VaR for a Liability with a Point Mass at Zero and Exponential Losses
Summary
The document analyzes a liability that is zero in most cases and otherwise follows an exponential loss distribution. Because the loss distribution has a point mass at zero, its cumulative distribution function jumps there; the usual continuous-distribution intuition for quantiles needs adjustment. The answer forms the cumulative probability by combining the probability of zero loss with the conditional exponential distribution for positive losses.
For confidence levels above the probability assigned to zero loss, the quantile can be found by solving the mixture cumulative distribution for the desired level. The stated example compares the resulting tail quantile with that of an exponential distribution without the zero-loss mass. This illustrates how a chance of no loss changes VaR. The discussion uses a particular convention for VaR and focuses on levels where the quantile lies in the positive-loss portion; care is needed with conventions and boundary levels, especially at the jump in the distribution.
Key ideas
- A zero-loss probability creates a jump in the loss cumulative distribution.
- For positive losses, the distribution is a scaled exponential component of the mixture.
- VaR is obtained by inverting the full mixture distribution at the selected confidence level.
- The zero-loss mass changes tail quantiles relative to a purely exponential loss model.
- Quantile conventions and confidence levels at the distribution's jump require care.
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Full text
# Calculate VaR for a liabilty taking a exponential distribution?
# Calculate VaR for a liabilty taking a exponential distribution?
An insurance company faces the liability loss off
$L = \begin{cases} 0, & \mbox{with probability } 0.75 \\ Z, & \mbox{with probability } 0.25\end{cases}$
where $Z\sim Exp(\mu)$. I want to calculate $\mbox{VaR}_{\alpha}$ for any $\alpha\in(0,1)$ when $R_0=1$.
I know that $\mbox{VaR}_{\alpha}(X)=F_L^{-1}(1-\alpha)$ (with $L=-X/R_0)$
However, I really don't know how I should intepret the liability since $Z$ is a random variable. Anyone who could help me out?
## Answer by Richi Wa (score 6, accepted)
https://quant.stackexchange.com/a/22628
The VaR of level $\alpha$ a loss random variable (the bigger the worse) is the quantity $q$ such that the loss is bigger with probability $1-\alpha$.
Thus we need a $q$ such that $$ P[L>q] = 1-\alpha, $$ where we can imagine $\alpha=99\%$ and thus we need the starting point of the $1\%$ tail. Because we have a probability of a loss of size $0$ of $75\%$ we have a discontinuity of the cdf there.
We have to analyze the cdf of $L$ in detail: Recalling that the density of an exponential rv is given by $\mu \exp(-\mu x)$ we get $$ P[L \le q] = 0.75+0.25 \int_0^q \mu \exp(-\mu x) dx = 0.75 + 0.25(1- \exp(-\mu q)). $$ For a given level $\alpha$ we can then solve $$ \alpha = 0.75 + 0.25(1- \exp(-\mu q)) $$ by rearranging terms and get $$ (\alpha-0.75)/0.25 = 1 - \exp(-\mu q) $$ and finally $$ q = -\frac{1}{\mu} \ln\left(1- (\alpha-0.75)/0.25 \right). $$
For example for $\mu=1$ and $\alpha=0.99$ we get a VaR of $3.2189$ in the mixed case, whereas in the exponential case without the mass at zero it would be $-\ln(1-\alpha) = 4.6052$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.