Why Mixtures Need Their Own Quantile Calculation
A loss distribution may combine several distinct risk sources or scenarios. For example, a hypothetical insurer might model losses differently depending on which of several mutually exclusive operating conditions occurs. The overall loss distribution is then a mixture: its component distributions are combined using scenario probabilities. Even if each component is simple, the mixture’s quantile is determined by the weighted CDFs together, not by averaging component quantiles.
This matters for Tail Value at Risk (TVaR). Once the mixture quantile is found, TVaR averages the losses in the upper tail, with only the necessary fraction of any probability mass at the quantile included. The calculation must therefore keep track of both the mixture weights and the tail probability. This tutorial is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.
For a continuous mixture, the quantile usually satisfies \(F_X(q_\alpha)=\alpha\). But the CDF can also reach or cross \(\alpha\) at a component boundary, remain constant across a gap in the possible losses, or jump past \(\alpha\) at an atom. The definition using the infimum handles all of these cases. For any integrable loss, the corresponding calculation is
The first term in the numerator is the expected loss contributed by outcomes strictly above the quantile. The second term adds just enough probability at the quantile to make the selected upper tail have probability \(1-\alpha\). For a continuous distribution there is no probability mass at the quantile, so \(F_X(q_\alpha)=\alpha\) and this second term is zero.
How to Locate and Average the Mixture Tail
The mixture CDF is a weighted sum, but its quantile is not a weighted sum of component quantiles. First determine which component intervals contain the solution, or identify the jump or boundary where the CDF reaches the target. Then solve the weighted CDF equation in that region. Once the quantile is known, calculate the upper-tail contribution component by component.
Combine the component CDFs using their scenario weights. Check the CDF at component support endpoints and at any atoms.
Find the first loss amount at which the mixture CDF is at least \(\alpha\). If it lies within a continuous interval, solve \(F_X(q_\alpha)=\alpha\); if the CDF jumps, use the jump location.
Sum the weighted expected losses strictly above \(q_\alpha\), then add \(q_\alpha(F_X(q_\alpha)-\alpha)\) to include any needed fraction of quantile mass.
Divide the selected tail’s expected loss amount by \(1-\alpha\). Check that the probability represented in the numerator is exactly \(1-\alpha\).
Why the Fractional-Mass Formula Works
The formula follows by considering the upper \(1-\alpha\) fraction of the loss distribution. All outcomes strictly greater than \(q_\alpha\) belong in that tail. Their probability is \(1-F_X(q_\alpha)\). If the CDF jumps at \(q_\alpha\), the probability at the quantile is \(F_X(q_\alpha)-F_X(q_\alpha-)\), where \(F_X(q_\alpha-)\) is the probability strictly below it. To fill the desired tail probability, include an amount of probability equal to
That selected probability has loss amount \(q_\alpha\), so its expected-loss contribution is \(q_\alpha(F_X(q_\alpha)-\alpha)\). Adding it to the expected loss strictly above the quantile gives the total loss amount in the upper \(1-\alpha\) tail. Dividing by \(1-\alpha\) gives the tail average. If \(F_X(q_\alpha)=\alpha\), no quantile mass is needed. This argument also shows why simply taking \(E[X\mid X>q_\alpha]\) can be wrong when the quantile is an atom: the event \(X>q_\alpha\) may have less than \(1-\alpha\) probability.
Worked Calculations for Mixtures
Worked Example: The Quantile Is Exactly at a Component Boundary
A hypothetical one-period loss \(X\), measured in thousands of dollars, comes from one of two scenarios. With probability \(0.60\), \(X\) is Uniform\((0,10)\); with probability \(0.40\), it is Uniform\((10,20)\). Find the 60% quantile and \(\operatorname{TVaR}_{0.60}(X)\). Assume these are the full loss distributions for the period; no interest or mortality adjustment applies.
Method. Form the weighted CDF on the component intervals, then calculate the upper tail. The common mistake to avoid is treating the 60% quantile as an average of the two component quantiles rather than locating it from the mixture CDF.
For \(0\leq x\leq10\), the second component has not accumulated probability below \(x\), so \(F_X(x)=0.60(x/10)\). At the endpoint, \(F_X(10)=0.60(10/10)=0.60\). Thus the first loss amount at which the mixture CDF reaches 0.60 is \(q_{0.60}=10\). This is a component boundary, not an atom: both uniform distributions are continuous.
The upper \(1-0.60=0.40\) probability consists of the entire second component. Its conditional mean is \((10+20)/2=15\) thousand dollars, and its mixture weight is 0.40. The expected-loss contribution strictly above 10 is therefore \(0.40(15)=6\) thousand dollars. Also, \(F_X(q)-\alpha=0.60-0.60=0\). Hence
Check: the selected tail has probability 0.40 and consists exactly of the second scenario, whose mean is 15 thousand dollars. The result is therefore consistent with the scenario interpretation.
Worked Example: The Quantile Lies Inside a Later Component
A hypothetical loss \(X\), in thousands of dollars, has distribution Uniform\((0,8)\) with probability 0.75 and Uniform\((16,24)\) with probability 0.25. Calculate \(q_{0.90}\) and \(\operatorname{TVaR}_{0.90}(X)\). Assume no discounting or other adjustment.
Method. Check the CDF across the gap between the component supports, then solve within the upper component. The common mistake to avoid is assuming that the 90th percentile is the 90th percentile of the second component alone.
At \(x=8\), the first component is fully accumulated and the second has not begun, so \(F_X(8)=0.75\). Throughout the gap from 8 to 16, the CDF remains 0.75. For \(16\leq x\leq24\),
Set this equal to 0.90. Then \(0.25((x-16)/8)=0.15\), so \((x-16)/8=0.60\), \(x-16=4.8\), and \(q_{0.90}=20.8\) thousand dollars. The mixture CDF there is 0.90, as required.
The upper tail has probability 0.10. It consists of losses above 20.8 in the second component. Within that uniform component, the interval from 20.8 to 24 has length \(3.2\) out of 8, so its conditional probability is \(3.2/8=0.40\). Its mixture probability is \(0.25(0.40)=0.10\). The mean loss over that interval is \((20.8+24)/2=22.4\), giving an expected-loss contribution of \(0.10(22.4)=2.24\) thousand dollars. Since the mixture is continuous at the quantile, the fractional-mass term is zero. Thus
Check: the upper-tail probability is 0.10, not 0.40; the 0.40 is the fraction of the upper component’s conditional distribution included. Dividing its weighted expected loss, 2.24, by 0.10 gives the conditional tail average of 22.4.
Worked Example: The Quantile Is an Atom of the Mixture
A hypothetical one-period loss \(X\), in thousands of dollars, equals 5 with probability 0.40. With probability 0.60, it is Uniform\((0,10)\). Find \(q_{0.60}\) and \(\operatorname{TVaR}_{0.60}(X)\). Assume the stated distribution is complete and no discounting applies.
Method. Evaluate the CDF immediately before and at the atom, then use fractional quantile mass. The common mistake to avoid is averaging only losses strictly above 5, which would omit part of the required upper tail.
Immediately below 5, the uniform component contributes \(0.60(5/10)=0.30\), and the point mass has not yet entered. At 5, the point mass adds 0.40, giving \(F_X(5)=0.30+0.40=0.70\). Since \(0.30<0.60\leq0.70\), the quantile is \(q_{0.60}=5\).
The probability strictly above 5 comes only from the uniform component. It is \(0.60(5/10)=0.30\), and the mean of that component conditional on exceeding 5 is \((5+10)/2=7.5\). Therefore, \(E[X\mathbf{1}_{\{X>5\}}]=0.30(7.5)=2.25\) thousand dollars. The amount of probability at the quantile needed for the upper 0.40 tail is \(F_X(5)-0.60=0.10\). Its expected-loss contribution is \(5(0.10)=0.50\) thousand dollars. Therefore,
Check: the selected tail contains probability 0.30 strictly above 5 and 0.10 at 5, totaling 0.40. Its expected loss is \(2.25+0.50=2.75\) thousand dollars, and \(2.75/0.40=6.875\) thousand dollars. The calculation uses only a quarter of the atom’s 0.40 probability, not the whole atom.
Interpretation and Common Traps
A mixture quantile depends on how the component probabilities accumulate at each loss amount. A low-weight component may determine the extreme upper tail, while another component may determine a lower quantile. Component boundaries and support gaps can make the CDF flat over an interval; in that case, the quantile definition selects the first point at which the CDF reaches the target. An atom can instead make the CDF jump past the target, requiring fractional use of the probability at that loss.
- Do not average component quantiles. Solve using \(F_X(x)=\sum_j w_jF_j(x)\), or use the quantile definition at a jump.
- Do not confuse a component weight with a tail probability. A component may contribute only part of its probability to the selected upper tail.
- Do not omit the fractional-mass term. When \(F_X(q_\alpha)>\alpha\), the excess probability at the quantile supplies part of the required tail.
- Do not divide by the wrong probability. TVaR at level \(\alpha\) divides by \(1-\alpha\), the selected upper-tail probability.
Check Your Understanding
For each hypothetical loss distribution, identify the mixture quantile and calculate the stated TVaR. Amounts are in thousands of dollars; assume no discounting or other adjustments.
- A loss is Uniform\((0,5)\) with probability 0.80 and Uniform\((5,15)\) with probability 0.20. Find \(q_{0.80}\) and \(\operatorname{TVaR}_{0.80}\). (Answer: \(q_{0.80}=5\); \(\operatorname{TVaR}_{0.80}=10\).)
- A loss is Uniform\((0,4)\) with probability 0.50 and Uniform\((8,12)\) with probability 0.50. Find \(q_{0.90}\) and \(\operatorname{TVaR}_{0.90}\). (Answer: \(q_{0.90}=11.2\); \(\operatorname{TVaR}_{0.90}=11.6\).)
- A loss equals 2 with probability 0.25; otherwise it is Uniform\((0,8)\). Find \(q_{0.50}\) and \(\operatorname{TVaR}_{0.50}\). (Answer: \(q_{0.50}=8/3\); \(\operatorname{TVaR}_{0.50}=16/3\), or approximately 5.3333.)
- A loss equals 10 with probability 0.40; otherwise it is Uniform\((0,20)\). Find \(q_{0.50}\) and \(\operatorname{TVaR}_{0.50}\), including any fractional mass at the quantile. (Answer: \(q_{0.50}=10\); \(\operatorname{TVaR}_{0.50}=13\).)
- A loss is Uniform\((0,5)\) with probability 0.20, Uniform\((5,15)\) with probability 0.30, and Uniform\((15,25)\) with probability 0.50. Find \(q_{0.50}\) and \(\operatorname{TVaR}_{0.50}\). (Answer: \(q_{0.50}=15\); \(\operatorname{TVaR}_{0.50}=20\).)