Let $Z$ be any non-degenerate positive random variable with pdf $% g(z)$. Let $a>0$ and $r\neq 0$ denote arbitrary real numbers. Define the "$r$ -mean" of $Z$ shifted by the constant $a$ as follows. (It can be assumed that the necessary moments are finite.):
$M_{r}(a)=E \left[ (a+Z)^{r}\right] ^\frac{1}{r}=\left[ \int\limits_{0}^{\infty }(a+z)^{r}g(z)dz\right] ^{1/r}$
Furthermore, for some given $s>0$, define
$k(a)=\dfrac{M_{-1}(a)}{M_{s}(a)}=\dfrac{\left[ \int\limits_{0}^{\infty }(a+z)^{-1}g(z)dz\right] ^{-1}}{\left[ \int\limits_{0}^{\infty }(a+z)^{s}g(z)dz\right] ^{1/s}}$
I was able to show that $0<k(a)<1$. However, it needs to be shown that $k(a)$ is non-decreasing, so if $a_{1}<a_{2}$, then $% k(a_{1})\leq k(a_{2})$. (In other words, $k'(a)\geq 0$. )
I played with specific distributions (lognormal, uniform, exponential) and it seems to work, but it would be nice to have some general proof. I'd appreciate any suggestions. (Note: I am just an economist, not a mathematician. I apologize if I used wrong terminology or I am unaware of some well-known result.)
Indeed the function $k$ is non-decreasing, here is a proof.
The derivative of $E((a+Z)^x)$ with respect to $a$ being $xE((a+Z)^{x-1})$, for every $x$, it suffices to show that, for $s\gt0$, $$ E((a+Z)^{-2})\,E((a+Z)^{s})\geqslant E((a+Z)^{-1})\,E((a+Z)^{s-1}). $$ More generally, it happens that, for every positive $X$ such that every moment involved exists, $$ E(X^{-2})\,E(X^s)\geqslant E(X^{-1})\,E(X^{s-1}). $$ To prove this, consider a random variable $Y$ independent of $X$ with the same distribution, and integrate the pointwise inequality $$ (X^{s+1}-Y^{s+1})\,(X-Y)\,X^{-2}\,Y^{-2}\geqslant0. $$ The expectation of each product $X^aY^b$ being $E(X^a)E(Y^b)=E(X^a)E(X^b)$, one is left with the result that twice the difference between the LHS and the RHS of the inequality one wants to prove, is nonnegative.