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I am currently dealing with a proof of Pauline Barrieu`s 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 can`t 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.

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    $\begingroup$ 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 $\alpha$. $\endgroup$
    – whuber ♦
    Nov 2, 2017 at 19:00
  • $\begingroup$ 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. $\endgroup$
    – Quastiat
    Nov 2, 2017 at 19:26
  • $\begingroup$ 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. $\endgroup$
    – whuber ♦
    Nov 2, 2017 at 19:28
  • $\begingroup$ 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... $\endgroup$
    – Quastiat
    Nov 2, 2017 at 19:34
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    $\begingroup$ 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 $1/2$ $\endgroup$
    – Quastiat
    Nov 3, 2017 at 23:08

1 Answer 1

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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 $$(F^{-1})^{\prime}(\alpha - \epsilon) = \dfrac{1}{F^{\prime}(F^{-1}(\alpha-\epsilon))} = \dfrac{1}{f(F^{-1}(\alpha-\epsilon))}\text{.}$$

I leave the rest to you.

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  • $\begingroup$ 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 $$ \frac{1/f(F^{-1}(\alpha))}{2\cdot 1/f(F^{-1}(\alpha))}=\frac{1}{2} $$ $\endgroup$
    – Quastiat
    Nov 3, 2017 at 23:04
  • $\begingroup$ @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. $\endgroup$
    – Clarinetist
    Nov 3, 2017 at 23:48

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