TVaR for Mixed Distributions

Why Mixed Distributions Need Care

A loss distribution is mixed when it has both a continuous component and one or more point masses. For example, a hypothetical benefit might vary continuously over a range of outcomes, while a separate event produces a fixed loss with positive probability. To calculate tail value at risk (TVaR), the upper tail may therefore contain a continuous range of losses, one or more complete point masses, and possibly only part of the mass at the VaR threshold.

The key is to select exactly the worst \(1-\alpha\) fraction of outcomes, where \(\alpha\) is the confidence level. If the cumulative distribution jumps at the VaR, the upper tail generally needs only part of that jump. Including the whole atom would select too much probability; omitting it would select too little.

Definition: Let \(X\) be a loss with finite expected value and let \(0<\alpha<1\). Define \(\operatorname{VaR}_{\alpha}(X)=\inf\{x:F_X(x)\geq\alpha\}\), where \(F_X\) is the cumulative distribution function. TVaR at level \(\alpha\) is the mean loss in the worst \(1-\alpha\) fraction of outcomes, with only the required fraction of any point mass at the VaR included.

Write \(q=\operatorname{VaR}_{\alpha}(X)\). Since \(F_X(q)\geq\alpha\), the probability strictly above \(q\) is \(1-F_X(q)\). The amount of probability to take from the atom at \(q\) is \(F_X(q)-\alpha\). These combine to give exactly the required upper-tail probability:

\bigl(1-F_X(q)\bigr)+\bigl(F_X(q)-\alpha\bigr)=1-\alpha.

If the distribution has continuous density \(f_c\) and atoms at values \(x_j\) with probabilities \(p_j\), then \(f_c\) is the unconditional density of the continuous component; its integral need not equal one. The TVaR calculation is

\operatorname{TVaR}_{\alpha}(X) = \frac{ \displaystyle \int_{q}^{\infty}x f_c(x)\,dx +\displaystyle\sum_{x_j>q}x_jp_j +q\bigl(F_X(q)-\alpha\bigr) }{1-\alpha}.

The integral and sum account for all losses strictly above \(q\); the final term adds the required fraction at \(q\). If \(q\) is not an atom, then \(F_X(q)=\alpha\) under the usual continuous-crossing case, and that final term is zero. If an atom occurs at \(q\), then \(F_X(q)-\alpha\) is between zero and the probability of that atom.

Why the Fractional-Atom Formula Holds

The formula follows by constructing a tail with probability exactly \(1-\alpha\). First take every outcome with loss strictly greater than \(q\). Its probability is \(1-F_X(q)\). The probability still needed is \((1-\alpha)-(1-F_X(q))=F_X(q)-\alpha\). Every outcome at the threshold has loss \(q\), so the expected-loss contribution from this selected part is \(q(F_X(q)-\alpha)\). Adding that contribution to the losses strictly above \(q\) gives the numerator. Dividing by the selected probability \(1-\alpha\) gives the average loss in that tail.

This reasoning also explains why TVaR for a mixed distribution is not generally the conditional mean \(E[X\mid X\geq q]\). That conditional mean includes the entire atom at \(q\), even when only a fraction of it belongs in the worst \(1-\alpha\) outcomes.

Find the quantile.
Use the cumulative distribution and its jumps to identify \(q=\operatorname{VaR}_{\alpha}(X)\).

Count the probability above the threshold.
Calculate \(1-F_X(q)\), including continuous mass and every atom strictly above \(q\).

Complete the tail at the threshold.
Add probability \(F_X(q)-\alpha\) at loss \(q\), not necessarily the whole atom.

Calculate and check the average.
Sum the loss contributions and divide by \(1-\alpha\). Confirm that the selected tail probabilities add to \(1-\alpha\).

Worked Calculations

Worked Example: A VaR Atom with a Continuous Tail Above It

A hypothetical loss \(X\) has probability \(0.7\) of following a uniform distribution on \([0,100]\), and probability \(0.3\) of being exactly \(60\). Calculate \(\operatorname{VaR}_{0.6}(X)\) and \(\operatorname{TVaR}_{0.6}(X)\).

Method. Find the jump at \(60\), then combine the continuous losses above \(60\) with only the fraction of the atom needed for a tail probability of \(1-0.6=0.4\).

For \(x<60\), \(F_X(x)=0.7x/100\), so \(F_X(60-)=0.7(60/100)=0.42\). At \(60\), the point mass raises the CDF to \(F_X(60)=0.42+0.3=0.72\). Since \(0.42<0.6\leq0.72\), \(q=60\).

The continuous probability strictly above \(60\) is \(0.7(100-60)/100=0.28\). The probability to take at \(60\) is \(F_X(60)-0.6=0.72-0.6=0.12\). Together, \(0.28+0.12=0.40\), as required.

The continuous density on \([0,100]\) is \(0.7/100=0.007\). Its tail loss contribution is

0.007\int_{60}^{100}x\,dx =0.007\left(\frac{100^2-60^2}{2}\right) =0.007(3{,}200) =22.4.

The threshold contribution is \(60(0.12)=7.2\). Thus the tail loss total is \(22.4+7.2=29.6\), and

\operatorname{TVaR}_{0.6}(X)=\frac{29.6}{0.4}=74.

Check. The selected tail has probability \(0.28+0.12=0.4\). A common mistake is to include the entire atom at \(60\): that gives \((22.4+60(0.3))/(0.28+0.3)=40.4/0.58\approx69.6552\), which averages a larger, incorrect portion of the distribution. The correct TVaR is \(74\).

Worked Example: An Atom at the Threshold and Another Atom Above It

A hypothetical loss has probability \(0.5\) of being uniform on \([0,10]\), probability \(0.3\) of being uniform on \([10,20]\), and probability \(0.2\) of being exactly \(12\). Calculate \(\operatorname{TVaR}_{0.7}(X)\).

Method. Locate the quantile by tracking the CDF through the continuous component and the jump at \(12\), then add losses strictly above the quantile and the needed part of its atom.

The first uniform component contributes total probability \(0.5\) by loss \(10\). Between \(10\) and \(12\), the second component contributes density \(0.3/10=0.03\), so \(F_X(12-)=0.5+0.03(12-10)=0.56\). The atom at \(12\) raises this to \(F_X(12)=0.56+0.2=0.76\). Since \(0.56<0.7\leq0.76\), \(q=12\).

The continuous probability strictly above \(12\) is \(0.03(20-12)=0.24\). The amount needed at \(12\) is \(0.76-0.7=0.06\). The total is \(0.24+0.06=0.30=1-0.7\).

The continuous tail contribution is

0.03\int_{12}^{20}x\,dx =0.03\left(\frac{20^2-12^2}{2}\right) =0.03(128) =3.84.

The threshold contributes \(12(0.06)=0.72\). The numerator is \(3.84+0.72=4.56\), giving

\operatorname{TVaR}_{0.7}(X)=\frac{4.56}{0.30}=15.2.

Check. The tail contains probability \(0.24\) from the continuous losses above \(12\) and \(0.06\) from the atom at \(12\). The atom at \(12\) is not fully included, and the lower uniform component contributes nothing to this upper tail.

Worked Example: A Continuous VaR with a Higher Point Mass

A hypothetical loss has probability \(0.8\) of being uniform on \([0,100]\) and probability \(0.2\) of being exactly \(80\). Calculate \(\operatorname{VaR}_{0.5}(X)\) and \(\operatorname{TVaR}_{0.5}(X)\).

Method. Since the quantile falls below the atom, first solve for the continuous quantile. Then include the full point mass above it in the tail.

For losses below \(80\), \(F_X(x)=0.8x/100=0.008x\). Solving \(0.008q=0.5\) gives \(q=62.5\). This is below \(80\), so the calculation is consistent with the CDF expression used.

The continuous probability above \(62.5\) is \(0.8(100-62.5)/100=0.3\). The atom at \(80\) adds probability \(0.2\), making the upper-tail probability \(0.3+0.2=0.5\). There is no atom at \(q\), so no threshold correction is needed.

The continuous tail contribution is

0.008\int_{62.5}^{100}x\,dx =0.004(100^2-62.5^2) =0.004(6{,}093.75) =24.375.

The point mass contributes \(80(0.2)=16\). Hence the tail loss total is \(24.375+16=40.375\), and

\operatorname{TVaR}_{0.5}(X)=\frac{40.375}{0.5}=80.75.

Check. The selected probability is exactly \(0.5\). A point mass need not be at the VaR to affect TVaR: here the entire atom at \(80\) lies above the threshold and must be included.

Common Checks and Interpretation

For a mixed distribution, calculating the loss contribution without checking the probability contribution is a frequent source of error. The integral may be evaluated correctly while the wrong amount of an atom is included. After identifying \(q\), explicitly verify

P(X>q)+\bigl(F_X(q)-\alpha\bigr)=1-\alpha.

Also distinguish an atom at the VaR from an atom strictly above it. At the VaR, include only \(F_X(q)-\alpha\) probability. Strictly above the VaR, include the full probability of each atom. In either case, compute the numerator using loss multiplied by the probability selected, then divide by the same total tail probability \(1-\alpha\).

Key takeaway: TVaR for a mixed loss is the expected loss of a precisely sized upper tail. Integrate the continuous part above VaR, add every full atom above VaR, and allocate only the required fraction of any atom at VaR.

Check Your Understanding

For each problem, find the VaR threshold and TVaR at the stated level. All uniform components are unconditional mixture components.

A loss is uniform on \([0,50]\) with probability \(0.8\) and equals \(70\) with probability \(0.2\). Find VaR and TVaR at \(\alpha=0.6\). (Answer: VaR \(=37.5\); TVaR \(=56.875\).)
A loss is uniform on \([0,10]\) with probability \(0.6\), equals \(12\) with probability \(0.25\), and equals \(20\) with probability \(0.15\). Find VaR and TVaR at \(\alpha=0.7\). (Answer: VaR \(=12\); TVaR \(=16\).)
A loss is uniform on \([0,20]\) with probability \(0.75\) and equals \(8\) with probability \(0.25\). Find VaR and TVaR at \(\alpha=0.6\). (Answer: VaR \(=9.3333\); TVaR \(=14.6667\).)
A loss is uniform on \([0,100]\) with probability \(0.7\) and equals \(80\) with probability \(0.3\). Find VaR and TVaR at \(\alpha=0.5\). (Answer: VaR \(=71.4286\); TVaR \(=82.2857\).)

What You'll Learn