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Let us consider a real-valued r.v. $X$ such that $\mathsf E X > 0$ and $X>x_0$ a.s. Denote $$ g(x,y) = \frac{x}{y(y+x)}. $$ It is right that for any $X$ there exists $y_0>-x_0$ such that for all $y>y_0$ one have $\mathsf E g(X,y)>0$?

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Here is a more elegant formulation of the question. Suppose that $X$ is a random variable such that ${\rm E}(X) > 0$ and $X > x_0$ a.s., $x_0 < 0$ fixed. Is it true that $$ {\rm E}\bigg[\frac{1}{{y + X}}\bigg] < \frac{1}{y} $$ for all sufficiently large $y$? – Shai Covo Jun 20 '11 at 20:04
I think it follows from the fact that $$\mathsf E g(X,y) = \frac{1}{y} - \mathsf E\left[\frac{1}{y+X}\right]$$. Thanks anyway. – Ilya Jun 21 '11 at 6:45
You are right - good observation. – Shai Covo Jun 21 '11 at 7:13
@Shai: it was in fact the first formulation of the problem but I posted here the one with a function $g$ since it seemed to me be more convenient to deal with. – Ilya Jun 21 '11 at 8:39
Thanks for noting this. (I deleted my second comment.) – Shai Covo Jun 21 '11 at 8:58
up vote 2 down vote accepted

For every $x$ and $z$ such that $1+zx$ is positive, let $h(x,z)=x/(1+zx)$. For every positive $z$ such that $z\le1/(2|x_0|)$, one has $1+zx_0\ge\frac12$, hence $|h(X,z)|\le2|X|$. Furthermore, $h(X,z)\to X$ when $z\to0^+$. By Lebesgue dominated convergence theorem, $E(h(X,z))\to E(X)$ when $z\to 0^+$.

Since $E(X)$ is positive, this implies that there exists a positive $z_0$ such that for every $z$ in $(0,z_0)$, $E(h(X,z))\ge E(X)-\frac12E(X)$ hence $E(h(X,z))$ is positive.

Note finally that $g(x,y)=h(x,1/y)/y^2$ hence for every positive $y$, $E(g(X,y))$ is positive if and only if $E(h(X,1/y))$ is. This happens at least for every $y\ge1/z_0$.

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Thank you, that was helpful for the question of bounds for submartingales. – Ilya Jun 20 '11 at 13:46

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