Mixed Severity Distributions With Atoms

Why Mixed Severity Distributions Need Special Care

A loss severity model need not be entirely discrete or entirely continuous. A hypothetical insurance portfolio might have a positive probability of no payment, while positive payments vary continuously. Another model might allow a fixed benefit amount as well as a range of continuously varying losses. Such models combine point masses, also called atoms, with a continuous component.

The key distinction is that an atom assigns positive probability to one exact value, whereas a continuous density assigns probability to intervals through integration. A density value at a single point is not itself a point probability. This distinction matters whenever a question asks for a probability at a particular loss amount or uses interval endpoints that may contain atoms.

Definition: A severity random variable $X$ has an atom at $a$ if $P(X=a)>0$. The probability $P(X=a)$ is the mass of that atom. A continuous component has a density whose integral over an interval gives the probability contributed by that component. A mixed distribution has at least one atom and a continuous component, or more than one kind of component.

Representing a Mixed Distribution

Suppose a model has atoms at distinct values $a_1,\ldots,a_k$, with masses $p_1,\ldots,p_k$, and a continuous component with (unconditional) density $g$. The word unconditional means that $g$ already incorporates the probability of entering the continuous component. The total mass must satisfy

\sum_{j=1}^{k}p_j+\int_{-\infty}^{\infty}g(t)\,dt=1.

The cumulative distribution function is $F_X(x)=P(X\le x)$. For this mixed model,

F_X(x)=\sum_{j:a_j\le x}p_j+\int_{-\infty}^{x}g(t)\,dt.

The sum accounts for atoms at or below $x$; the integral accounts for continuous probability up to $x$. If instead the model is described as a mixture that first selects a component, and the continuous component is selected with probability $w$ and has conditional density $f$, then its contribution to the unconditional density is $g(t)=wf(t)$. This weighting prevents accidentally counting the continuous component as though it had total probability one in addition to the atoms.

Key result: At any value $a$, the probability at the point is the jump of the CDF: $$ P(X=a)=F_X(a)-F_X(a-), $$ where $F_X(a-)=\lim_{x\uparrow a}F_X(x)$. A continuous density contributes no point mass at $a$; any jump there is due to an atom.

Why the CDF Jump Gives the Atom

For any $x<a$, the event $X\le x$ is contained in $X<a$. As $x$ increases toward $a$, these events approach $X<a$, not $X\le a$. By continuity of probability for an increasing sequence of events, their probabilities approach $P(X<a)=F_X(a-)$. The event $X\le a$ consists of the disjoint events $X<a$ and $X=a$. Therefore,

F_X(a)=P(X\le a)=P(X<a)+P(X=a)=F_X(a-)+P(X=a),

which rearranges to the jump formula. This also explains why the CDF is right-continuous and why atoms create jumps, while a continuous component alone creates no jump.

Calculating Probabilities on Intervals

For any $a<b$, the event $a<X\le b$ is the difference between $X\le b$ and $X\le a$. Hence,

P(a<X\le b)=F_X(b)-F_X(a) =\sum_{j:a<a_j\le b}p_j+\int_a^b g(t)\,dt.

The formula includes an atom at the upper endpoint $b$, but excludes one at the lower endpoint $a$. If the requested interval includes or excludes different endpoints, adjust the atom terms accordingly. The continuous integral is unchanged by including or excluding a single endpoint, because a single point has zero probability under a continuous component.

List the atoms in the requested event.
Check the interval notation before adding any point masses. For $a<X\le b$, include only atoms satisfying $a<a_j\le b$.

Find the continuous contribution.
Integrate the unconditional density over the interval. If the density is conditional on selecting the continuous component, multiply by that component's mixture weight.

Add the contributions and check.
The total is the sum of the relevant atom masses and the density integral. Check that it is between zero and one and is consistent with the model's total probability.

Worked Example: A No-Payment Atom and a Continuous Payment

A hypothetical policy payment $X$, measured in dollars, is zero with probability $0.30$. With probability $0.70$, the payment is uniform on $(0,100)$. Find $P(X\le 40)$, $P(X=0)$, and $P(20<X\le 60)$. State the CDF at $0$.

Method and model. This is a two-component mixture: an atom of mass $0.30$ at zero and a continuous uniform component with weight $0.70$. Conditional on selecting that continuous component, its density is $1/100$ on $(0,100)$. The unconditional density is therefore $g(x)=0.70/100=0.007$ on that interval and zero elsewhere.

The atom probability is immediately $P(X=0)=0.30$. Since the continuous component has no point mass at zero, the CDF at zero is

F_X(0)=P(X\le 0)=P(X=0)=0.30.

For $X\le 40$, include the atom at zero and the continuous probability from 0 to 40:

P(X\le 40)=0.30+\int_0^{40}0.007\,dx =0.30+0.007(40) =0.30+0.28 =0.58.

For $20<X\le 60$, the zero atom is outside the interval. The continuous contribution is

P(20<X\le 60)=\int_{20}^{60}0.007\,dx =0.007(60-20) =0.28.

Check. The full continuous component has probability $0.007(100)=0.70$, and the atom has probability $0.30$, so the model totals $1.00$. The interval probability $0.28$ is also no greater than the continuous component's total probability. The common mistake avoided here is treating the uniform density as if it had total weight one and then adding the atom on top of it.

Worked Example: An Atom Inside a Continuous Range

A hypothetical payment $X$ has an atom of mass $0.20$ at zero, an atom of mass $0.30$ at $50$, and a continuous component of weight $0.50$ that is uniform on $(0,100)$. Calculate $F_X(50-)$, $F_X(50)$, $P(X=50)$, and $P(40\le X\le 70)$.

Method and model. The conditional uniform density is $1/100$, so the unconditional density is $g(x)=0.50/100=0.005$ on $(0,100)$. The two atom masses and the continuous weight sum to $0.20+0.30+0.50=1.00$.

Immediately below 50, the CDF includes the atom at zero and the continuous mass accumulated from 0 to 50, but not the atom at 50:

F_X(50-)=0.20+\int_0^{50}0.005\,dx =0.20+0.25 =0.45.

At 50, include the atom there as well:

$$ F_X(50)=F_X(50-)+0.30=0.45+0.30=0.75. $$

Thus the jump is $0.75-0.45=0.30$, which agrees with $P(X=50)=0.30$.

The closed interval from 40 to 70 includes the atom at 50, but not the atom at zero. Its continuous contribution is

\int_{40}^{70}0.005\,dx=0.005(30)=0.15.

Therefore,

P(40\le X\le 70)=0.15+0.30=0.45.

Check. The continuous component contributes $0.50$ over its full support, so $0.15$ on this 30-dollar interval is consistent with its length being $30/100$ of the support. The interval probability also includes exactly the interior atom at 50. The common mistake avoided is calculating only the density integral and omitting the point mass.

Worked Example: Integrating a Nonuniform Density and Including an Endpoint Atom

A hypothetical nonnegative loss $X$ has mass $0.10$ at zero and mass $0.20$ at 2. With probability $0.70$, it has a continuous conditional density $f(x)=2x/9$ for $0<x<3$, and zero elsewhere. Find $P(1<X\le 2)$, $F_X(2-)$, and $F_X(2)$.

Method and model. First check the conditional density: $\int_0^3(2x/9)\,dx=[x^2/9]_0^3=1$. Thus the unconditional density is $g(x)=0.70(2x/9)=1.4x/9$ on $(0,3)$. The total mass is $0.10+0.20+0.70=1.00$.

On the interval $1<X\le2$, the atom at zero is excluded and the atom at 2 is included. The continuous probability is

\int_1^2 \frac{1.4x}{9}\,dx =\frac{1.4}{9}\left[\frac{x^2}{2}\right]_1^2 =\frac{1.4}{18}(4-1) =\frac{4.2}{18} =\frac{7}{30}.

Adding the atom at 2 gives

P(1<X\le2)=\frac{7}{30}+0.20 =\frac{7}{30}+\frac{1}{5} =\frac{13}{30} \approx 0.4333.

To find the CDF just below 2, include the atom at zero and integrate the continuous density only up to 2. Do not include the atom at 2:

F_X(2-)=0.10+\frac{1.4}{9}\left[\frac{x^2}{2}\right]_0^2 =0.10+\frac{1.4}{18}(4) =0.10+\frac{14}{45} =\frac{37}{90} \approx 0.4111.

At 2, add its point mass:

F_X(2)=F_X(2-)+0.20 =\frac{37}{90}+\frac{1}{5} =\frac{55}{90} =\frac{11}{18} \approx 0.6111.

Check. The jump is $11/18-37/90=10/90=1/9$, which is not $0.20$—so recheck the arithmetic: $0.20=18/90$, and $37/90+18/90=55/90$, giving jump $18/90=0.20$. The interval calculation is $7/30+1/5=7/30+6/30=13/30\approx0.4333$, as stated. This explicit jump check is useful because arithmetic errors can disguise an otherwise correct endpoint rule. The common mistake avoided is leaving out the atom at the upper endpoint while including it in a closed interval.

Common Exam Traps

Using the density value as a point probability. A density such as $g(2)$ is not $P(X=2)$. The point probability is the CDF jump, and a continuous component alone has zero probability at one exact point.
Forgetting the mixture weight. If $f$ is conditional on selecting the continuous component, its contribution to an interval is the component weight multiplied by the integral of $f$. Alternatively, integrate the already weighted unconditional density $g$.
Ignoring endpoint notation. In $a<X\le b$, an atom at $a$ is excluded and an atom at $b$ is included. For a closed interval, atoms at both endpoints are included.
Assuming every CDF change is a jump. The CDF can increase continuously over a range because of a density. Only the difference between the value at a point and its left limit identifies the atom at that point.
Failing to check total probability. The sum of all atom masses and the integral of the unconditional density must equal one.

Key takeaway: For mixed severity models, calculate point probabilities from CDF jumps and calculate continuous probabilities by integrating the correctly weighted density. Then handle each atom according to the exact interval endpoints.

Check Your Understanding

For each model, identify the relevant atom and continuous contributions before combining them.

A hypothetical payment is zero with probability $0.15$; otherwise it is uniform on $(0,200)$. Find $F_X(0)$ and $P(50<X\le150)$. (Answer: $F_X(0)=0.1500$; $P(50<X\le150)=0.4250$)
A loss has mass $0.10$ at zero, mass $0.25$ at 40, and a continuous component of weight $0.65$ uniform on $(0,100)$. Find $F_X(40-)$, $F_X(40)$, and $P(20\le X\le60)$. (Answer: $F_X(40-)=0.3600$; $F_X(40)=0.6100$; $P(20\le X\le60)=0.5100$)
A severity has mass $0.20$ at 1 and, with probability $0.80$, conditional density $f(x)=2x/16$ on $0<x<4$. Find $P(X=2)$ and $P(1<X\le3)$. (Answer: $P(X=2)=0$; $P(1<X\le3)=0.4000$)
A payment has mass $0.30$ at 5 and a continuous component of weight $0.70$ uniform on $(0,10)$. Find $P(X\le5)$, $P(X=5)$, and $P(X>5)$. (Answer: $P(X\le5)=0.6500$; $P(X=5)=0.3000$; $P(X>5)=0.3500$)
A loss has mass $0.12$ at zero, mass $0.18$ at 3, and a continuous component of weight $0.70$ uniform on $(0,6)$. Find $F_X(3-)$, $F_X(3)$, and $P(2<X\le4)$. (Answer: $F_X(3-)=0.4700$; $F_X(3)=0.6500$; $P(2<X\le4)=0.4133$)

What You'll Learn