Why the Distribution Differential Changes Sign in Hill Estimation

# Change of measure


# Change of measure












I am looking at the derivation of the Hill estimator. It is $ \bar{F}(x) = 1 - F(x)$ the right tail of the distribution. In the derivation they use the equation $$ \frac{1}{\bar{F}(u)}\int\limits_u^\infty (\log(x)-\log(u))dF(x) = -\frac{1}{\bar{F}(u)}\int\limits_u^\infty (\log\left(\frac{x}{u}\right)d\bar{F}(x)$$

I think it is $dF(x) = d(1 - \bar{F}(x)) = d(1) -d\bar{F}(x) = 0 - d\bar{F}(x)$. I kind of understand it how $dF(x)$ becomes $-d\bar{F}(x)$, but I am lacking a precise explaination. Is there an mathematical explanaition / theorem for this transform, I think even Ito's Lemma gives me the right intuition here.

## Answer by jacques (score 5, accepted)

https://quant.stackexchange.com/a/55258

You can't have a precise argument without a precise definition. In general, the appropriate notion of integral here is the Lebesgue-Stieltjes integral. In a fairly general setup, let $F: \mathbb R \to \mathbb R$ be a right-continuous function that is of locally bounded variation, that is $$V_F([a,b]) := \sup\lbrace \sum_{i=1}^n \vert F(x_{i+1}) - F(x_i ) \vert :\; a = x_0 < x_1 < \dots < x_n = b , n \in\mathbb N \rbrace < \infty\quad \forall a < b $$ Such a function can be written as $F = f - g$ for right-continuous, monotonously growing functions $f, g$. Then, $\mu^f ([a, b[) := f(b) - f(a)$ and $\mu^g$ analogously for $g$ define measures (in fact, they are premeasures on the ring generated by half-open intervals). By a standard procedure, this can be extended to an outer measure and hence a measure (this is known as the Caratheodory construction, see Rudin Real and Complex Analysis). This gives a decomposition $\mu^F = \mu^f - \mu^g$, where $\mu^f, \mu^g$ are genuine, positive measures and $\mu^F$ is a signed measure. In general, $f, g$ and $\mu^f, \mu^g$ are not unique, but it is possible to choose them such that $\mathbb R = A \cup B$ with $A \cap B = \emptyset$, and $\mu^f (A) = \mu^g (B) = 0$. Such a decomposition is called the Hahn-Jordan decomposition (then, the measure $\vert \mu^F \vert := \mu^f + \mu^g$ is called the absolute variation of $\mu^F$ and $V_F ([a,b]) = \vert \mu^F \vert ([a,b])$).

Now, if you have a probability variable $X$, $F(x) := P [X \leq x]$ defines a right-continuous function, and it has bounded variation indeed, and hence you can define an integral $$\int_U h\,d F = \int_U h\,d\mu^F = \int_U h\,d\mu^f - \int_U h\,d\mu^g$$ It is not important here that $\mu^F = \mu^f - \mu^g$ is the Hahn-Jordan decomposition, but any decomposition works and gives the same result (actually this is the key observation, and the proof should be in any text on signed or complex measures).

However, in the case of $\overline{F}$, one possible decomposition is $\overline F = 0 - (-1 + F), f = 0, g = -1 + F$. This gives you $$\int h\,d\overline F = -\int h\,d (-1 + F)$$ But now it is obvious that $(-1 + F)(b) - (-1 + F)(b) = F(b) - F(a)$, therefore $$\int h\,d(-1 + F) = \int h\,dF$$

EDIT: By the way, signed measures constitute a vector space and this is compatible with the vector space structure of BV functions, so $d(1-F) = d(1) - dF$ is indeed a valid argument. But I feel like it misses the main point.