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It is a standard fact in probability that almost sure convergence is stronger than convergence in probability. I can only see the differences in the proof. However, is there a way to view it intuitively? Is it true that almost sure convergence has a tighter hold on the tails of a sequence of random variables than convergence in probability does? The definition of convergence in probability I am using is that given $\epsilon >0$:

$$ \lim_{n \to \infty} P(|X_n-X|> \epsilon) = 0 $$

and the definition of almost sure convergence I am using is:

$$ P(\lim_{n \to \infty}X_n = X) = P\left(\omega \in \Omega: \lim_{n \to \infty}X_n(\omega) = X(\omega)\right) = 1 $$

The two above appear almost exactly the same to me, except that the limit on $n$ is outside of convergence in probability and within the probability measure for almost sure convergence. Is there an easy to understand intuitive difference here? Thanks!

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marked as duplicate by Did probability Feb 6 '16 at 22:33

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    $\begingroup$ I imagine that when you say weaker you mean stronger... I think the only way to get an intuition for the difference is to look at some examples of sequences that converge in probability but not almost surely. $\endgroup$ – David C. Ullrich Feb 5 '16 at 22:52
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    $\begingroup$ Consider $Y_{n}=|X_n-X|> \epsilon$. Convergence in probability means $P(Y_n)\to 0$. Convergence a.s. means that almost all $\omega\in\Omega$ are covered only by a finite number of $Y_n$. When $X_n$ converges in probability only, $Y_n$ can "travel around" $\Omega$ covering all points infinitely many times. $\endgroup$ – A.S. Feb 6 '16 at 0:06
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    $\begingroup$ Yes and Yes. Imagine $Y_n$'s staying the same size while "travelling around" $\Omega$ until it covers all of it (in finite number of steps). Then $Y_n$ decreases in size and continues its "travel" while staying the same size until all of $\Omega$ is covered. Then it further decreases in size - and the process repeats indefinitely while sizes converge to $0$. The the sequence converges in probability, but not a.s. $\endgroup$ – A.S. Feb 6 '16 at 0:54
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    $\begingroup$ Neither. Size of $Y_n$ is $P(Y_n)$ (all for a fixed $\epsilon$ of course), so it is effectively a function of $n$. $\endgroup$ – A.S. Feb 6 '16 at 1:07
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    $\begingroup$ Note that if $P(Y_n)$ is summable, then you automatically have a.s. convergence, since $Y_n$ does done fast enough that $Y_n$ doesn't get to cover any (non-zero) part of $\Omega$ infinitely many times. $\endgroup$ – A.S. Feb 6 '16 at 1:29
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For simplicity, consider the case where $X = 0$ and $X_n$ is the indicator function of an event $E_n$. "$X_n$ converges almost surely to $0$" says that with probability $1$, only finitely many of the events $E_n$ occur. "$X_n$ converges in probability to $0$" says that the probability of event $E_n$ goes to $0$ as $n \to \infty$.

Consider a case where for each $m$ you partition the sample space into $m$ events, each of probability $1/m$, and take all these events for all $m$ to form your sequence $E_n$. Then $X_n \to 0$ in probability because the probabilities of the individual events go to $0$, but each sample point is in infinitely many $E_n$ (one for each $m$) so $X_n$ does not go to $0$ almost surely.

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    $\begingroup$ Can I ask from your example how the $m$ are connected to $E_n$? Is it the case that each $E_n$ contains a larger and larger partition? $\endgroup$ – user136503 Feb 6 '16 at 0:08
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    $\begingroup$ $E_1$ corresponds to $m=1$, $E_2$ and $E_3$ to $m=2$, $E_4, E_5, E_6$ to $m=3$, etc. $\endgroup$ – Robert Israel Feb 6 '16 at 0:49

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