Zero-Modified Frequency Distributions

Why Modify the Probability of Zero?

A frequency model may describe the relative pattern of positive counts reasonably well but assign a probability of zero that does not suit a particular application. For example, a hypothetical portfolio might contain many policies known to have at least one reported claim, or its observed zero-claim frequency might differ from that predicted by a baseline model. A zero-modified model changes the probability of zero while preserving the baseline model’s proportions among positive counts.

This is different from simply discarding zero-count observations. In zero truncation, the count is conditioned to be positive, so the modified model assigns probability zero to a zero count. Zero modification instead allows any chosen zero probability below one and rescales the positive probabilities to make the total equal one.

Definition: Let $N$ have baseline probabilities $p_n=P(N=n)$, for $n=0,1,2,\ldots$, with $p_0<1$. Choose a new zero probability $\pi$, where $0\leq \pi<1$, and define the zero-modified count $N^*$ by $$ P(N^*=0)=\pi,\qquad P(N^*=n)=q_n=c\,p_n\quad(n\geq 1), \qquad c=\frac{1-\pi}{1-p_0}. $$ The factor $c$ rescales all positive-count probabilities by the same amount.

The restriction $p_0<1$ ensures that the baseline model has some positive-count probability to rescale. The restriction $\pi<1$ ensures that the modified model also has positive probability of a positive count. If $\pi=p_0$, then $c=1$ and the distribution is unchanged. If $\pi>p_0$, the positive probabilities shrink; if $\pi<p_0$, they increase.

Conditional Positive Counts and the Panjer Recursion

The defining rescaling leaves the distribution conditional on a positive count unchanged:

P(N^*=n\mid N^*>0) =\frac{q_n}{1-\pi} =\frac{p_n}{1-p_0}, \qquad n\geq 1.

Thus, zero modification changes how often the count is zero, but not the relative pattern of counts $1,2,\ldots$ among positive outcomes. This gives an immediate result for Panjer recursions.

Theorem: Suppose the baseline positive probabilities satisfy $$ p_n=\left(a+\frac{b}{n}\right)p_{n-1},\qquad n\geq 2. $$ Then the zero-modified probabilities satisfy the same recursion with the same $a$ and $b$: $$ q_n=\left(a+\frac{b}{n}\right)q_{n-1},\qquad n\geq 2. $$ Conversely, because $q_n=c\,p_n$ for every $n\geq1$ with $c>0$, the modified positive probabilities satisfy this recursion if and only if the baseline positive probabilities do.

This is precisely the positive-count recursion associated with the $(a,b,1)$ class. The boundary probability $q_1$ is specified separately in that class; the recursion links $q_2,q_3,\ldots$ to $q_1$, not $q_1$ to $q_0$. The result concerns positive-count recursion and does not, by itself, establish the $(a,b,0)$ boundary relation.

Why the Recursion Is Preserved

The justification is the cancellation of a common positive scaling factor. For $n\geq2$, both $q_n$ and $q_{n-1}$ are positive-count probabilities, so each is multiplied by $c$. Therefore:

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

In ratio form, when the denominator is positive, this says $q_n/q_{n-1}=p_n/p_{n-1}$. The altered probability $q_0=\pi$ does not enter this calculation. That independence from zero is why positive-count recursion can survive even when the probability at zero changes substantially.

The probability generating function also follows directly. If $G_N(s)=E[s^N]$ is the baseline probability generating function, then the zero-modified generating function is:

G_{N^*}(s)=\pi+c\bigl(G_N(s)-p_0\bigr).

Indeed, $G_N(s)-p_0$ contains exactly the terms for counts $n\geq1$, and multiplying them by $c$ produces the modified positive probabilities. Differentiating at $s=1$, when the relevant moments are finite, gives $E[N^*]=cE[N]$. More generally, for any integer $k\geq1$, $E[(N^*)^k]=cE[N^k]$, because the zero term contributes zero to that moment.

Worked Example: Modifying a Geometric Count

A hypothetical count $N$ has the geometric distribution on $0,1,2,\ldots$ with $P(N=n)=(1-r)r^n$ and $r=1/2$. Construct a zero-modified model with $P(N^*=0)=0.20$. Find $q_1,q_2,q_3$, identify the positive-count recursion, and calculate the modified mean and variance.

Method. First find the baseline zero probability and rescaling factor. Then apply the same factor to each positive probability. This avoids the common mistake of using the zero-truncated probabilities without adjusting them to the requested zero probability.

The baseline probabilities begin $p_0=1/2$, $p_1=(1/2)(1/2)=1/4$, $p_2=(1/2)(1/2)^2=1/8$, and $p_3=(1/2)(1/2)^3=1/16$. The positive baseline probability is $1-p_0=1/2$. Hence $c=(1-0.20)/(1-0.50)=0.80/0.50=1.60$.

Therefore $q_0=0.20$, $q_1=1.60(0.25)=0.40$, $q_2=1.60(0.125)=0.20$, and $q_3=1.60(0.0625)=0.10$. Each successive positive probability is half the preceding one, so $q_n=(1/2)q_{n-1}$ for $n\geq2$. The positive recursion has $a=1/2$ and $b=0$, the same as the baseline positive recursion. The probabilities shown sum to $0.20+0.40+0.20+0.10=0.90$; the remaining positive counts have total probability $0.10$, as also follows from $c(1-p_0)=1-0.20=0.80$.

For this geometric model, $E[N]=r/(1-r)=1$ and $\operatorname{Var}(N)=r/(1-r)^2=2$, so $E[N^2]=2+1^2=3$. Thus $E[N^*]=cE[N]=1.60$ and $E[(N^*)^2]=cE[N^2]=4.80$. It follows that $\operatorname{Var}(N^*)=4.80-(1.60)^2=2.24$. These results check the moment calculations against the zero-modification formulas.

Worked Examples with Other Baseline Models

Worked Example: Modifying a Negative Binomial Count

Let the baseline count have a negative binomial distribution with $r=2$ successes required and success probability $p=1/2$, using $p_n=\binom{n+r-1}{n}p^r(1-p)^n$. Set the modified zero probability to $\pi=0.40$. Find $q_0,\ldots,q_4$ and the recursion parameters for the positive counts.

Method. Compute the baseline probabilities, find the single scaling factor, and then check consecutive positive-probability ratios. This avoids trying to infer the recursion from $q_0$, which is not part of the $(a,b,1)$ recursion.

The baseline zero probability is $p_0=(1/2)^2=0.25$. The next four probabilities are $p_1=2(1/2)^2(1/2)=0.25$, $p_2=3(1/2)^2(1/2)^2=0.1875$, $p_3=4(1/2)^2(1/2)^3=0.125$, and $p_4=5(1/2)^2(1/2)^4=0.078125$. The rescaling factor is $c=(1-0.40)/(1-0.25)=0.60/0.75=0.80$.

Thus $q_0=0.40$, $q_1=0.80(0.25)=0.20$, $q_2=0.80(0.1875)=0.15$, $q_3=0.80(0.125)=0.10$, and $q_4=0.80(0.078125)=0.0625$. For this negative binomial parameterization, $$ \frac{p_n}{p_{n-1}} =\frac{n+1}{n}\left(\frac12\right) =\frac12+\frac{1}{2n},\qquad n\geq1. $$ For $n\geq2$, the modified probabilities have the same ratios, so $a=1/2$ and $b=1/2$. For example, $q_2/q_1=0.15/0.20=0.75=0.5+0.5/2$, and $q_3/q_2=0.10/0.15=2/3=0.5+0.5/3$. The construction assigns total probability $0.60$ to positive counts, as required, since $c(1-p_0)=0.80(0.75)=0.60$.

Worked Example: A Modified Poisson Count and Its Boundary

A baseline count is Poisson with mean $\lambda=1$. A hypothetical adjustment sets the probability of zero to $0.50$. Find $q_1,q_2,q_3$ to four decimal places, give the positive-count recursion parameters, and determine whether the modified probabilities satisfy the same $(a,b,0)$ relation as the baseline Poisson model.

Method. Use the zero-modification factor for the probabilities, then test the separate zero boundary condition. The common mistake is to assume that preserving the positive recursion also preserves the relation between the probabilities at zero and one.

For a Poisson count, $p_0=e^{-1}$ and $p_n=e^{-1}/n!$. The scaling factor is $c=0.50/(1-e^{-1})$. Therefore $q_1=ce^{-1}=0.50/(e-1)\approx0.2910$. The positive recursion for a Poisson model is $p_n=(1/n)p_{n-1}$; hence $a=0$ and $b=1$. It follows that $q_2=q_1/2\approx0.1455$ and $q_3=q_2/3\approx0.0485$, to four decimal places.

For the baseline Poisson distribution, the $(a,b,0)$ boundary relation is $p_1=(a+b)p_0=p_0$. With the same $a=0,b=1$, the modified relation would require $q_1=q_0=0.50$, but $q_1\approx0.2910$. It fails. The positive-count $(a,b,1)$ recursion is preserved, but the same $(a,b,0)$ boundary condition is not. As a probability check, the positive mass is $c(1-p_0)=0.50$, which combines with $q_0=0.50$ to give one.

When Does the Zero Boundary Relation Survive?

For a model in the $(a,b,0)$ class, the recursion includes the boundary relation $p_1=(a+b)p_0$, as well as the positive-count recursion for later terms. If zero modification changes only $p_0$ and rescales the positive probabilities, the inherited positive recursion still holds. But for the modified distribution to satisfy the same $(a,b,0)$ boundary relation, it must also have $q_1=(a+b)q_0$.

Using $q_1=cp_1=c(a+b)p_0$, this boundary condition reduces to $cp_0=\pi$. Since $c=(1-\pi)/(1-p_0)$, it holds exactly when $\pi=p_0$. Thus, for a baseline $(a,b,0)$ model, a genuine change in the zero probability generally leaves the positive $(a,b,1)$ recursion intact but breaks the same $(a,b,0)$ boundary relation. This statement concerns the inherited parameters; it does not rule out other possible parameter descriptions.

Check the baseline zero probability.
Record $p_0$ and verify that $p_0<1$.

Calculate the positive scaling factor.
Use $c=(1-\pi)/(1-p_0)$, not $1/(1-p_0)$ unless the modified zero probability is zero.

Apply the factor to positive probabilities.
For every $n\geq1$, set $q_n=cp_n$, while setting $q_0=\pi$.

Check recursion and boundary separately.
Ratios among positive probabilities determine the $(a,b,1)$ recursion; the relation involving $q_0$ must be checked independently for $(a,b,0)$.

Common Exam Trap

The principal trap is to confuse zero modification with zero truncation. For zero truncation, the positive probabilities are $p_n/(1-p_0)$, and the probability at zero is exactly zero. For zero modification with $\pi>0$, the correct positive probabilities are $(1-\pi)p_n/(1-p_0)$. For instance, in the geometric example above, the zero-truncated value of $P(N=1)$ would be $0.25/0.50=0.50$, but the zero-modified value is $0.80(0.25)=0.40$, because positive counts must total $0.80$, not one.

A second trap is to treat the preserved positive recursion as proof that the entire modified distribution remains in the original $(a,b,0)$ class. The recursion for $n\geq2$ says nothing by itself about $q_1$ relative to $q_0$. State which recursion is being checked and verify its boundary condition explicitly.

Key takeaway: Zero modification assigns $q_0=\pi$ and multiplies every positive baseline probability by $c=(1-\pi)/(1-p_0)$. Because this multiplier is common to all positive terms, their $(a,b,1)$ recursion is preserved exactly; the zero boundary must be checked separately.

Check Your Understanding

For each problem, identify the scaling or recursion step before giving the requested result.

A baseline geometric count has $p_n=(1-0.4)(0.4)^n$ for $n\geq0$. Its zero probability is modified to $0.30$. Find $c$, $q_1$, and $q_2$, and state the positive-count recursion parameters. (Answer: $c=1.75$, $q_1=0.42$, $q_2=0.168$, $a=0.4,\ b=0$.)
A baseline Poisson count has mean $2$, and the modified zero probability is $0.20$. Find the modified probability of exactly one count to four decimal places. (Answer: $q_1=1.6/(e^2-1)\approx0.2504$.)
A baseline negative binomial count has $r=3$ successes required and success probability $p=1/2$. Its zero probability is modified to $0.50$. Find $c$, $q_1$, $q_2$, and the positive-count recursion parameters. (Answer: $c=4/7$, $q_1=q_2=3/28\approx0.1071$, $a=1/2,\ b=1$.)
A baseline $(a,b,0)$ distribution has $p_0=0.30$, and its zero probability is modified to $0.60$. Does it satisfy the same $(a,b,0)$ boundary relation? Give $q_1/q_0$ in terms of $a+b$. (Answer: No; $q_1/q_0=(2/7)(a+b)$, not $a+b$ in general.)
A baseline $(a,b,0)$ distribution has $p_0=0.20$. What modified zero probability is required to preserve its same $(a,b,0)$ boundary relation after the positive probabilities are rescaled? (Answer: $\pi=0.20$, so the distribution is unchanged.)

What You'll Learn