TVaR Diversification for Discrete Portfolios

Why Discrete Portfolio TVaR Needs Care

For a discrete loss, positive probability can sit at the VaR threshold. The losses strictly above the threshold may therefore account for less than the tail probability used in TVaR. A correct calculation fills the remaining tail with only the required fraction of the probability at the threshold—not necessarily the whole atom.

For a portfolio, the same calculation is performed on the distribution of the aggregate loss. That distribution depends on the joint outcomes, not just on the marginal distributions of the individual losses. Comparing portfolio TVaR with the sum of individual TVaRs provides a direct numerical check of diversification through subadditivity.

Definition: Let $X$ be a finite-support loss with distribution function $F_X(x)=P(X\leq x)$, and let $0<\alpha<1$. Use the quantile convention $$ \operatorname{VaR}_{\alpha}(X)=\inf\{x:F_X(x)\geq\alpha\}. $$ The quantile-integral definition of TVaR is $$ \operatorname{TVaR}_{\alpha}(X)=\frac{1}{1-\alpha}\int_{\alpha}^{1}\operatorname{VaR}_{u}(X)\,du. $$ This definition includes the appropriate fraction of an atom at the VaR threshold.

Here $\alpha$ is the confidence level, and $1-\alpha$ is the probability mass in the upper tail being averaged. When the distribution has an atom at $q=\operatorname{VaR}_{\alpha}(X)$, TVaR is generally not the conditional mean $E[X\mid X\geq q]$. That conditional mean includes the entire atom at $q$, while the quantile-integral definition may need only part of it.

A Discrete Formula and Its Justification

The quantile-integral definition gives a convenient calculation formula. Write $q=\operatorname{VaR}_{\alpha}(X)$. By the quantile convention, $F_X(q)\geq\alpha$, while the probability strictly above $q$ is $1-F_X(q)$. Those losses do not, by themselves, fill a tail of probability $1-\alpha$. The additional probability required at $q$ is $F_X(q)-\alpha$.

Formula: For a finite-support loss $X$, $$ \operatorname{TVaR}_{\alpha}(X) = \frac{ E\!\left[X\,\mathbf{1}_{\{X>q\}}\right] +q\bigl(F_X(q)-\alpha\bigr) }{1-\alpha}, \qquad q=\operatorname{VaR}_{\alpha}(X). $$ The term $E[X\,\mathbf{1}_{\{X>q\}}]$ includes all probability strictly above $q$; the second term includes only the amount of the threshold atom needed to make the total tail probability $1-\alpha$.

To justify the formula, let $Q(u)=\operatorname{VaR}_{u}(X)$. For $u$ from $\alpha$ through $F_X(q)$, the quantile is $q$. For larger $u$, the quantile moves through the loss values above $q$, with each such value contributing its probability mass to the integral. Thus the part of the integral at the threshold is $q(F_X(q)-\alpha)$, and the part above the threshold is $E[X\mathbf{1}_{\{X>q\}}]$. Dividing their sum by $1-\alpha$ gives the stated formula.

Equivalently, imagine selecting a tail of total probability $1-\alpha$: include every outcome with loss greater than $q$, then select just enough probability at $q$ to reach that total. This description is a calculation device for a distribution with an atom; it does not require physically splitting an insured loss.

Worked Calculations

Worked Example: A Single Loss With a Threshold Atom

A hypothetical equipment policy has annual loss $X$ with the following distribution: $P(X=0)=0.50$, $P(X=4)=0.25$, $P(X=10)=0.20$, and $P(X=30)=0.05$, with losses measured in thousands of dollars. At confidence level $\alpha=0.80$, find VaR and TVaR.

Method. First accumulate the probabilities in increasing loss order: $F_X(0)=0.50$, $F_X(4)=0.75$, and $F_X(10)=0.95$. Since the cumulative probability first reaches $0.80$ at loss $10$, $q=\operatorname{VaR}_{0.80}(X)=10$.

The tail probability is $1-\alpha=0.20$. Losses strictly above $10$ contribute probability $0.05$ and loss amount $30(0.05)=1.50$. The probability to take at the threshold is $F_X(10)-\alpha=0.95-0.80=0.15$, contributing $10(0.15)=1.50$. Therefore, $$ \operatorname{TVaR}_{0.80}(X) =\frac{1.50+1.50}{0.20} =15 $$ thousand dollars.

Check. The selected tail has probability $0.05+0.15=0.20$, and its weighted loss total is $1.50+1.50=3.00$. Its average is $3.00/0.20=15$ thousand dollars. The common mistake is to average the entire event $X\geq10$, which gives $(10(0.20)+30(0.05))/0.25=14$ thousand dollars. That uses probability $0.25$, not the required tail probability $0.20$.

Worked Example: Strict Diversification for Mutually Exclusive Losses

A hypothetical two-coverage portfolio has four equally likely joint outcomes. The paired losses $(X,Y)$, in thousands of dollars, are $(10,0)$, $(0,10)$, $(0,0)$, and $(0,0)$. At $\alpha=0.75$, calculate each individual TVaR and the TVaR of the aggregate loss $Z=X+Y$. Check subadditivity.

Individual loss $X$. Its distribution is $P(X=0)=0.75$ and $P(X=10)=0.25$. Therefore, $\operatorname{VaR}_{0.75}(X)=0$. Using $q=0$, $F_X(0)-\alpha=0.75-0.75=0$, and $E[X\mathbf{1}_{\{X>0\}}]=10(0.25)=2.50$. Hence $\operatorname{TVaR}_{0.75}(X)=2.50/0.25=10$ thousand dollars. By symmetry, $\operatorname{TVaR}_{0.75}(Y)=10$ thousand dollars.

Aggregate loss. The joint outcomes give $Z=10$ in two states and $Z=0$ in two states. Thus $P(Z=0)=0.50$, $P(Z=10)=0.50$, and $\operatorname{VaR}_{0.75}(Z)=10$. At $q=10$, there is no loss strictly above the threshold, and $F_Z(10)-\alpha=1-0.75=0.25$. It follows that $$ \operatorname{TVaR}_{0.75}(Z) =\frac{0+10(0.25)}{0.25} =10 $$ thousand dollars.

Check. The subadditivity inequality is $10\leq10+10=20$, so it holds strictly. The joint outcomes matter: the two coverages never produce a loss of $10$ at the same time. A common error is to infer aggregate risk from marginal TVaRs alone; here the joint table is what determines the aggregate distribution.

Worked Example: A Portfolio With Different Threshold Atoms

A hypothetical portfolio has three possible states, with probabilities $0.50$, $0.30$, and $0.20$. Its paired losses $(X,Y)$, in thousands of dollars, are respectively $(2,6)$, $(8,4)$, and $(14,0)$. At $\alpha=0.60$, calculate the two individual TVaRs and aggregate TVaR, then check subadditivity.

For $X$. The losses are $2$ with probability $0.50$, $8$ with probability $0.30$, and $14$ with probability $0.20$. The cumulative probability at $2$ is $0.50$, and at $8$ it is $0.80$, so $q_X=8$. The tail probability is $0.40$: all $0.20$ probability at loss $14$, plus $F_X(8)-\alpha=0.80-0.60=0.20$ at loss $8$. The weighted amount is $14(0.20)+8(0.20)=2.80+1.60=4.40$. Thus $\operatorname{TVaR}_{0.60}(X)=4.40/0.40=11$ thousand dollars.

For $Y$. Its losses are $0$ with probability $0.20$, $4$ with probability $0.30$, and $6$ with probability $0.50$. The cumulative probability reaches $1$ at $6$, while it is only $0.50$ at $4$; hence $q_Y=6$. The threshold amount needed is $F_Y(6)-\alpha=1-0.60=0.40$. There are no losses above $6$, so $\operatorname{TVaR}_{0.60}(Y)=6(0.40)/0.40=6$ thousand dollars.

For the aggregate. The state totals are $8$, $12$, and $14$, with probabilities $0.50$, $0.30$, and $0.20$. Their cumulative probabilities are $0.50$, $0.80$, and $1$, so $q_Z=12$. The loss above $12$ contributes $14(0.20)=2.80$; the threshold contributes $12(0.80-0.60)=2.40$. Therefore $\operatorname{TVaR}_{0.60}(Z)=(2.80+2.40)/0.40=13$ thousand dollars.

Check. Portfolio TVaR is $13$, while the sum of individual TVaRs is $11+6=17$, all in thousands of dollars. Thus $13\leq17$, as required by subadditivity. The calculation also shows why each threshold must be found from its own distribution: the thresholds for $X$, $Y$, and $Z$ are $8$, $6$, and $12$, respectively.

Checking Diversification in a Discrete Portfolio

For losses $X$ and $Y$ on the same probability space, form the aggregate loss $Z=X+Y$ state by state. Then construct the distribution of $Z$, including any repeated totals by adding their probabilities. Calculate $\operatorname{TVaR}_{\alpha}(Z)$ from that aggregate distribution, and compare it with the sum of the separately calculated values:

\operatorname{TVaR}_{\alpha}(X+Y) \leq \operatorname{TVaR}_{\alpha}(X)+\operatorname{TVaR}_{\alpha}(Y).

This is the subadditivity comparison for TVaR. It states that the portfolio’s tail risk does not exceed the sum of its component tail risks. It does not say the inequality must be strict: some portfolios have equality. Nor does knowing the two marginal distributions alone determine portfolio TVaR; dependence between the losses affects the aggregate outcomes.

Build the joint outcomes.
List each possible pair of losses with its probability. Confirm that the probabilities sum to $1$.

Find marginal distributions.
Group outcomes by each component’s loss, adding probabilities where a loss value repeats.

Find the aggregate distribution.
Add the component losses in each joint outcome, then combine equal totals and their probabilities.

Calculate each TVaR at the same confidence level.
For every distribution, find its own VaR threshold and use only the required probability at that threshold.

Compare the results.
Check whether aggregate TVaR is no greater than the sum of the component TVaRs.

Common Errors to Avoid

Including the whole threshold atom. The calculation may require only $F_X(q)-\alpha$ probability at $q$. Including more changes the tail probability being averaged.
Using the wrong tail probability. The denominator is $1-\alpha$, not $F_X(q)$, $P(X\geq q)$, or $P(X>q)$.
Using a conditional mean without checking the atom. $E[X\mid X\geq q]$ generally averages a different event from the quantile-integral TVaR.
Adding losses without preserving their joint states. Aggregate losses must be formed from paired outcomes; adding quantiles or marginal TVaRs does not construct the aggregate distribution.
Comparing different confidence levels. The subadditivity check must use the same $\alpha$ for the aggregate and both components.

Key takeaway: For a discrete loss, TVaR averages exactly $1-\alpha$ probability from the upper tail. Include all probability above the VaR threshold and only the amount needed at the threshold itself. For a portfolio, apply this rule to the joint-state aggregate distribution before checking subadditivity.

Check Your Understanding

In each problem, use the quantile convention and the discrete TVaR formula stated above. Losses are in the units specified.

A loss is $0$ with probability $0.40$, $5$ with probability $0.40$, and $20$ with probability $0.20$. At $\alpha=0.75$, find VaR and TVaR. (Answer: VaR $=5$, TVaR $=17$.)
A loss is $1$ with probability $0.50$, $7$ with probability $0.30$, and $17$ with probability $0.20$. At $\alpha=0.80$, find TVaR and explain how much probability is selected at the VaR threshold. (Answer: TVaR $=17$; VaR $=7$, and $0$ probability at the threshold is needed.)
Four equally likely joint outcomes for $(X,Y)$ are $(8,0)$, $(0,8)$, $(0,0)$, and $(0,0)$. At $\alpha=0.50$, find both individual TVaRs and aggregate TVaR, and check subadditivity. (Answer: $\operatorname{TVaR}_{0.50}(X)=4$, $\operatorname{TVaR}_{0.50}(Y)=4$, and $\operatorname{TVaR}_{0.50}(X+Y)=8$; equality holds.)
Three joint outcomes have probabilities $0.40$, $0.35$, and $0.25$, with paired losses $(X,Y)=(0,9),(6,3),(12,0)$. At $\alpha=0.60$, calculate the two component TVaRs and aggregate TVaR. (Answer: $\operatorname{TVaR}_{0.60}(X)=9.75$, $\operatorname{TVaR}_{0.60}(Y)=9$, and $\operatorname{TVaR}_{0.60}(X+Y)=10.875$; subadditivity holds.)
A loss is $0$ with probability $0.70$, $100$ with probability $0.20$, and $500$ with probability $0.10$. At $\alpha=0.90$, find VaR and TVaR. (Answer: VaR $=100$, TVaR $=500$.)

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