The effect on expectation of biasing a distribution by a monotonic function Let 
$$ x \sim g(x)$$
where $g(x)$ denotes the pdf of $x$. Let the pdf of another variable $x^*$ be denoted by $f(x^*)$ and let 
$$f(x^*) \propto g(x^*) z(x^*) $$ 
where $z(x^*)$ is a monotonic increasing function. 
It seems to me intuitively that, provided they are well defined, $E(x^*) \geq E(x)$. How can one prove (or disprove) this? 
 A: This answer is motivated by the fact that the difference in expected values of two random variables is the (signed) area between their cumulative distribution functions (CDFs):

In this figure, the expected value of $F$ exceeds that of $G$ by the area between the curves.  This is proven in an appendix below.
The plan, then, is to demonstrate that the non-decreasing ratio of the densities of $X^{*}$ and $X$ implies the CDF of the first lies below the CDF of the second, implying the difference in expectations is positive.  
The crux of the matter is to show that initially $F$ increases more slowly than $G$ (at the left), where both begin (asymptotically) at the same level of $0.$  The same thing happens at the right, where $F$ drops more quickly than $G$ from right to left, although both begin (asymptotically) at the same height of $1.$  They meet somewhere in the middle, but along the way $G$ has never been lower than $F.$

Begin with the basic inequality
$$\int u(x)\,\mathrm{d}x \ge 0$$
for all almost everywhere non-negative functions $u.$ This inequality will be invoked several times--watch for it.
When $u$ and $v$ are functions with $u(x)\ge v(x)$ for all $x,$ applying the inequality to their difference shows
$$\int u(x)\,\mathrm{d}x \ge \int v(x)\,\mathrm{d}x\tag{*}$$
with equality only when $u(x)=v(x)$ almost everywhere.
Turning to the setting of the question, observe that since
$$\int g(x)\,\mathrm{d}x = 1 = \int z(x)g(x)\,\mathrm{d}x,$$
it is not possible for $z(x) \lt 1$ or $z(x)\gt 1$ for all $x$, because then $(*)$ would imply $1 \ne 1$ in either case.  Therefore there exists an $x_0$ at which $z$ crosses $1:$
$$z(x) \le 1\text{ for }x \le x_0\,\text{ and }\, z(x) \ge 1\text{ for }x \ge x_0.$$
Since $g(x)\ge 0$ everywhere, this implies
$$z(x)g(x) \le g(x)\text{ for }x \le x_0\,\text{ and }\, z(x)g(x) \ge g(x)\text{ for }x \ge x_0.$$
Define
$$G(x) = \int^x_{-\infty} g(t)\,\mathrm{d}t$$
and 
$$F(x) = \int^x_{-\infty} z(t)g(t)\,\mathrm{d}t.$$
It is immediate from the inequality $(*)$ applied to the functions $u=g$ and $v(x)=z(x)g(x)$ that for all $x\le x_0$
$$G(x)\ge F(x)$$
and for all $x \ge x_0$
$$1 - F(x) = \int_x^\infty z(t)g(t)\,\mathrm{d}t \ge \int_x^\infty g(t)\,\mathrm{d}t = 1 - G(x),$$
whence for all $x,$
$$G(x)\ge F(x).$$
But now
$$E[X^*] - E[X] = \int_{-\infty}^\infty (G(x)-F(x))\,\mathrm{d}x \ge 0,$$
QED.

If the final expression for the difference of expectations surprised you, note that (writing $f(x)=z(x)g(x)$) an integration by parts gives
$$\eqalign{
E[X^*] - E[X] &= \int x\,f(x)\,\mathrm{d}x -  \int x\,g(x)\,\mathrm{d}x \\
&=  \int x\,(f(x)-g(x))\,\mathrm{d}x \\
&= \left. x(F(x)-G(x))\right|_{-\infty}^\infty + \int_{-\infty}^\infty (G(x)-F(x))\,\mathrm{d}x
}$$
and 
$$\left. x(F(x)-G(x))\right|_{-\infty}^\infty = \lim_{x\to\infty,y\to-\infty} \left[x(F(x)-G(x)) - y(F(y)-G(y))\right] = 0 - 0 = 0$$
provided both expectations exist and are finite.

This demonstration applies more generally, and with no change at all, to the situation where $X^{*}$ has a distribution $\mathrm{d}\lambda$ that is absolutely continuous with respect to the distribution $\mathrm{d}\nu$ of $X$ and its Radon-Nikodym derivative $z = \mathrm{d}\lambda / \mathrm{d}\nu$ is a non-decreasing function.  This has applications to discrete distribution families like the Poisson distribution.  The concept goes by the name "Monotone Likelihood Ratio."
Another generalization applies to the expectations of any nondecreasing function of $X:$ at the end you arrive at the integral of the (generalized) derivative of that function, multiplied by $G(x)-F(x),$ which again must give a non-negative result.
