Moment Bounds for Stop-Loss Premiums

Why Bound a Stop-Loss Premium?

For an aggregate loss $S$ and an attachment point $d$, the stop-loss payment is $(S-d)_+=\max(S-d,0)$. Its expected value is the pure stop-loss premium. A full calculation may require the distribution of $S$, but sometimes the available information consists only of its mean and variance. Moment bounds then give a quick way to check the scale of the premium and to identify calculations that cannot be correct.

The lower bound uses convexity and requires a finite mean. The upper bound uses the second moment about the attachment point and requires a finite variance. Neither bound generally determines the exact premium. In particular, knowing the mean and variance does not usually determine how much probability lies above $d$, which is essential to the stop-loss payment.

Definition: For an aggregate loss $S$ and fixed attachment $d$, the stop-loss premium is $E[(S-d)_+]$, where $x_+=\max(x,0)$. If $E[S]$ is finite, then $$ E[(S-d)_+] \geq \max(E[S]-d,0). $$ If, in addition, $\operatorname{Var}(S)$ is finite, then $$ E[(S-d)_+] \leq \sqrt{\operatorname{Var}(S)+(E[S]-d)^2}. $$

Here $d$ is the attachment applied to the aggregate loss, not a per-claim deductible. The bounds concern the undiscounted expected payment at one time; no interest or mortality model is involved. Ordinarily $S$ is a nonnegative aggregate loss and $d\geq 0$, although the inequalities themselves hold for any real-valued $S$ and fixed real $d$ under the stated moment conditions.

Why the Bounds Hold

The lower bound follows by applying Jensen’s inequality to the convex function $g(x)=x_+$. For a random variable $Z$ with a finite mean, Jensen’s inequality gives $E[g(Z)]\geq g(E[Z])$. Set $Z=S-d$. Since $d$ is fixed, $E[Z]=E[S]-d$, so

E[(S-d)_+] \geq (E[S]-d)_+ = \max(E[S]-d,0).

This argument also explains why a positive expected loss above attachment is a guaranteed lower bound: the average of the positive parts cannot be less than the positive part of the average. If $E[S]\leq d$, the result is only the trivial lower bound zero; it does not mean the stop-loss premium is zero. Some possible outcomes may still exceed the attachment.

For the upper bound, write $Z=S-d$. Cauchy–Schwarz applied to $Z_+$ and the constant random variable 1 gives $E[Z_+]\leq \sqrt{E[Z_+^2]}$. Pointwise, $0\leq Z_+\leq |Z|$, so $Z_+^2\leq Z^2$. Therefore,

E[(S-d)_+] \leq \sqrt{E[(S-d)^2]} = \sqrt{\operatorname{Var}(S)+(E[S]-d)^2}.

The final equality uses the identity $E[(S-d)^2]=\operatorname{Var}(S)+(E[S]-d)^2$. The square root is the root-mean-square distance of $S$ from the attachment. It is an upper bound because the stop-loss payment discards outcomes below $d$, while the squared distance counts deviations on both sides.

Confirm the available moments.
Use the lower bound when the mean is finite. Use the upper bound only when the variance is also finite.

Measure the mean relative to attachment.
Calculate $E[S]-d$, then take its positive part for the lower bound.

Calculate the upper bound.
Add the variance to $(E[S]-d)^2$, then take the square root.

Check the result.
The lower bound must not exceed the upper bound. A calculated premium must lie between them.

Worked Applications

Worked Example: A Two-Point Aggregate Distribution

A hypothetical one-period portfolio has aggregate loss $S$ equal to 2,000 with probability 0.5 and 2,800 with probability 0.5. A stop-loss contract attaches at 2,200. Calculate the moment bounds and the exact premium as a check. This is a discrete distribution with the two stated outcomes and probabilities. There is no discounting or mortality assumption.

Step 1: Calculate the mean.

$$ E[S]=0.5(2{,}000)+0.5(2{,}800)=1{,}000+1{,}400=2{,}400. $$

Step 2: Calculate the variance. The deviations from the mean are $-400$ and $400$, so

\operatorname{Var}(S)=0.5(-400)^2+0.5(400)^2 =0.5(160{,}000)+0.5(160{,}000)=160{,}000.

Step 3: Apply the lower bound.

\max(E[S]-d,0)=\max(2{,}400-2{,}200,0)=200.

Step 4: Apply the upper bound.

\sqrt{160{,}000+(2{,}400-2{,}200)^2} =\sqrt{160{,}000+40{,}000} =\sqrt{200{,}000}\approx 447.21.

Step 5: Check against the exact payment. At $S=2{,}000$, the payment is zero. At $S=2{,}800$, it is $2{,}800-2{,}200=600$. Thus the exact premium is $0.5(0)+0.5(600)=300$, which satisfies $200\leq 300\leq 447.21$. The method is to calculate moments first and use the actual distribution only for the check. A common mistake is to report the lower bound, 200, as the exact premium.

Worked Example: A Mean Below the Attachment

A hypothetical aggregate loss has this discrete distribution: $S=0$ with probability 0.25, $S=800$ with probability 0.50, and $S=1{,}600$ with probability 0.25. The attachment is $d=1{,}000$. Find both bounds and check them against the exact premium. No discounting or mortality applies.

Step 1: Find the mean.

$$ E[S]=0.25(0)+0.50(800)+0.25(1{,}600)=0+400+400=800. $$

Step 2: Find the variance. The deviations from 800 are $-800$, $0$, and $800$.

\operatorname{Var}(S)=0.25(800^2)+0.50(0^2)+0.25(800^2) =160{,}000+0+160{,}000=320{,}000.

Step 3: Calculate the bounds. Because the mean is below attachment, the Jensen lower bound is zero. The upper bound is

\sqrt{320{,}000+(800-1{,}000)^2} =\sqrt{320{,}000+40{,}000} =\sqrt{360{,}000}=600.

Step 4: Check the exact premium. The first two outcomes pay zero. The outcome 1,600 pays $1{,}600-1{,}000=600$, so the exact premium is $0.25(600)=150$. It lies between zero and 600. The method correctly produces a valid interval, but the lower bound alone is uninformative here. The common mistake is to infer that a mean below attachment implies no payment; the 1,600 outcome shows why that inference fails.

Worked Example: A Binomial Claim Count with Fixed Severity

A hypothetical portfolio has claim count $N\sim\operatorname{Binomial}(4,0.25)$. Each claim has fixed severity 400, so $S=400N$, and the aggregate stop-loss attaches at 500. Find the moment bounds and verify them by evaluating the possible payments. Treat this as a one-period undiscounted model with no mortality assumption.

Step 1: Calculate the aggregate mean and variance. For a binomial count, $E[N]=np=4(0.25)=1$ and $\operatorname{Var}(N)=np(1-p)=4(0.25)(0.75)=0.75$. Multiplying a random variable by 400 multiplies its mean by 400 and its variance by $400^2$. Therefore,

E[S]=400(1)=400,\qquad \operatorname{Var}(S)=400^2(0.75)=160{,}000(0.75)=120{,}000.

Step 2: Calculate the bounds.

\max(400-500,0)=0, $$ $$ \sqrt{120{,}000+(400-500)^2} =\sqrt{120{,}000+10{,}000} =\sqrt{130{,}000}\approx 360.56.

Step 3: Verify using the count distribution. A payment occurs only when $N$ is 2, 3, or 4. The probabilities and payments are

P(N=2)=\binom{4}{2}(0.25)^2(0.75)^2=0.2109375, $$ $$ P(N=3)=\binom{4}{3}(0.25)^3(0.75)=0.046875,\qquad P(N=4)=(0.25)^4=0.00390625.

The respective payments are $800-500=300$, $1{,}200-500=700$, and $1{,}600-500=1{,}100$. Hence

E[(S-500)_+] =0.2109375(300)+0.046875(700)+0.00390625(1{,}100) =63.28125+32.8125+4.296875 =100.390625.

Rounded to the nearest cent, the exact premium is 100.39, within the bounds of 0 and 360.56. The common mistake is to use $\operatorname{Var}(N)$ directly as the aggregate variance; the fixed severity requires multiplying it by $400^2$.

Interpreting and Checking the Results

These bounds are most useful when the full aggregate distribution is unavailable or when a quick reasonableness check is needed. They depend only on the first two moments, so different loss distributions with the same mean and variance can have different exact stop-loss premiums while sharing these bounds. A wide interval is therefore not a calculation error; it may simply reflect how little information the moments provide about the upper tail.

Several checks help prevent common errors:

Do not omit the positive part in the lower bound. If $E[S]-d$ is negative, the lower bound is zero, not a negative premium.
Do not confuse the bounds with an exact premium. Use the aggregate distribution, a probability table, or an appropriate tail calculation when the exact value is requested and the information permits it.
Use variance, not standard deviation, in the formula. The radicand is $\operatorname{Var}(S)+(E[S]-d)^2$.
Keep the attachment and aggregate loss on the same scale. If losses are stated in thousands, express $d$ in thousands before calculating.
Check the ordering. The lower bound cannot exceed the upper bound, and any exact premium must lie between them.

For example, a result below $\max(E[S]-d,0)$ signals an error in the premium or the moments. A result above the square-root bound signals that the variance, attachment, or units should be checked. These bounds are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus because they connect aggregate moments to stop-loss pricing without requiring a complete distribution calculation.

Check Your Understanding

For each problem, calculate the requested moment bounds. Treat all losses as hypothetical, undiscounted aggregate losses; no mortality assumption applies.

An aggregate loss is 500 with probability 0.4 and 1,500 with probability 0.6. The attachment is 1,200. Find the lower and upper bounds, then calculate the exact premium. (Answer: lower bound 0; upper bound 500; exact premium 180.)
A nonnegative aggregate loss has mean 900 and variance 160,000. The attachment is 700. Find both moment bounds. (Answer: lower bound 200; upper bound $\sqrt{200{,}000}\approx 447.21$.)
A claim count has distribution $N\sim\operatorname{Binomial}(3,0.2)$, with fixed severity 250 per claim. The aggregate attachment is 400. Calculate the bounds using the aggregate mean and variance. (Answer: lower bound 0; upper bound $\sqrt{92{,}500}\approx 304.14$.)
An aggregate loss is 0 or 1,000, each with probability 0.5, and the attachment is 600. Calculate both bounds and the exact premium. (Answer: lower bound 0; upper bound $\sqrt{260{,}000}\approx 509.90$; exact premium 200.)
Two independent aggregate components each have mean 300 and variance 40,000. Their sum is subject to an attachment of 500. Find the bounds for the sum. (Answer: lower bound 100; upper bound 300.)

What You'll Learn