Separating Zero Claims from Positive Counts
A frequency model may need to represent two different features: whether any claims occur at all, and how many occur once the count is positive. A hurdle model makes that separation explicit. It assigns a probability to zero and specifies a separate distribution for counts of one or more. The positive-count distribution might have a convenient Panjer recursion, but that does not follow merely from having a hurdle.
The key membership question is whether the positive probabilities satisfy the required recursion at every count from two onward. The probability of zero and the probability of one are boundary probabilities in the (a,b,1) class; they do not determine the recursion for larger counts.
Testing Membership in the (a,b,1) Class
For a count distribution with probabilities \(p_n=P(N=n)\), the (a,b,1) recursion is \(p_n=(a+b/n)p_{n-1}\) for \(n\geq2\). The values \(p_0\) and \(p_1\) are boundary probabilities; this recursion does not impose a relation between them. If the probabilities at consecutive counts are positive, the same test can be written as a ratio: \(p_n/p_{n-1}=a+b/n\).
In a hurdle model, every positive probability is multiplied by the same factor \(1-\pi\). Consequently, for \(n\geq2\), the ratio of two consecutive unconditional positive probabilities is the same as the ratio of the corresponding conditional probabilities.
This criterion is stated as a recursion so it also applies when writing ratios would involve a zero probability. In the common case where the positive probabilities are all strictly positive, the ratio form is convenient. Membership requires one fixed pair \((a,b)\) to work at every applicable index—not a different pair for each count.
Why Conditioning a Panjer Model Can Preserve the Recursion
Suppose a nonnegative count distribution has probabilities \(f_n\) and belongs to the (a,b,0) class, so \(f_n=(a+b/n)f_{n-1}\) for \(n\geq1\). Let \(f_0<1\), and condition this distribution on being positive. Its conditional probabilities are \(q_n=f_n/(1-f_0)\) for \(n\geq1\). For \(n\geq2\), both terms in the recursion are divided by the same positive constant. Thus the recursion among positive counts is unchanged.
The conditioning step does not need to preserve the original recursion at \(n=1\): the (a,b,1) definition starts at \(n=2\), and \(q_1\) is a boundary probability for the positive-count distribution. Now assign any hurdle probability \(\pi<1\) to zero and scale these \(q_n\) by \(1-\pi\). The resulting hurdle distribution remains in the (a,b,1) class.
This is a sufficient construction, not a requirement that every hurdle model come from conditioning an (a,b,0) model. An unrestricted hurdle model may use any valid conditional positive-count distribution. It belongs to the (a,b,1) class only if that chosen distribution itself satisfies the recursion.
Worked Applications
Worked Example: A Hurdle Geometric Positive-Count Model
A hypothetical policy portfolio has probability \(\pi=0.30\) of no claims in a year. Conditional on at least one claim, the count follows a geometric distribution on \(1,2,\ldots\), with \(q_n=(1-r)r^{n-1}\) and \(r=0.40\). Find \(p_0,p_1,p_2,p_3\) and determine whether the resulting distribution belongs to an (a,b,1) class.
Method. First compute the conditional probabilities, then multiply each by \(1-\pi=0.70\) to obtain the unconditional positive-count probabilities. For the geometric law, \(q_1=1-r=0.60\), \(q_2=0.60(0.40)=0.24\), and \(q_3=0.24(0.40)=0.096\). The conditional probabilities sum to one because \(\sum_{n=1}^{\infty}(1-r)r^{n-1}=1\).
The zero probability is \(p_0=0.30\). For positive counts, \(p_1=0.70(0.60)=0.42\), \(p_2=0.70(0.24)=0.168\), and \(p_3=0.70(0.096)=0.0672\).
The positive-count ratios are \(p_2/p_1=0.168/0.42=0.40\) and \(p_3/p_2=0.0672/0.168=0.40\). In fact, \(q_n/q_{n-1}=r=0.40\) for every \(n\geq2\), so the recursion holds with \(a=0.40\) and \(b=0\). The probabilities also check at the boundary: \(p_0+p_1+p_2+p_3=0.9552\), leaving \(0.0448\) for counts of four or more; the full positive probability is \(0.70\) by construction.
Common mistake avoided. Do not apply the geometric recursion from \(n=1\) to relate \(p_1\) to \(p_0\). The hurdle assigns \(p_0\) separately, and (a,b,1) membership is tested starting at \(n=2\).
Worked Example: A Hurdle Model with a Zero-Truncated Poisson Component
A hypothetical service has probability \(\pi=0.25\) of zero events in a year. Conditional on a positive count, its event count follows a Poisson distribution with parameter \(\lambda=2\), conditioned on being positive. Find \(p_1,p_2,p_3\), identify the (a,b,1) parameters, and check the probability construction.
Method. For an unconditioned Poisson variable with parameter \(2\), the probability at \(n\) is \(f_n=e^{-2}2^n/n!\), and \(f_0=e^{-2}\). Therefore the probability of a positive count is \(1-e^{-2}\), and the conditional positive-count probabilities are \(q_n=2^n/[n!(e^2-1)]\) for \(n\geq1\). In particular, \(q_1=2/(e^2-1)\), \(q_2=2/(e^2-1)\), and \(q_3=4/[3(e^2-1)]\).
Here \(e^2-1\approx6.389056\). The positive probabilities are scaled by \(1-\pi=0.75\), giving \(p_1=1.5/6.389056\approx0.2348\), \(p_2=1.5/6.389056\approx0.2348\), and \(p_3=1/6.389056\approx0.1565\). Probabilities are rounded to four decimal places.
For \(n\geq2\), the consecutive conditional probabilities have ratio \(q_n/q_{n-1}=2/n\). Thus the same ratio holds for the unconditional probabilities, and the distribution belongs to the (a,b,1) class with \(a=0\) and \(b=2\). For example, \(p_2/p_1=1=2/2\), and \(p_3/p_2\approx0.6667=2/3\).
The zero probability is \(p_0=0.25\). The entire positive part has probability \(0.75\), since the conditional probabilities \(q_n\) sum to one; hence the distribution sums to one. The displayed first three probabilities sum to \(0.6261\), with the remaining \(0.1239\) in counts of four or more.
Common mistake avoided. The positive probabilities must be conditioned by dividing the Poisson probabilities by \(1-e^{-2}\), not by leaving them unchanged. The hurdle scaling is applied after that conditioning.
Worked Example: A Hurdle Model That Fails the Recursion Test
A hypothetical risk model has \(\pi=0.20\). Its conditional positive-count distribution is an equal mixture of two geometric distributions on \(1,2,\ldots\): one has \(r=0.50\), and the other has \(r=0.25\). Determine whether the hurdle distribution belongs to any (a,b,1) class.
Method. Compute consecutive conditional probabilities and their ratios. For a geometric distribution on positive integers, the first four probabilities are \((1-r), (1-r)r, (1-r)r^2, (1-r)r^3\). Averaging the two component probabilities with equal weights gives \(q_1=5/8\), \(q_2=7/32\), \(q_3=11/128\), and \(q_4=19/512\). Each component is a probability distribution, so their equal mixture is also normalized.
The first three consecutive ratios are \(q_2/q_1=7/20\), \(q_3/q_2=11/28\), and \(q_4/q_3=19/44\). If a single pair \((a,b)\) worked, the first two ratios would require \(a+b/2=7/20\) and \(a+b/3=11/28\). Subtracting the second equation from the first gives \(b/6=-3/70\), so \(b=-9/35\). Substitution gives \(a=67/140\).
Those parameters predict a ratio at \(n=4\) of \(a+b/4=67/140-9/140=29/70\). But the actual ratio is \(19/44\), and \(29/70\ne19/44\). Therefore no single pair \((a,b)\) satisfies the recursion at all these indices, so this hurdle distribution is not in the (a,b,1) class.
For a boundary check, \(p_0=0.20\) and \(p_1=0.80(5/8)=0.50\); likewise, \(p_2=0.80(7/32)=0.175\), \(p_3=0.80(11/128)=0.06875\), and \(p_4=0.80(19/512)=0.0296875\). Scaling has not changed any of the ratios used in the test.
Common mistake avoided. Matching two ratios only determines a candidate pair \((a,b)\). It does not establish membership; the candidate must satisfy the recursion at every subsequent count as well.
Distinguishing Hurdle and Zero-Modified Models
A zero-modified model starts with a specified base distribution and changes its zero probability while rescaling the positive probabilities to preserve total probability. In that construction, the conditional distribution given a positive count is fixed by the base distribution. A hurdle model instead treats the zero probability and the conditional positive-count distribution as separately specified components. A zero-modified model can therefore be represented as a particular hurdle model, but an arbitrary hurdle model need not have the positive-count shape inherited from a chosen base distribution.
For Panjer membership, this distinction leads to a direct workflow:
Write \(p_0=\pi\) and \(p_n=(1-\pi)q_n\) for \(n\geq1\), verifying that the \(q_n\) form a probability distribution on the positive integers.
Compute \(q_n/q_{n-1}\), where defined, and check whether one pair \((a,b)\) gives \(a+b/n\) at every index \(n\geq2\).
If the test succeeds, report the (a,b,1) parameters and keep \(p_0\) and \(p_1=(1-\pi)q_1\) as boundary probabilities. The recursion itself does not determine them.
The main exam trap is to infer class membership from the name of the positive-count model or from only one or two matching ratios. A hurdle probability changes the level of every positive probability but not their ratios. The shape of the conditional positive-count distribution decides membership.
Check Your Understanding
For each problem, test the positive-count recursion rather than extending it to the zero probability.
- A hurdle model has \(\pi=0.20\), and its positive count is geometric on \(1,2,\ldots\) with \(r=0.30\). Find \(p_1,p_2\) and the (a,b,1) parameters. (Answer: \(p_1=0.56\), \(p_2=0.168\), \(a=0.30\), \(b=0\).)
- A hurdle model has \(\pi=0.20\); conditional on a positive count, \(N\) has a \(\operatorname{Binomial}(4,0.5)\) distribution conditioned on being positive. Identify its (a,b,1) parameters and find \(p_1\). (Answer: \(a=-1\), \(b=5\), \(p_1=16/75\approx0.2133\).)
- A hurdle model has \(\pi=0.10\) and conditional probabilities \(q_1=0.40,q_2=0.24,q_3=0.16,q_4=0.10\). Decide whether it can belong to an (a,b,1) class. (Answer: No; the first two ratios imply \(a=0.8,b=-0.4\), which predict \(q_4/q_3=0.7\), not \(0.625\).)
- A hurdle model has \(\pi=0.35\), \(q_1=0.25\), and ratios \(q_2/q_1=1.2\), \(q_3/q_2=0.8\), and \(q_4/q_3=0.55\). Find \(p_1\) and determine membership. (Answer: \(p_1=0.1625\); no, the first two ratios imply \(a=0,b=2.4\), predicting \(q_4/q_3=0.6\).)
- A hurdle model has \(\pi=0.50\); its positive-count component is a Poisson distribution with parameter \(\lambda=1\), conditioned on being positive. Find \(a,b,p_1,p_2\). (Answer: \(a=0,b=1,p_1=1/[2(e-1)]\approx0.2910,p_2=1/[4(e-1)]\approx0.1455\).)