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$X_n$ is a sequence of independent random variables with $\mathbb{P}[X_n=n^2-1]=n^{-2},\mathbb{P}[X_n=-1]=1-n^{-2}$ and $Var[X_n]$ is unbounded. Set $S_n=X_1+...+X_n$. Prove that $\frac{S_n}{n} \rightarrow -1$ in probability.

Till now, I have proved that $\mathbb{E}[X_n]=0$. But then I am confused, no matter by weak law of large numbers or strong law of large numbers, this doesn't make sense to me that $\frac{S_n}{n}$ converges to -1. Can anyone explain to me what happened? Thanks!

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Doesn't the weak law only hold for i.i.d random variables? (ie, they need to be identically distributed). –  Andrew D Apr 11 '13 at 23:05
    
I don't think so. Weak law only need those random variables uncorrelated and the variance is bounded. –  Cancan Apr 11 '13 at 23:06
    
Going off MathWorld (mathworld.wolfram.com/WeakLawofLargeNumbers.html) seems to suggest otherwise; having had a quick look around elsewhere, it appears that the strong law holds for non-identically distributed random variables provided they follow some stricter conditions: (mathworld.wolfram.com/StrongLawofLargeNumbers.html) –  Andrew D Apr 11 '13 at 23:10
    
This is is also true, because strong law only requires them to be independent but not necessarily identically distributed. :) –  Cancan Apr 11 '13 at 23:12

1 Answer 1

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The series $\sum\limits_nP[X_n\ne-1]$ converges hence $X_n=-1$ for every $n$ large enough, almost surely, say for every $n\geqslant N$. Thus, $S_n=S_N+N-n$ for every $n\geqslant N$, in particular $S_n/n\to-1$ almost surely (hence also in probability). The independence hypothesis is not needed.

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This is a cool explanation! but could you please help me with the other question? From Wikipedia, I got to know that the fundamental difference between weak and strong law of large number is that, for weak law, sample average \bd{converges} to the expected value and, for strong law, sample \bd{converges almost surely} to the expected value. Could you please briefly show me the difference? Perhaps, can you show me an example that weak law holds but strong law fails to demonstrate what "almost surely convergence" really means? Thanks! –  Cancan Apr 11 '13 at 23:17
    
And in addition, why do you write $S_n=S_N+N-n$. Why do you get the $N-n$ in the equation? :) –  Cancan Apr 11 '13 at 23:26
    
Please check your definition of the weak LLN, the statement in your comment is odd. For an example of weak LLN without strong LLN, see here. –  Did Apr 11 '13 at 23:30
    
About $N-n$: say, if $X_k=-1$ for every $k\geqslant N$, what is $S_n-S_N$ for $n\geqslant N$? –  Did Apr 11 '13 at 23:31
    
Sorry, I thought the equation $S_n= S_N+N-n$ again and again and I couldn't figure out why is so? Till now can conclude that $X_n \displaystyle \rightarrow^{a.s.} -1$ but how did you get the conclusion from here? still not clear to me. –  Cancan Apr 11 '13 at 23:38

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