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Zero-Truncated Poisson Counts

Learn how conditioning a Poisson count to be positive changes its probabilities and boundary value while preserving its recursive ratios.

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What You'll Learn

  • Define a Poisson count conditioned on being positive and state its support.
  • Derive the zero-truncated Poisson probability at one.
  • Obtain the recursive probabilities for all positive counts.
  • Identify the distribution’s parameters in the Panjer \((a,b,1)\) class.
  • Compute event probabilities using the correct truncated normalization.
  • Recognize errors involving the zero-count probability and the recursion’s starting index.

Why Zero-Truncated Poisson Counts Need a Boundary Probability

A Poisson model can describe the number of claims reported in a fixed period, but some data sets include only periods in which at least one claim occurred. For example, a hypothetical insurer might examine only policy-months with a reported claim. The resulting count is not an ordinary Poisson count: its probability at zero is removed, and the remaining probabilities must be renormalized.

The zero-truncated Poisson model is useful in the Panjer frequency framework because its positive-count probabilities retain a simple recursive relationship. To use that relationship correctly, however, one must distinguish the probability at one, which is a boundary value, from the recursion that generates probabilities at two and above.

Definition: Let \(N_0\sim\operatorname{Poisson}(\lambda)\), where \(\lambda>0\), so that \(P(N_0=n)=e^{-\lambda}\lambda^n/n!\) for \(n=0,1,2,\ldots\). A zero-truncated Poisson count \(N\) has the conditional distribution of \(N_0\) given \(N_0>0\). Write \(p_n=P(N=n)\). Its support is \(n=1,2,\ldots\), and \(p_0=0\).

The condition \(\lambda>0\) matters: when \(\lambda=0\), a Poisson count is zero with probability one, so conditioning on a positive count is not defined. For positive \(\lambda\), \(P(N_0>0)=1-e^{-\lambda}\), and conditioning divides each positive Poisson probability by this value.

Deriving the Probability at One and the Recursion

For any \(n\geq1\), apply the conditional probability definition and then simplify by multiplying numerator and denominator by \(e^\lambda\). This gives the probability mass function in a form convenient for both calculation and recursion.

$$ p_n=P(N_0=n\mid N_0>0) =\frac{e^{-\lambda}\lambda^n/n!}{1-e^{-\lambda}} =\frac{\lambda^n}{n!(e^\lambda-1)},\qquad n\geq1. $$

Setting \(n=1\) gives the boundary probability. It is not the ordinary Poisson probability \(e^{-\lambda}\lambda\), because it is conditional on a positive count.

Boundary probability: For a zero-truncated Poisson count with parameter \(\lambda>0\), $$ p_1=\frac{\lambda}{e^\lambda-1}. $$

For \(n\geq2\), divide the expression for \(p_n\) by the expression for \(p_{n-1}\). The common normalizing factor \(e^\lambda-1\) cancels, as does the corresponding power of \(\lambda\), leaving a ratio that depends on \(n\).

$$ \frac{p_n}{p_{n-1}} =\frac{\lambda^n}{n!(e^\lambda-1)} \frac{(n-1)!(e^\lambda-1)}{\lambda^{n-1}} =\frac{\lambda}{n}, \qquad n\geq2. $$

Thus, once \(p_1\) is known, every subsequent probability follows from \(p_n=(\lambda/n)p_{n-1}\). The recurrence does not generate \(p_1\) from \(p_0\): since \(p_0=0\), trying to apply a recurrence at \(n=1\) would not produce the positive boundary probability.

Connection to the Panjer \((a,b,1)\) Class

In the Panjer \((a,b,1)\) class, probabilities satisfy \(p_n=(a+b/n)p_{n-1}\) for \(n\geq2\), with a separate boundary probability \(p_1\). Comparing this defining ratio with the zero-truncated Poisson recursion identifies its parameters directly.

Theorem: A zero-truncated Poisson count with parameter \(\lambda>0\) belongs to the Panjer \((a,b,1)\) class with \(a=0\), \(b=\lambda\), and boundary probability \(p_1=\lambda/(e^\lambda-1)\). The recursion is \(p_n=(\lambda/n)p_{n-1}\) for \(n\geq2\).

The distribution’s normalization can also be checked directly. Summing its mass function over the positive integers gives \(\sum_{n=1}^{\infty}\lambda^n/[n!(e^\lambda-1)]\). The exponential series gives \(\sum_{n=1}^{\infty}\lambda^n/n!=e^\lambda-1\), so the probabilities sum to one. This confirms why the denominator is \(e^\lambda-1\), rather than \(e^\lambda\).

1
Set the model and parameter.
Use a Poisson count with \(\lambda>0\), observed conditional on being positive.
2
Calculate the boundary value.
Find \(p_1=\lambda/(e^\lambda-1)\) from the conditional probability formula.
3
Generate the remaining probabilities.
For each \(n\geq2\), calculate \(p_n=(\lambda/n)p_{n-1}\).
4
Identify the Panjer parameters.
Match the ratio to \(a+b/n\): here \(a=0\) and \(b=\lambda\).

Worked Applications

Worked Example: Probabilities and a Tail Event

A hypothetical portfolio has a zero-truncated Poisson claim count \(N\) with parameter \(\lambda=2\). Find \(p_1,p_2,p_3,p_4\), then calculate \(P(N\geq5)\). Give probabilities to four decimal places.

Method and boundary value. Since the model is conditioned on \(N>0\), begin with the truncated boundary probability: \(p_1=2/(e^2-1)\). Using \(e^2-1=6.3890561\), \(p_1=2/6.3890561=0.3130353\).

Recursion. Apply \(p_n=(2/n)p_{n-1}\) successively: \(p_2=(2/2)(0.3130353)=0.3130353\); \(p_3=(2/3)(0.3130353)=0.2086902\); and \(p_4=(2/4)(0.2086902)=0.1043451\).

The event \(N\geq5\) is the complement of \(N=1,2,3,4\), since \(N=0\) is impossible under this model. Therefore, \(P(N\geq5)=1-(0.3130353+0.3130353+0.2086902+0.1043451)=0.0608941\). To four decimal places, \(P(N\geq5)=0.0609\).

Check. The four displayed probabilities sum to \(0.9391059\); adding the tail \(0.0608941\) gives \(1.0000000\), up to rounding. The common mistake avoided is using the ordinary, unconditioned Poisson probability at one as the boundary value.

Worked Example: Recognizing the Panjer Parameters

A hypothetical reporting system records a positive count with a zero-truncated Poisson distribution having parameter \(\lambda=3\). Find \(p_1,p_2,p_3\), and state the Panjer \((a,b,1)\) parameters. Give probabilities to four decimal places.

Boundary value. Here \(e^3-1=19.0855369\), so \(p_1=3/19.0855369=0.1571871\), which rounds to \(0.1572\).

Recursive probabilities. First, \(p_2=(3/2)(0.1571871)=0.2357807\), which rounds to \(0.2358\). Then, \(p_3=(3/3)(0.2357807)=0.2357807\), also \(0.2358\).

Class parameters. The ratio is \(p_n/p_{n-1}=3/n=0+3/n\) for \(n\geq2\). Therefore \(a=0\) and \(b=3\), with \(p_1=0.1572\) as the boundary probability.

Check. The first three probabilities sum to \(0.1571871+0.2357807+0.2357807=0.6287485\), which is less than one, as expected because counts \(4,5,\ldots\) retain positive probability. A common mistake is to set \(p_3=(3/3)p_2\) and then substitute \(p_1\) for \(p_2\); the recurrence must be applied one count at a time.

Worked Example: A Cumulative Probability and Its Trap

For a hypothetical zero-truncated Poisson count with \(\lambda=2\), calculate \(P(N\geq3)\). Give the result to four decimal places.

Method. Use the complement \(P(N\geq3)=1-P(N\leq2)\). The relevant positive-count probabilities are \(p_1=2/(e^2-1)=0.3130353\) and \(p_2=(2/2)p_1=0.3130353\). Hence \(P(N\geq3)=1-(0.3130353+0.3130353)=0.3739294\), or \(0.3739\) to four decimal places.

Common mistake and correction. A tempting but incorrect calculation is \(1-(1+2)/(e^2-1)=0.5304\). Its numerator includes the unconditioned Poisson term for zero while its denominator is the zero-truncated normalizer; those terms do not represent \(P(N\leq2)\) under the conditional model. The correct numerator for the truncated cumulative probability is \(2+2\), corresponding to counts one and two, and the correct tail probability is \(0.3739\).

Check. From the first example, \(p_3=0.2086902\), so the tail from three onward must exceed \(0.2086902\). The result \(0.3739294\) passes this check.

Common Interpretation and Calculation Errors

The main practical distinction is between the Poisson model before conditioning and the positive-count model after conditioning. The latter has no probability at zero, and all positive probabilities are rescaled by the same factor. That common factor cancels from adjacent probability ratios, which is why the recursion remains simple even though the boundary value changes.

  • Using the untruncated boundary. The value \(e^{-\lambda}\lambda\) is \(P(N_0=1)\), not \(P(N=1\mid N_0>0)\). Divide by \(1-e^{-\lambda}\), or use \(\lambda/(e^\lambda-1)\).
  • Starting the recursion at zero. The zero-truncated count has \(p_0=0\). Calculate \(p_1\) from its boundary formula, then recurse for \(n\geq2\).
  • Using the wrong recursion index. The factor is \(\lambda/n\) when calculating \(p_n\) from \(p_{n-1}\). For example, \(p_2=(\lambda/2)p_1\), not \(\lambda p_1\).
  • Mixing normalizers. A truncated probability uses the denominator \(e^\lambda-1\) in the simplified mass function. Do not include the zero-count term in a numerator while retaining this positive-count normalizer.
  • Reading a recursion as a complete distribution. The ratio alone does not supply the probability at one. A boundary value is required to determine the sequence.
Key takeaway: For a zero-truncated Poisson count with \(\lambda>0\), set \(p_1=\lambda/(e^\lambda-1)\), then use \(p_n=(\lambda/n)p_{n-1}\) for \(n\geq2\). Its Panjer parameters are \(a=0\) and \(b=\lambda\); the zero probability is exactly zero.

Check Your Understanding

For each problem, use the positive-count boundary probability and apply the recursion from \(n=2\) onward. Answers are rounded to four decimal places where appropriate.

  1. A hypothetical count is zero-truncated Poisson with \(\lambda=1\). Find \(p_1\). (Answer: \(p_1=0.5820\))
  2. A zero-truncated Poisson count has \(\lambda=2\). Find \(p_1\) and \(p_2\). (Answer: \(p_1=0.3130,\ p_2=0.3130\))
  3. For a zero-truncated Poisson count with \(\lambda=3\), find \(p_1,p_2,p_3\). (Answer: \(p_1=0.1572,\ p_2=0.2358,\ p_3=0.2358\))
  4. A hypothetical reporting period has a zero-truncated Poisson count with \(\lambda=2\). Find \(P(N\geq3)\). (Answer: \(0.3739\))
  5. A zero-truncated Poisson count has \(\lambda=1.5\). Find \(p_1,p_2,p_3\), and identify its Panjer parameters. (Answer: \(p_1=0.4308,\ p_2=0.3231,\ p_3=0.1616;\ a=0,\ b=1.5\))

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