Compound Bernoulli Models

Why a Bernoulli Claim Count Produces a Mixture

Some hypothetical coverages can have at most one claim during a specified period. For example, a contract might cover one scheduled piece of equipment against a single specified event. If a claim occurs, its amount is random; if no claim occurs, the aggregate loss is zero. The resulting aggregate distribution combines a probability mass at zero with the severity distribution for the claim.

The key distinction is that the claim probability scales the probabilities of the positive claim amounts. It does not change their relative proportions. If the severity is strictly positive whenever a claim occurs, the entire probability of zero aggregate loss comes from no claim occurring.

Definition: Let $N\sim\operatorname{Bernoulli}(p)$, where $0\le p\le1$, so $P(N=1)=p$ and $P(N=0)=1-p$. Let $X$ be the claim severity, independent of $N$, and assume $X\ge0$. The aggregate loss is $S=\sum_{j=1}^{N}X_j$, with an empty sum defined as zero. Since $N$ can only be zero or one, $S=0$ when $N=0$ and $S=X_1$ when $N=1$.

Here $F_X(x)=P(X\le x)$ denotes the severity distribution function. The severity is assumed to have the same distribution whenever a claim occurs; because there is at most one claim, no assumption about dependence among multiple severities is needed. If severity is strictly positive, then $P(X=0)=0$. If zero severity is possible, it must be accounted for separately because a claim can occur without producing a positive aggregate loss.

Distribution of the Aggregate Loss

Conditioning on the two possible claim counts gives the aggregate distribution. For $s<0$, $P(S\le s)=0$, since aggregate losses are nonnegative. For $s\ge0$, the event $S\le s$ occurs either because $N=0$, or because $N=1$ and $X\le s$. These mutually exclusive cases yield

F_S(s)=P(S\le s)= \begin{cases} 0, & s<0,\\ (1-p)+pF_X(s), & s\ge0. \end{cases}

This expression includes the probability of no claim and the probability of a claim whose severity is at most $s$. In particular, at zero it gives

$$ P(S=0)=F_S(0)=1-p+pF_X(0). $$

When $X$ is strictly positive, $F_X(0)=0$, so $P(S=0)=1-p$. For any positive severity value $x>0$, the aggregate has the same point probability as the severity multiplied by the probability of a claim:

P(S=x)=pP(X=x),\qquad x>0.

If severity is continuous and strictly positive, the aggregate distribution has an atom of size $1-p$ at zero and a continuous component for positive losses. Where $f_X$ exists, the density on $s>0$ is $f_S(s)=pf_X(s)$. The density integrates to $p$, not one: the remaining probability, $1-p$, is concentrated at zero. Consequently, this is a mixed distribution, not an ordinary continuous distribution over all possible aggregate losses.

Theorem: If $X>0$ with probability one, then the aggregate distribution consists of an atom $1-p$ at zero and a positive-loss component with total probability $p$. Conditional on $S>0$, the distribution of $S$ is exactly the severity distribution $F_X$.

The conditional statement follows because $S>0$ is equivalent to $N=1$ when severity is strictly positive. Given that a claim occurred, the aggregate loss is the claim severity itself. Thus, for $s\ge0$, $P(S\le s\mid S>0)=F_X(s)$.

Why the Mixture Formula Holds

The formula can be derived directly by partitioning on the claim count. For $s\ge0$, the two events $N=0$ and $N=1$ are disjoint and together cover all possibilities. When $N=0$, the aggregate is zero and therefore no greater than $s$. When $N=1$, the aggregate equals $X$.

Split by claim count.
Write $P(S\le s)=P(S\le s,N=0)+P(S\le s,N=1)$.

Evaluate the no-claim case.
For $s\ge0$, $S=0$ whenever $N=0$, so $P(S\le s,N=0)=P(N=0)=1-p$.

Evaluate the claim case.
When $N=1$, $S=X$. Independence gives $P(S\le s,N=1)=P(X\le s,N=1)=F_X(s)p$.

Combine the cases.
Adding the disjoint probabilities gives $F_S(s)=(1-p)+pF_X(s)$ for $s\ge0$.

This derivation also shows why the zero-loss probability needs care. If $X$ can equal zero, the claim case can contribute to $S=0$ as well as the no-claim case. In that setting the zero probability is $1-p+pP(X=0)$, not just $1-p$.

Worked Examples

Worked Example: A Discrete Severity Distribution

A hypothetical one-event coverage has probability $p=0.40$ of a claim during the year. Conditional on a claim, the payment $X$ is \$1,000 with probability $0.50$, \$3,000 with probability $0.30$, and \$5,000 with probability $0.20$. Find the aggregate distribution and $P(S\le \$3{,}000)$.

Method. Multiply each claim-severity probability by $0.40$, and assign the no-claim probability $1-0.40$ to zero. The severity values are all positive, so the no-claim case is the only way to have zero aggregate loss.

The probability of no claim is $1-0.40=0.60$. The positive aggregate probabilities are

\begin{aligned} P(S=\$1{,}000)&=0.40(0.50)=0.20,\\ P(S=\$3{,}000)&=0.40(0.30)=0.12,\\ P(S=\$5{,}000)&=0.40(0.20)=0.08. \end{aligned}

Therefore the distribution assigns probabilities $0.60, 0.20, 0.12,$ and $0.08$ to aggregate losses of \$0, \$1,000, \$3,000, and \$5,000, respectively. The requested probability includes the zero, \$1,000, and \$3,000 outcomes:

P(S\le \$3{,}000)=0.60+0.20+0.12=0.92.

Check. The probabilities sum to $0.60+0.20+0.12+0.08=1.00$. A common mistake is to report $0.30$, the severity probability at \$3,000, as the aggregate probability at that amount. The correct aggregate probability is $0.40(0.30)=0.12$, because a claim must first occur.

Worked Example: A Continuous Exponential Severity

A hypothetical policy has probability $p=0.25$ of one claim. Given a claim, severity $X$ is exponential with mean \$1,000, so its rate is $\lambda=1/1000$ per dollar and $F_X(x)=1-e^{-x/1000}$ for $x\ge0$. Find the aggregate probability of no loss, $P(S\le \$1{,}200)$, and $P(\$500<S\le \$2{,}000)$.

Method. Use the zero atom for no claim and scale the exponential severity probabilities by $p$ for positive losses. Since exponential severity is strictly positive with probability one, $P(S=0)=1-p=0.75$.

At \$1,200, the aggregate CDF is

\begin{aligned} P(S\le \$1{,}200) &=0.75+0.25F_X(1200)\\ &=0.75+0.25(1-e^{-1.2})\\ &=0.75+0.25(0.6988058)\\ &=0.9247015\\ &\approx 0.9247. \end{aligned}

For the interval from \$500 to \$2,000, zero is not included. The probability is the claim probability multiplied by the severity probability in that interval:

\begin{aligned} P(\$500<S\le \$2{,}000) &=0.25\bigl(F_X(2000)-F_X(500)\bigr)\\ &=0.25\bigl((1-e^{-2})-(1-e^{-0.5})\bigr)\\ &=0.25(e^{-0.5}-e^{-2})\\ &=0.25(0.6065307-0.1353353)\\ &=0.1177989\\ &\approx 0.1178. \end{aligned}

Check. The probability below \$1,200 includes the entire zero atom, while the interval probability does not. The zero probability is $0.7500$, less than the CDF value $0.9247$, as required. Probabilities are rounded to four decimal places.

Worked Example: Severity Can Equal Zero

A hypothetical reporting system records a claim with probability $p=0.60$. Conditional on a recorded claim, the reported severity is \$0 with probability $0.20$, \$2,000 with probability $0.50$, and \$5,000 with probability $0.30$. Find the aggregate distribution and the probability of a positive aggregate loss.

Method. Include both ways to produce zero aggregate loss: no claim, and a claim whose reported severity is zero. The no-claim probability is $1-0.60=0.40$. The claim-with-zero-severity probability is $0.60(0.20)=0.12$. Thus

$$ P(S=0)=0.40+0.12=0.52. $$

For the positive outcomes, scale their conditional severity probabilities by $0.60$:

\begin{aligned} P(S=\$2{,}000)&=0.60(0.50)=0.30,\\ P(S=\$5{,}000)&=0.60(0.30)=0.18. \end{aligned}

The probability of a positive aggregate loss is therefore $0.30+0.18=0.48$, equivalently $1-P(S=0)=1-0.52=0.48$. Check. The three aggregate probabilities sum to $0.52+0.30+0.18=1.00$. The common mistake is to use $1-p=0.40$ as the zero-loss probability even though the severity can itself be zero; the correction is to add $pP(X=0)$.

Common Interpretation Traps

The Bernoulli model is simple, but its atom and mixture weights create several recurring errors:

Confusing severity probabilities with aggregate probabilities. For a positive discrete amount $x$, use $P(S=x)=pP(X=x)$, not $P(X=x)$.
Omitting the no-claim atom from a CDF. When $X>0$, every $s\ge0$ includes probability $1-p$ from no claim. It must be added to $pF_X(s)$.
Treating the continuous density as a full probability density. If severity is continuous and positive, the aggregate density integrates to $p$. The remaining probability $1-p$ is an atom at zero.
Assuming zero aggregate loss always means no claim. This is valid only when $P(X=0)=0$. Otherwise a zero-severity claim also contributes probability mass at zero.
Using the model when more than one claim can occur. A Bernoulli count permits only zero or one claim. If the count can exceed one, the aggregate can involve sums of severities and this distribution no longer describes it.

Key takeaway: For a Bernoulli claim count with probability $p$ and independent nonnegative severity $X$, the aggregate is zero when no claim occurs and has severity distribution $F_X$ when a claim occurs. If severity is strictly positive, the zero atom is $1-p$, and every positive severity probability or density is multiplied by $p$.

Check Your Understanding

For each problem, identify the zero-loss probability before calculating any positive-loss probabilities.

A hypothetical coverage has claim probability $0.30$. Given a claim, payment is \$100 with probability $0.70$ and \$400 with probability $0.30$. Find the aggregate probabilities at \$0, \$100, and \$400. (Answer: $0.70,\ 0.21,\ 0.09$, respectively.)
A policy has claim probability $0.20$, and conditional severity is exponential with mean \$500. Find $P(S\le \$500)$. (Answer: $0.9264$.)
A reporting model has claim probability $0.50$; conditional severity is \$0 with probability $0.10$, \$2,000 with probability $0.60$, and \$10,000 with probability $0.30$. Find the aggregate distribution. (Answer: $P(S=0)=0.55,\ P(S=\$2{,}000)=0.30,\ P(S=\$10{,}000)=0.15$.)
A coverage has claim probability $0.40$, and strictly positive severity is \$1, \$2, or \$4 with probabilities $0.20,\ 0.50,$ and $0.30$. Find the probability that aggregate loss is at least \$2. (Answer: $0.32$.)
A policy has claim probability $0.35$, and conditional severity is exponential with rate $1/1000$ per dollar. Find $P(0<S\le \$1{,}000)$. (Answer: $0.2212$.)

What You'll Learn