Why Zero Truncation Preserves the Recursion

Why Conditioning Changes the Boundary

A frequency model may assign positive probability to no claims, even when an analysis concerns only policies or periods with at least one claim. Conditioning on a positive count removes the zero-count outcome and rescales every positive-count probability. The rescaling is the same for all positive counts, so their ratios remain unchanged. The important qualification is that the recursion for the original model links count one to count zero; after conditioning, the new probability at zero is zero, so that first link no longer holds.

This distinction matters when using a Panjer recursion: the interior probability ratios can be carried over, but the first positive probability must be recalculated as a boundary probability. The derivation below makes the valid range explicit.

Definition: Let \(N\) be a nonnegative integer-valued count with probabilities \(p_n=P(N=n)\). An \((a,b,0)\) distribution has a probability sequence satisfying \(p_n=(a+b/n)p_{n-1}\) for \(n\geq 1\), wherever the recursion applies. Assume the distribution is admissible and \(P(N>0)>0\). Its zero-truncated probabilities are \(q_n=P(N=n\mid N>0)\) for \(n\geq 1\), with \(q_0=0\).

The parameters \(a\) and \(b\) belong to the original count distribution. Zero truncation does not change them in the ratios between positive-count probabilities. It does change the probability sequence: \(q_n\) is conditional, whereas \(p_n\) is unconditional.

The Recursion After Zero Truncation

Write \(c=P(N>0)=1-p_0\). By the definition of conditional probability, for every \(n\geq1\),

q_n=P(N=n\mid N>0) =\frac{P(N=n)}{P(N>0)} =\frac{p_n}{c}.

For \(n\geq2\), both \(n\) and \(n-1\) are positive counts. Thus both probabilities in their ratio are divided by the same positive constant \(c\). Substituting the original recursion gives

\frac{q_n}{q_{n-1}} =\frac{p_n/c}{p_{n-1}/c} =\frac{p_n}{p_{n-1}} =a+\frac{b}{n}, \qquad n\geq2,

whenever the ratio is defined. Equivalently, the recurrence can be written without division:

q_n=\left(a+\frac{b}{n}\right)q_{n-1}, \qquad n\geq2.

That equation is the central result: the positive-count probabilities retain the original recursion from \(n=2\) onward. At \(n=1\), however, \(q_0=0\) while \(q_1=p_1/c\) can be positive. The original recursion at \(n=1\) was \(p_1=(a+b)p_0\); it does not become the corresponding recursion \(q_1=(a+b)q_0\). That would force \(q_1=0\), contradicting positive probability of a count of one.

Theorem: If an admissible \((a,b,0)\) count \(N\) satisfies \(P(N>0)>0\), then conditioning on \(N>0\) gives \(q_0=0\), \(q_n=p_n/(1-p_0)\) for \(n\geq1\), and \(q_n=(a+b/n)q_{n-1}\) for \(n\geq2\) wherever the recursion applies. In particular, the new first-count boundary is \(q_1=p_1/(1-p_0)\), not a value obtained by applying the recursion at \(n=1\).

Derivation and Interpretation

The result follows from one feature of conditioning: each retained outcome is multiplied by the same factor. Specifically, \(P(N>0)=c\), so every positive-count probability is multiplied by \(1/c\). A ratio between two such probabilities cancels this factor. At the boundary, one member of the original ratio, \(p_0\), is not retained at all; the conditional sequence instead assigns zero probability to count zero.

Find the conditioning probability.
Set \(c=P(N>0)=1-p_0\). The condition \(c>0\) is necessary to define the conditional distribution.

Rescale positive probabilities.
For each \(n\geq1\), set \(q_n=p_n/c\), and set \(q_0=0\).

Take ratios only between positive counts.
For \(n\geq2\), the factors \(1/c\) cancel in \(q_n/q_{n-1}\), leaving \(a+b/n\).

Set the first-count boundary directly.
Use \(q_1=p_1/c\). Do not extend the ratio argument to \(n=1\), where the lower count is zero.

The theorem does not say that every probability in the conditional distribution is unchanged. In general, \(q_n\neq p_n\), because conditioning changes their scale. Nor does it say that ratios involving the zero-count probability are preserved. It says that the relation among successive positive-count probabilities is unchanged.

For a distribution with finite support, the same qualification about support applies as before conditioning. If a probability at the endpoint is zero, a ratio with that probability in the denominator is not defined; the recursion can still be stated as a probability equation where valid. Zero truncation does not create positive probability beyond the original maximum count.

Worked Example: Rescaling an Infinite Positive-Count Sequence

A hypothetical count model has \(p_n=0.4(0.6)^n\) for \(n\geq0\). Condition on at least one claim. Find \(q_1,q_2,q_3\), verify two successive ratios, and state the boundary probability.

Method. First find the probability of the conditioning event. The original zero-count probability is \(p_0=0.4\), so \(P(N>0)=1-0.4=0.6\).

The original probabilities needed are \(p_1=0.4(0.6)=0.24\), \(p_2=0.4(0.6)^2=0.144\), and \(p_3=0.4(0.6)^3=0.0864\). Divide each by \(0.6\):

q_1=\frac{0.24}{0.6}=0.4,\qquad q_2=\frac{0.144}{0.6}=0.24,\qquad q_3=\frac{0.0864}{0.6}=0.144.

The sequence has ratio \(p_n/p_{n-1}=0.6\), which is the \((a,b,0)\) recursion with \(a=0.6\) and \(b=0\). The conditional ratios confirm the theorem: \(q_2/q_1=0.24/0.4=0.6\), and \(q_3/q_2=0.144/0.24=0.6\). The boundary is \(q_1=0.4\), calculated by conditioning rather than from \(q_0\).

As a check, the conditional probabilities sum to one: \(\sum_{n=1}^{\infty}0.4(0.6)^{n-1}=0.4/(1-0.6)=1\). The common mistake this avoids is treating the original \(p_0\) as the conditional zero probability. The conditional model assigns \(q_0=0\).

Worked Example: Checking a Finite-Support Recursion

A hypothetical count \(N\) has probabilities \(p_0=1/8\), \(p_1=3/8\), \(p_2=3/8\), and \(p_3=1/8\), with no probability above three. Its recursion has \(a=-1\) and \(b=4\). Find the zero-truncated probabilities and check the recursion for the positive counts where it applies.

Method. The conditioning probability is \(P(N>0)=1-1/8=7/8\). Therefore,

q_1=\frac{3/8}{7/8}=\frac{3}{7},\qquad q_2=\frac{3/8}{7/8}=\frac{3}{7},\qquad q_3=\frac{1/8}{7/8}=\frac{1}{7}.

For \(n=2\), the original ratio factor is \(a+b/2=-1+4/2=1\); the conditional ratio is \(q_2/q_1=(3/7)/(3/7)=1\). For \(n=3\), the factor is \(a+b/3=-1+4/3=1/3\); the conditional ratio is \(q_3/q_2=(1/7)/(3/7)=1/3\). Thus the unchanged ratios agree with the original recursion. The probabilities also check: \(3/7+3/7+1/7=1\).

At \(n=1\), the original factor would be \(-1+4=3\), and indeed \(p_1/p_0=3\). But applying that factor to the conditional boundary would give \(q_1=3q_0=0\), which is wrong. The conditional first probability must instead be obtained by dividing \(p_1\) by \(7/8\).

Worked Example: A Boundary and an Interior Probability

A hypothetical finite-support count model follows \(p_n=(a+b/n)p_{n-1}\), with \(a=-1/2\), \(b=3/2\), \(p_0=4/9\), and no probability above two. Condition on \(N>0\). Find the conditional distribution and verify its interior ratio.

Method. At \(n=1\), the factor is \(-1/2+3/2=1\), so \(p_1=p_0=4/9\). At \(n=2\), the factor is \(-1/2+(3/2)/2=-1/2+3/4=1/4\), so \(p_2=(1/4)(4/9)=1/9\). The probabilities sum to \(4/9+4/9+1/9=1\); the stated finite support is consistent because the next factor is \(-1/2+(3/2)/3=0\), giving \(p_3=0\).

The probability of a positive count is \(1-p_0=5/9\). Hence \(q_1=(4/9)/(5/9)=4/5\) and \(q_2=(1/9)/(5/9)=1/5\), with \(q_0=0\). The conditional interior ratio is \(q_2/q_1=(1/5)/(4/5)=1/4\), matching \(a+b/2=1/4\). The conditional probabilities sum to \(4/5+1/5=1\).

The common mistake this calculation avoids is assuming that unchanged ratios imply unchanged probabilities. The ratio \(p_2/p_1\) is retained, but both probabilities have been rescaled to form a distribution conditional on a positive count.

Common Mistakes and Practical Checks

A reliable calculation separates the boundary from the interior. Compute \(1-p_0\), use it to obtain \(q_1\), and then use the ratio recursion only for \(n\geq2\). For a requested list of conditional probabilities, each positive \(p_n\) may be divided by the same conditioning probability. For a recursion calculation, start from \(q_1\) and move upward.

Applying the recursion at count one. Since \(q_0=0\), this incorrectly forces \(q_1=0\). Calculate \(q_1=p_1/(1-p_0)\) instead.
Forgetting the normalization. Reusing \(p_n\) as \(q_n\) generally gives probabilities summing to \(1-p_0\), not one.
Changing \(a\) or \(b\) after conditioning. The ratio for successive positive counts is unchanged, so the same parameters apply there.
Using a ratio beyond the support. If the denominator probability is zero, the ratio is undefined. Respect the original support and any endpoint where the probability recurrence yields zero.
Conditioning when \(P(N>0)=0\). If \(p_0=1\), the conditional distribution given \(N>0\) does not exist.

These checks are useful whenever a positive-count model is built from a distribution that includes zero. The key is not to confuse the preserved positive-count ratios with the changed boundary probabilities.

Key takeaway: Zero truncation multiplies every positive-count probability by \(1/(1-p_0)\). That common factor cancels between two positive counts, preserving \(q_n=(a+b/n)q_{n-1}\) for \(n\geq2\). Count one is a new boundary: \(q_1=p_1/(1-p_0)\).

Check Your Understanding

For each problem, distinguish the conditional boundary from the ratios between positive counts.

A count model has \(p_0=0.25\), \(p_1=0.30\), and recursion parameters \(a=0.4\), \(b=0.2\). Find \(q_1\) and \(q_2\). (Answer: \(q_1=0.4000,\ q_2=0.2000\).)
A model has \(p_0=0.20\), \(p_1=0.16\), and \(a=0.5,\ b=0.3\). Find \(q_1\) and the ratio \(q_2/q_1\). (Answer: \(q_1=0.2000,\ q_2/q_1=0.6500\).)
A finite count distribution has probabilities \(1/8,3/8,3/8,1/8\) at counts zero through three, respectively. Find its zero-truncated probabilities. (Answer: \(q_1=3/7,\ q_2=3/7,\ q_3=1/7\).)
A count model has \(P(N>0)=0.8\), \(q_1=0.25\), and recursion parameters \(a=0.5,\ b=0.5\). Find the unconditional \(p_1\) and conditional \(q_2\). (Answer: \(p_1=0.2000,\ q_2=0.1875\).)
A model has \(p_0=0.4\), \(a=0.2\), and \(b=0.3\). A proposed calculation uses \(q_1=(a+b)q_0\). Explain why it fails and find the correct \(q_1\). (Answer: The recursion cannot be applied at \(n=1\) after truncation; \(q_1=p_1/(1-p_0)=1/3\).)

What You'll Learn