2
$\begingroup$

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?

$\endgroup$

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.