5

I'm pretty sure it's a limit but I haven't been able to find any page explaining this notation (see below).

It's from a paper on block maxima. 3 out of 5 occurences:

$V=(-1/logF)^\leftarrow$ (p.4)

$x_n=F^\leftarrow(1-p_n)$ (p.7)

$k_0^{(i)}\sim \frac{n}{(1/s)^\leftarrow (n)}(\frac{\sigma_i^2}{B_i^2})^{1/(1-2\rho)}$ (p.10, most important)

2 more on page 10 related to the last one.

Richard Löwenström
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  • This doesn't look like a limit to me, because on page 7 you have $x_n = F^\leftarrow(1-p_n) = V(1/(−\log(1−p_n)))$, where $V$ is defined in the occurrence on page 4. What do you think it is a limit of? – A.P. Apr 14 '15 at 11:39
  • Maybe you're right and it would explain why I'm so confused. I thought it could be a limit related to inputting an argument to a function. If you look at $(1/s)^\leftarrow (n)$ then my guess was that the brackets around $(1/s)$ implied it was a function maybe and you input n into this? Another idea, it might be related to regularly varying functions and the brackets tell us that the arrow affects everything within? – Richard Löwenström Apr 14 '15 at 11:46

1 Answers1

6

Sometimes you see $f^{\leftarrow}$ as an alternate notation for an inverse function $f^{-1}$. This was proposed to reserve the $-1$ exponent for $1/f$ only. But this has not caught on.

So let's see if this works. What is $V=(-1/\log F)^\leftarrow$? $$ V(\theta)=(-1/\log F)^\leftarrow(\theta)\qquad\Longleftrightarrow\qquad \theta = \frac{-1}{\log(F(V(\theta)))} $$ If so, then $$ \log(F(V(\theta))) = \frac{-1}{\theta} \\ F(V(\theta)) = e^{-1/\theta} \\ V(\theta) = F^{\leftarrow}\big(e^{-1/\theta}\big) $$ Now if $1-p_n = e^{-1/\theta}$, so $-1/\theta=\log(1-p_n)$, and $\theta=-1/\log(1-p_n)$, then we have $$ F^{\leftarrow}(1-p_n) = V\left(-1/\log(1-p_n)\right) $$ as they said.

GEdgar
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