Assume $f$ and $g$ are nonnegative with
$$\int\_0^\infty f(x)dx=1=\int\_0^\infty g(x)dx
$$
and
$$\int\_0^\infty xf(x)dx<\infty > \int\_0^\infty xg(x)dx
$$
Is it true for nonnegative numbers $p$, $q$ with $p+q=1$, and $b\ge 0$ that



> 
> $$
> p\int\_c^\infty xf(x)dx + q\int\_d^\infty xg(x)dx \le p\int\_b^\infty xf(x)dx + q\int\_b^\infty xg(x)dx
> $$
> 
> 
> 


where $c$ and $d$ are defined by
$$
\int\_c^\infty f(x)dx = \int\_d^\infty g(x)dx = p\int\_b^\infty f(x)dx + q\int\_b^\infty g(x)dx ?
$$



An inequality concerning convexity and expectation
Yes, this works. Let's say $c\le b$, so $b\le d$. Then (slightly rearranging) we want to show that
$$
p\int\_c^b xf \le q \int\_b^d xg \quad\quad\quad (1).
$$
Rearranging the definition of $c,d$, we see that
$$
\int\_c^b f = q\int\_b^{\infty}(g-f),\quad\quad
\int\_b^d g = p \int\_b^{\infty}(g-f) ,
$$
and this produces the identity $p\int\_c^b f= q\int\_b^d g$, which implies (1).


