The risk measures of valueatrisk and tailvalueatrisk are discussed in the preceding post. This post extends the preceding post with an algorithm on evaluating the tailvalueatrisk of a mixture distribution with discrete mixing weights.
The preceding post introduces the notions of valueatrisk (VaR) and tailvalueatrisk (TVaR). These are two particular examples of risk measures that are useful for insurance companies and other enterprises in a risk management context. For , VaR at the security level gives the threshold that the probability of a loss more adverse than the threshold is at most . Thus in our context VaR is a percentile of the loss distribution. TVaR is a conditional expected value. At the security level , TVaR is the expected value of the losses given that the losses exceed the threshold VaR.
The preceding post gives several representations of TVaR. It also gives the formula for TVaR for several distributions – exponential, Pareto, normal and lognormal. We now discuss TVaR of a mixture distribution. If a distribution is the mixture of two distributions and if each of the individual distributions has a clear formulation of TVaR, can we mix the two TVaR's? The answer is that we can provided some adjustments are made. The following gives the formula.
Suppose that the loss is a mixture of two distributions represented by the random variables and , with weights and , respectively. Let be the th percentile of the loss , i.e. . Then the tailvalueatrisk at the percent security level is:

The comparison is with formula (3) in the preceding post. That formula shows that TVaR is . In other words, TVaR is VaR plus the mean excess loss function evaluated at . The content within the squared brackets in Formula (a) is a weighted average of the two individual mean excess loss functions with the adjustment of multiplying with the probabilities and . This formula is useful if the (VaR) can be calculated and if the mean excess loss functions are accessible.
We give an example and then show the derivation of Formula (a).
Example 1
The mean excess loss function of an exponential distribution is constant. Let’s consider the mixture of two exponential distributions. Suppose that losses follow a mixture of two exponential distributions where one distribution has mean 5 (75% weight) and the other has mean 10 (25% weight). Determine the VaR and TVaR at the security level 99%.
First, calculate the 99th percentile of the mixture, which is the solution to the following equation.
By letting , we solve the following equation.
Use the quadratic formula to solve for . Then solve for . The following is the 99th percentile of the loss .
The following gives the TVaR.
Note that the mean excess loss for the first exponential distribution is 5 and for the second one is 10 (the unconditional means). The survival functions and are also easy to evaluate. As long as the percentile of the mixture is calculated, the formula is very useful. In this example, the two exponential parameters are set so that the calculation of percentiles uses the quadratic formula. If the parameters are set differently, then we can use software to evaluate the required percentile.
Deriving the formula
Suppose that is the mixture of , with weight , and , with weight . The density function for is and the density function for is . The density function of is then . We derive from the basic definition of TVaR. Let be the th percentile of .
The formula derived here is for mixtures for two distributions. It is straightforward to extend it for mixtures of any finitemixture.
Practice Problems
Practice problems are available in the companion blog to reinforce the concepts of valueatrisk and tailvalueatrisk. Practice Problems 10G and 10H in that link are for TVaR of mixtures.
actuarial
math
Daniel Ma
mathematics
2018 – Dan Ma
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Hi! I have a doubt. Why you put the mixture of density functions equal to the quantile?
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I assume you refer to the equation . This is taking the weighted average of the two survival functions. Equivalently, you can take the weighted average of the two CDFs.
Doing this will give the same result. Thank you for your question and for your detailed reading.
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