How to find a survival function in terms of a mean residual life function? I found in a book that one way to express the survival function is by using the mean residual life; that is
$$
S(x) =\dfrac{mrl(0)}{mrl(x)}\,\exp\left\{ -\int_{0}^{x}\dfrac{du}{mrl(u)} \right\}
$$
where $S(x)$ and $mrl(x)$ are the surival function and the mean residual life at time $x,$ respectively.
How can I prove that?
 A: The mean residual life function for a non-negative variable $X$ with survival function $S(x) = \Pr(X \gt x)$ is its expectation conditional on $X$ exceeding the value $x \ge 0,$
$$M(x) = E\left[X \mid X \gt x\right] = \frac{1}{S(x)} \int_x^\infty S(t)\,\mathrm dt$$
for all $x$ where $S(x) \gt 0.$  (This implicitly assumes the expectation of $X$ is finite.)
Your goal is to find $S$ in terms of $M.$
Multiplying both sides by $S(x)$ gives an identity for all $x\ge 0$ and differentiating that using the Product Rule (on the left) and the Fundamental Theorem of Calculus (on the right) yields

$$M^\prime(x)S(x) + M(x)S^\prime(x) = -S(x).$$

This is a linear first-order ordinary differential equation for $S.$ Because it is non-singular ($M$ is nonzero throughout the region of interest), the implicit initial condition $S(0)=1$ (which automatically holds when the distribution of $X$ is continuous at $0$) uniquely determines the solution.  Its solution is routine, but for those unfamiliar with solving ODEs, here are the details.

Again supposing $S(x)\gt 0,$ necessarily $M(x) \ne 0$ and so we may divide both sides by $S(x)M(x)$ and rearrange to separate the functions $S$ and $M$ on either side of the equation:
$$\frac{\mathrm d}{\mathrm dx} \log(S(x)) = \frac{S^\prime(x)}{S(x)} = -\frac{M^\prime(x)}{M(x)} - \frac{1}{M(x)} = -\frac{\mathrm d}{\mathrm dx} \log(M(x)) -\frac{1}{M(x)}.$$
Integrating both sides (starting at the natural origin of $0$) shows that
$$\log(S(x)) - \log(S(0)) = - \left(\log(M(x)) - \log(M(0))\right) - \int_0^x \frac{\mathrm du}{M(u)}.$$
Exponentiating this and recognizing $\log(S(0)) = \log(1) = 0$ yields the equivalent equality,
$$S(x) = \frac{M(0)}{M(x)}\, \exp \left(-\int_0^x \frac{\mathrm du}{M(u)}\right).$$
If we define $M(0)$ to be any positive value for $S(x) = 0,$ the equality continues to hold for all $x \ge 0,$ QED.
A: It is not very elegant, but I wanted to show what I managed to do.
We will start from
$S(t) = \dfrac{vmr(0)}{vmr(t)}\,\exp\left\{-\int_{0}^{t} \frac{du}{vmr(u)}\right\},$
then
$-log[S(t)] = -log\left[\int_{0}^{\infty}S(u)\,du\right] +log[vmr(t)]-\int_{0}^{t} \frac{du}{vmr(u)}, $
because $vmr(0) =\int_{0}^{\infty}S(u)\,du$. Now, deriving respect $t$
$-\dfrac{\partial}{\partial t}log[S(t)] = \dfrac{\partial}{\partial t}\left( -log\left[\int_{0}^{\infty}S(u)\,du\right] \right) + \dfrac{\partial}{\partial t}\left( log[vmr(t)] \right) - \dfrac{\partial}{\partial t}\left( \int_{0}^{t} \frac{du}{vmr(u)} \right).$
Using Leibniz rule for differentiation under the integral sign, and the fact that
$\dfrac{\partial}{\partial t}\left( -log\left[\int_{0}^{\infty}S(u)\,du\right] \right) =0$ (because it is not a function that depends on $t$), we have that
$-\dfrac{\partial}{\partial t}log[S(t)] = \dfrac{\dfrac{\partial}{\partial t}vmr(t)}{vmr(t)} +\dfrac{1}{vmr(t)}$
$h(t) = \dfrac{\dfrac{\partial}{\partial t}vmr(t) +1}{vmr(t)},$
because $h(t) =-\dfrac{\partial}{\partial t}log[S(t)]$. Now, using 
$
\dfrac{\partial}{\partial t}vmr(t) = \dfrac{\partial}{\partial t}\left( \dfrac{\int_{t}^{\infty}S(u)\,du}{S(t)} \right)$
$ = \dfrac{S(t)[-S(t)] -\int_{t}^{\infty}S(u)\,du\left( \dfrac{\partial}{\partial t}S(t) \right)}{[S(t)]^{2}}$
$ = \dfrac{-[S(t)]^{2} - \dfrac{\partial}{\partial t}S(t)\int_{t}^{\infty}S(u)\,du}{[S(t)]^{2}},$
we can check that
$h(t) = \left( \dfrac{\partial}{\partial t}vmr(t) +1 \right) \left( \dfrac{1}{vmr(t)} \right)$
$h(t) = \left( \dfrac{-[S(t)]^{2} - \dfrac{\partial}{\partial t}S(t)\int_{t}^{\infty}S(u)\,du +[S(t)]^{2}}{[S(t)]^{2}}\right)  \left( \dfrac{S(t)}{\int_{t}^{\infty}S(u)\,du} \right)$
$h(t) = \dfrac{-\dfrac{\partial}{\partial t}S(t)}{S(t)}$
$h(t) = -\dfrac{\partial}{\partial t}log[S(t)]$
That we know to be true. And being just a series of equalities, we can return, obtaining the relationship that we want to demonstrate.
