Deriving TVaR from Stop-Loss Expectations

Why Stop-Loss Expectations Give TVaR

VaR identifies a loss threshold at a selected confidence level, but it does not describe how large losses tend to be once that threshold is exceeded. Tail value-at-risk (TVaR) adds that information. A useful way to calculate it is through the expected amount by which a loss exceeds the VaR threshold: the stop-loss expectation.

The connection is especially direct for a continuous loss distribution. The conditional mean above the VaR threshold equals the threshold plus the average excess over it, conditional on exceeding it. The stop-loss expectation supplies that excess in unconditional dollars, so dividing by the probability of entering the tail converts it to a conditional amount.

Definition: Let $X$ be a nonnegative, integrable loss with a continuous distribution, and let $0<\alpha<1$. Define $\operatorname{VaR}_{\alpha}(X)=q_{\alpha}$, where $F_X(q_{\alpha})=\alpha$. In this continuous setting, $\operatorname{TVaR}_{\alpha}(X)=E[X\mid X>q_{\alpha}]$. The stop-loss expectation at threshold $t$ is $E[(X-t)^{+}]$, where $(z)^{+}=\max(z,0)$.

The continuity condition matters: it ensures there is no probability mass at the threshold and $P(X>q_{\alpha})=1-\alpha$. For a distribution with an atom at the quantile, the conditional mean strictly above the quantile need not agree with every standard TVaR convention. The identity below is stated for the continuous case.

Deriving the Stop-Loss Identity

Set $q=q_{\alpha}$. On the event $X>q$, the excess over the threshold is $X-q$; otherwise the positive part is zero. Thus $E[(X-q)^{+}]=E[(X-q)\mathbf{1}_{\{X>q\}}]$. The conditional expectation formula then gives $E[X\mid X>q]=E[X\mathbf{1}_{\{X>q\}}]/P(X>q)$.

To relate the numerator to the stop-loss expectation, write $X=q+(X-q)$ on the tail event. Since $P(X>q)=1-\alpha$, the resulting equation is

\begin{aligned} E[X\mathbf{1}_{\{X>q\}}] &=qP(X>q)+E[(X-q)^{+}]\\ &=q(1-\alpha)+E[(X-q)^{+}]. \end{aligned}

Dividing by $1-\alpha$ establishes the identity:

\operatorname{TVaR}_{\alpha}(X) =q_{\alpha}+\frac{E[(X-q_{\alpha})^{+}]}{1-\alpha}.

This derivation also explains the denominator. The stop-loss expectation averages the excess over all outcomes, counting zero excess for losses at or below the threshold. Dividing by the tail probability $1-\alpha$ gives the mean excess among tail outcomes.

Result: For a nonnegative continuous loss and a threshold $t$, the stop-loss expectation can also be written $E[(X-t)^{+}]=\int_t^\infty S_X(x)\,dx$. Consequently, at the VaR threshold, $\operatorname{TVaR}_{\alpha}(X)=q_{\alpha}+\frac{1}{1-\alpha}\int_{q_{\alpha}}^\infty S_X(x)\,dx$, provided the tail expectation is finite.

The integral result follows by writing $(X-t)^{+}=\int_t^\infty \mathbf{1}_{\{X>x\}}\,dx$. Taking expectations and interchanging the expectation and integral for a nonnegative integrand gives $E[(X-t)^{+}]=\int_t^\infty P(X>x)\,dx=\int_t^\infty S_X(x)\,dx$. This is useful when the survival function is easier to integrate than the loss density is to condition on.

Worked Calculations

Worked Example: Exponential Loss and a 90% Confidence Level

A hypothetical equipment policy has loss $X$ with an exponential distribution of mean $\theta=\$1{,}000$, so $S_X(x)=e^{-x/1000}$ for $x\geq 0$. Calculate $\operatorname{TVaR}_{0.90}(X)$, using the stop-loss identity.

Method and assumptions. The exponential model is continuous, has finite mean, and has scale (mean) $\theta=1000$. Its quantile satisfies $F_X(q)=0.90$, or $S_X(q)=0.10$.

Find VaR. Solve $e^{-q/1000}=0.10$. Taking logarithms gives $q=-1000\ln(0.10)=1000\ln(10)=\$2{,}302.5851$.

Find the stop-loss expectation. Using the survival integral, $E[(X-q)^{+}]=\int_q^\infty e^{-x/1000}\,dx=1000e^{-q/1000}$. Since $e^{-q/1000}=0.10$, this is $\$100$.

Calculate TVaR. The tail probability is $1-\alpha=0.10$, so $\operatorname{TVaR}_{0.90}(X)=2302.5851+100/0.10=\$3{,}302.5851$. Rounded to the nearest cent, TVaR is \$3,302.59.

Check. The exponential distribution is memoryless, so the mean excess above any threshold is $\$1{,}000$. The result is $q+\$1{,}000$, matching the calculation. The method avoids mistakenly adding the unconditional stop-loss amount of $\$100$ directly to VaR.

Worked Example: Uniform Loss with an Upper Endpoint

For a hypothetical coverage, let $X$ be uniformly distributed from $\$0$ to $\$12{,}000$. Find TVaR at confidence level $75\%$, using the stop-loss expectation.

Method and assumptions. The density is $f_X(x)=1/12000$ for $0\leq x\leq12000$, and zero otherwise. The loss is continuous, and its upper endpoint makes the stop-loss integral finite.

Find VaR. For a uniform loss on $[0,12000]$, $F_X(x)=x/12000$ within the interval. Thus $q_{0.75}=0.75(12000)=\$9{,}000$.

Calculate the stop-loss expectation. Losses above $q$ range from 9,000 to 12,000:

\begin{aligned} E[(X-9000)^{+}] &=\int_{9000}^{12000}(x-9000)\frac{1}{12000}\,dx\\ &=\frac{1}{12000}\left[\frac{(x-9000)^2}{2}\right]_{9000}^{12000}\\ &=\frac{3000^2}{24000}=\$375. \end{aligned}

Calculate TVaR. The tail probability is $1-0.75=0.25$. Therefore $\operatorname{TVaR}_{0.75}(X)=9000+375/0.25=9000+1500=\$10{,}500$.

Check. Conditional on exceeding 9,000, the uniform loss remains uniform on $[9000,12000]$. Its conditional mean is $(9000+12000)/2=\$10{,}500$, confirming the stop-loss result. The method avoids treating the stop-loss expectation of $\$375$ as the conditional mean excess.

Worked Example: Pareto Loss with a Heavy Tail

A hypothetical catastrophe loss $X$ has a Pareto Type I distribution with minimum loss $x_m=\$1{,}000$ and shape parameter $a=2$: $S_X(x)=(1000/x)^2$ for $x\geq1000$. Find $\operatorname{TVaR}_{0.90}(X)$.

Method and assumptions. This is a continuous Pareto model. Its mean is finite because $a=2>1$; a finite tail mean is required for TVaR. The desired tail probability is $0.10$.

Find VaR. Solve $S_X(q)=0.10$: $(1000/q)^2=0.10$, so $q=1000/\sqrt{0.10}=\$3{,}162.2777$.

Calculate the stop-loss expectation. Integrating the survival function from $q$ to infinity gives

\begin{aligned} E[(X-q)^{+}] &=\int_q^\infty \left(\frac{1000}{x}\right)^2 dx\\ &=1{,}000{,}000\left[-\frac{1}{x}\right]_q^\infty\\ &=\frac{1{,}000{,}000}{q} =\$316.2278. \end{aligned}

Calculate and check TVaR. Divide by the tail probability and add VaR: $\operatorname{TVaR}_{0.90}(X)=3162.2777+316.2278/0.10=\$6{,}324.5553$. Rounded to the nearest cent, the answer is \$6,324.56. As a check, the Pareto conditional tail mean for $a>1$ is $aq/(a-1)=2q$, which here is $\$6{,}324.5553$.

The method makes the finiteness condition visible: for Pareto shape $a\leq1$, the survival integral from any finite threshold diverges, so the tail mean is infinite and this finite-TV aR calculation does not apply.

Calculation Workflow and Exam Traps

A reliable calculation separates the quantile from the expected excess. First identify the loss distribution and verify its assumptions. Next find $q_{\alpha}$ from $F_X(q_{\alpha})=\alpha$, or equivalently $S_X(q_{\alpha})=1-\alpha$. Then calculate $E[(X-q_{\alpha})^{+}]$, either from the positive-part definition or by integrating $S_X$. Finally, divide that expectation by $1-\alpha$ and add $q_{\alpha}$.

Confirm the tail definition.
Use a continuous loss model with $F_X(q_{\alpha})=\alpha$, so the probability above the threshold is $1-\alpha$.

Calculate the quantile.
Invert the distribution function or solve the survival equation $S_X(q_{\alpha})=1-\alpha$.

Calculate the stop-loss amount.
Evaluate $E[(X-q_{\alpha})^{+}]$, not an expectation that includes negative differences below the threshold.

Convert to a tail conditional amount.
Divide by $1-\alpha$, then add $q_{\alpha}$. Check units and whether the result is at least the VaR threshold.

A frequent error is to replace the positive part with an ordinary difference. For the uniform example, $E[X]=\$6{,}000$ and $q=\$9{,}000$. The incorrect calculation $E[X-q]=6000-9000=-\$3{,}000$ includes losses below the threshold as negative excesses. But the stop-loss payment is zero below the threshold, so the correct quantity is $E[(X-q)^{+}]=\$375$, as calculated by integrating only from 9,000 to 12,000. A negative stop-loss expectation is impossible.

A second trap is dividing by $\alpha$ instead of $1-\alpha$. The stop-loss expectation is an unconditional average over the entire distribution, while the desired mean excess is conditional on being in the upper tail. Its conditioning event has probability $1-\alpha$, not $\alpha$. The result should also satisfy $\operatorname{TVaR}_{\alpha}(X)\geq q_{\alpha}$; if it does not, recheck the positive part, tail probability, and arithmetic.

Key takeaway: For a continuous loss with a finite tail mean, TVaR is the VaR threshold plus the stop-loss expectation at that threshold divided by the tail probability. The positive-part operator and the factor $1-\alpha$ are essential to the identity.

Check Your Understanding

For each problem, calculate the requested quantity using the stop-loss identity. Answers are rounded to the nearest cent when needed.

A hypothetical loss is uniform on $[0,\$20{,}000]$. Find TVaR at the 80% confidence level. (Answer: \$18,000.00)
A hypothetical loss is exponential with mean $\$500$. Find TVaR at the 95% confidence level. (Answer: \$1,997.87)
A hypothetical loss follows a Pareto Type I distribution with $x_m=\$200$ and $a=3$. Find TVaR at the 87.5% confidence level. (Answer: \$600.00)
A hypothetical exponential loss has mean $\$800$. At threshold $t=\$1{,}600$, its survival probability is $e^{-2}$. Calculate the stop-loss expectation at $t$ and the conditional mean loss given $X>t$. (Answer: stop-loss expectation $\$108.27$; conditional mean $\$2,400.00$)
A hypothetical loss has survival function $S_X(x)=(1+x/1000)^{-2}$ for $x\geq0$. Find TVaR at the 90% confidence level. (Answer: \$5,324.56)

What You'll Learn