Conditional Claim Counts Given At Least One Claim

Why Condition on a Positive Claim Count?

A portfolio analysis may focus only on policies that produced at least one claim. The resulting claim count is no longer distributed like the original count: zero-claim policies have been excluded, and probabilities for positive counts must be rescaled. This tutorial develops that adjustment for binomial, Poisson, and negative binomial models.

Let $N$ be a nonnegative integer-valued claim count. The conditioning event is $N\geq 1$, which has probability $1-P(N=0)$. The conditional distribution describes the count among the cases for which this event occurred. It is sometimes called a zero-truncated distribution.

Definition: For any count model with $P(N=0)<1$, the count conditional on at least one claim has probabilities $$ P(N=k\mid N\geq 1)=\frac{P(N=k)}{1-P(N=0)},\qquad k=1,2,\ldots. $$ The denominator is the probability that the conditioning event occurs. The conditional distribution assigns probability zero to $N=0$.

The same adjustment applies to events involving several positive counts. For any integer $m\geq 1$,

P(N\geq m\mid N\geq 1)=\frac{P(N\geq m)}{1-P(N=0)}.

The numerator is already contained in the conditioning event, so no additional subtraction is needed. For an interval of positive counts, divide the original probability of that interval by $1-P(N=0)$.

Conditional Mean: A Short Derivation

The conditional mean has a particularly direct form. Because $N=0$ contributes nothing to the unconditional mean, the expected count on the positive-count cases is the unconditional expected count divided by the chance of being in those cases.

\begin{aligned} E[N\mid N\geq 1] &=\sum_{k=1}^{\infty} kP(N=k\mid N\geq 1)\\ &=\sum_{k=1}^{\infty} k\frac{P(N=k)}{1-P(N=0)}\\ &=\frac{\sum_{k=1}^{\infty} kP(N=k)}{1-P(N=0)} =\frac{E[N]}{1-P(N=0)}. \end{aligned}

This argument requires a finite mean and $P(N\geq 1)>0$. It also explains why conditioning on at least one claim raises the mean: when $P(N=0)>0$, the denominator is less than one, while the numerator remains the original mean.

Formulas for the Three Frequency Models

Use the model’s own parameters to find both its unconditional mean and its zero-count probability. Then substitute those values into the conditional formulas. The models below use these conventions.

Binomial: $N\sim\operatorname{Bin}(n,\pi)$ counts successes in $n$ independent trials, each with success probability $\pi$. Thus $E[N]=n\pi$ and $P(N=0)=(1-\pi)^n$.
Poisson: $N\sim\operatorname{Poisson}(\lambda)$, where $\lambda>0$ is the mean count. Thus $E[N]=\lambda$ and $P(N=0)=e^{-\lambda}$.
Negative binomial: Here $N$ counts failures before the $r$th success in independent trials with success probability $p$, where $r$ is a positive integer and $0<p<1$. Thus $P(N=k)=\binom{k+r-1}{k}p^r(1-p)^k$, $P(N=0)=p^r$, and $E[N]=r(1-p)/p$.

For each model, the conditional point probability is its ordinary point probability divided by $1-P(N=0)$. In particular, the conditional means are

\begin{aligned} N\sim\operatorname{Bin}(n,\pi):\quad &E[N\mid N\geq 1]=\frac{n\pi}{1-(1-\pi)^n},\\ N\sim\operatorname{Poisson}(\lambda):\quad &E[N\mid N\geq 1]=\frac{\lambda}{1-e^{-\lambda}},\\ N\text{ negative binomial}(r,p):\quad &E[N\mid N\geq 1]=\frac{r(1-p)/p}{1-p^r}. \end{aligned}

These formulas apply only when the positive-count event has positive probability. For the binomial model this requires $n\geq 1$ and $\pi>0$; the stated Poisson and negative binomial parameter ranges also give positive probability of a claim.

Worked Example: Binomial Counts Among Policies With Claims

A hypothetical book has a fixed group of 8 independent policy exposures. Each exposure has probability $\pi=0.20$ of producing one claim, and can produce at most one claim in the period. Let $N\sim\operatorname{Bin}(8,0.20)$. Find (a) the probability of exactly one claim given at least one, (b) the probability of at least two claims given at least one, and (c) the conditional mean.

Method. Calculate the zero probability first, then use it as the common conditioning denominator. The model has $n=8$, $\pi=0.20$, and $1-\pi=0.80$.

The probability of no claims and the probability of at least one claim are

P(N=0)=0.80^8=0.16777216,\qquad P(N\geq 1)=1-0.16777216=0.83222784.

For exactly one claim,

P(N=1)=\binom{8}{1}(0.20)(0.80)^7 =8(0.20)(0.2097152)=0.33554432.

Therefore,

P(N=1\mid N\geq 1)=\frac{0.33554432}{0.83222784}=0.4032.

The unconditional probability of at least two claims is one minus the zero- and one-claim probabilities:

P(N\geq 2)=1-0.16777216-0.33554432=0.49668352.

Since $N\geq 2$ implies $N\geq 1$,

P(N\geq 2\mid N\geq 1)=\frac{0.49668352}{0.83222784}=0.5968.

Finally, $E[N]=n\pi=8(0.20)=1.6$. Hence

E[N\mid N\geq 1]=\frac{1.6}{0.83222784}=1.9226.

As a check, the conditional probabilities of exactly one and at least two claims sum to $0.4032+0.5968=1.0000$. The conditional mean exceeds the unconditional mean of 1.6, as expected after zero counts are removed. The common mistake avoided here is using the unconditional probability $P(N\geq 2)$ as the conditional answer without dividing by $P(N\geq 1)$.

Worked Example: A Poisson Count Conditioned on a Claim

Suppose a hypothetical portfolio segment has a Poisson claim count with mean $\lambda=1$ per period. Given that at least one claim occurs, calculate the probability of three or more claims and the expected claim count.

Method. Use $N\sim\operatorname{Poisson}(1)$, so $P(N=k)=e^{-1}/k!$, $P(N=0)=e^{-1}$, and $E[N]=1$. The conditioning probability is

P(N\geq 1)=1-e^{-1}=1-0.36787944=0.63212056.

To find the probability of three or more claims, first calculate the probabilities for zero, one, and two claims:

P(N=0)=0.36787944,\quad P(N=1)=0.36787944,\quad P(N=2)=\frac{e^{-1}}{2}=0.18393972.

Thus the unconditional tail probability is

P(N\geq 3)=1-(0.36787944+0.36787944+0.18393972)=0.08030140.

Because $N\geq 3$ is contained in $N\geq 1$, divide by $0.63212056$:

P(N\geq 3\mid N\geq 1)=\frac{0.08030140}{0.63212056}=0.1270.

The conditional mean is

E[N\mid N\geq 1]=\frac{1}{0.63212056}=1.5820.

Both results are plausible: some positive-count cases have exactly one claim, so the conditional mean is above one, while three-or-more claims account for a smaller portion of the positive-count cases. The method avoids treating the unconditional Poisson mean as the mean after zero-count cases have been excluded.

Worked Example: Negative Binomial Counts With a Claim

In a hypothetical model, $N$ counts failures before the second success in independent trials, with success probability $p=0.50$. Given $N\geq 1$, find the probability of at least two failures and the expected number of failures.

Method. This is the negative binomial convention with $r=2$ and $p=0.50$. In particular, $N=0$ means both required successes occur without any failures. The zero-count probability and positive-count probability are

P(N=0)=p^r=0.50^2=0.25,\qquad P(N\geq 1)=1-0.25=0.75.

The probability of exactly one failure is

P(N=1)=\binom{1+2-1}{1}(0.50)^2(0.50)^1 =2(0.25)(0.50)=0.25.

Therefore, the unconditional probability of at least two failures is

P(N\geq 2)=1-P(N=0)-P(N=1)=1-0.25-0.25=0.50.

Conditioning gives

P(N\geq 2\mid N\geq 1)=\frac{0.50}{0.75}=\frac{2}{3}=0.6667.

The unconditional mean is $r(1-p)/p=2(0.50)/(0.50)=2$. Consequently,

E[N\mid N\geq 1]=\frac{2}{0.75}=2.6667.

As a check, $P(N=1\mid N\geq 1)=0.25/0.75=1/3$, and $1/3+2/3=1$. The calculation depends on the stated negative binomial convention: here the count is failures before $r$ successes, not the number of trials needed to reach the $r$th success.

Common Exam Trap: The Denominator Must Match the Condition

A conditional probability is not generally the same as its unconditional numerator. For example, in the Poisson example above, a tempting but incorrect calculation is

P(N\geq 3\mid N\geq 1)\ \color{#000}{\text{(incorrect)}}=P(N\geq 3)=0.0803.

The event $N\geq 3$ is a subset of $N\geq 1$, but the probability must still be measured relative to the positive-count cases. The correct calculation is

P(N\geq 3\mid N\geq 1) =\frac{P(N\geq 3)}{P(N\geq 1)} =\frac{0.08030140}{0.63212056} =0.1270.

A related error is subtracting the zero probability from the numerator even when the target event already excludes zero. For $N\geq 3$, $P(N=0)$ is not part of the event, so the numerator is $P(N\geq 3)$, not $P(N\geq 3)-P(N=0)$.

Using the Conditional Model Correctly

Conditioning changes the distribution, not just the wording of a probability. When calculating any positive-count probability, apply the same conditioning denominator to the relevant original probability. When calculating the mean, divide the original mean by the probability of a positive count. These operations are valid for each of the three frequency models, but their zero probabilities and means differ.

This distinction matters when a question describes a group selected because it had at least one claim. Using the original mean would include zero-claim cases that are absent from the selected group. Similarly, an unconditioned tail probability answers a question about the full portfolio, not about the selected positive-count cases. This tutorial is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, condition on at least one claim only where the question specifies that condition. Answers are rounded to four decimal places.

A hypothetical group has $N\sim\operatorname{Bin}(5,0.10)$. Find $E[N\mid N\geq 1]$. (Answer: 1.2210)
A claim count has a Poisson distribution with $\lambda=2$. Find $P(N=1\mid N\geq 1)$. (Answer: 0.3130)
Let $N$ count failures before the third success, with success probability $p=0.60$. Find $E[N\mid N\geq 1]$. (Answer: 2.5510)
A hypothetical portfolio has $N\sim\operatorname{Bin}(6,0.25)$. Find $P(N\geq 2\mid N\geq 1)$. (Answer: 0.5670)
A claim count is Poisson with $\lambda=0.50$. Find $E[N\mid N\geq 1]$. (Answer: 1.2707)

What You'll Learn