Why Condition a Binomial Count on Being Positive?
A binomial count describes the number of successes in a fixed number of independent trials, each with the same success probability. In some applications, the group being modelled has already been selected because it experienced at least one success. The relevant count distribution is then not the original binomial distribution: it is the binomial distribution conditioned on a positive count.
Conditioning changes every positive-count probability by the same normalizing factor. It does not change the ratios between successive positive-count probabilities. That distinction is useful in the Panjer framework: the zero-truncated binomial probabilities retain a recursion of the \((a,b,1)\) form, but their boundary probability must be recalculated for the conditioned model.
This treatment is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
Conditioning and the Boundary Probability
The original binomial probability at \(k\) is \(P(N=k)=\binom{n}{k}p^kq^{n-k}\). The probability of the conditioning event is \(P(N>0)=1-P(N=0)=1-q^n\). Therefore, for each \(k=1,\ldots,n\),
In particular, the first positive probability, called the boundary probability, is
The denominator is the probability that the count is positive, not the probability that it is zero. The conditioning event removes the zero count; it does not shift each count up by one.
Worked Example: Calculating the Boundary Probability
A hypothetical screening program performs \(n=5\) independent tests on each participant. Each test has probability \(p=2/5\) of a positive result. Among participants with at least one positive result, find the probability of exactly one positive result. Give the final probability to four decimal places.
Method. Use the conditional probability formula, dividing the original probability of exactly one success by the probability of at least one success.
Here \(q=1-p=3/5\). The probability of no positive results is \[ q^5=(3/5)^5=243/3125. \] Thus the conditioning probability is \[ P(N>0)=1-243/3125=2882/3125. \] The original probability of exactly one positive result is \[ P(N=1)=\binom{5}{1}(2/5)(3/5)^4 =5(2/5)(81/625)=810/3125. \] Consequently, \[ g_1=\frac{810/3125}{2882/3125} =\frac{810}{2882} =\frac{405}{1441} \approx 0.2811. \]
Check. The denominator \(2882/3125\) is less than 1, so conditioning on a positive result raises the probability of exactly one result from \(810/3125=0.2592\) to approximately \(0.2811\). A common mistake is to report \(0.2804\); the exact ratio \(405/1441\) gives \(0.2811\) to four decimal places.
Deriving the Positive-Count Recursion
The earlier material on the Panjer probability recursion uses the form \(p_k=(a+b/k)p_{k-1}\). For a zero-truncated count, the recursion starts at \(k=2\), because its first probability \(g_1\) is supplied separately. To find the parameters, begin with the ratio of two successive probabilities of the original binomial count.
For positive \(k\), both binomial probabilities in this ratio are divided by the same conditioning probability \(1-q^n\). That common factor cancels. Hence the ratio also holds for \(g_k/g_{k-1}\) when \(k=2,\ldots,n\), giving the recursion below.
This is an \((a,b,1)\) recursion: the positive probabilities are generated from the boundary probability \(g_1\), rather than from a positive probability at zero. There is also a useful endpoint check. Extending the recursion one step, the multiplier at \(k=n+1\) is \[ a+\frac{b}{n+1}=-\frac{p}{q}+\frac{p}{q}=0. \] Thus it produces \(g_{n+1}=0\), consistent with the binomial model having no counts above \(n\). This endpoint does not mean the recursion generates positive probabilities beyond the binomial support.
Record \(n\), \(p\), and \(q=1-p\), and confirm that \(0<p<1\).
Use \(P(N>0)=1-q^n\).
Calculate \(g_1=npq^{n-1}/(1-q^n)\), then use \(a=-p/q\) and \(b=(n+1)p/q\).
Generate probabilities only through \(k=n\), verify they are nonnegative, and check that they sum to 1.
Worked Recursions and Probability Checks
Worked Example: Generating the Full Positive-Count Distribution
A hypothetical inspection process makes four independent checks, each with probability \(p=1/2\) of finding a defect. Given that at least one defect is found, calculate the entire distribution of the number of defects and then find the probability of at least three. Use the recursion and check its normalization.
Method. Find \(g_1\) from conditioning, then generate the remaining probabilities with the Panjer recursion. Here \(n=4\), \(p=q=1/2\), so \[ P(N>0)=1-(1/2)^4=15/16 \] and \[ g_1=\frac{4(1/2)(1/2)^3}{15/16} =\frac{4/16}{15/16}=\frac{4}{15}. \] The parameters are \[ a=-\frac{p}{q}=-1,\qquad b=\frac{(n+1)p}{q}=5. \] For \(k=2\), \[ g_2=\left(-1+\frac{5}{2}\right)\frac{4}{15} =\frac{3}{2}\frac{4}{15}=\frac{2}{5}. \] For \(k=3\), \[ g_3=\left(-1+\frac{5}{3}\right)\frac{2}{5} =\frac{2}{3}\frac{2}{5}=\frac{4}{15}. \] For \(k=4\), \[ g_4=\left(-1+\frac{5}{4}\right)\frac{4}{15} =\frac{1}{4}\frac{4}{15}=\frac{1}{15}. \] The distribution is therefore \(g_1=4/15\), \(g_2=2/5\), \(g_3=4/15\), and \(g_4=1/15\). The requested probability is \[ P(N\geq 3\mid N>0)=g_3+g_4=4/15+1/15=1/3. \]
Check. The probabilities sum to \(4/15+6/15+4/15+1/15=1\). A common mistake is to use \(g_1=1/4\), the unconditioned probability of exactly one defect. The required boundary is \(g_1=(1/4)/(15/16)=4/15\).
Worked Example: Using the Recursion for an Upper-Tail Probability
A hypothetical device has six independent components, each with probability \(p=1/3\) of requiring replacement in a particular period. Given that at least one component requires replacement, find the probability that five or six components require replacement. Also verify the probability sequence.
Method. Calculate the boundary probability and recursion parameters, generate each probability through the finite endpoint, and add the requested tail.
Here \(n=6\), \(p=1/3\), and \(q=2/3\). The probability of no replacements is \[ q^6=(2/3)^6=64/729, \] so \(P(N>0)=665/729\). The boundary probability is \[ g_1=\frac{6(1/3)(2/3)^5}{665/729} =\frac{64/243}{665/729}=\frac{192}{665}. \] The recursion parameters are \(a=-1/2\) and \(b=7/2\). Applying the recursion gives \[ g_2=\left(-\frac12+\frac{7}{4}\right)\frac{192}{665} =\frac54\frac{192}{665}=\frac{240}{665}, \] \[ g_3=\left(-\frac12+\frac{7}{6}\right)\frac{240}{665} =\frac23\frac{240}{665}=\frac{160}{665}, \] \[ g_4=\left(-\frac12+\frac{7}{8}\right)\frac{160}{665} =\frac38\frac{160}{665}=\frac{60}{665}, \] \[ g_5=\left(-\frac12+\frac{7}{10}\right)\frac{60}{665} =\frac15\frac{60}{665}=\frac{12}{665}, \] and \[ g_6=\left(-\frac12+\frac{7}{12}\right)\frac{12}{665} =\frac1{12}\frac{12}{665}=\frac{1}{665}. \] Therefore, \[ P(N\geq5\mid N>0)=g_5+g_6=\frac{12+1}{665} =\frac{13}{665}\approx 0.0195. \]
Check. The numerators sum to \(192+240+160+60+12+1=665\), so the conditional probabilities sum to 1. The endpoint probability is small but positive at \(k=6\); setting it to zero would incorrectly discard the possibility that all six components require replacement.
Interpretation and Exam Traps
Zero truncation changes the distribution by conditioning, not by changing the number of trials or the success probability. The recursion parameters \(a\) and \(b\) describe the ratios of successive positive probabilities and are unchanged by conditioning. The boundary probability is different because it must be scaled by \(1/(1-q^n)\).
| Quantity | Original binomial count | Zero-truncated binomial count |
|---|---|---|
| Support | \(0,\ldots,n\) | \(1,\ldots,n\) |
| Probability at zero | \(q^n\) | 0 |
| Boundary probability | \(P(N=1)=npq^{n-1}\) | \(g_1=npq^{n-1}/(1-q^n)\) |
| Positive-count recursion | \(P(N=k)=(a+b/k)P(N=k-1)\) | \(g_k=(a+b/k)g_{k-1}\), for \(2\leq k\leq n\) |
Two checks are especially useful. First, \(g_1\) must be at most 1 and, because \(N>0\), the probabilities \(g_1,\ldots,g_n\) must sum to 1. Second, the recursion is applied only to positive counts starting at \(k=2\); using it to calculate \(g_1\) from \(g_0=0\) would produce zero, not the required boundary probability.
- Do not confuse conditioning with shifting. The conditional count has support \(1,\ldots,n\); it is not the original count plus one.
- Keep the event in the denominator clear. The denominator is \(P(N>0)=1-q^n\), not \(P(N=0)\).
- Do not alter the recursion parameters after conditioning. The common conditioning factor cancels in each ratio of positive probabilities.
- Respect the finite support. Generate only through \(n\); the recursion's zero multiplier at \(n+1\) confirms the endpoint.
Check Your Understanding
Work through these original problems in order. Each uses a binomial count conditioned on being positive.
- A hypothetical batch has three independent items, each defective with probability \(p=0.2\). Find the conditional probability of exactly one defective item. (Answer: \(48/61\approx0.7869\))
- A count is zero-truncated \(\operatorname{Binomial}(4,1/2)\). State its boundary probability \(g_1\) and its recursion parameters \(a,b\). (Answer: \(g_1=4/15,\ a=-1,\ b=5\))
- A hypothetical unit undergoes five independent tests with positive-test probability \(p=1/4\). Find \(g_2\) using the boundary probability and the recursion. (Answer: \(g_2=270/781\))
- A zero-truncated \(\operatorname{Binomial}(6,1/3)\) count models component replacements. Find the conditional probability that at least five components require replacement. (Answer: \(13/665\approx0.0195\))
- A hypothetical count is zero-truncated \(\operatorname{Binomial}(4,1/3)\). Calculate \(g_3\) using the boundary probability and the recursion, and check that your recursion parameters correspond to this \(n,p\). (Answer: \(g_3=8/65\), with \(a=-1/2,\ b=5/2\))