# Positive expectation

Let us consider a real-valued r.v. $X$ such that $\mathsf E X > y$ and $X\geq x_0$ a.s. Denote $$g(x,y,r) = \frac{1}{y+r} - \frac{1}{x+r}.$$

It is right that for any $X$ there exists $r_0>-x_0$ such that for all $r>r_0$ one have $\mathsf E g(X,y,r)>0$?

The question is similar to this one: Estimation of an expectation

I guess it can be done through the expansion in series. Anyway if there will be another idea - I appreciate it since expansion will give $r_0$ which increases with $y$.

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If you call $h(x,y) = 1/y - 1/(x+y)$ the function from your previous question, then $g(x,y,r)=h(x-y,r+y)$, so this is actually the same question (replacing $X$ by $X-y$). – pgassiat Jun 21 '11 at 21:06

The method is similar to the one used to solve your previous question: you want to prove that $E(h(X,r))$ is positive for $r$ large enough, with $$h(x,r)=r(x-y)/(x+r).$$ You know that $|h(x,r)|\le r|x-y|/(r+x_0)$ for every $r>-x_0$, that $|X-y|$ is integrable, and that $h(x,r)\to x-y$ when $r\to+\infty$. Lebesgue dominated convergence theorem yields that $E(h(X,r))\to E(X)-y$. This limit is positive hence $E(h(X,r))$ is positive for every $r$ large enough.
By the way, methods using series expansions are doomed unless $X$ satisfies some strong integrability conditions.
Doing calculations formally we have $$\mathsf E g(X,y,r)=\frac{\mathsf EX-y}{r^2}+o(r^{-2})$$ as $r\to+\infty$. To justify them seem not so difficult, but I can be mistaken. Condition $\mathsf EX<\infty$ seems to be required though. Besides the proof is already given by Didier Piau.