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I've got a question regarding a step in a proof, the situation is following:

Let $X_{1},\dots,X_{n}$ be independent, symmetric stochastic variables so that $\sum\limits_{n=1}^{\infty}X_{n}$ exists in probability. If I use the fact that convergent in probability implies the same convergence of the corresponding Cauchy sequence i can write: $$\lim_{n,m\to\infty}P(|S_{m}-S_{n}|>t)=0$$ where $S_{n}=\sum\limits_{i=1}^{n}X_{i}$. My textbook then says that from the triangle inequality we can write that as: $$\lim_{n\to\infty}\sup_{m\geq n+1}P(|S_{m}-S_{n}|>t)=0$$ Could someone please help me fill in the blanks?

I hope I didn't leave anything relevant out, let me know if you feel i did.

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This uses the property of Cauchy sequences. One has that for every $\epsilon>0$, there exists an $N$ such that for all $n,m\geq N$, $P(|S_n-S_m|>t)<\epsilon$. So if we temporarly anchor $n>N$, the inequality holds even after taking the supremum over $m\geq n+1$, so long as $n\geq N$.

As far as triangle inequality goes, likely the book is referring to the fact that $|S_n-S_m|\leq |S_n-S_N|+|S_m-S_N|$ which implies that if $|S_n-S_m|>t$ then either $|S_n-S_N|>t/2$ or $|S_m-S_N|>t/2$, so that one can write:

$P(|S_n-S_m|>t) \leq P(|S_n-S_N|>t/2) + P(|S_m-S_N|>t/2)$

and so if we again think of $N$ as any large anchored quantity, then the two terms on the right can be made uniformly small for $n,m\geq N$. In other words if you take the supremum in $m$ over $m\geq n+1$, then the right hand side can be made arbitrarely small by picking a large enough $n$ and $N$. Essentially this "trick" decouples $n$ and $m$ from eachother.

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Thanks! That helped - I should of course have thought of the strict definition of Cauchy instead. I'm still wondering how what you write gives the exact result? (It's one of the subjects to an impending oral exam to do this proof - so I kind of want to be absolute precise) –  Henrik Jun 13 '12 at 6:58
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