Conditional Excess Loss Versus Stop-Loss Payment

Why the Two Excess Measures Differ

An aggregate loss above an attachment can be described in two useful ways. The stop-loss payment averages the insurer’s payment over every outcome, including outcomes in which nothing is paid. The conditional excess averages the amount above the attachment only among outcomes that exceed it. The first is an unconditional expected payment; the second is an average conditional on the cover being triggered.

These quantities are related, but they are not interchangeable. In particular, the stop-loss payment is not the average amount paid after a claim triggers the cover. To convert between the measures, the probability of exceeding the attachment is essential.

Definition: Let $S$ be a nonnegative aggregate loss and let $d\geq 0$ be an attachment. The stop-loss payment is $(S-d)^+=\max(S-d,0)$, and its expected value is $E[(S-d)^+]$. If $P(S>d)>0$, the conditional excess is $E[S-d\mid S>d]$, the expected amount by which aggregate loss exceeds $d$, given that it does exceed $d$.

The conditioning event is strictly $S>d$. An outcome with $S=d$ produces no payment, and it does not satisfy this event. If there is a probability mass at $d$, including that mass in the conditioning event changes the conditional mean, even though those outcomes contribute zero to the payment.

The Conversion Formula

The positive-part payment can be written using an indicator: it equals $S-d$ when $S>d$, and zero otherwise. This gives the conversion directly.

(S-d)^+=(S-d)\mathbf{1}_{\{S>d\}}.

Taking expectations and conditioning on whether $S>d$ gives the key result. The contribution from $S\leq d$ is zero; on $S>d$, the average excess is the conditional expectation.

\begin{aligned} E[(S-d)^+] &=E\left[(S-d)\mathbf{1}_{\{S>d\}}\right]\\ &=P(S>d)\,E[S-d\mid S>d]. \end{aligned}

Theorem: If $P(S>d)>0$, then $$ E[(S-d)^+]=P(S>d)\,E[S-d\mid S>d], \qquad E[S-d\mid S>d]=\frac{E[(S-d)^+]}{P(S>d)}. $$ The expected stop-loss payment averages over all outcomes. The conditional excess divides that expected payment by the probability that a positive payment occurs. If $P(S>d)=0$, the stop-loss payment is zero, but the conditional excess given $S>d$ is undefined.

The quantities have the same monetary units, but different interpretations. If losses are measured in dollars, both are dollar amounts. The exceedance probability is dimensionless. A useful consistency check is that the conditional excess must be at least the unconditional stop-loss payment when the exceedance probability is positive and at most one.

Calculating from a Discrete Loss Distribution

For a discrete aggregate loss, list each possible value of $S$, its probability, and its excess above $d$. Values below or equal to $d$ contribute zero to the stop-loss payment. Values strictly above $d$ contribute their excess multiplied by their probability. Sum those contributions to obtain $E[(S-d)^+]$, then divide by $P(S>d)$ if the conditional excess is required.

Worked Example: A Discrete Aggregate Loss with an Attachment-Point Atom

A hypothetical portfolio has the following aggregate loss distribution, in dollars. The attachment is $d=\$250$. Find the exceedance probability, the expected stop-loss payment, and the conditional excess given exceedance.

Method and model. Treat the stated table as the complete discrete distribution of $S$. The listed probabilities sum to $0.10+0.20+0.25+0.25+0.20=1.00$.

Aggregate loss $S$	Probability	Payment $(S-250)^+$
$0	0.10	$0
$100	0.20	$0
$250	0.25	$0
$350	0.25	$100
$550	0.20	$300

Step 1: Find the exceedance probability. Strict exceedance occurs at losses of $350 and $550, so $P(S>250)=0.25+0.20=0.45$. The probability at exactly $250 is not included.

Step 2: Find the stop-loss payment. The expected payment is the sum of each payment times its probability: $E[(S-250)^+]=0.25(\$100)+0.20(\$300)=\$25+\$60=\$85.00$.

Step 3: Find the conditional excess. Divide the expected payment by the probability of a positive payment: $E[S-250\mid S>250]=\$85.00/0.45=\$188.888\ldots$, or $188.89 to the nearest cent.

Check and mistake avoided. The conversion gives $0.45(\$188.888\ldots)=\$85.00$. A common mistake is to use $P(S\geq250)=0.70$ in the denominator. That would give $\$85/0.70=\$121.43$, which is the mean excess conditional on $S\geq250$, not on $S>250$. The $250 outcomes add no payment but do affect that different conditional mean.

Continuous Models and Conditional Excess

For a continuous aggregate-loss model with density $f_S$, the same distinction applies. The stop-loss payment is an unconditional expectation, obtained by integrating the payment over the part of the distribution above the attachment. Dividing by the exceedance probability gives the conditional excess, provided that probability is positive.

E[(S-d)^+]=\int_d^\infty (s-d)f_S(s)\,ds, \qquad P(S>d)=\int_d^\infty f_S(s)\,ds.

A continuous distribution has zero probability at any single point, so $P(S\geq d)=P(S>d)$. That equality does not generally hold for a discrete or mixed distribution with an atom at $d$. The strict conditioning event remains important regardless of the model.

Worked Example: Exponential Aggregate Loss

Suppose a hypothetical aggregate loss $S$ has an exponential distribution with mean $200. Its density is $f_S(s)=\frac{1}{200}e^{-s/200}$ for $s\geq0$. The attachment is $d=\$300$. Find the exceedance probability, conditional excess, and expected stop-loss payment.

Method and assumptions. Use the exponential model with scale (and mean) $200, and apply the density only on its support $s\geq0$. The exponential distribution is memoryless: for $x\geq0$, $P(S>d+x\mid S>d)=P(S>x)$.

Step 1: Find the exceedance probability. $P(S>300)=e^{-300/200}=e^{-1.5}=0.2231$ to four decimals.

Step 2: Find the conditional excess. For $x\geq0$, $P(S-300>x\mid S>300)=P(S>300+x)/P(S>300)=e^{-(300+x)/200}/e^{-300/200}=e^{-x/200}$. Thus, conditional on exceeding $300, the excess has the same exponential distribution with mean $200. Therefore $E[S-300\mid S>300]=\$200.00$.

Step 3: Find the expected payment. Multiply the conditional excess by the exceedance probability: $E[(S-300)^+]=e^{-1.5}(\$200)=\$44.626\ldots$, or $44.63 to the nearest cent.

Check and mistake avoided. The rounded check is $0.2231(\$200)\approx\$44.62$; using the unrounded probability gives $44.63. Do not report $200 as the expected payment: it is the mean payment conditional on exceedance. The unconditional payment is smaller because no payment occurs in the other $1-0.2231$ of outcomes.

Using a Normal Approximation

A normal model can be used as an approximation for aggregate loss when its use is justified by the application. It is not a nonnegative-loss model: a normal distribution assigns some probability to negative values. For an attachment in the upper part of the distribution, the usual normal stop-loss calculation still evaluates the modeled excess above that attachment. State clearly when the normal distribution is an approximation rather than an exact description.

If $S\sim N(\mu,\sigma^2)$, with $\sigma>0$, define $z=(d-\mu)/\sigma$. Let $\phi$ and $\Phi$ denote the standard normal density and distribution functions. Then $P(S>d)=1-\Phi(z)$, and integration of the excess over the normal upper tail gives the expected stop-loss payment.

E[(S-d)^+]=\sigma\phi(z)+(\mu-d)\bigl[1-\Phi(z)\bigr].

The conditional excess is this expected payment divided by $1-\Phi(z)$. The formula is useful only when the exceedance probability is positive; in a nondegenerate normal model it is positive for every finite $d$.

Worked Example: Conditional Excess under a Normal Approximation

A hypothetical aggregate loss is approximated by $S\sim N(500,100^2)$, with losses and the attachment in dollars. Take $d=\$600$. Find the exceedance probability, expected stop-loss payment, and conditional excess. Round probabilities to four decimals and dollar amounts to the nearest cent.

Method and assumptions. Use a normal approximation with mean $\mu=\$500$ and standard deviation $\sigma=\$100$. The variance is $\$10{,}000$, not $\$100$. The calculation is for the stated approximation.

Step 1: Standardize the attachment. $z=(600-500)/100=1$. Standard normal values are $1-\Phi(1)=0.1586553$ and $\phi(1)=0.2419707$. Thus $P(S>600)=0.1587$ to four decimals.

Step 2: Calculate the expected payment. Substitute into the normal stop-loss formula: $E[(S-600)^+]=100(0.2419707)+(500-600)(0.1586553)$. The two terms are $24.1971 and $-\$15.8655$, so the payment is $\$8.3316$, or $8.33.

Step 3: Calculate the conditional excess. $E[S-600\mid S>600]=\$8.3316/0.1586553=\$52.513\ldots$, or $52.51.

Check and mistake avoided. Using unrounded values, $0.1586553(\$52.513\ldots)=\$8.3316$, as required. The conditional mean is not the stop-loss payment: dividing by the exceedance probability raises the $8.33 unconditional average to the $52.51 average among exceedances. Do not use the variance, rather than the standard deviation, in standardizing or in the formula.

Common Interpretation and Calculation Traps

The conversion formula is short, but its use depends on identifying the correct quantity and event. In a problem, first determine whether the requested result averages payments across all outcomes or only across exceedances. Then check the attachment convention, particularly when the distribution has an atom at that point.

Using the conditional mean as the expected payment. $E[S-d\mid S>d]$ is an average among exceedances. Multiply it by $P(S>d)$ to recover $E[(S-d)^+]$.
Conditioning on $S\geq d$ instead of $S>d$. When $P(S=d)>0$, the extra outcomes have zero excess but change the conditional denominator. Follow the event specified in the question.
Including losses below the attachment as negative payments. A stop-loss payment uses the positive part, $(S-d)^+$, not $S-d$ over all outcomes.
Dividing by zero. If $P(S>d)=0$, the stop-loss payment is zero and the conditional excess given $S>d$ is undefined.
Rounding too early. Keep the exceedance probability and expected payment unrounded through the division; round only the requested final values.

The distinction matters whenever an expected payment is compared with the average severity of a triggered payment. The former includes the chance that the payment is zero; the latter describes the payment size once the attachment is exceeded. Keeping both the conditioning event and the averaging population explicit prevents an apparently plausible answer from having the wrong interpretation.

Key takeaway: For $P(S>d)>0$, the unconditional stop-loss payment equals the probability of exceeding the attachment times the conditional excess. Identify the strict exceedance event, account for any mass at the attachment, and use unrounded values in the conversion.

Check Your Understanding

For each problem, distinguish the expected payment across all outcomes from the conditional excess among strict exceedances. Dollar amounts are hypothetical.

A discrete aggregate loss takes values $50, $100, $180, and $300 with respective probabilities 0.20, 0.30, 0.30, and 0.20. At attachment $d=\$100$, find $P(S>d)$, $E[(S-d)^+]$, and $E[S-d\mid S>d]$. (Answer: $0.50$, $\$64.00$, and $\$128.00$.)
An aggregate loss is exponential with mean $150, and the attachment is $150. Find the exceedance probability, conditional excess, and expected stop-loss payment. (Answer: $0.3679$, $\$150.00$, and $\$55.18$.)
A discrete aggregate loss is $0 with probability 0.20, $200 with probability 0.30, $400 with probability 0.30, and $700 with probability 0.20. At $d=\$200$, find the exceedance probability and both excess measures. (Answer: $0.50$, $\$160.00$, and $\$320.00$.)
A normal approximation is $S\sim N(1000,200^2)$, in dollars, with attachment $d=\$1200$. Find the exceedance probability, expected stop-loss payment, and conditional excess. Round the probability to four decimals and dollar amounts to cents. (Answer: $0.1587$, $\$16.66$, and $\$105.03$.)
A discrete loss is $100 with probability 0.40, $200 with probability 0.40, and $500 with probability 0.20. For $d=\$100$, find the strict exceedance probability, the conditional excess given $S>100$, and the conditional excess given $S\geq100$. (Answer: $0.60$, $\$200.00$, and $\$120.00$, respectively.)

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