# Tail-value-at-risk of a mixture

The risk measures of value-at-risk and tail-value-at-risk are discussed in the preceding post. This post extends the preceding post with an algorithm on evaluating the tail-value-at-risk of a mixture distribution with discrete mixing weights.

The preceding post introduces the notions of value-at-risk (VaR) and tail-value-at-risk (TVaR). These are two particular examples of risk measures that are useful for insurance companies and other enterprises in a risk management context. For $0, VaR at the security level $p$ gives the threshold that the probability of a loss more adverse than the threshold is at most $1-p$. Thus in our context VaR is a percentile of the loss distribution. TVaR is a conditional expected value. At the security level $p$, 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 $X$ is a mixture of two distributions represented by the random variables $X_1$ and $X_2$, with weights $w$ and $1-w$, respectively. Let $\pi_p$ be the $100p$th percentile of the loss $X$, i.e. $\pi_p=\text{VaR}_p(X)$. Then the tail-value-at-risk at the $100p$ percent security level is:

\displaystyle \begin{aligned} \text{TVaR}_p(X)&=\pi_p+\frac{1}{1-p} \biggl[w \times P(X_1>\pi_p) \times e_{X_1}(\pi_p)\\& \ \ +(1-w) \times P(X_2>\pi_p) \times e_{X_2}(\pi_p)\biggr] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (a) \end{aligned}

The comparison is with formula (3) in the preceding post. That formula shows that TVaR is $\pi_p+e(\pi_p)$. In other words, TVaR is VaR plus the mean excess loss function evaluated at $\pi_p$. 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 $P(X_1>\pi_p)$ and $P(X_2>\pi_p)$. This formula is useful if the $\pi_p$ (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.

$\displaystyle 0.75 e^{-x/5}+0.25 e^{-x/10}=1-0.99=0.01$

By letting $y=e^{-x/10}$, we solve the following equation.

$\displaystyle 0.75 y^2+0.25 y-0.01=0$

Use the quadratic formula to solve for $y$. Then solve for $x$. The following is the 99th percentile of the loss $X$.

$\displaystyle \pi_p=-10 \times \text{ln} \biggl(\frac{-1+\sqrt{1.48}}{6} \biggr)=33.2168$

The following gives the TVaR.

\displaystyle \begin{aligned} \text{TVaR}_p(X)&=\pi_p+\frac{1}{1-0.99} \biggl[0.75 \times e^{-\pi_p/5} \times 5 +0.25 \times e^{-\pi_p/10} \times 10 \biggr] \\&=42.7283 \end{aligned}

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 $P(X_1>\pi_p)$ and $P(X_2>\pi_p)$ are also easy to evaluate. As long as the percentile $\pi_p$ 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 $X$ is the mixture of $X_1$, with weight $w$, and $X_2$, with weight $1-w$. The density function for $X_1$ is $f_1(x)$ and the density function for $X_2$ is $f_2(x)$. The density function of $X$ is then $f(x)=w f_1(x)+(1-w) f_2(x)$. We derive from the basic definition of TVaR. Let $\pi_p$ be the $100p$th percentile of $X$.

\displaystyle \begin{aligned} \text{TVaR}_p(X)&=\frac{\int_{\pi_p}^\infty x f(x) \ dx}{1-p}\\&=\pi_p+\frac{\int_{\pi_p}^\infty (x-\pi_p) f(x) \ dx}{1-p} \\&=\pi_p+\frac{\int_{\pi_p}^\infty (x-\pi_p) (w f_1(x)+(1-w) f_2(x)) \ dx}{1-p} \\&=\pi_p+\frac{1}{1-p} \biggl[w \int_{\pi_p}^\infty (x-\pi_p) f_1(x) \ dx +(1-w) \int_{\pi_p}^\infty (x-\pi_p) f_2(x) \ dx\biggr] \\&=\pi_p+\frac{1}{1-p} \biggl[w \ P(X_1>\pi_p) \ \frac{\int_{\pi_p}^\infty (x-\pi_p) f_1(x) \ dx}{P(X_1>\pi_p)}\\& \ \ +(1-w) \ P(X_2>\pi_p) \ \frac{\int_{\pi_p}^\infty (x-\pi_p) f_2(x) \ dx}{P(X_2>\pi_p)} \biggr] \\&=\pi_p+\frac{1}{1-p} \biggl[w \ P(X_1>\pi_p) \ e_{X_1}(\pi_p) +(1-w) \ P(X_2>\pi_p) \ e_{X_2}(\pi_p) \biggr] \end{aligned}

The formula derived here is for mixtures for two distributions. It is straightforward to extend it for mixtures of any finite-mixture.

Practice Problems

Practice problems are available in the companion blog to reinforce the concepts of value-at-risk and tail-value-at-risk. Practice Problems 10-G and 10-H in that link are for TVaR of mixtures.

actuarial
math

Daniel Ma
mathematics

$\copyright$ 2018 – Dan Ma

## 3 thoughts on “Tail-value-at-risk of a mixture”

1. Hi! I have a doubt. Why you put the mixture of density functions equal to the quantile?

Like

• I assume you refer to the equation $\displaystyle 0.75 e^{-x/5}+0.25 e^{-x/10}=1-0.99=0.01$. This is taking the weighted average of the two survival functions. Equivalently, you can take the weighted average of the two CDFs.

$\displaystyle 0.75 (1-e^{-x/5})+0.25 (1-e^{-x/10})=0.99$

Doing this will give the same result. Thank you for your question and for your detailed reading.

Like