Tag Info

New answers tagged

4

If the loss distribution is normal with mean $\mu$ and variance $\sigma^2$, then the Value-at-Risk and Expexted Shortfall (or CVaR) at level $\alpha \in (0, 1)$ are \begin{align*} \mbox{VaR}_\alpha & = \mu + \sigma \Phi^{-1}(\alpha) , \\ \mbox{ES}_\alpha & = \mu + \sigma \frac{\phi\{\Phi^{-1}(\alpha)\}}{1 - \alpha} , \end{align*} where $\phi$ ...


2

I don't know what you did when you tried pulling out $1-\alpha$, the correct expression would be $\lim_{\alpha \to 1} \frac{\mu(1-\alpha) + \sigma {\phi^{-1}(\alpha)}}{(1-\alpha)(\mu + \sigma \phi^{-1}(\alpha))}$. Anyhow, you can try using the substitution $\Phi^{-1}(\alpha) = x$, $x \to \infty$ and $\alpha = \Phi(x)$. Then the expression becomes ...


1

You can use the either, as both necessarily are symmetric positive definite; covariance is a personal preference. It's really just a matter of scaling, as $\mathcal{N}(0,\Sigma)$ is distributionally $\sqrt{\Sigma} \mathcal{N}(0,1) $. Correlation would require additional scaling (i.e. multiplication of every $\mathcal{N}(0,\rho)$ element by its respective ...



Top 50 recent answers are included