I have to compute the mean value, the variance, the value-at-risk $\mathrm{V@R}_{\alpha}$ and the expected shortfall $\mathrm{ES}$ of a random variable with CDF
$$ F(x) = \begin{cases} 1- (\frac{3}{2x})^{-3}, &\text{if } x\geq 3\\ 0, &\text{otherwise} \end{cases} $$
I have computed the mean and variance, but what about computing the value at risk and expected shortfall if I am not given a confidence level $\alpha$?
Let us denote a random variable which has the above CDF by $X$ (there are more than one!).
Notice that $F(3)=\frac{7}{8}$. The $\mathrm{V@R}_{\alpha}$ is defined as $\mathrm{V@R}_{\alpha}[X]=F^{-1}(1-\alpha)$. If $1-\alpha\geq \frac{7}{8}$ then $\mathrm{V@R}_{\alpha}[X]$ is determined by
$$ 1-\left(\frac{3}{2\mathrm{V@R}_{\alpha}[X]}\right)^{-3}=1-\alpha \Rightarrow \mathrm{V@R}_{\alpha}[X] = \frac{2}{\alpha^3}. $$
If $x<3$ then, clearly, $\mathrm{V@R}_{\alpha}=-\infty$.
$\mathrm{ES}_{\alpha}$ is given by
$$ \mathrm{ES}_{\alpha}[X] = \mathbb{E}[X\mid X\geq \mathrm{V@R}_{\alpha}[X]]. $$
Now special care is required because your CDF is discontinuous. Recall first that the above expectation is given by the Lebesgue integral
$$ \mathrm{ES}_{\alpha}[X] = \int_{\mathrm{V@R}_{\alpha}[X]}^{+\infty}X\mathrm{dP} $$
Then, you have to distinguish cases: for $\alpha < \frac{1}{8}$, for $\alpha=\frac{1}{8}$ and for $\alpha>\frac{1}{8}$ and you need to take into consideration that the discontinuity contributes to the expectation. For details you may refer to various SO answers.
