Stop-Loss Premiums Under a Normal Aggregate Model

Why the Normal Model Is Useful for Stop-Loss Calculations

A stop-loss contract pays when aggregate loss exceeds an attachment point. Even when an aggregate distribution is complicated, a normal model can make the expected payment straightforward to calculate using a standard normal density and survival probability. The central task is to standardize the attachment point correctly and retain precision until the final rounding.

This tutorial treats the normal distribution as the assumed model for aggregate loss. In applications, it may be used as an approximation to an aggregate distribution. A normal variable can take negative values, so the approximation is most credible when that feature is not materially misleading for the contract being priced. The calculations below are expected payments, not expense-loaded premiums.

Definition: Let aggregate loss $X\sim N(\mu,\sigma^2)$, with $\sigma>0$. For an attachment point $d$, the unlimited stop-loss payment is $(X-d)_+=\max(X-d,0)$. Its pure stop-loss premium is $E[(X-d)_+]$. Write $\phi(z)$ for the standard normal density and $\overline{\Phi}(z)=1-\Phi(z)$ for the standard normal survival function.

The density and survival function are

\phi(z)=\frac{1}{\sqrt{2\pi}}e^{-z^2/2}, \qquad \overline{\Phi}(z)=\int_z^\infty \phi(t)\,dt.

Standardize the attachment point using $z=(d-\mu)/\sigma$. The value $z$ counts how many standard deviations the attachment lies above or below the mean. The survival probability $\overline{\Phi}(z)$ is the probability that loss exceeds the attachment.

Unlimited Stop-Loss Premium Under a Normal Model

Theorem: If $X\sim N(\mu,\sigma^2)$, $\sigma>0$, and $z=(d-\mu)/\sigma$, then $$ E[(X-d)_+]=\sigma\phi(z)+(\mu-d)\overline{\Phi}(z). $$

The formula has two terms. The probability of exceeding $d$ is $\overline{\Phi}(z)$, while the density term accounts for the amount of loss above the attachment, not merely the chance of exceeding it. Since an unlimited stop-loss payment is nonnegative, the full expression must be nonnegative, even when $\mu-d$ is negative.

Why the Formula Holds

For a normal variable, write $X=\mu+\sigma Z$, where $Z$ has the standard normal distribution. The payment is positive only when $Z>z$. Integrating the positive payment over that tail gives

\begin{aligned} E[(X-d)_+] &=\int_z^\infty (\mu+\sigma t-d)\phi(t)\,dt\\ &=(\mu-d)\overline{\Phi}(z) +\sigma\int_z^\infty t\phi(t)\,dt. \end{aligned}

Because $\phi'(t)=-t\phi(t)$, the remaining integral is

\int_z^\infty t\phi(t)\,dt =-\int_z^\infty \phi'(t)\,dt =\phi(z),

where $\phi(t)$ tends to zero as $t$ tends to infinity. Substitution yields the stated formula. This derivation also explains why using only the exceedance probability would omit part of the expected payment.

Worked Example: Unlimited Stop-Loss on a Normal Aggregate

A hypothetical benefit plan models annual aggregate claims as $X\sim N(1000,200^2)$, with amounts measured in thousands of dollars. Calculate the pure stop-loss premium for an unlimited contract attaching at $d=1200$ thousand dollars. Use $\phi(1)=0.2419707$ and $\overline{\Phi}(1)=0.1586553$.

Method. Standardize the attachment, then apply the normal tail formula. The model has mean $\mu=1000$, standard deviation $\sigma=200$, and attachment $d=1200$, all in thousands of dollars.

First, $z=(1200-1000)/200=1$. Then,

\begin{aligned} E[(X-1200)_+] &=200\phi(1)+(1000-1200)\overline{\Phi}(1)\\ &=200(0.2419707)+(-200)(0.1586553)\\ &=48.39414-31.73106\\ &=16.66308. \end{aligned}

The premium is $16.66308$ thousand dollars, or $16,663.08. The result is positive, as required for an expected payment, and is expressed in the same loss units as the model.

Limited Stop-Loss Layers

A limited stop-loss contract with attachment $d$ and layer limit $L>0$ pays $\min((X-d)_+,L)$. Equivalently, it pays the loss in the layer from $d$ to $d+L$: zero below $d$, the amount above $d$ between the two thresholds, and $L$ above $d+L$.

Theorem: Define $G(c)=E[(X-c)_+]$ for $X\sim N(\mu,\sigma^2)$, where $$ G(c)=\sigma\phi\left(\frac{c-\mu}{\sigma}\right) +(\mu-c)\overline{\Phi}\left(\frac{c-\mu}{\sigma}\right). $$ Then the expected payment for a layer attaching at $d$ with limit $L$ is $$ E[\min((X-d)_+,L)]=G(d)-G(d+L). $$ If the insurer pays a fixed coinsurance share $r$, with $0\le r\le1$, its expected payment is $r[G(d)-G(d+L)]$.

To justify the subtraction, consider any realized loss $x$. The upper stop-loss payment $(x-(d+L))_+$ is zero until loss passes the top of the layer. Above that point it is exactly the amount by which loss exceeds the top. Subtracting it from $(x-d)_+$ leaves zero below the attachment, the layer loss within the layer, and exactly $L$ above the top. Taking expectations gives the formula.

Identify the thresholds.
The lower attachment is $d$; the upper threshold is $d+L$, not $L$ by itself.

Calculate both unlimited premiums.
Evaluate $G(d)$ and $G(d+L)$ using the normal formula, standardizing each threshold separately.

Subtract, then apply any share.
The layer expectation is $G(d)-G(d+L)$. Multiply by coinsurance only if the stated contract pays a fixed share of the layer loss.

Worked Example: Limited Layer and Precision at the Final Step

A hypothetical pool has aggregate claims $X\sim N(800,150^2)$, in dollars. A stop-loss contract attaches at $d=800$ and has a limit of $L=150$, so its upper threshold is $950$. Calculate the expected payment. Use $\phi(0)=0.3989422804$, $\phi(1)=0.2419707245$, and $\overline{\Phi}(1)=0.1586552539$.

Method. Calculate the unlimited stop-loss premium at each end of the layer and subtract without rounding either amount to cents in the intermediate steps.

At the lower threshold, $z_d=(800-800)/150=0$, so

\begin{aligned} G(800) &=150\phi(0)+(800-800)\overline{\Phi}(0)\\ &=150(0.3989422804)+0\\ &=59.84134206. \end{aligned}

At the upper threshold, $z_{950}=(950-800)/150=1$, and therefore

\begin{aligned} G(950) &=150\phi(1)+(800-950)\overline{\Phi}(1)\\ &=150(0.2419707245)-150(0.1586552539)\\ &=36.295608675-23.798288085\\ &=12.497320590. \end{aligned}

Thus the expected limited payment is

$$ G(800)-G(950)=59.84134206-12.497320590 =47.344021470. $$

Rounded to the nearest cent, the expected payment is $47.34. The intermediate values were kept unrounded; rounding each unlimited premium to cents before subtraction is an avoidable source of small discrepancies.

Worked Example: Coinsurance on a Symmetric Layer

A hypothetical reinsurer models aggregate loss as $X\sim N(1200,300^2)$, in thousands of dollars. It covers the layer from $900$ thousand to $1500$ thousand dollars and pays 80% of covered layer loss. Calculate its expected payment.

Method. The attachment is $d=900$, and the limit is $L=600$, so the upper threshold is $d+L=1500$. First calculate the full layer expectation, then multiply by the coinsurance share $r=0.80$.

For the lower threshold, $z_1=(900-1200)/300=-1$. Using $\phi(-1)=\phi(1)=0.2419707245$ and $\overline{\Phi}(-1)=0.8413447461$,

\begin{aligned} G(900) &=300(0.2419707245)+(1200-900)(0.8413447461)\\ &=72.59121735+252.40342383\\ &=324.99464118. \end{aligned}

For the upper threshold, $z_2=(1500-1200)/300=1$, so

\begin{aligned} G(1500) &=300(0.2419707245)+(1200-1500)(0.1586552539)\\ &=72.59121735-47.59657617\\ &=24.99464118. \end{aligned}

The full layer expectation is $324.99464118-24.99464118=300$ thousand dollars. The reinsurer's expected payment is $0.80(300)=240$ thousand dollars, or $240,000.00. As a check, the layer is centered on the normal mean and spans one standard deviation on either side; the calculated full-layer expectation is positive and does not exceed the $600$-thousand-dollar limit.

Common Exam Trap: Confusing Exceedance Probability With Expected Payment

A frequent error is to use the probability of exceeding the attachment as though it were the expected amount paid. For example, in the first worked example, the calculation

(\mu-d)\overline{\Phi}(z) =(-200)(0.1586553)=-31.73106

is not the expected stop-loss payment. It is only one term in the formula, and by itself it is negative, which cannot be a valid expected payment. The missing term is $\sigma\phi(z)=48.39414$. Including it gives $48.39414-31.73106=16.66308$ thousand dollars.

For a limited layer, another threshold error is to calculate $G(L)$ rather than $G(d+L)$. The upper stop-loss in the subtraction begins at the top of the layer, not at the size of the limit measured from zero. Always write down both $d$ and $d+L$ before standardizing.

Practical Checks Before Reporting a Premium

Check the model and units. Confirm that $\mu$, $\sigma$, $d$, and $L$ use the same units, and that $\sigma$ is the standard deviation rather than the variance.
Check the standardized thresholds. Use $z=(c-\mu)/\sigma$ at each threshold $c$; reversing the subtraction reverses the tail.
Check the layer construction. For a layer with attachment $d$ and limit $L$, subtract $G(d+L)$ from $G(d)$. The result should be nonnegative and no greater than $L$.
Keep precision until the end. Use adequate precision for normal density and survival values, perform the subtraction and any coinsurance multiplication, and round the final dollar amount only.

Key takeaway: Under a normal aggregate model, an unlimited stop-loss expectation is $\sigma\phi(z)+(\mu-d)\overline{\Phi}(z)$. A limited layer is the difference between the unlimited expectations at its attachment and its upper threshold. Standardize both thresholds, preserve precision through the final calculation, and apply coinsurance to the layer payment.

Check Your Understanding

For each problem, assume the stated normal model and calculate expected payments in the units specified. Round monetary answers to the nearest cent.

A hypothetical pool has aggregate loss $X\sim N(500,100^2)$ dollars and an unlimited attachment of $600$ dollars. Using $\phi(1)=0.2419707$ and $\overline{\Phi}(1)=0.1586553$, find the expected payment. (Answer: $8.33)
A hypothetical plan has $X\sim N(1500,250^2)$ dollars. An unlimited contract attaches at $1250$ dollars. Use $\phi(1)=0.2419707$ and $\overline{\Phi}(-1)=0.8413447$ to calculate its expected payment. (Answer: $270.83)
For aggregate loss $X\sim N(2000,400^2)$ dollars, calculate the expected payment for a layer attaching at $2000$ dollars with a limit of $400$ dollars. Use $\phi(0)=0.3989423$, $\phi(1)=0.2419707$, and $\overline{\Phi}(1)=0.1586553$. (Answer: $126.25)
A hypothetical insurer models annual aggregate loss as $X\sim N(1000,200^2)$ thousands of dollars. It covers the layer attaching at $900$ with a limit of $300$. Use $\phi(-0.5)=0.3520653$, $\overline{\Phi}(-0.5)=0.6914625$, $\phi(1)=0.2419707$, and $\overline{\Phi}(1)=0.1586553$. Find the expected layer payment. (Answer: $122.90 thousand)
A hypothetical reinsurer models aggregate loss as $X\sim N(1000,200^2)$ dollars and pays 75% of a layer attaching at $1100$ dollars with a limit of $500$ dollars. Use $\phi(0.5)=0.3520653$, $\overline{\Phi}(0.5)=0.3085375$, $\phi(3)=0.00443185$, and $\overline{\Phi}(3)=0.00134990$. What is its expected payment? (Answer: $29.61)

What You'll Learn