$X$ is a random variable, $E[\mid X\mid ]<\infty$ and $F_X$ is its cumulative distribution function.
I would like to proof that $E[X\mid X\geq b]\geq E[X \mid X \geq a]$ for $b\geq a$ and $F(b)<1$.
I started with writing
$E[X \mid X \geq b] = \frac{1}{P[X\geq b]}\int_b^{\infty}xf_X(x)dx$
and
$E[X \mid X \geq a] = \frac{1}{P[X\geq a]}\int_a^{\infty}xf_X(x)dx$.
I think the term before the integral grows faster than the integral decreases. But I do not know how to proof that. I would be glad if somebody could help me with that. I think I am missing something elementary.
Or is there a more elegant way to proof it?
The idea is that $E[X \mid X \ge a]$ is a weighted average of $E[X \mid X \ge b]$ and something smaller than it.
Specifically, write $$E[X \mid X \ge a] = \frac{E[X \mid X \ge b] P[X \ge b] + E[X \mid a \le X < b] P[a \le X < b]}{P[X \ge a]}$$ and now observe that $$E[X \mid X \ge b] \ge b \ge E[X \mid A \le X < b].$$