Show that $\frac{\int_t^\infty (u-t)f(u)du}{S(t)}=\frac{\int_t^\infty S(u)du}{S(t)}$

Question

Show that $$vmr(t)=\frac{\int_t^\infty (u-t)f(u)du}{S(t)}=\frac{\int_t^\infty S(u)du}{S(t)}$$ Suggestion: Use integration by parts knowing that $f(u)du=-dS(u)$

First some explanations,

Let $T$ a continous non-negative random variable that represents a failure time, where $S(t)=P(T\geq t)$ with $f(t)$ probability density function of $T$.

What I tried:

Let $a=(u-t)$ and $db=f(u)du$, then $da=du$ and $b=\int_t^\infty f(u)du=S(u)$

$$ab-\int_t^\infty b\space da$$

so

$$vmr(t)=\frac{\int_t^\infty (u-t)f(u)du}{S(t)}=\frac{1}{S(t)}\Big((u-t)S(u)-\int_t^\infty S(u)du\Big)$$

EDIT:

$(u-t)S(u)\Big|_t^\infty =0$ since $\lim_{u\rightarrow \infty}S(u)\rightarrow 0$ because $P(T\geq \infty)\rightarrow 0$.

But I still don't get the result because the $-$ signal.

$$vmr=-\frac{\int_t^\infty S(u)du}{S(t)}$$

I probably need to use the fact that $f(u)du=-dS(u)$ but I can't figure out what this $-dS(u)$ mean.

Answer 1

Let $I(t)=\int_t^\infty (t'-t)f(t')\,dt'$. Integrating $I(t)$ by parts with $u=t'-t$ and $v=\int^{t'}f(t'')\,dt''$, so that $du=dt'$ and $dv=-S(t')$, yields

$$\begin{align} I(t)&=\int_t^\infty (t'-t)f(t')\,dt'\\\\ &=\left.\left(-(t'-t)S(t')\right) \right|_t^\infty +\int_t^\infty S(t')\,dt'\\\\ &=\int_t^\infty S(t')\,dt' \end{align}$$

where we used $\lim_{t'\to \infty}t'S(t')=0$.