+1.25/3)(0.1536)=\tfrac16(0.1536)=0.0256,\\ p_4&=(-0.25+1.25/4)(0.0256)=0.0625(0.0256)=0.0016. \end{aligned}

At the next count, \(a+b/5=-0.25+1.25/5=0\), so \(p_5=0\). Hence the probabilities for counts above four are zero as well.

Check. Adding the probabilities gives \(0.4096+0.4096+0.1536+0.0256+0.0016=1.0000\). The values are to four decimal places, and the total confirms that the probability mass has been fully accounted for. The method avoids repeatedly evaluating the binomial coefficient in the mass function.

Worked Example: A Tail Probability From the Recursion

A hypothetical system runs six independent tests, each of which detects a specified condition with probability \(0.5\). Let \(N\) be the number of detections. Use the recursion to calculate \(P(N\geq4)\), and verify the complete probability sequence.

Method and parameters. Here \(n=6\), \(p=q=0.5\), so \(a=-1\), \(b=(7)(0.5)/(0.5)=7\), and \(p_0=(0.5)^6=1/64\).

Generate the probabilities. The successive multipliers are \(-1+7/k\):

\begin{aligned} p_1&=6p_0=6/64,\\ p_2&=(5/2)(6/64)=15/64,\\ p_3&=(4/3)(15/64)=20/64,\\ p_4&=(3/4)(20/64)=15/64,\\ p_5&=(2/5)(15/64)=6/64,\\ p_6&=(1/6)(6/64)=1/64. \end{aligned}

At \(k=7\), the multiplier is \(-1+7/7=0\), giving \(p_7=0\). The sequence from \(k=0\) through \(k=6\) has numerators \(1,6,15,20,15,6,1\), whose sum is \(64\); thus its probabilities sum to one.

Calculate the requested tail. Counts of at least four are four, five, or six, so \(P(N\geq4)=(15+6+1)/64=22/64=11/32=0.34375\). To four decimal places, the answer is \(0.3438\).

The important distinction is between the requested tail and the endpoint: the tail includes the maximum count \(6\), but not the zero probability at count \(7\). Omitting the probability at six would give the wrong event.

Worked Example: Recovering the Trial Count and Success Probability

A frequency sequence is reported to satisfy \(p_k=(a+b/k)p_{k-1}\), with \(a=-0.25\) and \(b=1.5\). Assume it comes from a nondegenerate binomial count. Identify the number of trials and the success probability, then use the resulting model to find the probability of at least three successes.

Method and parameter recovery. For a binomial count, \(a=-p/q\). Since \(q=1-p\), solving \(a=-p/(1-p)\) gives \(p=-a/(1-a)\). Thus \(p=0.25/1.25=0.2\), and \(q=0.8\). Also \(b=(n+1)p/q\), so \(n+1=bq/p=1.5(0.8)/0.2=6\). Therefore \(n=5\).

Compute the requested probability. The starting value is \(p_0=0.8^5=0.32768\). The recursion coefficients are \(a=-0.25\) and \(b=1.5\), giving

\begin{aligned} p_1&=(-0.25+1.5)(0.32768)=0.40960,\\ p_2&=(-0.25+1.5/2)(0.40960)=0.20480,\\ p_3&=(-0.25+1.5/3)(0.20480)=0.05120,\\ p_4&=(-0.25+1.5/4)(0.05120)=0.00640,\\ p_5&=(-0.25+1.5/5)(0.00640)=0.00032. \end{aligned}

At \(k=6\), \(-0.25+1.5/6=0\), confirming the maximum count is five. Therefore \(P(N\geq3)=p_3+p_4+p_5=0.05120+0.00640+0.00032=0.05792\), or \(0.0579\) to four decimal places.

Check. The recovered parameters are \(n=5\) and \(p=0.2\), and the zero multiplier occurs at \(n+1=6\), as required. This inversion method avoids treating \(b\) as a direct multiple of \(n\) rather than \(n+1\).

Common Error: Using the Wrong Value of \(b\)

A frequent mistake is to set \(b=np/q\). The ratio derivation shows why that is incorrect: the numerator is \(n-k+1=(n+1)-k\), so the coefficient of \(1/k\) contains \(n+1\), not \(n\).

For example, with \(n=4\) and \(p=0.2\), the incorrect choice would be \(b=4(0.2)/0.8=1\). Keeping the correct \(a=-0.25\), it gives the wrong first probability: \(p_1=(-0.25+1)(0.4096)=0.3072\). But directly from the binomial model, \(p_1=\binom41(0.2)(0.8)^3=0.4096\). The correct value \(b=5(0.2)/0.8=1.25\) yields the correct multiplier \(-0.25+1.25=1\).

Another trap is to continue the calculation without checking support. For a binomial count, the last positive probability is \(p_n\), and the recurrence must produce zero at \(k=n+1\). A nonzero multiplier at that step signals an incorrect parameter or an incorrect value of \(n\).

Key takeaway: For \(N\sim\operatorname{Bin}(n,p)\), set \(q=1-p\), \(a=-p/q\), \(b=(n+1)p/q\), and start from \(p_0=q^n\). The multiplier becomes zero at \(k=n+1\), which verifies the finite upper endpoint.

Check Your Understanding

For each problem, identify the relevant recursion values and check the finite support where requested.

A hypothetical count has \(n=5\) independent trials and success probability \(p=0.2\). Find \(a\), \(b\), and \(P(N=1)\) using \(p_0\) and the recursion. (Answer: \(a=-0.25,\ b=1.5,\ P(N=1)=0.4096\))
Let \(N\sim\operatorname{Bin}(4,0.3)\). Find \(p_0\), use the recursion to find \(p_2\), and give the result to four decimal places. (Answer: \(p_0=0.2401,\ p_2=0.2646\))
A binomial recursion has \(a=-0.5\) and \(b=4.5\). Recover the success probability and number of trials. (Answer: \(p=1/3,\ n=8\))
A hypothetical count follows \(\operatorname{Bin}(6,0.5)\). Use the probability sequence to find the probability of at least five successes, to four decimal places. (Answer: \(7/64=0.1094\))
For \(N\sim\operatorname{Bin}(5,0.4)\), find the recursion multiplier at \(k=6\) and state the resulting probability. (Answer: \(a=-2/3,\ b=4,\ a+b/6=0,\ p_6=0\))

What You'll Learn