Why the First Positive Probability Matters
When a frequency model is observed only when at least one event occurs, its positive-count probabilities must be recalculated. For example, a hypothetical service plan might record only months in which a customer makes a claim. The probability of exactly one claim among these recorded months is not the original probability of one claim: it is conditional on the claim count being positive.
A recursion can then generate the remaining positive-count probabilities efficiently. But the recursion needs a starting value, and the usual \((a,b,1)\) form begins with a boundary probability at count one. A correct calculation therefore needs three parts: identify the original probability of zero, calculate the probability of one, and condition the positive probabilities on a nonzero count.
Here \(p_k\) always refers to the original count model, while \(q_k\) refers to the zero-truncated model. Keeping those symbols separate helps prevent a common error: using an original probability as though it were already conditional.
From the Boundary Probability to the Recursion
In the \((a,b,1)\) frequency class, the positive probabilities satisfy a recursion of the form below for \(k\geq 2\). The value at one is a boundary probability; the recursion does not, by itself, specify it.
For a zero-truncated sequence, the same recursion applies to \(q_k\). This follows directly by conditioning: every positive probability is divided by the same factor, \(1-p_0\).
This identity explains why the recursion is preserved, but not why its starting value may be ignored. The conditional boundary is \(q_1=p_1/(1-p_0)\). To calculate it, use the original model’s boundary probability \(p_1\) and its zero probability \(p_0\), or use a stated conditional boundary if the problem supplies one.
A Reliable Calculation Procedure
The following steps separate conditioning from recursion. This is especially useful when the model’s zero probability and positive-count recursion are presented in different forms.
Record the parameters, support, \(p_0\), \(p_1\), and the applicable recursion. Do not assume the recursion determines \(p_1\).
Use \(P(N>0)=1-p_0\). It must be positive for zero truncation to be defined.
Set \(q_1=p_1/(1-p_0)\), keeping sufficient precision for later steps.
Apply \(q_k=(a+b/k)q_{k-1}\) for each required \(k\geq2\), unless the model has a different stated boundary or finite endpoint.
Sum all probabilities on the positive support, using an appropriate series identity if the support is infinite. The total must be one.
Worked Example: Zero-Truncated Negative Binomial Counts
A hypothetical monthly count \(N\) follows a negative binomial model counting failures before the second success, with success probability \(1/2\). Find the zero-truncated probabilities \(q_1,q_2,q_3,q_4\), give each to four decimal places, and verify the full sequence sums to one.
Method. For this model, \(p_k=(k+1)(1/2)^{k+2}\), \(k\geq0\). In particular, \(p_0=1/4\) and \(p_1=1/4\). Its positive-count ratio is \(\frac{p_k}{p_{k-1}}=\frac{k+1}{2k}=\frac12+\frac{1}{2k}\), for \(k\geq2\). Thus \(a=1/2\) and \(b=1/2\).
Condition on a positive count. The probability of a positive count is \(1-p_0=1-1/4=3/4\). Therefore \(q_1=p_1/(1-p_0)=(1/4)/(3/4)=1/3\).
Apply the recursion. At \(k=2\), the multiplier is \(1/2+1/4=3/4\), so \(q_2=(3/4)(1/3)=1/4\). At \(k=3\), it is \(1/2+1/6=2/3\), so \(q_3=(2/3)(1/4)=1/6\). At \(k=4\), it is \(1/2+1/8=5/8\), so \(q_4=(5/8)(1/6)=5/48\).
To verify the entire infinite sequence, the formula for the original probabilities gives \(q_k=(k+1)/(3\cdot 2^k)\) for \(k\geq1\). Using \(\sum_{k=1}^{\infty}2^{-k}=1\) and \(\sum_{k=1}^{\infty}k2^{-k}=2\), \(\sum_{k=1}^{\infty}q_k=(1/3)(1+2)=1\). The requested values are \(q_1=0.3333\), \(q_2=0.2500\), \(q_3=0.1667\), and \(q_4=0.1042\), to four decimal places.
Check and trap avoided. The first four terms sum to \(41/48\); the remaining positive-count tail is \(7/48\), so those four terms should not be mistaken for the full normalization check. Dividing \(p_1\) by \(1-p_0\) before recursing avoids mixing original and conditional probabilities.
Worked Example: A Finite-Support Zero-Truncated Count
A hypothetical count \(N\) is binomial with \(n=3\) trials and success probability \(1/2\). Find the zero-truncated probabilities and check their total. State where the recursion stops.
Find the original boundary probabilities. The model gives \(p_0=(1/2)^3=1/8\) and \(p_1=\binom31(1/2)(1/2)^2=3/8\). Hence \(P(N>0)=1-1/8=7/8\), and \(q_1=(3/8)/(7/8)=3/7\).
Determine the recursion. For a binomial count with \(n=3\) and success probability \(1/2\), the ratio for \(k=2,3\) is \((4-k)/k\). This is \(a+b/k\) with \(a=-1\) and \(b=4\). At \(k=2\), the multiplier is \(-1+4/2=1\), so \(q_2=q_1=3/7\). At \(k=3\), it is \(-1+4/3=1/3\), so \(q_3=(1/3)(3/7)=1/7\).
The support ends at three; there is no probability beyond it. The sum is \(q_1+q_2+q_3=3/7+3/7+1/7=1\). Thus the zero-truncated probabilities are \(3/7,3/7,1/7\), approximately \(0.4286,0.4286,0.1429\), to four decimal places.
Check and trap avoided. The zero multiplier occurs at \(k=4\), not at \(k=3\): \(-1+4/4=0\). Stopping one step early would incorrectly omit the positive probability at three.
Worked Example: A Zero-Truncated Poisson Sequence
A hypothetical count \(N\) is Poisson with mean \(\lambda=1\). Calculate \(q_1,q_2,q_3,q_4\) to four decimal places and verify normalization.
Find the boundary. For a Poisson count, \(p_0=e^{-1}\) and \(p_1=e^{-1}\). Therefore \(q_1=e^{-1}/(1-e^{-1})=1/(e-1)\), approximately \(0.5820\).
Recurse. The Poisson probabilities satisfy \(p_k=(1/k)p_{k-1}\). Thus \(a=0,b=1\), and \(q_k=q_{k-1}/k\) for \(k\geq2\). It follows that \(q_2=q_1/2\approx0.2910\), \(q_3=q_2/3\approx0.0970\), and \(q_4=q_3/4\approx0.0242\).
In general, \(q_k=1/[k!(e-1)]\) for \(k\geq1\). Since \(\sum_{k=1}^{\infty}1/k!=e-1\), the probabilities sum to one. The computed values are \(0.5820,0.2910,0.0970,0.0242\), each rounded to four decimal places.
Check and trap avoided. The recursion generates ratios, not the starting probability. Starting at \(q_1=1\), for example, would produce the wrong sequence even though every later ratio was applied correctly.
Normalization and Exam Checks
A recursion can generate a plausible-looking sequence without proving that it is a probability distribution. The boundary probability sets the scale; the recursion sets the relative sizes of later probabilities. For an infinite support, a complete check must account for the whole infinite sum, not merely the terms requested in a calculation. A probability-generating formula or a known series identity can provide that check.
For finite support, include every positive mass through the endpoint and confirm that the next probability is zero or that the model otherwise specifies no mass beyond the endpoint. Never continue a finite-support recurrence as if it generated additional positive counts. Conversely, do not stop merely because a term is small.
A frequent exam error is to condition only the first probability and then continue with original, unconditioned probabilities. That mixes two scales. Because each positive original probability is divided by the same \(1-p_0\), the recursion can be applied directly to the conditional probabilities after the correct \(q_1\) has been found.
Check Your Understanding
For each problem, identify the conditioning probability before using the recursion. Answers are rounded to four decimal places where appropriate.
- A hypothetical Poisson count has mean \(3\). After conditioning on a positive count, calculate \(q_1\) and \(q_2\). (Answer: \(q_1=3/(e^3-1)=0.1572\), \(q_2=0.2358\).)
- A hypothetical binomial count has \(n=4\) trials and success probability \(1/4\). Find \(q_1\) and \(q_2\). (Answer: \(q_1=108/175=0.6171\), \(q_2=54/175=0.3086\).)
- A hypothetical negative binomial count records failures before three successes, with success probability \(0.4\). Find the zero-truncated \(q_1,q_2,q_3\), using \(p_k=\binom{k+2}{2}(0.6)^k(0.4)^3\). (Answer: \(q_1=0.1231\), \(q_2=0.1477\), \(q_3=0.1477\).)
- A hypothetical binomial count has \(n=5\) trials and success probability \(0.2\). Calculate \(q_1\) and \(q_2\) using \(p_0=0.8^5\), \(p_1=5(0.2)(0.8)^4\), and the positive-count ratio \((6-k)/(4k)\). (Answer: \(q_1=1280/2101=0.6092\), \(q_2=640/2101=0.3046\).)
- A finite-support count has zero-truncated boundary \(q_1=2/5\) and recursion \(q_k=(3-k)q_{k-1}/k\) for \(k\geq2\). Calculate its positive probabilities and verify the endpoint and total. (Answer: \(q_2=1/5\), \(q_3=0\); the sequence ends at two and sums to \(3/5\), so the stated boundary and recursion do not define a normalized zero-truncated distribution.)