Mixed Severity Models and Coverage Atoms

Why Mixed Severity Models Need Special Care

A severity model can assign positive probability to particular loss amounts while also assigning probability across a continuous range. A coverage modification then transforms every loss into a payment. Some loss values remain distinct payments, while many different losses can map to the same payment. That many-to-one mapping creates or changes probability atoms in the payment distribution.

For example, a deductible can map every loss at or below the deductible to a payment of zero. A payment limit can map every sufficiently large loss to the same maximum payment. If the loss model already has atoms, an atom can also map to one of these amounts or coincide with an atom created by the coverage terms. The correct procedure is to track the entire set of losses producing each payment, not just transform the continuous density.

Definition: A severity distribution is mixed if it has both one or more atoms and a continuous component. An atom at $x_j$ with probability $p_j$ means $P(X=x_j)=p_j>0$. If the continuous component has density $f_c(x)$, its probability over a set $B$ is $\int_B f_c(x)\,dx$. The density $f_c$ need not integrate to one: its integral is the total probability assigned to the continuous component.

Transforming a Mixed Loss Distribution

Let $g(x)$ be a policy’s payment function and let $Y^L=g(X)$ be the payment per loss. For a deductible $d$ followed by a payment limit $u$, the standard payment function is

Y^L=\min\bigl((X-d)_+,u\bigr), \qquad (z)_+=\max(z,0).

Thus losses at or below $d$ produce zero payment, losses between $d$ and $d+u$ produce $X-d$, and losses at or above $d+u$ produce payment $u$. Atoms exactly at a boundary must be assigned by evaluating the payment function itself. For this particular function, $X=d$ pays zero and $X=d+u$ pays $u$.

Key result: For any payment amount $y$, its probability is the probability of all losses that map to it: $$ P(Y^L=y)=\sum_{j:g(x_j)=y}p_j+\int_{\{x:g(x)=y\}}f_c(x)\,dx. $$ For a continuous payment range, probabilities are obtained from the loss intervals mapping into that range. An isolated loss value has zero probability under a continuous component, but a whole interval of losses can map to one payment and create an atom.

For $Y^L=\min((X-d)_+,u)$, the zero-payment atom includes any original atoms at $x_j\le d$ and the continuous probability over $x\le d$. The payment-limit atom includes original atoms at $x_j\ge d+u$ and the continuous probability over $x\ge d+u$. An original atom strictly between those thresholds maps to its shifted payment $x_j-d$. If several atoms map to the same payment, add their probabilities.

Why Coverage Creates Atoms

The probability statement $P(Y^L=y)=P(g(X)=y)$ follows directly from the event definition: the payment equals $y$ precisely when the loss lies in the preimage $\{x:g(x)=y\}$. If $g$ is constant on an interval, that preimage contains an interval, not just one point. A continuous loss component can assign positive probability to the interval, even though it assigns probability zero to any one loss value.

For the deductible-and-limit payment, $g(x)=0$ throughout $x\le d$, and $g(x)=u$ throughout $x\ge d+u$. Therefore these regions can each contribute positive mass. In the interior, $g(x)=x-d$, so the payment varies with the loss; the continuous component remains continuous there, while any original atoms remain atoms at their transformed values.

List the loss components.
Record every original atom and its probability, then specify the continuous density and its total probability.

Partition losses by payment behavior.
For a deductible and limit, identify $x\le d$, $d<x<d+u$, and $x\ge d+u$. Include boundary atoms in the relevant regions by evaluating $g(x)$.

Collect payment atoms.
Add the probabilities of every original atom and continuous interval that maps to each fixed payment.

Transform the remaining component.
On a one-to-one part of the payment function, transform the loss interval or density and keep track of its probability mass.

Check and summarize.
Verify that all payment probabilities, including continuous probability, total one. Compute expectations by adding the atom and continuous contributions.

Worked Examples

Worked Example: A Limit Applied to a Mixed Severity

A hypothetical loss $X$ has probability $0.25$ at $2$, probability $0.15$ at $5$, and its remaining probability $0.60$ is uniformly distributed over $(0,6)$. A policy pays $Y^L=\min(X,3)$. Find $P(Y^L=3)$, $P(Y^L=2)$, and $E[Y^L]$.

Method. Map each atom through the payment function, then separately integrate the continuous component. The density of the continuous component is $0.60/6=0.10$ on $(0,6)$.

Payment-limit atom. The atom at $X=5$ maps to payment $3$. The continuous losses from $3$ through $6$ also map to payment $3$, with probability $0.10(6-3)=0.30$. Hence

$$ P(Y^L=3)=0.15+0.30=0.45. $$

Payment of 2. The atom at $X=2$ maps to 2. The continuous component has no atom at a single loss value, so

$$ P(Y^L=2)=0.25. $$

Expectation. The two original atoms contribute $0.25(2)=0.50$ and $0.15(3)=0.45$. The continuous component contributes

0.10\left(\int_0^3x\,dx+\int_3^6 3\,dx\right) =0.10\left(\frac{9}{2}+9\right) =1.35.

Therefore $E[Y^L]=0.50+0.45+1.35=2.30$. As a check, the payment masses are $0.25$ at 2 and $0.45$ at 3, while the remaining $0.30$ is continuous over payments between 0 and 3; the total probability is $1$.

Common mistake avoided: The continuous component cannot be omitted from the expectation just because it has no individual point masses. It contributes $1.35$, not zero.

Worked Example: A Deductible and Limit Combine Several Sources of Mass

A hypothetical loss $X$ has probability $0.10$ at $0$, probability $0.20$ at $4$, probability $0.10$ at $9$, and its remaining probability $0.60$ is uniformly distributed over $(0,10)$. A policy has deductible $d=2$, payment limit $u=4$, and pays $Y^L=\min((X-2)_+,4)$. Find the atoms of $Y^L$, its continuous density, and $E[Y^L]$.

Zero payment. The atom at zero pays zero. The continuous component over $0<X\le2$ also pays zero and has probability $0.60(2/10)=0.12$. Thus $P(Y^L=0)=0.10+0.12=0.22$.

Interior atom. The atom at loss $4$ maps to payment $4-2=2$, so $P(Y^L=2)=0.20$. The continuous component produces no atom at this particular payment.

Payment-limit atom. The limit applies when $X\ge d+u=6$. The atom at $9$ contributes $0.10$; the continuous component over $6\le X<10$ contributes $0.60(4/10)=0.24$. Therefore $P(Y^L=4)=0.34$.

Continuous payment density. For losses between 2 and 6, the payment is $X-2$, so the continuous density $0.60/10=0.06$ carries to payments between 0 and 4. The continuous part of the payment distribution has density $0.06$ on $(0,4)$, with total probability $0.06(4)=0.24$. The total probability is $0.22+0.20+0.34+0.24=1$.

Expectation. The atoms contribute $0.20(2)+0.34(4)=0.40+1.36=1.76$. The continuous contribution is

\int_0^4 y(0.06)\,dy=0.06\frac{4^2}{2}=0.48.

Thus $E[Y^L]=1.76+0.48=2.24$. As a direct check on the continuous contribution, losses from 2 to 6 contribute $0.60(1/10)\int_2^6(x-2)\,dx=0.48$, and losses from 6 to 10 contribute $0.60(4/10)(4)=0.96$. Together they give $1.44$, matching the payment-distribution calculation $0.48+0.24(4)=1.44$. Adding the original atoms’ contributions $0.40+0.40$ gives $2.24$.

Common mistake avoided: The atom at loss 9 does not disappear when the limit applies. Its probability is part of the payment-limit atom, along with the continuous probability above the threshold.

Worked Example: Coinsurance Changes Atom Locations, Not the Mapping Principle

A hypothetical loss $X$ has probability $0.20$ at $1$, probability $0.30$ at $6$, probability $0.10$ at $10$, and its remaining probability $0.40$ is uniformly distributed over $(0,8)$. The insurer pays $Y^L=\min(0.5(X-2)_+,4)$: there is a deductible of $2$, a coinsurance factor of $0.5$, and a payment limit of $4$. Find $P(Y^L=0)$, $P(Y^L=2)$, $P(Y^L=4)$, and $E[Y^L]$.

The payment is zero for $X\le2$. The atom at 1 contributes $0.20$, and the uniform component contributes $0.40(2/8)=0.10$, so $P(Y^L=0)=0.30$.

The atom at $X=6$ maps to $0.5(6-2)=2$, giving $P(Y^L=2)=0.30$. The atom at $X=10$ would produce $0.5(10-2)=4$, exactly the limit, so it contributes $0.10$ to $P(Y^L=4)$. The continuous component has losses only up to 8, and even at 8 the uncapped payment is $0.5(8-2)=3$; it contributes no payment-limit atom. Thus $P(Y^L=4)=0.10$.

For the continuous component, the loss density is $0.40/8=0.05$. Losses from 2 to 8 produce payments from 0 to 3, with $y=0.5(x-2)$, so $dx/dy=2$. The payment density on $(0,3)$ is $0.05(2)=0.10$, with total probability $0.10(3)=0.30$. The remaining $0.10$ of the continuous probability is in the zero-payment atom, and the full payment probabilities sum to $0.30+0.30+0.10+0.30=1$.

The expectation from the continuous part is $\int_0^3 0.10y\,dy=0.45$. The two positive-payment loss atoms contribute $0.30(2)=0.60$ and $0.10(4)=0.40$. Hence $E[Y^L]=0.45+0.60+0.40=1.45$.

Common mistake avoided: Applying the coinsurance factor does not scale every payment atom by $0.5$ after the limit. The atom at loss 10 has uncapped payment 4 and therefore maps to the limit payment 4.

Common Errors and Useful Checks

A particularly important exam trap is to calculate probabilities correctly but omit part of the expectation. In the first worked example, the atoms contribute $0.25(2)+0.15(3)=0.95$. Stopping there, or adding an incorrect continuous contribution, gives the wrong mean. The continuous component must be weighted by its total probability, or equivalently integrated using its unconditional density $0.10$. Its correct contribution is $1.35$, so the mean is $0.95+1.35=2.30$.

Other reliable checks are to list the loss intervals mapped to each fixed payment and to add every probability component exactly once. In particular, distinguish the probability of an original atom from the continuous probability on a threshold interval. The former is a point mass; the latter may also create a mass in the payment distribution when the payment function is constant on that interval.

Do not treat a mixed model as purely discrete or purely continuous. Compute contributions from both components.
Do not assume an atom stays at its original loss value. Evaluate the payment function at that value.
Do not count a capped loss twice. Original atoms and continuous probability in the capped region are added to one payment-limit atom.
Check probability accounting. Atoms plus continuous payment probability must sum to one.

Key takeaway: A coverage modification transforms a mixed severity by mapping loss values to payments. To find any payment atom, collect every original atom and every continuous loss interval that maps to that same amount; to find the mean, add the expected contributions from all atoms and the continuous component.

Check Your Understanding

For each problem, identify the relevant loss regions before calculating. Answers are provided at the end of each question for checking.

A hypothetical loss has probability $0.30$ at 1, probability $0.20$ at 5, and remaining probability $0.50$ uniformly distributed over $(0,4)$. A policy pays $Y^L=\min(X,2)$. Find $P(Y^L=2)$, $P(Y^L=1)$, and $E[Y^L]$. (Answer: $0.4500,\ 0.3000,\ 1.45$.)
A hypothetical loss has probability $0.20$ at 0, probability $0.25$ at 6, and remaining probability $0.55$ uniformly distributed over $(0,10)$. A policy pays $Y^L=\min((X-2)_+,3)$. Find $P(Y^L=0)$, $P(Y^L=3)$, and $E[Y^L]$. (Answer: $0.3100,\ 0.5250,\ 1.8225$.)
A hypothetical loss has probability $0.20$ at 3, probability $0.30$ at 8, and remaining probability $0.50$ uniformly distributed over $(0,10)$. A policy pays $Y^L=\min(0.4(X-1)_+,2)$. Find $P(Y^L=0)$, $P(Y^L=2)$, and $E[Y^L]$. (Answer: $0.0500,\ 0.5000,\ 1.41$.)
A hypothetical loss has probability $0.10$ at 2, probability $0.20$ at 7, and remaining probability $0.70$ uniformly distributed over $(0,8)$. A policy pays $Y^L=\min((X-2)_+,3)$. Find $F_{Y^L}(1)$, $P(Y^L=0)$, and $P(Y^L=3)$. (Answer: $0.3625,\ 0.2750,\ 0.4625$.)
A hypothetical loss has probability $0.15$ at 2, probability $0.25$ at 5, probability $0.10$ at 9, and remaining probability $0.50$ uniformly distributed over $(0,10)$. A policy pays $Y^L=\min((X-1)_+,4)$. Find $P(Y^L=0)$, $P(Y^L=4)$, and $E[Y^L]$. (Answer: $0.0500,\ 0.6000,\ 2.95$.)

What You'll Learn