TITLE: Proof that Expected Lifetime is longer than Remaining Lifetime if the Hazard Rate is increasing.
QUESTION [6 upvotes]: Let $X$ be a positive, continuous random variable. Denote the density of $X$ by $f(x)$ and its CDF by $F(x)$. Let $\bar{F}(x) = 1- F(x)$ be the survival function of $X$. Given that the Hazard Rate,
\begin{align}
\lambda(x) &= \frac{f(x)}{\bar{F}(x)}
\end{align}
is increasing, i.e. $\lambda'(x) \geq 0$, I want to prove that
\begin{align}
\mathbf{E}\left[ X-c \mid X>c \right] &\leq \mathbf{E}\left[ X \right] 
\end{align}
for any constant $c >0$.
Here's what I have tried so far:
Let
\begin{align}
\Lambda(x) &= \int_0^x \lambda(s) \mathop{}\!\mathrm{d} s.
\end{align}
This implies the following
\begin{align}
\lambda(x) &= \Lambda'(x),
\\
-\Lambda(x) &= \log \left(\bar{F}(x) \right),
\\
\bar{F}(x) &= e^{-\Lambda(x) }.
\end{align}
Writing the expected values as integrals (using the Darth Vader rule) and using the above, I can rewrite the inequality as
\begin{align}
\int_c^\infty e^{-\int_c^s \lambda(x) \mathop{}\!\mathrm{d}x } \mathop{}\!\mathrm{d}s &\leq
\int_0^\infty e^{
-\int_0^s \lambda(x) \mathop{}\!\mathrm{d}x } \mathop{}\!\mathrm{d}s.
\end{align}
However, I do not see how this holds given $\lambda$ in increasing, and I can't think of a way using that assumption.

REPLY [3 votes]: Since $\lambda$ is increasing, we have $\lambda(y+c) \geq \lambda(y)$, hence $\int_0^t \lambda(y + c) \,dy \geq \int_0^t \lambda(y)\,dy$, and thus $e^{-\int_0^t \lambda(y + c) \,dy} \leq e^{-\int_0^t \lambda(y) \,dy}$. Changing variables and using this inequality gives
$$\int_c^\infty e^{-\int_c^s \lambda(x) \,dx} \,ds = \int_0^\infty e^{-\int_c^{t+c} \lambda(x) \,dx} \,dt = \int_0^\infty e^{-\int_0^t \lambda(y+c) \,dy} \,dt \leq \int_0^\infty e^{-\int_0^t \lambda(y) \,dy} \,dt $$
as desired.