CVaR/VaR Ratio as alpha goes to 1

Question

I am having trouble taking the following limit of CVaR/VaR for a normal distribution as alpha approaches 1:

$\lim_{\alpha \to 1} \frac{\mu + \sigma \frac{\phi^{-1}(\alpha)}{1-\alpha}}{\mu + \sigma \phi^{-1}(\alpha)}$

First I tried pulling the $(1-\alpha)$ out of the CVaR denominator to get:

$\lim_{\alpha \to 1} \frac{\mu(1-\alpha) + \sigma {\phi^{-1}(\alpha)}}{(1-\alpha)(\mu + \sigma \phi^{-1}(\alpha))}$

Then I thought maybe I need to use L'Hopital's rule, but I have no idea how to do that with an inverse normal imbedded in my function. I feel that I'm probably missing something simple (and my days of calculus are too far behind me). Any hints for how to compute this limit?

Many thanks.

Answer 1

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 $\phi$ denotes the density function of the standard normal distribution, and $\Phi$ its distribution function.

Recall that the derivative of the density is $\phi'(z) = -z\phi(z)$. Then, setting $x = \Phi^{-1}(\alpha)$ and by l'Hopital's rule, the limit of the ratio is $$ \lim_{\alpha \to 1} \frac{\mbox{ES}_\alpha}{\mbox{VaR}_\alpha} = \lim_{x \to \infty} \frac{\mu \{1 - \Phi(x)\} + \sigma \phi(x)}{(\mu + \sigma x) \{1 - \Phi(x)\} } = \lim_{x \to \infty} \frac{1}{1 - \sigma \frac{1 - \Phi(x)}{(\mu + \sigma x)\phi(x)}}, $$ and by l'Hopital's rule $$ \lim_{x \to \infty} \frac{1 - \Phi(x)}{(\mu + \sigma x)\phi(x)} = \lim_{x \to \infty} \frac{1}{(\mu + \sigma x)x - \sigma} = 0. $$ Thus, $$ \lim_{\alpha \to 1} \frac{\mbox{ES}_\alpha}{\mbox{VaR}_\alpha} = 1 . $$

Answer 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

$\lim\limits_{x \to \infty} \frac{\mu + \sigma x/(1-\Phi(x))}{\mu + \sigma x}$

Then perhaps you can you L'Hospitals from hereon. It becomes a bit messy though, but with some effort you might be able to do it.

Do you have the answer?