Mean residual life: (improper) integration by parts involving p.d.f

Question

Let a random variable $X$ be distributed with p.d.f $f(\cdot)$.

I want to derive the following equality for the mean residual life: $$m(x)\equiv\frac{\int_{x}^{\infty}(z-x) f(z)dz}{1-F(x)}=\frac{\int_{x}^{\infty}1-F(z)dz}{1-F(x)}$$

Clearly, I can start by splitting the left-hand-side to get $\frac{\int_{x}^{\infty}z f(z)dz-x\int_{x}^{\infty} f(z)dz}{1-F(x)}=\frac{\left(\int_{x}^{\infty}z f(z)dz\right)-x(1-F(x))}{1-F(x)}$. From this it looks like all I need to do is integration by parts for $\int_{x}^{\infty}z f(z)dz$ with $u(z)=z$ and $v(z)=dF(z)$, but I don't see how this converges to the desired result.

Answer 1

Define $Z-x \equiv h(Z)$. $x$ is treated as fixed here. Then you have

$$m(x)\equiv\frac{\int_{x}^{\infty}(z-x) f(z)dz}{1-F(x)}= \frac {E\Big(h(Z)\cdot \mathbf 1_{\{Z \ge x\}}\Big)}{1-F(x)} \qquad [1]$$

Now the expected value can be written $$E\Big(h(Z)\cdot \mathbf 1_{\{Z \ge x\}}\Big) = h(x) + \int_{x}^\infty h'(x)P(Z\ge z) dz $$ since $P(h(Z)\ge h(x)) =1$.

But $h(x) = x-x = 0$, $h'(x) = 1$ and $P(Z\ge z) = 1-F(z)$.

So $$E\Big(h(Z)\cdot \mathbf 1_{\{Z \ge x\}}\Big) = \int_{x}^\infty [1-F(z)] dz$$

which leads to the desired result.

Answer 2

The mean residual life $m(x)$ at time $x$ of a system with lifetime $X$, a nonnegative random variable, is the expected lifetime beyond $x$ of a system that has survived till time $x$, that is, $$m(x) = E[X\mid X > x] - x.$$ Now, the conditional pdf of $X$ given that $X > x$ is just $\displaystyle \frac{f(z)}{P\{X > x\}}\mathbf 1_{\{z: z>x\}} = \frac{f(z)}{1-F(x)}\mathbf 1_{\{z: z>x\}}$ and so $$\begin{align} m(x) &= E[X\mid X > x] -x\\ &= \int_x^{\infty}z\frac{f(z)}{1-F(x)}\mathrm dz - x\\ &= \int_x^{\infty}z\frac{f(z)}{1-F(x)}\mathrm dz - x\int_x^{\infty}\frac{f(z)}{1-F(x)}\mathrm dz\\ &= \int_x^{\infty}\frac{(z-x)f(z)}{1-F(x)}\mathrm dz. \end{align}$$ The second equality uses the result that for a (integrable) nonnegative random variable $Y$ with CDF $F_Y(y)$, $$E[Y] = \int_0^\infty [1-F_Y(y)]\,\mathrm dy.$$ If you are unfamiliar with this result, see, for example, this question for a discussion of the intuition behind this result and then read the complete proof. The conditional CDF of $X$ given that $X > x$ is given by $$F_{X\mid X>x}(z\mid X > x) = \frac{F(z)-F(x)}{1-F(x)}\mathbf 1_{\{z: z>x\}}$$ so that the complementary CDF has value $\displaystyle \mathbf 1_{\{z: 0 < z < x\}} + \frac{1-F(z)}{1-F(x)} \mathbf 1_{\{z: z>x\}}$ on the positive real line. The integral of the complementary CDF over the positive real line gives $$E[X \mid X > x] = x + \int_x^\infty \frac{1-F(z)}{1-F(x)}\mathrm dz$$ and since $m(x) = E[X \mid X > x] - x$, the desired equality follows.