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This is a problem after the section "Strong Law of Large Numbers" of Shiryaev's Probability:

Let $\xi_1,\xi_2,...$ denote independent and identically distributed random variables such thatt $E|\xi_1|=\infty$. Show that $$\limsup_{n\to\infty}\left|\frac{S_n}{n}-a_n\right|=\infty\text{ (P-a.s.)}$$ for every sequence of constants $\{a_n\}$.

I have no idea about it. Any hint please. Thanks!

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2  
1. as a tail event it occurs with probability 0 or 1, 2. symmetrize $X_i \rightarrow X_i - \tilde X_i$ so the $X_i$ are symmetric and $a_i = 0$. ${ }{ }$ 3. show can't be $ < k$ on $X_i > ik$ i.o., which occurs by lack of first moment –  mike Jan 10 '13 at 14:36
    
@mike Thanks! But I can't catch the hint 2 clearly. Does it meam let $Z_i=X_i-Y_i$ where $Y_i$ are also iid?If so how can I prove $Z_i$ lack of first moment? –  Danielsen Jan 10 '13 at 15:47
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yes, and lack of first moment is a standard application of fubini: $\infty > \mathbb E(|X - Y|) = \int \int (|x-y| \mu(dx)\mu(dy) = \int \mathbb E(|X - y|) \mu(dy) $ which implies that $\infty > \mathbb E(|X - y|) $ a.e. y which implies $\infty > \mathbb E(|X |)$ –  mike Jan 10 '13 at 15:55
    
@mike I get it. Thank you very much! –  Danielsen Jan 10 '13 at 16:02
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@Danielsen You might wish to write down a full solution based on mike's hints, to post it here as an answer and even, after a while and if no other solution appears that you would prefer, to accept it. –  Did Jan 11 '13 at 18:20

1 Answer 1

Let $Y_i:=X_i-X'_i$, where $(X'_i)$ is an i.i.d. copy of $(X_i)_i$. Define $$A:=\left\{\limsup_{n\to \infty}\left|\frac 1n\sum_{i=1}^nX_i-a_n\right|<\infty\right\}$$ $$A':=\left\{\limsup_{n\to \infty}\left|\frac 1n\sum_{i=1}^nX'_i-a_n\right|<\infty\right\}.$$ The goal is to show that $\mathbb P(A)=0$. Since $A$ and $A'$ are independent and of equal probability, we are reduced to show that $\mathbb P(A\cap A')=0$. Notice that $$A\cap A'\subset \left\{\limsup_{n\to \infty}\left|\frac 1n\sum_{i=1}^nY_i\right|<\infty\right\},$$ and the sequence $(Y_i)_i$ is i.i.d., with $\mathbb E|Y_1|=\infty$. Indeed, by independence of $X_1$ and $X'_1$, $$\mathbb E|X_1-X'_1|=\int_\Omega\mathbb E|X_1-x'|\mathrm d\mathbb P_{X'_1}(x')=\infty.$$ Using the Borel-Cantelli lemma, we have that for each $R$, $\mathbb P(\limsup_i \{|Y_i|>iR\})=1$.

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How do you conclude from infinite mean that the series $P(|Y_i|>iR)$ diverges? –  Lost1 Sep 5 at 23:01
    
It's an application of Tonnelli's theorem. Alternatively, you can bound $\mathbb E[|Y_1|\chi(k\lt |Y_1|\leqslant k+1)]$ to conclude that the series $\sum k\mu(k\lt |Y_1|\leqslant k+1)$ is divergent, then sum by parts. –  Davide Giraudo Sep 6 at 9:25
    
@DavideGiraudo I think you have followed mike's hint, am I right? If so, how to prove his hint 3? Thanks! –  Danielsen Sep 6 at 15:14

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