# A simple distance inequality in probability

I was going through Durrett's probability when I was stuck in this problem. If $X_1 \ldots X_n$ are independent RVs, $S_n=\sum_{i=1}^n X_i$, $M\in \mathbb{Z}$ and $\epsilon>0$, there is this inequality:

$$P(\sup_{m,n\in M} |S_m-S_n|> 2\epsilon) \le P(\sup_{m\ge M} |S_m-S_M| > \epsilon)$$

I have been toying with triangle inequality to establish this, but unfortunately I cannot see this. Could someone help me please? Looks simple at first sight.

-
It folows from a simple observation that, if we have $a+b > 2c$, then either $a>c$ or $b>c$. Thus for $m,n\geq M$, the inequality $|S_m - S_n|>2\epsilon$ implies (by triangle inequality) $|S_m - S_M|+|S_n - S_M|>2\epsilon$, which again implies either $|S_m - S_M|>\epsilon$ or $|S_n - S_M|>\epsilon$, which finally implies $\sup_{m\geq M}|S_m - S_M|>\epsilon$. Then the last observation you need is that $\sup_{m,n\geq M}|S_m - S_n|>2\epsilon$ implies the existence of some $m,n>M$ satisfying $|S_m - S_n|>2\epsilon$. – Sangchul Lee Mar 30 '12 at 8:13

Since $|a-b|\leqslant|a-c|+|b-c|$ for every $a$, $b$ and $c$, if $|S_m-S_n|\gt2\epsilon$ for some $m$ and $n$, then either $|S_m-S_M|\gt\epsilon$ or $|S_n-S_M|\gt\epsilon$, for every $M$ (choose $a=S_m$, $b=S_n$ and $c=S_M$). In short, $$[|S_m-S_n|\gt2\epsilon]\subseteq[|S_m-S_M|\gt\epsilon]\cup[|S_n-S_M|\gt\epsilon],$$ hence $$\bigcup_{m,n\geqslant M}[|S_m-S_n|\gt2\epsilon]\subseteq\bigcup_{m\geqslant M}[|S_m-S_M|\gt\epsilon].$$ Finally, thanks to the strict inequalities, the events in the LHS and in the RHS of this inclusion coincide respectively with the events $$\left[\sup\limits_{m,n\geqslant M}|S_m-S_n|\gt2\epsilon\right]\quad\text{and}\quad\left[\sup\limits_{m\geqslant M}|S_m-S_M|\gt\epsilon\right].$$