On page 19 of this freely available book by Emil Gumbel (in English) the following two equations are printed:

$$\Phi_n(x_n)=F^n(x_n)\tag{3.1}$$

which stands for the probability distribution that the largest value of a sample of $n$ independent observations from the probability $F$ – i.e. the probability that the largest value is less than $x_n,$

as well as its derivative

$$\phi_n(x_n) = nF^{(n-1)}(x_n)f(x_n)\tag{3.2}$$

which would correspond to the pdf.

This all seems clear.

The question is about the equations on page 21:

The asymptotic distribution of the largest value for the exponential

type may be derived in the following way: The mode $\bar x_n$ of the

largest value is obtained from equation $( 3.2)$ as the solution of

$$\frac{n-1}{F(x)}f(x) =-\frac{f'(x)}{f(x)}$$

If $x$ becomes very large, the density of probability $f (x )$ becomes

very small , and the same holds for the probability $1 -F(x)$ of a

value surpassing $x.$ If the variate is unlimited, the derivative $f '

> ( x )$ also converges toward zero . If it is legitimate from a certain

value $x$ onward to apply l'Hôpital's rule and obtain the relation

$$\frac{f(x)}{1-F(x)} \sim-\frac{f'(x)}{f(x)}$$

How do these two last equations follow from $(3.1)$ and $(3.2)?$

## Best Answer

A mode of a continuous probability distribution is a value at which the probability density function (pdf) attains its maximum value. Assuming that the mode $\bar x_n$ is unique, it is therefore the solution of $$ \phi_n'(\bar x_n)=0\Rightarrow n [\phi_{n-1}(\bar x)f(\bar x)+F^{n-1}(\bar x)f'(\bar x)]=0 . $$ Substituting $\phi_{n-1}(\bar x)=(n-1)F^{n-2}(\bar x)f(\bar x)$, and dividing by $F^{n-1}(\bar x)f(\bar x)$, we get the equation $$ (n-1)F^{-1}(\bar x)f(\bar x)+\frac{f'(\bar x)}{f(\bar x)}=0 $$ that is equivalent to the first equation you sought.

The second equation you are after [NOTE: there is a typo in what you wrote, the numerator is $f(x)$, not $x$] does not follow (I think) from (3.1) or (3.2). You just want to estimate for large $x$ how the ratio $$ \frac{f(x)}{1-F(x)} $$ behaves (where both numerator and denominator quite clearly go to zero). You can estimate the ratio by using L'Hopital, which gives $$ \frac{f(x)}{1-F(x)}\sim -\frac{f'(x)}{f(x)}\ , $$ which (coincidentally) - if evaluated for $x=\bar x$ - is equal to the right hand side of the first equation you were after. From this fact, some conclusions may be drawn (I did not look more in detail into the paper you linked) - but again I don't think the asymptotic relation follows from any of the previous statements.