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Suppose $U_1,...,U_n$ are independent random variable with $\mathbb{E}[U_i]=0$.

Define $Z_k:=\sum_{i=1}^k U_i$. Set $T:=\inf \lbrace k \in N \mid |Z_k|>2\alpha \rbrace$.

Clearly $\lbrace T=k \rbrace$ is independent of $(Z_n-Z_k)=\sum_{i=k+1}^n U_i$.

I do not get why:

$$\Pr(|Z_n|>\alpha)\geq \Pr(|Z_n-Z_T|\leq \alpha, T\leq n)$$

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By the definition of $T$ we have $Z_T>2\alpha$ whenever $T<\infty$. If, in addition, $|Z_n-Z_T|\leq \alpha$ we get

$$|Z_n|-\alpha=\alpha + |Z_n|-2\alpha > \alpha + |Z_n|-|Z_T| \geq \alpha - |Z_n-Z_T|\geq 0.$$

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  • $\begingroup$ thank you very much, it is very clear. However I don't get the last inequality. Why should it be: $|Z_n|-|Z_T|\geq -|Z_N-Z_T|$ ? I see it is similar to the reverse triangular inequality but in that I have $\mid |Z_n|-|Z_T|\mid$ $\endgroup$ – mastro Jul 26 '15 at 8:41
  • $\begingroup$ It is just the triangle inequality. Work on it for awhile... :) $\endgroup$ – user940 Jul 26 '15 at 14:10

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