Why Shifting a Count Is Not Zero Truncation

Why the Two Operations Are Different

Suppose \(N\) is a claim count with probability function \(p_n=P(N=n)\). There are two natural ways to obtain a count that is always positive. One is to condition on the event \(N>0\), keeping the realized positive count unchanged. The other is to add one to every realized count, including realizations where \(N=0\). Both resulting counts are positive, but they need not have the same distribution or probability ratios.

This distinction matters when identifying a frequency distribution for a Panjer recursion. The recursion is about probabilities at particular integer indices. Conditioning rescales the probabilities at the existing positive indices; shifting relocates each probability to the next index. That relocation changes which original probability appears in a ratio.

Definition: Let \(N\) have probabilities \(p_n=P(N=n)\), for integers \(n\geq 0\), with \(P(N>0)>0\). The zero-truncated count \(N\mid N>0\) has probabilities \(r_0=0\) and \(r_n=p_n/(1-p_0)\) for \(n\geq1\). The shifted count \(N+1\) has probabilities \(q_0=0\) and \(q_n=p_{n-1}\) for \(n\geq1\).

For zero truncation, every positive probability is divided by the same normalizing constant. For shifting, the probability originally at zero becomes the probability at one, the original probability at one becomes the probability at two, and so forth. The formulas give, for indices where the denominators are positive,

\frac{r_n}{r_{n-1}}=\frac{p_n}{p_{n-1}}, \qquad \frac{q_n}{q_{n-1}}=\frac{p_{n-1}}{p_{n-2}}, \qquad n\geq2.

The second ratio uses the original ratio at index \(n-1\), not at index \(n\). That one-index difference is the essential reason a shifted count may fail to retain the original Panjer form.

Panjer Recursions and the Index Change

A nonnegative integer distribution is in the \((a,b,0)\) Panjer class when its probabilities satisfy \(p_n=(a+b/n)p_{n-1}\) for \(n\geq1\), along with the conditions that the resulting probabilities are nonnegative and sum to one. In the \((a,b,1)\) form, the recursion is required for \(n\geq2\); the probabilities at zero and one are specified consistently with a normalized probability distribution.

p_n=\left(a+\frac{b}{n}\right)p_{n-1}.

If the original count satisfies this recursion, zero truncation retains it at indices \(n\geq2\). Indeed, \(r_n=p_n/(1-p_0)\) and \(r_{n-1}=p_{n-1}/(1-p_0)\), so their common normalizing factor cancels. The recursion at \(n=1\) is not inherited: \(r_0=0\), while \(r_1\) is positive. This is why the relevant form for a zero-truncated sequence is generally the \((a,b,1)\) recursion.

Result: If \(p_n=(a+b/n)p_{n-1}\) holds for the original count at \(n\geq2\), then the zero-truncated probabilities satisfy \(r_n=(a+b/n)r_{n-1}\) at \(n\geq2\). By contrast, the shifted probabilities satisfy \(q_n=(a+b/(n-1))q_{n-1}\) at \(n\geq2\), whenever these equations are defined.

The shift formula follows directly from the definitions: \(q_n=p_{n-1}\) and \(q_{n-1}=p_{n-2}\). Applying the original recursion at index \(n-1\) gives \(p_{n-1}=(a+b/(n-1))p_{n-2}\), hence the stated formula. The denominator is \(n-1\) because the adjacent shifted probabilities came from adjacent original probabilities one index earlier.

For an infinite-support sequence with positive adjacent probabilities, a shifted ratio of \(a+b/(n-1)\) generally cannot be written as \(A+B/n\) for fixed constants \(A\) and \(B\). If equality held for all \(n\geq2\), taking limits as \(n\) increases would give \(A=a\). The remaining equality \(b/(n-1)=B/n\) would imply \(bn=B(n-1)\) for every \(n\), which is impossible for \(b\ne0\). This argument explains why shifting often leaves the class, but it does not say that every shift must do so: special cases, such as a geometric count, can retain the form.

Worked Comparisons

Worked Example: Shifting a Poisson Count

A hypothetical portfolio has claim count \(N\sim\operatorname{Poisson}(2)\). Compare the ratios for \(N\mid N>0\) and \(N+1\), and determine whether the shifted count satisfies any \((A,B,1)\) recursion.

Method. First write the Poisson probabilities, then calculate adjacent ratios at their actual indices. For \(N\sim\operatorname{Poisson}(\lambda)\), \(p_n=e^{-\lambda}\lambda^n/n!\), so \(p_n/p_{n-1}=\lambda/n\) for \(n\geq1\). Here \(\lambda=2\), giving \(p_n/p_{n-1}=2/n\).

The probability of a positive count is \(1-p_0=1-e^{-2}\). Thus the first two zero-truncated probabilities are

r_1=\frac{2e^{-2}}{1-e^{-2}}=\frac{2}{e^2-1}, \qquad r_2=\frac{2e^{-2}}{1-e^{-2}}=\frac{2}{e^2-1}.

Therefore \(r_2/r_1=1\), in agreement with the original Poisson ratio at index \(2\), namely \(2/2=1\). For the shifted count, \(q_1=p_0=e^{-2}\), \(q_2=p_1=2e^{-2}\), \(q_3=p_2=2e^{-2}\), and \(q_4=p_3=(4/3)e^{-2}\). Its ratios are

\frac{q_2}{q_1}=2,\qquad \frac{q_3}{q_2}=1,\qquad \frac{q_4}{q_3}=\frac{2}{3}.

Suppose these ratios came from \(q_n=(A+B/n)q_{n-1}\) for \(n\geq2\). The first two equations would be \(A+B/2=2\) and \(A+B/3=1\). Subtracting gives \(B/6=1\), so \(B=6\) and \(A=-1\). At \(n=4\), that recursion predicts \(A+B/4=-1+6/4=1/2\), but the actual ratio is \(2/3\). The shifted count therefore does not satisfy an \((A,B,1)\) recursion. The calculation avoids the mistake of using the original Poisson ratio \(2/n\) at shifted index \(n\): the shifted ratio is \(2/(n-1)\).

Worked Example: A Shifted Binomial Count and Its Endpoint

Let a hypothetical batch have \(N\sim\operatorname{Binomial}(3,1/2)\), and set \(M=N+1\). Test whether \(M\) satisfies an \((A,B,1)\) recursion. Include the equation at the first index beyond the support.

Method. The original probabilities are \(p_0=1/8\), \(p_1=3/8\), \(p_2=3/8\), and \(p_3=1/8\). Shifting gives \(q_1=1/8\), \(q_2=3/8\), \(q_3=3/8\), \(q_4=1/8\), and \(q_5=0\). The ratios within the positive support are \(q_2/q_1=3\), \(q_3/q_2=1\), and \(q_4/q_3=1/3\).

If the first two ratios followed \(A+B/n\), then \(A+B/2=3\) and \(A+B/3=1\). Subtraction gives \(B/6=2\), so \(B=12\) and \(A=-3\). At \(n=4\), the predicted coefficient is \(-3+12/4=0\). The recursion would imply \(q_4=0\cdot q_3=0\), contradicting the actual value \(q_4=1/8\). Thus the shift already fails within its positive support.

There is also an essential endpoint check. Since the last positive probability is \(q_4\) and \(q_5=0\), a claimed recursion must hold at \(n=5\):

0=q_5=\left(A+\frac{B}{5}\right)q_4.

For \(A=-3\) and \(B=12\), the right side is \((-3+12/5)(1/8)=(-3/5)(1/8)=-3/40\), not zero. In a finite-support sequence ending at \(m\), it is not enough to check ratios only where both adjacent probabilities are positive. The recursion must also hold at \(m+1\), requiring \(A+B/(m+1)=0\) when \(q_m>0\).

Worked Example: A Geometric Shift That Does Remain in the Class

Suppose a hypothetical claim count has \(P(N=n)=2^{-(n+1)}\) for \(n\geq0\), a geometric distribution on the nonnegative integers. Compare \(N\mid N>0\) with \(N+1\), and identify the recursion for the shifted count.

Method. Here \(p_0=1/2\), so \(P(N>0)=1/2\). For each \(n\geq1\), zero truncation gives

r_n=\frac{2^{-(n+1)}}{1/2}=2^{-n}.

Shifting gives \(q_n=p_{n-1}=2^{-n}\) for \(n\geq1\) as well. Thus the two resulting distributions agree in this special case. Their probabilities sum to one because \(\sum_{n=1}^{\infty}2^{-n}=1\), and each adjacent ratio is \(q_n/q_{n-1}=1/2\) for \(n\geq2\). The shifted count satisfies the \((A,B,1)\) recursion with \(A=1/2\) and \(B=0\). This special outcome follows from the constant original ratios; it is not a general property of shifting.

Finite Support, Traps, and Practical Checks

For a finite-support count, recursion membership is an equation check, not merely a comparison of ratios. Suppose probabilities are positive through index \(m\) and zero thereafter. A proposed \((a,b,0)\) recursion must hold from \(n=1\) through \(n=m+1\); a proposed \((A,B,1)\) recursion must hold from \(n=2\) through \(n=m+1\). At the endpoint, the equation is

0=p_{m+1}=\left(a+\frac{b}{m+1}\right)p_m \quad\text{or}\quad 0=p_{m+1}=\left(A+\frac{B}{m+1}\right)p_m,

as appropriate. Since \(p_m>0\), the coefficient at that endpoint must be zero. For \(n>m+1\), both adjacent probabilities are zero, so the recursion equation is automatically satisfied. Checking only ratios between two positive probabilities omits the endpoint and can falsely accept a finite-support sequence.

A common exam trap is to write \(q_n/q_{n-1}=a+b/n\) immediately after defining \(q_n=p_{n-1}\). That expression uses the wrong original index: the ratio is \(p_{n-1}/p_{n-2}\), so the correct coefficient is \(a+b/(n-1)\). Another trap is to conclude that a sequence is outside the class only because its positive-support ratios do not fit, while overlooking the separate endpoint equation in a finite-support test.

Write the new probabilities.
Use \(r_n=p_n/(1-p_0)\) for zero truncation and \(q_n=p_{n-1}\) for shifting.

Form ratios with their indices visible.
For \(n\geq2\), zero truncation uses the original ratio at \(n\); shifting uses it at \(n-1\).

Test one fixed pair of recursion parameters.
Check every required index, not just a selected pair of adjacent probabilities.

Check the finite endpoint.
If support ends at \(m\), substitute \(n=m+1\) into the recursion and verify that the resulting probability is zero.

The distinction is useful whenever a model conditions on observing at least one claim, or when a count is deliberately increased by one. Conditioning changes the probability scale but leaves each positive count at its original index. Adding one changes the count value and shifts every probability to a new index. Recognizing which operation the scenario describes prevents errors in both the probabilities and any subsequent Panjer classification.

Key takeaway: Zero truncation preserves the original adjacent-probability ratios at indices \(n\geq2\). Shifting replaces the original ratio at \(n\) with the one at \(n-1\), so a Panjer form need not be preserved. For finite support, always test the recursion one index beyond the last positive probability.

Check Your Understanding

For each problem, distinguish the operation, calculate the requested ratios, and check any stated support endpoint.

A count has a Poisson distribution with mean \(3\). Find the ratio of the second to the first positive probability after zero truncation, then find \(q_2/q_1\) and \(q_3/q_2\) for the shifted count. (Answer: \(r_2/r_1=3/2\); \(q_2/q_1=3\) and \(q_3/q_2=3/2\).)
Let \(N\sim\operatorname{Binomial}(4,1/3)\), and define \(q_n=P(N+1=n)\). Give \(q_1\), the ratios \(q_2/q_1,\ q_3/q_2,\ q_4/q_3,\ q_5/q_4\), and the endpoint condition at \(n=6\). (Answer: \(q_1=16/81\); the ratios are \(2,\ 3/4,\ 1/3,\ 1/8\); since \(q_5=1/81\) and \(q_6=0\), a recursion requires \(A+B/6=0\).)
Let \(N\sim\operatorname{Binomial}(2,1/2)\) and \(M=N+1\). Using \(q_2/q_1=2\) and \(q_3/q_2=1/2\), solve for \(A,B\) if \(M\) were to satisfy \(q_n=(A+B/n)q_{n-1}\). Then check the recursion at the endpoint \(n=4\). (Answer: \(A=-5/2,\ B=9\); the endpoint coefficient is \(A+B/4=-1/4\), so the recursion predicts \(q_4=-1/16\), not \(0\), and the shifted count is not in the \((A,B,1)\) class.)
A count has probabilities \(p_n=(1-r)r^n\) for \(n\geq0\), where \(0<r<1\). State the adjacent ratio for the shifted count at \(n\geq2\), and identify one valid \((A,B,1)\) representation. (Answer: \(q_n/q_{n-1}=r\); \(A=r,\ B=0\).)
An original count has positive adjacent ratios \(p_n/p_{n-1}=a+b/n\) for every \(n\geq1\). For its shift, state the ratio at index \(n\geq2\), and explain why it cannot equal \(A+B/n\) at every such index when \(b\ne0\). (Answer: \(q_n/q_{n-1}=a+b/(n-1)\); equality for all \(n\) would require \(bn=B(n-1)\) for all \(n\), which is impossible when \(b\ne0\).)

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