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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?

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  • $\begingroup$ You should find the CDF of $L$, otherwise calculate VaR by simulation. $\endgroup$ – Egodym Jan 6 '16 at 21:03
  • $\begingroup$ How do I intepretate the CDF of $L$? It is a discrete distribution with an output $Z$ that has a continious distribution $\endgroup$ – Elekko Jan 6 '16 at 21:16
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    $\begingroup$ It is a "mixture" of two distributions, an exponential and a trivial X=0 distribution, with mixing weights 0.25 and 0.75. $\endgroup$ – noob2 Jan 6 '16 at 21:19
  • $\begingroup$ Thats the keypoint I actually don't really 100% understand. $\endgroup$ – Elekko Jan 6 '16 at 23:08
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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 $-\ln(1-\alpha) = 4.6052$.

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