TITLE: A Variance-Tail Description for Continuous Probability Distributions
QUESTION [8 upvotes]: Start with a continuous probability distribution  given by a density function f(x). Let X be a real random variable whose distribution is given by the probability distribution.
I would like to ask about the following specific way to describe the distribution in terms of the variance of tails.
Given a number t between 0 and 1 let s be such that the probability that $X\ge s$ equals t. (In other words, $\int_s^{\infty}f(x)=t$.) Let $W_X(t)$ be the variance of $X$ conditioned on $X\ge s$.
The function $W_X(t)$ seems an interesting way to describe the original probability distribution, and its tail behavior. I wonder if there is an explicit useful way to move from $W$ back to $f$; Can you give a dictionary between the tail behavior of $f$ and of that of $W$; And an explicit description of $W$ for some "famous" probability distributions. (E.g. the normal distribution, and the Tracy Widom distribution.) And of course if this kind of "transform" is considered in the literature.  
UPDATE: Many thanks to Robert and Brendan to their answers. Robert's  asymptotic expansion seem amazing. I will be happy to understand the asymptotics as t goes to 0.
Looking at various distribution one sees that when the distribution is exponential, W is constant; when the distribution is normal and thus decays doubly exponentially, W(t) behaves like $\log^{-1}(1/t)$. I am curious how the asymptotic decay of $W(t)$ translates to the tail behavior of the probability distribution. For the tracy Widom distribution when you ask what is the probability that $X \ge M+t \sigma$ you get an expression like $e^{-t^{3/2}}$ and for the probability that $X \le M-t \sigma$ you get something like $e^{-t^3}$. I wonder how these tails behavior can be described in terms of $W_X(t)$ and $W'_X(t)$ respectively, where $W'_X(t)$ deals with conditioning on $X \le s$ rather than on $X \ge s$. So let me form a simple question:
Question: If $W_X(t)$ behaves like $\log^a (1/t)$, for some real $a>0$, what does it say about the tail behavior of $X$. (More generally, how the decay of $W$ translates to the decay of the distribution.)

REPLY [2 votes]: Expression for the quantile function (the inverse CDF) for the case $E(X) = 0$:
Let $\overline{F}(s) = 1 - F(s)$. Then, $t = \overline{F}(s)$ and
$$\frac{d}{ds} E(X^2|X \geq s) = \frac{d}{ds} \Big( \frac{\int_s^{+\infty}x^2f(x)dx}{\int_s^{+\infty}f(x)dx} \Big) =$$
$$ = \frac{-s^2 f(s) \int_s^{+\infty}f(x)dx+f(s) \int_s^{+\infty}x^2f(x)dx}{\big(\int_s^{+\infty}f(x)dx\big)^2} = $$
$$ = \frac{-s^2 f(s) + f(s) W(\overline{F}(s))}{\overline{F}(s)}$$
On the other hand,
$$ \frac{d}{ds} E(X^2|X \geq s) = \frac{d}{ds} W(\overline{F}(s)) = -W'(\overline{F}(s)) f(s)$$
where $W'$ is the derivative of $W$. Therefore,
$$ \frac{s^2 - W(\overline{F}(s))}{\overline{F}(s)} = W'(\overline{F}(s))$$
or, using $t$,
$$ \frac{(\overline{F}^{-1}(t))^2 - W(t)}{t} = W'(t) $$
and 
$$ \overline{F}^{-1}(t) = \sqrt{tW'(t) + W(t)}$$
($\overline{F}^{-1}(t)$ should be positive in some neighborhood of $t = 0$).
Then, since $\overline{F}^{-1}(t) = F^{-1}(1-t)$,
$$ F^{-1}(x) = \sqrt{(1-x)W'(1-x) + W(1-x)} $$
I hope that it's correct and that it can tell something about the decay of the distribution when $t$ tends to $+0$. Maybe it can also be extended to the case $E(X) \neq 0$.