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I have to contruct a sequence of random variables that follow Weak Law of Large Numbers, but don't follow Strong Law of Large Numbers. Can canyone give me any hint please? Basically i need to choose such a sequence of integrable r.v-s that:

1) $$(S_n-\mathbb{E}S_n)/n \rightarrow 0 \qquad \mbox{in probability}$$

and

2) $$\mathbb{P}(\{\omega:(S_n-\mathbb{E}S_n)/n \rightarrow 0 \})<1$$

$S_n:=X_1+\ldots+X_n$

The only thing I know is that if $X_1,X_2,\ldots$ were identically distributed and independent, then its impossible.

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    $\begingroup$ Pretty sure there's a typo or a misunderstanding here - LLN says that $(S_n-\mathbb ES_n)/n\to0$ in some sense or other, where $S_n=X_1+\dots+X_n$. $\endgroup$ – David C. Ullrich Jul 19 '15 at 18:14
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    $\begingroup$ Are you sure you're including all the requirements? Simply taking $X_n=X_1$ seems like a solution to the problem as you stated it... $\endgroup$ – David C. Ullrich Jul 19 '15 at 18:16
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    $\begingroup$ @DavidC.Ullrich : it would then follow the strong law too. But I agree, the question is badly asked $\endgroup$ – Tryss Jul 19 '15 at 18:18
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    $\begingroup$ @Tryss I don't see why. Yes, I realized before you spoke up that my second comment was totally wrong. But the reason is that then $(X_n)$ does not satisfy the weak LLN... $\endgroup$ – David C. Ullrich Jul 19 '15 at 18:20
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    $\begingroup$ Whoever made that close vote: How the heck is this question "not about math, as defined in the help center"??? $\endgroup$ – David C. Ullrich Jul 19 '15 at 18:26
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A counterexample to the question exactly as stated is very simple. Let $A_n:[0,1]\to\mathbb C$ be any sequence of functions that tends to $0$ in measure but not pointwise. Ok, also say $\int_0^1 A_n=0$. Let $S_n=nA_n$, then let $X_1=S_1$ and $X_n=S_n-S_{n-1}$ for $n>1$.

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  • $\begingroup$ Could you explain this a bit more? I just don't understand how this sequence "works" for weak law but not for strong law... $\endgroup$ – luka5z Jul 23 '15 at 18:15
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    $\begingroup$ Do you believe that there exist functions $A_n$ that tend to $0$ in measure but not pointwise? If so: trace through the equations I gave and you see $A_n=(1/n)(X_1+\dots+X_n)$. So $(1/n)(X_1+\dots+X_n)\to0$ in measure but not almost surely. $\endgroup$ – David C. Ullrich Jul 23 '15 at 18:41
  • $\begingroup$ Such functions can be $A_n(x)=x^{n}$ on $[0,1]$? $\endgroup$ – luka5z Jul 23 '15 at 19:19
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    $\begingroup$ I can't tell exactly what you're missing here, but if you really think those functions do not tend to $0$ almost everywhere it follows that you're missing something much more basic than the details of the example I gave. My advice is find a book on measure theory and read up on almost everywhere convergence and convergence in measure... $\endgroup$ – David C. Ullrich Jul 23 '15 at 19:27

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