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.

  • $\begingroup$ The expression of CVaR should read $\mu + \sigma \phi\{\Phi^{-1}(\alpha)\} /(1-\alpha)$, and the expression of VaR should read $\mu + \sigma \Phi^{-1}(\alpha)$. $\endgroup$
    – QuantIbex
    Commented Apr 12, 2014 at 21:33

2 Answers 2


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, $$ \lim_{\alpha \to 1} \frac{\mbox{ES}_\alpha}{\mbox{VaR}_\alpha} = 1 . $$

  • $\begingroup$ I don't see why $$\lim_{x \to \infty} \frac{\mu \{1 - \Phi(x)\} + \sigma \phi(x)}{(\mu + \sigma x) \{1 - \Phi(x)\} } = \lim_{x \to \infty} \frac{1}{1 - \sigma \frac{1 - \Phi(x)}{(\mu + \sigma x)\phi(x)}}, $$. I've been trying that equality for a few hours but I just don't get it $\endgroup$
    – BlueRedem1
    Commented Jul 12, 2021 at 20:51

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?

  • $\begingroup$ Thanks, Mike. I missed my $(1-\alpha)$ in the denominator by mistake. In any case I do know that the correct answer is 1. I'm just struggling in the details of how to get there. $\endgroup$
    – AmethystJ
    Commented Apr 11, 2014 at 21:23
  • $\begingroup$ Using your hints, then we would have: $\lim\limits_{x \to \infty} \frac{1}{1-\Phi(x)}$ but unfortunately it seems like that goes to infinity not 1. As a reference, this is something called the "shortfall-to-quantile ratio". $\endgroup$
    – AmethystJ
    Commented Apr 11, 2014 at 21:31
  • $\begingroup$ Well, is that the expression you really want to compute the limit of? If the denominator is the CVaR then you have the wrong expression I believe. It should be something like $\phi(\Phi^{-1}(p))/p$. $\endgroup$ Commented Apr 12, 2014 at 9:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.