proof non-decreasing conditional expectation I have to show that, 
$$g(x)=E(\Pi\mid \Pi>h(x))= 
\frac{\int\limits_{ h(x)}^1 \pi f_\Pi(\pi) \, d\pi}{ \int\limits_{h(x)}^1 f_\Pi(\pi) \, d\pi},$$
is a non-decreasing function;
where $h(x)$ is decreasing, $f(\pi)$ is non-decreasing density function, and $\Pi$ is between $0$ and $1$.
Intuitively, if we increase $x$, then, as $h(x)$ is decreasing, we let $\Pi$ take smaller values , and so expected value tends to be lower. However, I'm not able to prove it (Honestly, I'm not sure if the condition for a non-decreasing $f(\pi)$ is necessary).
I have tried something like splitting the expected value into two parts:
$$\frac{\int\limits_{ h(x)}^1 \pi f(\pi) \, d\pi}{ \int\limits_{h(x)}^1 f(\pi) \, d\pi} \geq \frac{\int\limits_{ h(\pi+\epsilon)}^{h(x)} \pi f(\pi) \, d\pi + \int\limits_{ h(x)}^1 \pi f(\pi) \, d\pi}{ \int\limits_{h(x+\epsilon)}^{h(x)} f(\pi) \, d\pi + \int\limits_{h(x)}^1 f(\pi) \, d\pi} . $$
for any $\epsilon>0$ and see what happens, but I'm not able to see any clear way to proof that $g(x) \geq g(x+\epsilon)$ since both the numerator and denominator move on the same direction.
 A: One can also show that $\Gamma(y):=E[\Pi\mid\Pi>y]$ is non-decreasing. If the density $f_\pi$ is continuous, it is easy to compute $\Gamma'(y)$ and check that it is non-negative. More generally, even if $\Pi$ doesn't have a density, the function $y\mapsto \Gamma(y)$ is of bounded variation, being the ratio of two such functions.  The associated (possibly signed)   measure satisfies
$$
d\Gamma(y) = {\Gamma (y)-y\over \overline F(y)}{1\over 1+\Delta F(y)/\overline F(y)}\,dF(y),
$$
where $F$ is the cdf of $\Pi$, $\overline F(y) =1-F(y)$, and $\Delta F(y) = F(y)-F(y-)$. Clearly $\Gamma(y)\ge y$, and so $d\Gamma$ is a positive measure, and $\Gamma$ is non-decreasing.
A: To find the conditional expectation, first find the conditional distribution, then find the expectation of that distribution in the same way in which you find the expectation of any distribution.
Suppose $A\subseteq [w,\infty)$.  Then
$$
\Pr(\Pi \in A \mid \Pi > w) = \frac{\Pr(\Pi\in A\ \&\ \Pi>w)}{\Pr(\Pi>w)} = \frac{\Pr(\Pi\in A)}{\Pr(\Pi>w)} = \frac{\int_A f(\pi)\,d\pi}{\int_w^\infty f(\pi)\,d\pi}.
$$
Letting $c= \dfrac 1 {\int_w^\infty f(\pi)\,d\pi}$, this says
$$
\Pr(\Pi\in A\mid \Pi>w) = \int_A cf(\pi)\,d\pi.
$$
That means $cf(\pi)\,d\pi$ is the conditional distribution given that $\Pi>w$.  Therefore the conditional expectation is
$$
\operatorname{E}(\Pi\mid \Pi>w) = \int_w^\infty \pi c f(\pi)\,d\pi = \frac{\int_w^\infty \pi f(\pi)\,d\pi}{\int_w^\infty f(\pi)\,d\pi}.
$$
That takes care of what your first paragraph says you want to show, and it does not depend on the decreasing nature of your function $h$.
If you want to show that as $w$ decreases, so does $\operatorname{E}(\Pi\mid \Pi>w)$, that still doesn't depend on anything about your function $h$. Suppose $w_1<w_2$.  Then
$$
\operatorname{E}(\Pi\mid \Pi>w_1) = \frac{\int_{w_1}^\infty \pi f(\pi)\,d\pi}{\int_{w_1}^\infty f(\pi)\,d\pi} = \frac{\int_{w_1}^{w_2} \pi f(\pi)\,d\pi + \int_{w_2}^\infty \pi f(\pi)\,d\pi}{\int_{w_1}^{w_2} f(\pi)\,d\pi + \int_{w_2}^\infty f(\pi)\,d\pi}
$$
This last expression has the form $\dfrac{A+B}{C+D}$, where $\dfrac A C$ is the conditional expected value of $\Pi$ given that $w_1<\Pi<w_2$, and $\dfrac B D$ is the conditional expected value of $\Pi$ given that $\Pi>w_2$.  Thus $\dfrac A C < w_2 < \dfrac B D.$  Now it remains to prove this:
$$
\text{If } C,D>0 \text{ and } \frac A C < \frac B D \text{ then } \frac{A+B}{C+D} < \frac B D.
$$
