2
$\begingroup$

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.

$\endgroup$
6
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ Nov 3, 2017 at 23:48

Your Answer

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

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