# How to simplify the derivation of the inverse cdf yielded from l'Hospital rule?

I am currently dealing with a proof of Pauline Barrieus Paper "Assessing Financial Model Risk" (page 19).

At one point she applys the l'Hospital rule on a limit equation. We have some cumulative distribution function $$F$$ and its derivation $$f$$ as a density function given. I just cant follow that step:

$$\begin{equation} \lim_{\varepsilon \rightarrow 0}\frac{F^{-1}(\alpha)-F^{-1}(\alpha-\varepsilon)}{F^{-1}(\alpha+\varepsilon)-F^{-1}(\alpha-\varepsilon)} \end{equation}$$ after applying l'Hospital rule it should be $$\begin{equation} \lim_{\varepsilon \rightarrow 0}\frac{1/f(\alpha-\varepsilon)}{1/f(\alpha+\varepsilon)+1/f(\alpha-\varepsilon)} \end{equation}$$

But why? I really don't get it.

My approach would be applying $$[F^{-1}(\alpha)]^{\prime}=\frac{1}{f(F^{-1}(\alpha))}$$

But from there I won't get any further. Also the fact that $$F^{-1}(\alpha)=q_{\alpha}$$ doesn't seem to help.

Maybe it's obvious, but I can't see it, so if there is someone who know how to deal with a such an equation, your help would be very much appreciated.

• You're quite correct. Consider, after all, a distribution $F$ that is not supported on a neighborhood of the interval $[0,1]$. In that case both $f(\alpha-\varepsilon)$ and $f(\alpha+\varepsilon)$ would be zero for sufficiently small $\varepsilon$, making the limit undefined. Evaluating the original limit depends on whether $f$ is continuous at $\alpha$.
– whuber
Nov 2, 2017 at 19:00
• Thank you for your quick response, I get that, but I don't under stand how to get from $\frac{F^{-1}(\alpha)-F^{-1}(\alpha-\varepsilon)}{F^{-1}(\alpha+\varepsilon)-F^{-1}(\alpha-\varepsilon)}$ to $\frac{1/f(\alpha-\varepsilon)}{1/f(\alpha+\varepsilon)+1/f(\alpha-\varepsilon)}$ regardless of the limit. Nov 2, 2017 at 19:26
• My point is that this is obviously wrong. Therefore you need to return to the first expression and perform the calculation correctly. You haven't given enough information about $F$ and $\alpha$ to allow anyone to go any further than that in terms of deriving a definite answer.
– whuber
Nov 2, 2017 at 19:28
• Oh okay, so this is not true in general? Guess I missinterpreted your first answer. Actually $F$ could be any distribtution function for example standard normal or student t, but I guess that doesn't change a thing. Also we have $\alpha \in (0,1)$, $\varepsilon<\alpha$ and $VaR_{\alpha}(X)=F^{-1}(\alpha)$ if that helps anyhow... Nov 2, 2017 at 19:34
• Alright yeah you're right thanks a lot, so I don't actually need the form $1/f(\alpha+\varepsilon)$, but $1/f(F^{-1}(\alpha+\varepsilon))$ would work as well to get the limit of $1/2$ Nov 3, 2017 at 23:08

Hint: This is a limits problem. Since you're taking the limit with respect to $\epsilon$, you need to take the derivative with respect to $\epsilon$, treating $\alpha$ as if it were a constant (assuming $\alpha$ is not a function of $\epsilon$).
For the numerator, the $F^{-1}(\alpha)$, for instance, goes to $0$, since $\alpha$ is regarded as a constant.
The term $-F^{-1}(\alpha - \epsilon)$ has derivative $-(F^{-1})^{\prime}(\alpha - \epsilon) \cdot (-1) = (F^{-1})^{\prime}(\alpha - \epsilon)$, where $\prime$ denotes the derivative. Now, by this page, we have $$(F^{-1})^{\prime}(\alpha - \epsilon) = \dfrac{1}{F^{\prime}(F^{-1}(\alpha-\epsilon))} = \dfrac{1}{f(F^{-1}(\alpha-\epsilon))}\text{.}$$
• Thanks a lot! I actually ignored the fact that I have to derive with respect to $\varepsilon$ and not $\alpha$. So it could work like that: $$\frac{[F^{-1}]'(\alpha)+[-F^{-1}]'(\alpha-\varepsilon)}{[F^{-1}]'(\alpha+\varepsilon)+[-F^{-1}]'(\alpha-\varepsilon)} = \frac{0+[F^{-1}]'(\alpha-\varepsilon)}{[F^{-1}]'(\alpha+\varepsilon)+[F^{-1}]'(\alpha-\varepsilon)}= \frac{1/f(F^{-1}(\alpha-\varepsilon))}{1/f(F^{-1}(\alpha+\varepsilon))+1/f(F^{-1}(\alpha-\varepsilon))}$$ Which with respect to limit tends to be $$\frac{1/f(F^{-1}(\alpha))}{2\cdot 1/f(F^{-1}(\alpha))}=\frac{1}{2}$$ Nov 3, 2017 at 23:04
• @Quastiat Yeah, note that there are some implicit assumptions made here. Namely, that $f$ is continuous at $F^{-1}(\alpha)$, and $F^{-1}$ is continuous at $\alpha$. Otherwise, you can't bring the limit inside the functions. Nov 3, 2017 at 23:48