Stop-Loss Pricing for Mixed Aggregate Distributions

Why Mixed Distributions Need Separate Calculations

An aggregate loss may have a continuous range of possible values and still assign positive probability to particular amounts. For example, a hypothetical portfolio might produce no claims with positive probability, while positive claim totals vary continuously in other cases. Such a distribution is mixed: it has both point masses and a continuous component.

For an aggregate loss $X$ and attachment $d$, the stop-loss payment is $(X-d)_+=\max(X-d,0)$. Its expected value depends on both parts of the distribution. Each atom above the attachment contributes its probability multiplied by its excess above the attachment. The continuous part contributes the integral of excess loss times its density. Keeping these contributions separate prevents a common error: treating a density that describes only one component as if it already included that component’s probability.

Definition: Suppose $X$ has point masses $p_j=P(X=x_j)$ at values $x_j$, and a continuous component with unconditional density $g(x)$. The density $g$ is a subdensity: its integral is the total probability assigned to the continuous component, which can be less than 1. If $E[(X-d)_+]$ is finite, the stop-loss premium at attachment $d$ is $$ \Pi(d)=E[(X-d)_+] =\sum_j p_j(x_j-d)_+ + \int_d^\infty (x-d)g(x)\,dx. $$

The sum includes every atom, but atoms at or below $d$ contribute zero. In particular, an atom at exactly $d$ does not generate a payment: its excess is $d-d=0$. The integral uses the unconditional subdensity. If a problem instead gives a conditional density $f_c$ for the continuous component and states that the component has probability $w$, use $g(x)=wf_c(x)$. Equivalently, multiply the integral calculated using $f_c$ by $w$.

All amounts in the worked examples are measured in thousands of dollars. For each example, the stated probabilities and component weights define the full distribution; no interest or mortality assumption is needed because the calculation is an expected payment on aggregate loss.

Why the Formula Holds

The formula follows by splitting the expectation according to whether $X$ comes from a point mass or from the continuous component. On the event that $X=x_j$, the payment is the fixed amount $(x_j-d)_+$, and that event has probability $p_j$. Its contribution to the expectation is therefore $p_j(x_j-d)_+$. On the continuous part, the payment is zero for $x\le d$ and is $x-d$ for $x>d$, so its expectation is the integral of $x-d$ against the subdensity above $d$. Adding the contributions accounts for the entire probability distribution.

E[(X-d)_+] =\sum_j E[(X-d)_+\mathbf{1}_{\{X=x_j\}}] +E[(X-d)_+\mathbf{1}_{\{X\text{ is in the continuous component}\}}].

For each atom, the first expectation equals $p_j(x_j-d)_+$. For the continuous component, the second equals $\int_d^\infty (x-d)g(x)\,dx$. Thus the decomposition gives the stated formula. It requires a finite expected stop-loss payment; if the positive tail is sufficiently heavy that this expectation is infinite, the premium is not finite.

Separate the Atoms and the Density

A reliable calculation starts by listing the atoms and the continuous component separately. Then compare every atom with the attachment, and integrate only over the portion of the continuous support above it. The following sequence makes the bookkeeping explicit.

Check the distribution.
Add the atom probabilities and continuous-component probability. The total must be 1.

Calculate atom payments.
For each atom $x_j$, include $p_j(x_j-d)$ only when $x_j>d$. An atom at $d$ contributes zero.

Calculate the continuous payment.
Integrate $(x-d)$ times the unconditional subdensity over its support above $d$. If using a conditional density, include its component probability.

Add and check.
Add the atom and continuous contributions, retain the stated loss units, and check that no probability weight or payment was counted twice.

Worked Example: A Uniform Component and Three Atoms

A hypothetical aggregate loss $X$, in thousands of dollars, has probability $0.20$ of being 0, probability $0.25$ of being 4, and probability $0.15$ of being 10. With probability $0.40$, $X$ is conditionally uniform on $[2,8]$. Find the stop-loss premium at attachment $d=5$.

Method and distribution check. The atom probabilities sum to $0.20+0.25+0.15=0.60$, and the continuous component has probability $0.40$, so the total is $1.00$. Its conditional density on $[2,8]$ is $1/6$. Therefore its unconditional subdensity is $0.40(1/6)=1/15$ on that interval.

Atom contributions. The atom at 0 is below 5 and contributes $0.20(0)=0$. The atom at 4 is also below 5 and contributes $0.25(0)=0$. The atom at 10 is above 5 and contributes $0.15(10-5)=0.15(5)=0.75$ thousand dollars.

Continuous contribution. Only the part from 5 to 8 can pay. Using the unconditional subdensity, $$ \int_5^8 (x-5)\frac{1}{15}\,dx =\frac{1}{15}\left[\frac{(x-5)^2}{2}\right]_5^8 =\frac{1}{15}\left(\frac{9}{2}-0\right) =0.30 $$ thousand dollars.

The total is $0.75+0.30=1.05$ thousand dollars, or $1,050.00. As a check, the continuous component’s probability above 5 is $0.40(3/6)=0.20$; the atom at 10 adds $0.15$, so $P(X>5)=0.35$. Both the continuous and atom contributions are positive, as expected.

Common mistake avoided. Using the conditional density $1/6$ directly would give a continuous contribution of $(1/6)(9/2)=0.75$ thousand dollars and an incorrect total of $1.50$ thousand dollars. The conditional density describes the distribution given that the continuous component occurred; its integral must be weighted by $0.40$. The correct continuous contribution is $0.40(0.75)=0.30$.

Worked Example: An Exponential Tail With a Large Atom

A hypothetical aggregate loss, in thousands of dollars, has atoms $P(X=0)=0.25$, $P(X=7)=0.20$, and $P(X=10)=0.05$. With probability $0.50$, $X=6+E$, where $E$ has an exponential distribution with rate $1$ per thousand-dollar unit. Find the stop-loss premium at $d=8$, rounded to the nearest cent.

Method and distribution check. The atom probabilities sum to $0.25+0.20+0.05=0.50$; together with the continuous component’s probability $0.50$, they sum to 1. Conditional on the continuous component, $X$ has density $e^{-(x-6)}$ for $x>6$. Its unconditional subdensity is therefore $0.50e^{-(x-6)}$.

Atom contributions. The atoms at 0 and 7 are below 8 and contribute zero. The atom at 10 contributes $0.05(10-8)=0.10$ thousand dollars.

Continuous contribution. In the integral, set $y=x-8$. Then $x-6=y+2$, and the lower limit $x=8$ becomes $y=0$: $$ \int_8^\infty (x-8)0.50e^{-(x-6)}\,dx =0.50e^{-2}\int_0^\infty ye^{-y}\,dy =0.50e^{-2}(1) =0.0676676 $$ thousand dollars. Here $\int_0^\infty ye^{-y}\,dy=1$, as follows by integration by parts.

Adding the contributions gives $0.10+0.0676676=0.1676676$ thousand dollars, or $167.6676 before rounding. Rounded to the nearest cent, the premium is $167.67. As a check, $P(X>8)=0.05+0.50e^{-2}=0.1176676$; the atom at 10 and the exponential tail are the only parts above the attachment.

Common mistake avoided. The atom at 7 contributes nothing even though the continuous component starts at 6. The continuous density does not turn a separate point mass into a continuous payment; each atom must be handled using its own probability and loss value.

Worked Example: An Atom at Attachment and a Gap in the Density

A hypothetical aggregate loss $X$, in thousands of dollars, has atoms of probability $0.10$ at 4, $0.10$ at 8, and $0.20$ at 13. Its continuous part has unconditional density $0.10$ on each of $[5,8]$ and $[10,13]$, and zero elsewhere. Find the stop-loss premium at $d=8$.

Method and distribution check. The atoms total $0.10+0.10+0.20=0.40$. The continuous density integrates to $0.10(8-5)+0.10(13-10)=0.30+0.30=0.60$. Thus the full probability is $0.40+0.60=1.00$.

Atom contributions. The atom at 4 contributes zero. The atom at 8 is exactly at the attachment, so its payment is $0.10(8-8)=0$. The atom at 13 contributes $0.20(13-8)=1.00$ thousand dollars.

Continuous contribution. The density on $[5,8]$ lies at or below the attachment, so it contributes zero. On $[10,13]$, integrate the excess against the density: $$ \int_{10}^{13}(x-8)(0.10)\,dx =0.10\left[\frac{(x-8)^2}{2}\right]_{10}^{13} =0.10\left(\frac{25-4}{2}\right) =1.05 $$ thousand dollars.

The premium is $1.00+1.05=2.05$ thousand dollars, or $2,050.00. A probability check gives $P(X>8)=0.20+0.10(3)=0.50$: the atom at 13 and the continuous mass on $[10,13]$ are the only outcomes strictly above 8.

Common mistake avoided. Counting the atom at 8 as a positive payment would be wrong: its excess is exactly zero. Likewise, the density on $[5,8]$ cannot contribute merely because its interval touches the attachment.

Exam Checks for Mixed Stop-Loss Calculations

Mixed distributions require more than applying an integral mechanically. The distribution may be presented by component probabilities and conditional distributions, or by an already weighted subdensity. Read that distinction carefully. A useful check is to verify the total probability before calculating, then confirm that only outcomes strictly above the attachment have positive payments. This topic is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Do not omit a point mass. An atom can make a substantial contribution even when the rest of the distribution is continuous.
Do not double-weight a subdensity. A density whose integral is already the continuous component’s probability is unconditional; do not multiply it by that probability again.
Do not confuse probability with payment. The probability above $d$ is a useful check, but the stop-loss premium also depends on how far losses exceed $d$.
Keep units explicit. If loss values are in thousands of dollars, the calculated expectation is also in thousands of dollars until converted.

Key takeaway: For a mixed aggregate loss, add the payment from each atom above attachment to the integral of excess loss against the continuous subdensity above attachment. An atom exactly at attachment contributes zero, and conditional component densities must be weighted by their component probabilities.

Check Your Understanding

For each problem, verify the component probabilities before calculating. All loss amounts are in thousands of dollars.

$X$ has atoms $P(X=0)=0.40$, $P(X=5)=0.30$, and $P(X=11)=0.30$. Find the stop-loss premium at $d=6$. (Answer: $1.5000$ thousand dollars.)
$X$ has atoms of probability $0.25$ at 2 and $0.25$ at 9; with probability $0.50$, it is conditionally uniform on $[4,10]$. Find the premium at $d=7$. (Answer: $0.8750$ thousand dollars.)
$X$ has atoms $P(X=0)=0.20$ and $P(X=8)=0.20$. With probability $0.60$, $X=4+E$, where $E$ is exponential with rate $0.5$ per thousand-dollar unit. Find the premium at $d=6$, to four decimal places in thousands. (Answer: $0.8415$ thousand dollars.)
$X$ has atoms of probability $0.15$ at 6, $0.10$ at 9, and $0.25$ at 12. Its remaining probability is split equally between uniform distributions on $[7,9]$ and $[11,13]$. Find the premium at $d=9$. (Answer: $1.5000$ thousand dollars.)
$X$ has atoms of probability $0.10$ at 5, $0.20$ at 10, and $0.20$ at 14. Its unconditional density is $0.10$ on $[6,10]$, $0.10$ on $[12,13]$, and zero elsewhere. Find the premium at $d=8$. (Answer: $2.2500$ thousand dollars.)

What You'll Learn