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Let $ (X_n) $ be a series of real random variables. To show:

$X_n \xrightarrow{\text{ $\mathbb{P} $ }} 0 , $ ( $n \rightarrow \infty) $

is only valid, if and only if there exists an $n_0$ for all $ \epsilon > 0$, so that the inequation $ \mathbb{P} (|X_n| > \epsilon ) < \epsilon $ is fullfilled for all $n_0 \leq n$

my idea: convergence in probability means : $ \mathbb{P} (|X_n| > \epsilon) \rightarrow 0, n \rightarrow \infty$. Let $\omega \in \Omega$ $\Rightarrow$ there exists an $n_0 \in \mathbb{N} $, so that for all $ n_0 \leq n$ : $\mathbb{P} (|X_n| > \epsilon) < \mathbb{P} ({ \omega_i})$. Therefore $\omega_i$ can't be an Element of ${|X_n| > \epsilon}$ and this implies that $ X_n (\omega_i) < \epsilon $.

Is this a solid idea to proof it? Help and hints are much appreciated :-)

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  • $\begingroup$ You seem to be very confused about notation. The expression $\mathbb P(\omega_i)$ is meaningless, for example. $\endgroup$ – Math1000 Jun 13 '18 at 14:03
  • $\begingroup$ No, it is not a solid idea to prove it. $\endgroup$ – drhab Jun 13 '18 at 19:53
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Based on: $$\forall\epsilon>0\exists n_{0}\forall n\geq n_{0}P\left(\left|X_{n}\right|>\epsilon\right)<\epsilon\tag1$$ it must be shown that for every $\epsilon>0$ we have $\lim_{n\to\infty}P\left(\left|X_{n}\right|>\epsilon\right)=0$ (the other side is trivial).

Assume that this is not the case, or equivalently that some $\epsilon>0$ exists together with a $\delta>0$ and a subsequence $\left(X_{n_{k}}\right)$ such that $P\left(\left|X_{n_{k}}\right|>\epsilon\right)\geq\delta$ for every $k$.

This implies that $\delta<\epsilon$ because for $k$ large enough we have $P\left(\left|X_{n_{k}}\right|>\epsilon\right)<\epsilon$ according to $(1)$.

But then $P\left(\left|X_{n_{k}}\right|>\delta\right)\geq P\left(\left|X_{n_{k}}\right|>\epsilon\right)\geq\delta$ for every $k$, contradicting $(1)$.

This proves that our assumption was wrong, and we are ready.

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