Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $y$.

share|improve this question
2  
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
add comment

2 Answers

up vote 2 down vote accepted

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.