An inequality concerning convexity and expectation

Question

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 $$ \int_c^\infty f(x)dx = \int_d^\infty g(x)dx = p\int_b^\infty f(x)dx + q\int_b^\infty g(x)dx ? $$

Answer 1

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).