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The problem:

Let $\xi,\eta$ be two independent integrable r.v. such that $\mathsf P\{\xi> 0\} = 1$ and $\mathsf P\{\eta\geq 0\} = 1$ and $ a = \mathsf E[\xi-\eta]>0. $ Check if $ \lim\limits_{k\to\infty}\mathsf P\{(\xi-\eta)+k\xi\geq y\} = 1 $ for any fixed $y$.

I can neither prove that the limit is $1$ nor find a counterexample, so any help is appreciated.

What I've tried so far: since $a>0$ then $p = \mathsf P\{\xi-\eta\geq a\}>0$. Then for a fixed $y$ we have: $ \mathsf P\{(\xi-\eta)+k\xi\geq y\} \geq p\cdot \mathsf P\{a+k\xi\geq y|\xi-\eta\geq a\} $ but even if $\mathsf P\{a+k\xi\geq y|\xi-\eta\geq a\}\to1$ with $k\to\infty$ it woulnd't be sufficient for the original problem.

3 Answers 3

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I don't think independence is necessary. Since $E[\xi-\eta]>0$, $P(\eta=\infty)=0$. So, for almost every $\omega$, by archimedean property we can find an integer $k$ such that $(k+1)\xi(\omega)\ge \eta(\omega)+y$.

Hence $P\left(\bigcup_{k=1}^{\infty}\{(\xi-\eta)+k\xi\ge y\}\right)=1$ Since the events in the union are increasing in $k$, we have $\lim_{k\to \infty}P\{(\xi-\eta)+k\xi\ge y\}=1$

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    nope, I meant bounds on the difference $1-\mathsf P$ as I presented in my answer below. I would appreciate if you could take a look on it and tell me if there are any mistakes or there are better bounds - though I cannot ask you to do it because your answer is cool and clear, good job ;)2011-11-24
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Since $\mathsf{P}\{\xi> 0\} = 1$, we have $ \lim_{k\to\infty}\mathsf{P}\left\{\xi>\frac{y+\frac{2}{\epsilon}\mathsf{E}|\xi-\eta|}{k}\right\}=1\tag{1} $ for any $\epsilon>0$. Furthermore, Markov's Inequality implies that $ \mathsf{P}\left\{\xi-\eta<-\frac{2}{\epsilon}\mathsf{E}|\xi-\eta|\right\}\le\mathsf{P}\left\{|\xi-\eta|>\frac{2}{\epsilon}\mathsf{E}|\xi-\eta|\right\}\le\frac{\epsilon}{2}\tag{2} $ Note that $ \begin{align} \mathsf{P}(x+y\ge A+B) &\ge\mathsf{P}(x\ge A\wedge y\ge B)\\ &=\mathsf{P}(x\ge A)-\mathsf{P}(x\ge A\wedge y Therefore, given $\epsilon>0$, $(1)$ insures that there is a $k$ so that $ \mathsf{P}\left\{\xi>\frac{y+\frac{2}{\epsilon}\mathsf{E}|\xi-\eta|}{k}\right\}\ge1-\frac{\epsilon}{2}\tag{4} $ then $(2)$, $(3)$, and $(4)$ yield $ \begin{align} \mathsf{P}\{(\xi-\eta)+k\xi>y\} &\ge\mathsf{P}\{k\xi>y+\frac{2}{\epsilon}\mathsf{E}|\xi-\eta|\}-\mathsf{P}\{\xi-\eta<-\frac{2}{\epsilon}\mathsf{E}|\xi-\eta|\}\\ &\ge1-\epsilon\tag{5} \end{align} $ Therefore, $(5)$ tells us that $ \lim_{k\to\infty}\mathsf{P}\{(\xi-\eta)+k\xi>y\}=1 $

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    nice point, but it can hold as well if a<0. Nevertheless, it seems to me now that it's not so important what is greater, $\xi$ or $\eta$ because we mostly use $k\xi$ to obtain bounds2011-11-26
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Inspired by the answer by Ashok, I also derived bounds for the convergence which I need as well. Just for the case I put it here as an answer.

Let us take $F_\xi$ to be a c.d.f. of $\xi$. Consider $ 1-\mathsf P\{(\xi-\eta)+k\xi\geq y\} = \mathsf P\{\eta > (k+1)\xi -y\}. $ We have then: $ \mathsf P\{\eta > (k+1)\xi -y\} = \int\limits_0^\infty \mathsf P\{\eta > (k+1)x -y\}\,dF_\xi(x)$ $ = \int\limits_0^{\frac{y+\sqrt{k+1}}{k+1}} \mathsf P\{\eta > (k+1)x -y\}\,dF_\xi(x)+\int\limits_{\frac{y+\sqrt {k+1}}{k+1}}^\infty \mathsf P\{\eta > (k+1)x -y\}\,dF_\xi(x). $ The first term is $\displaystyle{F_\xi\left(\frac{y+\sqrt{k+1}}{k+1}\right)}$ and the second we bound by Markov inequality: $ \int\limits_{\frac{y+\sqrt {k+1}}{k+1}}^\infty \mathsf P\{\eta > (k+1)x -y\}\,dF_\xi(x)\leq \mathsf E\eta\int\limits_{\frac{y+\sqrt {k+1}}{k+1}}^\infty \frac{dF_\xi(x)}{(k+1)x-y}. $

Since for the denominator we have: $(k+1)x-y\geq \sqrt{k+1}$ for all $x$ in the domain of integration, $ \int\limits_{\frac{y+\sqrt {k+1}}{k+1}}^\infty \mathsf P\{\eta > (k+1)x -y\}\,dF_\xi(x)\leq \frac{\mathsf E\eta}{\sqrt{k+1}} $ so $ \mathsf P\{\eta > (k+1)\xi -y\}\leq F_\xi\left(\frac{y+\sqrt{k+1}}{k+1}\right)+\frac{\mathsf E\eta}{\sqrt{k+1}} $ which in particular means that $\mathsf P\{(\xi-\eta)+k\xi\geq y\}\to 1$ with $k\to\infty$.