I was watching a YouTube video about finance and in the video, the lecturer mentioned excess distribution function and mean excess function.
\begin{align} F_u(x)&=P(X\le u+x|X>u)\\ e(u)&=E(X-u|X>u) \end{align}
where $X$ is a random variable with distribution function $F(x)=P(X\le x)$.
However, I don't really know how to compute these quantities. My background is computer science, and I'm doing some programming about finance. What I've searched and learned is that,
Let's say random variable $X$ follows a uniform distribution on [0, 1], then the $x_F$, which is the upper end-point of $X$, equals to 1.
The Wikipedia page about Pickands–Balkema–de Haan theorem gives a formula, $$F_u(y)=P(X-u\le y|X>u)=\frac{F(u+y)-F(u)}{1-F(u)}$$
The cumulative distribution function for standard uniform random variable $X$ is $F(x)=x$
For mean excess function, I found some materials on this page, on hu-berlin.de, $$e(x)=E(X-x|X>x)=\frac{\int_x^\infty(1-F(u))du}{1-F(x)}$$
Does 2. 3. and 4. suggest that the $F_u(x)$ and $e(u)$ can be computed as the following?
\begin{align} F_u(x)&=\frac{F(u+x)-F(u)}{1-F(u)}=\frac{u+x-u}{1-u}=\frac{x}{1-u}\\ e(u)&=\frac{\int_x^\infty(1-F(u))du}{1-F(x)}=\frac{\int_x^\infty(1-u)du}{1-x} \end{align}
And if we let $u\rightarrow x_F$, then is the following right?
\begin{align} \lim_{u\rightarrow x_F}F_u(x)&=\frac{x}{1-u}=\frac{x}{1-1}=\infty\\ \lim_{u\rightarrow x_F}e(u)&=\frac{\int_x^\infty(1-u)du}{1-x}=\frac{\int_x^\infty(1-1)du}{1-x}=0 \end{align}
I'm now totally lost, could anyone shed some light on this?
Update 1:
\begin{align} F_u(x)&=\frac{F(u+x)-F(u)}{1-F(u)}\\ &\not=\frac{u+x-u}{1-u} \end{align}
$F(u)$ can be approximated by $\frac{n-N_u}{n}$