# 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?

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 $$M(x)$$ to be any positive value for $$S(x) = 0,$$ the equality continues to hold for all $$x \ge 0,$$ QED.

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.