Stop-Loss Expectations and Tail Risk

Why Stop-Loss Expectations Matter

A loss amount and the portion of that loss above a specified threshold answer different questions. For a hypothetical insurer, the expectation of the excess above a retention can quantify the average amount by which losses exceed that retention. The same quantity also gives a direct route to tail value-at-risk (TVaR) when the retention is a value-at-risk quantile.

This connection is useful because a stop-loss expectation can often be calculated directly from a loss distribution's survival function or from an excess-loss formula. Care is needed, especially for discrete losses: the upper tail used in TVaR may include only part of the probability mass at the quantile.

Definition: For a loss $X$ and threshold $d$, the stop-loss expectation at $d$ is $E[(X-d)^+]$, where $(X-d)^+=\max(X-d,0)$. It is the expected excess of the loss above $d$; losses at or below $d$ contribute zero.

Let $F_X(x)=P(X\leq x)$ be the distribution function and $S_X(x)=P(X>x)=1-F_X(x)$ the survival function. Using the lower quantile convention, define $\operatorname{VaR}_p(X)=\inf\{x:F_X(x)\geq p\}$, for $0<p<1$. In this tutorial, TVaR uses the integral-quantile definition:

\operatorname{TVaR}_p(X)=\frac{1}{1-p}\int_p^1 \operatorname{VaR}_q(X)\,dq.

Assume $X$ has a finite expectation when using this formula. Unlike the conditional mean $E[X\mid X>\operatorname{VaR}_p(X)]$, the integral-quantile definition applies consistently when $X$ has discrete probability masses.

Calculating the Stop-Loss Expectation

For any real-valued loss $X$ and finite threshold $d$, the stop-loss expectation can be calculated from the survival function:

Theorem: If $E[(X-d)^+]$ is finite, then

E[(X-d)^+]=\int_d^\infty S_X(x)\,dx.

This identity says that the expectation of the excess is the area under the survival curve to the right of the threshold. For a loss distribution with atoms, the survival function still gives the correct integral; no separate adjustment for an atom at $d$ is needed.

Set the threshold.
Identify the value $d$ above which excess loss is being measured.

Write the survival function.
Use $S_X(x)=P(X>x)$, taking care that this is a strict exceedance probability.

Integrate the tail.
Evaluate $\int_d^\infty S_X(x)\,dx$, splitting the integral wherever the survival function changes form.

Check units and interpretation.
The result is an expected loss amount, not a probability and not a conditional mean by itself.

A related excess-loss formula follows when $P(X>d)>0$. Conditional on exceeding $d$, the excess is $X-d$, so

E[(X-d)^+]=P(X>d)E[X-d\mid X>d].

This formula is valid for discrete as well as continuous distributions. It calculates the stop-loss expectation by multiplying the probability of exceeding the threshold by the mean excess conditional on exceeding it. It does not, on its own, calculate TVaR.

Deriving the Link to Tail Value-at-Risk

The key result is that the stop-loss expectation at a value-at-risk quantile determines TVaR. Let $d=\operatorname{VaR}_p(X)$, and write $Q(u)=\operatorname{VaR}_u(X)$ for the lower quantile function. The quantile convention implies

F_X(d-) \leq p \leq F_X(d),

where $F_X(d-)=P(X<d)$. Thus, if there is probability mass at $d$, the probability level $p$ can fall within that mass. For quantile levels from $p$ through $F_X(d)$, the quantile equals $d$. At higher levels, the quantile is above $d$. The stop-loss expectation can be expressed as

E[(X-d)^+]=\int_{F_X(d)}^1 \bigl(Q(u)-d\bigr)\,du.

To see why, a uniform quantile level $u$ generates a loss with the distribution of $X$. Quantile levels up to $F_X(d)$ correspond to losses no greater than $d$ and contribute no positive excess. The remaining levels contribute $Q(u)-d$. Now split the integral defining TVaR at $F_X(d)$:

\begin{aligned} (1-p)\operatorname{TVaR}_p(X) &=\int_p^{F_X(d)}d\,du+\int_{F_X(d)}^1 Q(u)\,du\\ &=d\bigl(F_X(d)-p\bigr) +E[(X-d)^+]+d\bigl(1-F_X(d)\bigr)\\ &=d(1-p)+E[(X-d)^+]. \end{aligned}

Therefore the general quantile-threshold relationship is

Theorem: For $0<p<1$, $d=\operatorname{VaR}_p(X)$, and finite $E[(X-d)^+]$,

\operatorname{TVaR}_p(X) =d+\frac{E[(X-d)^+]}{1-p}.

The derivation explains how the formula handles atoms. The part of the atom at $d$ needed to reach probability level $p$ contributes $d$ to the quantile integral, while the stop-loss expectation counts only amounts strictly above $d$. Together these pieces give the full upper-tail integral. If $X$ is continuous at $d$, then $F_X(d)=p$, and the same result follows without a partial-atom segment.

Worked Calculations

Worked Example: A Continuous Loss with a Uniform Distribution

A hypothetical equipment policy has loss $X\sim\operatorname{Uniform}(0,100)$, with amounts measured in dollars. Calculate the 80% value-at-risk, the stop-loss expectation at that threshold, and the 80% TVaR.

Method. Find the quantile first, integrate the survival function above it, and then apply the quantile-threshold TVaR formula. This order avoids confusing the expected excess with the tail mean.

For $0\leq x\leq100$, the distribution function is $F_X(x)=x/100$. Solving $F_X(d)=0.80$ gives

d=\operatorname{VaR}_{0.80}(X)=100(0.80)=80.

On this range, $S_X(x)=1-x/100$. Therefore

\begin{aligned} E[(X-80)^+] &=\int_{80}^{100}\left(1-\frac{x}{100}\right)\,dx\\ &=\left[x-\frac{x^2}{200}\right]_{80}^{100}\\ &=(100-50)-(80-32)\\ &=50-48=2. \end{aligned}

The tail probability is $1-p=0.20$, so

\operatorname{TVaR}_{0.80}(X) =80+\frac{2}{0.20} =80+10 =90.

The result is consistent with the upper 20% of a uniform distribution on $[0,100]$: that portion runs from 80 to 100 and has mean 90. The stop-loss expectation is $2$ dollars, while TVaR is $90$ dollars; they are different quantities.

Worked Example: A Discrete Loss with an Atom at the Quantile

A hypothetical warranty portfolio has loss $X$ with $P(X=0)=0.60$, $P(X=40)=0.25$, and $P(X=100)=0.15$, with amounts in dollars. Find the 80% value-at-risk, stop-loss expectation at that threshold, and 80% TVaR.

Method. Use the lower quantile convention, then calculate the excess above the quantile. The important check is whether the probability level falls inside an atom.

The cumulative probabilities are $F_X(0)=0.60$, $F_X(40)=0.85$, and $F_X(100)=1$. Since $0.60<0.80\leq0.85$,

\operatorname{VaR}_{0.80}(X)=40.

Only the loss of 100 exceeds 40. Its excess is $100-40=60$, so

$$ E[(X-40)^+] =0.60(0)+0.25(0)+0.15(60) =9. $$

The TVaR is consequently

\operatorname{TVaR}_{0.80}(X) =40+\frac{9}{1-0.80} =40+45 =85.

Check this using the quantile integral. The interval from 0.80 to 0.85 has quantile 40, and the interval from 0.85 to 1 has quantile 100:

\operatorname{TVaR}_{0.80}(X) =\frac{(0.85-0.80)(40)+(1-0.85)(100)}{0.20} =\frac{2+15}{0.20} =85.

Both calculations agree. The quantile integral includes just 0.05 of probability at the loss amount 40, not the entire 0.25 atom.

Worked Example: Integrating a Piecewise Survival Function

A hypothetical loss $X$ has survival function

S_X(x)= \begin{cases} 1, & x<0,\\ 1-\dfrac{x}{80}, & 0\leq x\leq40,\\ \dfrac{100-x}{120}, & 40<x\leq100,\\ 0, & x>100. \end{cases}

Calculate the 50% value-at-risk, the stop-loss expectation at that threshold, and the 50% TVaR. Assume amounts are in dollars.

Method. First locate the quantile from $F_X(x)=1-S_X(x)$. Then integrate only the part of the survival function above the threshold, splitting at the point where its formula changes.

At $x=40$, $S_X(40)=1-40/80=0.50$, so $F_X(40)=0.50$. The survival function is continuous there, and hence

d=\operatorname{VaR}_{0.50}(X)=40.

For $40\leq x\leq100$, the relevant survival function is $(100-x)/120$. Thus

\begin{aligned} E[(X-40)^+] &=\int_{40}^{100}\frac{100-x}{120}\,dx\\ &=\frac{1}{120}\left[100x-\frac{x^2}{2}\right]_{40}^{100}\\ &=\frac{1}{120}\left((10000-5000)-(4000-800)\right)\\ &=\frac{1}{120}(5000-3200) =\frac{1800}{120} =15. \end{aligned}

Finally,

\operatorname{TVaR}_{0.50}(X) =40+\frac{15}{0.50} =40+30 =70.

As a check, $P(X>40)=0.50$. Conditional on exceeding 40, the survival probability of the excess at $y$, for $0\leq y\leq60$, is $S_X(40+y)/0.50=1-y/60$. Its integral is $\int_0^{60}(1-y/60)\,dy=30$, giving an expected excess of $0.50(30)=15$, as calculated.

Common Exam Trap: Confusing Excess with Tail Mean

A stop-loss expectation is not itself a TVaR. It is the expected excess above the threshold across all outcomes, including zero contribution from outcomes at or below the threshold. Dividing it by a tail probability gives a conditional mean excess in some settings, not the tail mean loss.

For the discrete example, a tempting but incorrect calculation is

\frac{E[(X-40)^+]}{P(X>40)} =\frac{9}{0.15} =60,

and then to report 60 as $\operatorname{TVaR}_{0.80}(X)$. The value 60 is $E[X-40\mid X>40]$, the mean excess conditional on a strict exceedance. It is not the mean loss in the upper 20% of probability. The correct calculation is

\operatorname{TVaR}_{0.80}(X) =40+\frac{9}{0.20} =85.

The denominator is $1-p$, as specified by the integral-quantile definition. In this example, $F_X(40)=0.85$, which is greater than $p=0.80$. Using only the probability strictly above 40 would omit the 0.05 of probability at the quantile that belongs to the upper 20% tail.

Key takeaway: At $d=\operatorname{VaR}_p(X)$, $\operatorname{TVaR}_p(X)=d+E[(X-d)^+]/(1-p)$. Calculate the stop-loss expectation from the survival function or an excess-loss calculation, and retain the denominator $1-p$ even when the distribution has an atom at $d$.

Check Your Understanding

For each problem, use the stated lower-quantile convention and give monetary answers in dollars.

A hypothetical loss is uniform on $[0,60]$. Find its 75% value-at-risk, the stop-loss expectation at that threshold, and its 75% TVaR. (Answer: VaR = $45, stop-loss expectation = $1.875, TVaR = $52.50)
A hypothetical claim has probabilities $P(X=0)=0.40$, $P(X=30)=0.35$, and $P(X=90)=0.25$. Find the 60% value-at-risk and the 60% TVaR. (Answer: VaR = $30, TVaR = $67.50)
A continuous hypothetical loss has $P(X>25)=0.30$ and $E[X-25\mid X>25]=18$. If 25 is its 70% value-at-risk, find the stop-loss expectation there and its 70% TVaR. (Answer: stop-loss expectation = $5.40, TVaR = $43.00)
A hypothetical loss is uniform on $[0,100]$. Use the survival-function integral to find the stop-loss expectation at the 90% value-at-risk, then calculate 90% TVaR. (Answer: VaR = $90, stop-loss expectation = $0.50, TVaR = $95.00)
A hypothetical discrete loss has $P(X=10)=0.20$, $P(X=50)=0.50$, and $P(X=110)=0.30$. Find the 60% value-at-risk, the stop-loss expectation there, and the 60% TVaR. (Answer: VaR = $50, stop-loss expectation = $18, TVaR = $95.00)

What You'll Learn