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This question has been already asked a few times, and, for instance, the accepted answer here gives an excellent intuitive explanation between the difference of the two, but now I'm trying to establish an intuitive link between similar explanations and the formal definitions of the respective convergences.

So, written formally, $X_n \rightarrow_{a.s.} X$ is equivalent to $$ \forall \xi : \forall \epsilon \exists n_\epsilon : \forall n > n_\epsilon : P \{ \exists m \geq n : | X_m - X | > \xi \} < \epsilon. $$ On the other hand, $X_n \rightarrow_{p} X$ is equivalent to $$ \forall \xi : \forall \epsilon \exists n_\epsilon : \forall n > n_\epsilon : P \{| X_n - X | > \xi \} < \epsilon. $$ Factoring out the common predicates, it boils down to the difference between $P \{ \exists m \geq n : | X_m - X | > \xi \} < \epsilon$ and $P \{| X_n - X | > \xi \} < \epsilon$.

In other words, if I get it right, it boils down to whether the probability of a deviation too big of any (for the almost sure convergence) or each individual (for the convergence in probability) of the members of the sequence starting at a sufficiently large $n$ will be small enough.

It's immediately obvious why almost sure convergence includes convergence in probability. But what is not obvious to me is why the converse isn't true: why each individual member of the sequence being unlikely to get far from the limiting random variable doesn't imply that any member (starting perhaps at a much further point) won't be far too?

I know some counter-examples of random variables converging in probability but not almost surely, but they don't really help to grok what exactly breaks when trying to make the above implication. So, perhaps this is more of real analysis or logical question.

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  • $\begingroup$ do you know the example with the moving indicator functions already? Otherwise I am going to start composing an answer with them $\endgroup$ – Felix B. Nov 5 '17 at 21:43
  • $\begingroup$ @FelixB. do you mean independent $X_n : P\{X_n = 1\} = 1 - P\{X_n = 0\} = \frac{1}{n}$? That's indeed a very illustrative case for these types of convergence themselves (since the probability of the union of the corresponding events diverges, so there is no a.s. convergence), but it doesn't help much in building the intuition. $\endgroup$ – 0xd34df00d Nov 5 '17 at 22:03
  • $\begingroup$ yes - well I wrote it down now anyway. Why do you think it doesn't build intuition? $\endgroup$ – Felix B. Nov 5 '17 at 22:19
  • $\begingroup$ The common "natural language" explanation of the difference between these types of convergence is that a.s. convergence allows for only finite number of sequence members to be too far from the limiting random variable, while the convergence in probability allows for infinitely many such members, albeit getting more and more rare. I fail to learn a gut feel why the fact that every single sequence member's probability of getting too far doesn't imply that too. $\endgroup$ – 0xd34df00d Nov 5 '17 at 22:29
  • $\begingroup$ It is less about you finding a special random variable in the sequence which is itself too far and more about the fact, that they can be far in different ways/places. If they are far on the same set then it is something like the first example. But if the problem shifts then everything is affected. Which may not happen in the a.s. case. It is a union of sets in the end. While every bit of the union might be small all of them together can be a big set again. $\endgroup$ – Felix B. Nov 5 '17 at 22:43
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Okay so $\Omega=[0,1]$ and our probability measure is just the Lebesgue measure on the interval. Consider $X_k=\mathbb{1}_{[0,\frac{1}{k}]}$ now these converge almost surely and in probability. As it converges pointwise except for the 0. And $P(\{0\})=0$. So let's make those a bit more clever.

Now the issue is to write down a formula for this, but what happens if you move those intervals over the [0,1] interval before making them smaller? You could properly define this recursively. But it is probably easier to see from example:

$$1_{[0,1]},1_{[0,\frac12]},1_{[\frac12,1]},1_{[0,\frac13]},1_{[\frac23,\frac13]},...$$

Now the interval gets smaller or stays equal with every step. Which means that $\forall\varepsilon\ P(X_k>\varepsilon)$ converges to zero which is convergence in probability. But you can not find a zero set which is the only affected part. Because for every natural number you have a positive interval size which gets shifted over every part of the [0,1] intervall. Which means that actually the entire [0,1] intervall fulfills: $\exists m>k:X_m(x)>\varepsilon$ thus $P(\exists m>k:X_m>\varepsilon)=P([0,1])=1 \ \forall\varepsilon>0$

Note the difference between these two sets: $$\{x\in[0,1]:X_k(x)>\varepsilon\}$$ which is a smaller and smaller set for increasing k. And: $$\{x\in[0,1]:\exists m>k:X_m(x)>\varepsilon\} =[0,1]$$

Now there is actually a theorem which states that every sequence of random variables which converges in probability has a subsequence which converges almost surely. And you can easily construct such a sequence in our example. It is the sequence we started with.

So the difference is, that in one case you just say that the probability that it doesn't converge goes to zero. In the other case you can actually find a set which has probability zero, which is the only part in which it doesn't converge pointwise. This is a lot stronger, as the bad stuff is actually contained in a sense.

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  • $\begingroup$ I'm not sure the original sequence converges almost surely too. Assume for all small $\xi, \epsilon > 0$ there exists such $n$ that $P\{ \cup_{m = n}^\infty \{ X_m > \xi \}\} < \epsilon$. But if those are independent, then $P\{ U_{m = n}^\infty \{ X_m > \xi \}\} = \sum_{m = n}^\infty P \{ X_m > \xi \} = \sum_{m = n}^\infty \frac{1}{m}$, which diverges. $\endgroup$ – 0xd34df00d Nov 5 '17 at 22:26
  • $\begingroup$ $P(\bigcup\{X_m>\epsilon\})<\sum P(X_m>\epsilon)$ The union is not disjoint. $\bigcup_{m=n}\{X_m>\epsilon\}=[0,\frac 1n]$ $\endgroup$ – Felix B. Nov 5 '17 at 22:30
  • $\begingroup$ Oh, right, you're not considering the independent ones, as opposed to the example I was considering, thanks! $\endgroup$ – 0xd34df00d Nov 5 '17 at 22:32
  • $\begingroup$ Also independence doesn't have anything to do with it (at least as far as I know). You are confusing this with estimated values of products of random variables maybe. $\endgroup$ – Felix B. Nov 5 '17 at 22:35

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