The Severity Distribution After a Payment Transformation

Why a Payment Has Its Own Distribution

A ground-up loss is not necessarily the amount an insurer pays. A deductible can remove small losses, coinsurance can scale payments, and a limit can cap large payments. Consequently, even when the ground-up severity has a continuous distribution, the payment may have probability atoms—for example, a positive probability of zero payment or of reaching the policy limit. To model the payment correctly, first apply the contract’s payment function to the loss, then derive the distribution of the resulting random variable.

Definition: Let $X$ be a nonnegative ground-up loss, and let $g(x)$ be the amount paid under the coverage for a loss of size $x$. The claim payment is the random variable $Y=g(X)$. Its cumulative distribution function is \[ F_Y(y)=P(Y\le y)=P(g(X)\le y). \] The set of ground-up losses satisfying $g(x)\le y$ is the preimage of $(-\infty,y]$ under $g$. In general, $F_Y(y)$ is found from the probability of that set—not by assuming that $Y$ has the same distributional form as $X$.

The basic distribution formula remains valid whether the ground-up severity is continuous, discrete, or mixed. What changes is how the probability of the preimage is calculated. A continuous part of $X$ can produce a continuous part of $Y$, while values or regions of $X$ that the coverage maps to one payment amount can produce atoms.

Atoms and Continuous Payment Values

An atom at $y_0$ is positive probability concentrated at exactly one value: \[ P(Y=y_0)>0. \] For any payment map, the event $Y=y_0$ is precisely the event $g(X)=y_0$. Thus, if $X$ has density $f_X$ and may also have point masses,

P(Y=y_0) = \int_{\{x:g(x)=y_0\}} f_X(x)\,dx + \sum_{x:g(x)=y_0}P(X=x),

where the integral accounts for continuous probability on the level set and the sum accounts for atoms of $X$. In particular, a flat part of $g$ over an interval with positive continuous probability creates an atom in $Y$. An atom of $X$ also contributes an atom at its mapped payment amount.

For example, a deductible may map every loss up to the deductible to zero. A payment limit may map all sufficiently large losses to the same capped payment. These are different ground-up loss regions, but each can create an atom in the payment distribution. A continuous transformation on a smooth, strictly monotone region generally contributes a density instead.

Theorem: Suppose $X$ has density $f_X$, and on a region of interest the payment map $g$ is differentiable and strictly monotone, with $g'(x)\ne 0$. At a payment value $y$ whose inverse $x=g^{-1}(y)$ lies in that region, the density contribution is \[ f_Y(y)=\frac{f_X(x)}{|g'(x)|}. \] If multiple smooth, strictly monotone branches map to the same $y$, add their density contributions: \[ f_Y(y)=\sum_{x:g(x)=y}\frac{f_X(x)}{|g'(x)|}. \] These formulas describe the continuous density contributions only; atoms must be calculated separately.

The absolute derivative in the denominator is the change-of-variables adjustment. If a small interval of payments has width $dy$, its corresponding interval of losses has width approximately $dx=dy/|g'(x)|$. The probability in that loss interval is approximately $f_X(x)\,dx$, giving the stated density. The formula does not apply across flat regions, where $g'(x)=0$, or at jumps and branch boundaries; those locations require direct probability calculations.

Worked Payment Transformations

Worked Example: Deductible, Coinsurance, and a Payment Cap

A hypothetical policy has a deductible of $2 and pays 50% of the loss above the deductible, subject to a maximum payment of $3. The ground-up loss $X$ is uniform on $[0,10]$, in dollars. Derive the full distribution of the payment $Y$, including its atoms and continuous density.

Method. Write the coverage map first. Below the deductible the payment is zero; above it, the uncapped payment is $0.5(X-2)$. The cap is reached when $0.5(X-2)=3$, or $X=8$. Therefore,

Y=g(X)= \begin{cases} 0, & 0\le X\le 2,\\ 0.5(X-2), & 2<X<8,\\ 3, & X\ge 8. \end{cases}

The density of $X$ is $f_X(x)=1/10=0.1000$ for $0<x<10$. Every loss in $[0,2]$ maps to zero, so \[ P(Y=0)=P(X\le2)=\frac{2}{10}=0.2000. \] Every loss in $[8,10]$ maps to $3, so \[ P(Y=3)=P(X\ge8)=\frac{2}{10}=0.2000. \] The endpoints have probability zero under this continuous ground-up model, so their inclusion in these intervals does not change the results.

For $0<y<3$, the inverse payment map is $x=2+2y$, and $dx/dy=2$. Thus the continuous density is \[ f_Y(y)=f_X(2+2y)\left|\frac{dx}{dy}\right| =0.1000(2)=0.2000,\qquad 0<y<3. \] Its total probability is $0.2000(3)=0.6000$. The full distribution function is

F_Y(y)= \begin{cases} 0, & y<0,\\ 0.2000, & y=0,\\ 0.2000+0.2000y, & 0<y<3,\\ 1, & y\ge3. \end{cases}

The continuous component contributes 0.6000, and the atoms contribute $0.2000+0.2000=0.4000$, for a total probability of 1.0000. In particular, the left-hand limit at $3 is 0.8000, followed by a jump of 0.2000 at the cap. This check confirms that neither the capped losses nor the zero-payment losses have been lost from the distribution.

Worked Example: A Mixed Severity with a Capped Payment

For a hypothetical portfolio, a ground-up loss $X$ is zero with probability 0.1000, is uniform on $[0,4]$ with probability 0.6000, and equals 2 with probability 0.3000. The three components are mutually exclusive and exhaustive. The insurer pays $Y=\min(X,2)$, with all amounts in thousands of dollars. Find the payment atoms and continuous distribution.

Method. Treat each component of the mixed ground-up distribution separately and combine the probabilities that map to the same payment. The atom $X=0$ maps to $Y=0$, so $P(Y=0)=0.1000$. The uniform component has conditional density $1/4$ on $(0,4)$, and its unconditional density contribution is $0.6000/4=0.1500$.

For $0<y<2$, the payment equals the loss, so this density contribution remains 0.1500. Its total probability over that interval is $0.1500(2)=0.3000$. At $Y=2$, the uniform component contributes all its probability from $X\ge2$: the conditional probability is $(4-2)/4=0.5000$, so the unconditional contribution is $0.6000(0.5000)=0.3000$. The point mass $P(X=2)=0.3000$ also maps to $Y=2$. Therefore, \[ P(Y=2)=0.3000+0.3000=0.6000. \]

The payment distribution is consequently

F_Y(y)= \begin{cases} 0, & y<0,\\ 0.1000, & y=0,\\ 0.1000+0.1500y, & 0<y<2,\\ 1, & y\ge2. \end{cases}

The continuous probability is $0.1500(2)=0.3000$, and the atoms total $0.1000+0.6000=0.7000$. Together they sum to 1.0000. The ground-up atom at 2 is not the only source of the payment atom at 2: losses above the cap also contribute to it.

Worked Example: Mapping a Discrete Severity

A hypothetical severity model assigns the following probabilities to ground-up losses $X$, measured in thousands of dollars: $P(X=0)=0.1000$, $P(X=1)=0.2000$, $P(X=3)=0.2500$, $P(X=5)=0.2500$, and $P(X=8)=0.2000$. A policy pays $Y=\min(0.5X,2)$. Derive the payment distribution and identify the losses that combine at the cap.

Method. Evaluate the payment map at each possible severity, then add probabilities whenever different severities produce the same payment:

Ground-up loss $x$	Payment $g(x)=\min(0.5x,2)$	Probability
0	0	0.1000
1	0.5	0.2000
3	1.5	0.2500
5	2	0.2500
8	2	0.2000

Thus $P(Y=0)=0.1000$, $P(Y=0.5)=0.2000$, $P(Y=1.5)=0.2500$, and $P(Y=2)=0.2500+0.2000=0.4500$. The last atom combines losses of 5 and 8, because both reach or exceed the loss level at which the payment cap binds. The cumulative distribution is 0 below zero, 0.1000 on $[0,0.5)$, 0.3000 on $[0.5,1.5)$, 0.5500 on $[1.5,2)$, and 1.0000 on $[2,\infty)$. The probabilities sum to $0.1000+0.2000+0.2500+0.4500=1.0000$, as required.

Common Traps and Distribution Checks

A common error is to transform the payment values but forget that a change of scale also changes the density. In the first example, a mistaken calculation would use $f_Y(y)=f_X(2+2y)=0.1000$ for $0<y<3$, omitting the inverse slope. This assigns only $0.1000(3)=0.3000$ continuous probability. Including the two correct endpoint atoms then gives $0.3000+0.2000+0.2000=0.7000$, not 1.0000. The correction is to include $|dx/dy|=2$, giving density 0.2000 and continuous probability 0.6000.

Another trap is to report a capped payment as though it were only the endpoint of a continuous density. If a positive-probability range of losses is capped at the same amount, the payment distribution has a jump there. Similarly, a deductible does not merely shift a continuous distribution: if all losses below the deductible pay zero, it creates a zero-payment atom. Always check the jumps of $F_Y$ and verify that all continuous probability plus all atom probabilities equals one.

Key takeaway: Start with $Y=g(X)$, use $F_Y(y)=P(g(X)\le y)$, and identify every level set $g(x)=y$ that can carry positive probability. Use change of variables only on smooth, strictly monotone branches, and account separately for flat regions, caps, and ground-up atoms.

Check Your Understanding

For each problem, derive the requested payment distribution from the stated ground-up model and coverage map.

A hypothetical loss $X$ is uniform on $[0,6]$, in thousands of dollars. The policy pays $Y=\min((X-1)_+,2)$, where $(z)_+=\max(z,0)$. Give the zero-payment atom, the cap atom, and the density on the continuous payment interval. (Answer: $P(Y=0)=0.1667$, $P(Y=2)=0.5000$; $f_Y(y)=0.1667$ for $0<y<2$.)
A hypothetical ground-up loss is uniform on $[0,8]$, and the policy pays $Y=0.5X$ up to a maximum payment of 2. Find the payment density and the probability atom at the cap. (Answer: $f_Y(y)=0.2500$ for $0<y<2$; $P(Y=2)=0.5000$.)
A hypothetical severity has $P(X=0)=0.2000$; with probability 0.5000 it is uniform on $[0,4]$; and with probability 0.3000 it equals 3. The payment is $Y=\min(X,1)$. Find the zero atom, cap atom, and continuous density on $(0,1)$. (Answer: $P(Y=0)=0.2000$, $P(Y=1)=0.6750$, and $f_Y(y)=0.1250$ for $0<y<1$.)
A hypothetical discrete severity takes values $1,2,4,7$, with respective probabilities $0.1500,0.2500,0.3500,0.2500$. The payment is $Y=\min(0.5(X-1)_+,2)$. Give the payment values and their probabilities. (Answer: $P(Y=0)=0.1500$, $P(Y=0.5)=0.2500$, $P(Y=1.5)=0.3500$, and $P(Y=2)=0.2500$.)
A hypothetical continuous severity has density $f_X(x)=0.1000$ on $(0,10)$. The payment is zero for $X\le2$, equals $0.25(X-2)$ for $2<X<10$, and is capped at 2 for $X\ge10$. Give the continuous payment density, its interval, and both atoms. (Answer: $f_Y(y)=0.4000$ for $0<y<2$; $P(Y=0)=0.2000$, $P(Y=2)=0.0000$.)

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