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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.

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3 Answers 3

up vote 6 down vote accepted

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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Thanks, seems to be right. Do you think it is possible to find how $\mathsf P\{(\xi-\eta)+k\xi\geq y\}$ is close to $1$ for $k$ being large? –  Ilya Nov 24 '11 at 11:56
    
$\{(\xi-\eta)+k\xi\ge y\}\uparrow \bigcup_{k=1}^{\infty}\{(\xi-\eta)+k\xi\ge y\}$ right? I hope I understood your doubt correctly. –  Ashok Nov 24 '11 at 12:22
    
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 ;) –  Ilya Nov 24 '11 at 14:35

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 $$ \begin{align} \mathsf{P}(x+y\ge A+B) &\ge\mathsf{P}(x\ge A\wedge y\ge B)\\ &=\mathsf{P}(x\ge A)-\mathsf{P}(x\ge A\wedge y<B)\\ &\ge\mathsf{P}(x\ge A)-\mathsf{P}(y<B)\tag{3} \end{align} $$ 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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cool answer, thanks a lot. The only thing I wonder about now - is that no one of us used the fact that $a = \mathsf E(\xi-\eta)>0$ and maybe bounds would be better in that case. –  Ilya Nov 26 '11 at 19:32
    
It is $a=\mathsf{E}|\xi-\eta|>0$ (absolute values) which is simply a way of saying that $a=\mathsf{E}|\xi-\eta|\not=0$, and it was used with Markov's Inequality. –  robjohn Nov 26 '11 at 19:58
    
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 bounds –  Ilya Nov 26 '11 at 20:02

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 $k\to\infty$.

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