Retained and Ceded Severity Distributions

How Excess-of-Loss Changes a Claim Severity

A per-claim excess-of-loss treaty divides each individual loss between the insurer and the reinsurer. To describe the resulting severity distributions, it is not enough to transform the continuous part of the original loss distribution: a treaty can also concentrate a range of original losses at one retained amount or one ceded recovery. Those concentrated probabilities are atoms, or probability masses.

Let $X$ be a nonnegative claim loss. Suppose the treaty attaches at $d$ and has a maximum recovery of $u$ per claim. Thus the reinsurer pays nothing up to $d$, pays the excess above $d$ until the recovery reaches $u$, and pays at most $u$ on any one claim. The exhaustion point is $d+u$. In this tutorial, $u$ denotes the maximum recovery, not the exhaustion point.

Definition: The ceded recovery per original loss is $Y^L=\min((X-d)_+,u)$, where $(z)_+=\max(z,0)$. The retained loss is $R=X-Y^L$. Consequently, $X=R+Y^L$ for every claim. The payment per payment, $Y^P$, is the conditional distribution of $Y^L$ given $Y^L>0$, provided $P(X>d)>0$.

For the main formulas below, assume $X$ has a continuous distribution on the nonnegative real line, with cumulative distribution function $F_X$, survival function $S_X=1-F_X$, and density $f_X$. In particular, $X$ has no atoms at the attachment or exhaustion point. If the original severity has atoms, their probabilities must also be carried through the transformations; the general allocation $X=R+Y^L$ remains valid.

Distribution of the Retained Loss

The retained amount has three possible forms. A loss no greater than $d$ is retained in full. A loss above $d$ but no greater than $d+u$ produces the same retained amount, $d$. A loss above $d+u$ leaves the insurer responsible for the loss less the maximum recovery $u$.

R= \begin{cases} X, & 0\le X\le d,\\ d, & d<X\le d+u,\\ X-u, & X>d+u. \end{cases}

For $0\le r<d$, only losses $X\le r$ produce retained losses at most $r$, so $F_R(r)=F_X(r)$. At $r=d$, all losses through the exhaustion point have retained amount at most $d$, giving $F_R(d)=F_X(d+u)$. For $r>d$, the condition $R\le r$ is equivalent to $X\le r+u$, so the distribution function is $F_X(r+u)$.

F_R(r)= \begin{cases} 0, & r<0,\\ F_X(r), & 0\le r<d,\\ F_X(d+u), & r=d,\\ F_X(r+u), & r>d. \end{cases}

The jump at $d$ is the probability of an original loss in the layer, $d<X\le d+u$. Away from that jump, the density is inherited from the original severity, with the upper portion shifted left by $u$.

P(R=d)=F_X(d+u)-F_X(d),\qquad f_R(r)= \begin{cases} f_X(r), & 0<r<d,\\ f_X(r+u), & r>d. \end{cases}

The formula for the density describes only the continuous portions. If $X$ has an atom at zero, $R$ also has that atom at zero. Under the continuous-severity assumption, there is no such mass. In particular, the interval of original losses in the layer is not spread over retained values: it all becomes the single retained amount $d$.

Distribution of the Ceded Recovery

The recovery $Y^L$ is zero when the loss does not exceed the attachment, increases one-for-one with the loss within the layer, and equals $u$ once the layer is exhausted. Thus the recovery distribution has a mass at zero, a continuous part between zero and $u$, and a mass at $u$.

F_{Y^L}(y)= \begin{cases} 0, & y<0,\\ F_X(d), & y=0,\\ F_X(d+y), & 0<y<u,\\ 1, & y\ge u. \end{cases}

For $0<y<u$, the event $Y^L\le y$ is equivalent to $X\le d+y$. The endpoint masses and the density on the open interval are therefore

P(Y^L=0)=F_X(d),\qquad P(Y^L=u)=S_X(d+u),\qquad f_{Y^L}(y)=f_X(d+y)\quad(0<y<u).

These pieces account for the entire distribution: $F_X(d)$ is the probability below attachment, $F_X(d+u)-F_X(d)$ is the probability of a recovery strictly between zero and $u$, and $S_X(d+u)$ is the probability of reaching the maximum recovery. Their sum is one.

F_X(d)+\bigl(F_X(d+u)-F_X(d)\bigr)+S_X(d+u)=1.

The payment per payment $Y^P$ excludes claims with no recovery. For a continuous severity and $P(X>d)>0$, its distribution is the positive part of the recovery distribution divided by the probability of a positive recovery:

P(Y^P\le y)=\frac{F_X(d+y)-F_X(d)}{S_X(d)} \quad(0\le y<u),\qquad P(Y^P=u)=\frac{S_X(d+u)}{S_X(d)}.

The denominator is $S_X(d)$, not one: $Y^P$ is conditional on $X>d$. Its probabilities, including its continuous part and its mass at $u$, sum to one.

Worked Examples

Worked Example: A Uniform Severity With a Limited Layer

A hypothetical insurer has claim severity $X\sim\operatorname{Uniform}(0,200)$ dollars. A per-claim treaty attaches at $d=\$50$ and has maximum recovery $u=\$100$. Find the retained and ceded distribution components, then calculate their means as a check.

Method. Here $F_X(x)=x/200$ for $0\le x\le200$, and $f_X(x)=1/200=0.005$ on that interval. The exhaustion point is $d+u=50+100=\$150$.

For retained loss, the continuous part below $d$ has density $0.005$ on $(0,50)$, with probability $50/200=0.25$. The mass at $R=50$ is $F_X(150)-F_X(50)=150/200-50/200=0.50$. Above $d$, the retained loss is $X-100$, so its continuous part has density $f_X(r+100)=0.005$ for $50<r<100$, with probability $50/200=0.25$. The probabilities sum to $0.25+0.50+0.25=1$.

For ceded recovery, $P(Y^L=0)=F_X(50)=50/200=0.25$, and $P(Y^L=100)=S_X(150)=50/200=0.25$. The density is $0.005$ on $(0,100)$, giving continuous probability $100(0.005)=0.50$. Again, $0.25+0.50+0.25=1$.

The mean ceded recovery is the integral over the continuous part plus the contribution from the mass at $100$: $\int_0^{100} y(0.005)\,dy+100(0.25)=0.005(100^2/2)+25=25+25=\$50$. Since $E[X]=100$, the mean retained loss is $E[R]=E[X]-E[Y^L]=100-50=\$50$.

Check and trap avoided. The retained mean and ceded mean sum to the original mean, $50+50=100$. A common error is to spread the probability $0.50$ from losses in the layer across retained amounts from $50$ to $150$. Those losses all leave exactly $50$ retained, so that probability is an atom at $R=50$.

Worked Example: Conditional Payments With Exponential Severity

Suppose a hypothetical portfolio has $X\sim\operatorname{Exponential}(\lambda=0.1)$ in thousands of dollars, so $S_X(x)=e^{-0.1x}$ for $x\ge0$. The treaty attaches at $d=10\ln 2$ thousand dollars and has maximum recovery $u=10\ln 2$ thousand dollars. Find the probability of no recovery, the probability of the maximum recovery, and the mean recovery per payment.

The exhaustion point is $d+u=20\ln 2$. Since $e^{-0.1(10\ln 2)}=e^{-\ln 2}=1/2$, the probability of no recovery is $F_X(d)=1-S_X(d)=1/2$. Also $S_X(d+u)=e^{-0.1(20\ln 2)}=e^{-2\ln2}=1/4$, which is the per-loss probability of a recovery equal to $u$. The continuous recovery portion therefore has probability $1-1/2-1/4=1/4$.

To find the conditional distribution, divide the positive recovery probabilities by $P(X>d)=1/2$. For $0\le y<u$, $P(Y^P\le y)=[F_X(d+y)-F_X(d)]/S_X(d)$. Using the exponential survival function gives $P(Y^P\le y)=1-e^{-0.1y}$. At the limit, $P(Y^P=u)=S_X(d+u)/S_X(d)=(1/4)/(1/2)=1/2$.

The conditional mean follows by integrating the survival probability of the capped payment: $E[Y^P]=\int_0^u P(Y^P>y)\,dy=\int_0^u e^{-0.1y}\,dy=(1-e^{-0.1u})/0.1$. Because $e^{-0.1u}=1/2$, this is $(1-1/2)/0.1=5$ thousand dollars.

As a further check, the unconditional expected recovery is $P(X>d)E[Y^P]=(1/2)(5)=2.5$ thousand dollars. Directly, it is $\int_0^u S_X(d+y)\,dy=(1/2)\int_0^u e^{-0.1y}\,dy=(1/2)(5)=2.5$ thousand dollars.

Check and trap avoided. The maximum-recovery probability is $1/4$ per original loss but $1/2$ per positive payment. Using $1/4$ as the atom in the conditional distribution would fail to normalize the payment-per-payment probabilities.

Worked Example: Translating Retained and Ceded Events

A hypothetical risk has $X\sim\operatorname{Uniform}(0,300)$ dollars. A per-claim layer attaches at $d=\$60$ and has maximum recovery $u=\$90$. Find $P(R\le60)$, the mass at $R=60$, $P(Y^L=0)$, $P(Y^L=90)$, $P(30<Y^L<70)$, and $P(R>100)$.

The exhaustion point is $60+90=\$150$, and $F_X(x)=x/300$ on the support. Since $R\le60$ exactly when $X\le150$, $P(R\le60)=150/300=0.50$. The mass at $R=60$ is $P(60<X\le150)=(150-60)/300=90/300=0.30$. The remaining $0.20$ of $P(R\le60)$ is the continuous retained portion below $60$.

No recovery occurs when $X\le60$, so $P(Y^L=0)=60/300=0.20$. The maximum recovery occurs when $X\ge150$, so $P(Y^L=90)=(300-150)/300=0.50$. For a recovery strictly between $30$ and $70$, the original loss must be strictly between $60+30=90$ and $60+70=130$. Thus $P(30<Y^L<70)=(130-90)/300=40/300=2/15$.

For $R>100$, the claim must exceed the exhaustion point and satisfy $X-90>100$, or $X>190$. Therefore $P(R>100)=(300-190)/300=110/300=11/30$. The values are consistent with the transformed distributions: retained loss has probabilities $0.20$ below $60$, $0.30$ at $60$, and $0.50$ above $60$; ceded recovery has probabilities $0.20$ at zero, $0.30$ strictly inside the layer, and $0.50$ at the limit.

Check and trap avoided. For $R>100$, the correct original-loss threshold is $100+u=190$, not $100+d=160$. Above exhaustion, retained loss is $X-u$, so the limit $u$, not the attachment $d$, determines the shift.

Derivation and Exam Checks

The transformations follow directly from the treaty payment rule. For $d<X\le d+u$, the ceded amount is $X-d$, and subtracting it from $X$ gives $R=X-(X-d)=d$. For $X>d+u$, the recovery is capped at $u$, so $R=X-u$. These two cases explain both why retained losses have a mass at $d$ and why the upper continuous retained distribution is shifted by $u$. For ceded recovery inside the layer, $Y^L=X-d$, so a recovery in $(0,u)$ comes from an original loss shifted right by $d$; this yields $f_{Y^L}(y)=f_X(d+y)$.

A reliable check is to account for probability separately: below attachment, within the layer, and above exhaustion. For continuous severity, the retained distribution has continuous probability $F_X(d)$, a mass $F_X(d+u)-F_X(d)$, and upper-tail continuous probability $S_X(d+u)$. The ceded distribution has those same three probabilities assigned respectively to zero, the interior of the layer, and the limit. This matching reflects the one-to-one allocation of each original claim between retained loss and ceded recovery.

Keep the two meanings of “limit” distinct. Here $u$ is the maximum recovery; exhaustion is at $d+u$.
Do not confuse the per-loss and per-payment distributions. $Y^L$ includes claims with zero recovery; $Y^P$ conditions on recovery being positive.
Do not omit atoms. A continuous original severity can produce discrete masses after the treaty transformation.
Check the allocation. For each loss, retained loss plus ceded recovery must equal $X$.

Understanding these transformed severities is useful when assessing how a treaty redistributes claim sizes before considering the portfolio of claims. The distribution describes more than the average payment: it shows how much probability is concentrated at no recovery, at the retained attachment, or at the maximum cession. This treatment is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, identify the relevant threshold before calculating the probability.

A hypothetical severity is uniform on $(0,120)$, and a layer attaches at $d=30$ with maximum recovery $u=30$. Find the probability of zero recovery, the probability of maximum recovery, and the mass at retained loss $30$. (Answer: $0.2500,\ 0.5000,\ 0.2500$, respectively.)
A hypothetical severity is exponential with rate $\lambda=0.05$ per thousand dollars. A layer attaches at $d=10$ thousand dollars and has maximum recovery $u=20$ thousand dollars. Find the per-loss probabilities of zero recovery, an interior recovery, and maximum recovery. (Answer: $0.3935,\ 0.3834,\ 0.2231$, respectively.)
A hypothetical severity is uniform on $(0,250)$, with $d=50$ and $u=75$. Find $P(R\le50)$, the mass at $R=50$, and $P(Y^L=75)$. (Answer: $0.5000,\ 0.3000,\ 0.5000$.)
A hypothetical exponential severity has rate $\lambda=0.1$, with attachment $d=5$ and maximum recovery $u=10$. Conditional on a positive payment, find $P(Y^P\le5)$ and $P(Y^P=10)$. (Answer: $1-e^{-0.5}=0.3935,\ e^{-1}=0.3679$.)
A hypothetical severity is uniform on $(0,400)$, and the treaty has $d=80$, $u=120$. Find $P(40<Y^L<100)$ and $P(R>150)$. (Answer: $0.1500,\ 0.3250$.)

What You'll Learn