# A problem about strong law of large numbers of Shiryaev's Probability

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 $\{a_n\}$.

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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
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
@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

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 $R$, $\mathbb P(\limsup_i \{|Y_i|>iR\})=1$.
How do you conclude from infinite mean that the series $P(|Y_i|>iR)$ diverges? – Lost1 Sep 5 '14 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 '14 at 9:25
Let $X_i=\xi_i-\tilde{\xi}_i$, where $\tilde{\xi}_i$ is a independent copy of $\xi_i$. By the comment of mike, we have $E|X_i|=\infty$. And let $H_n=\sum_{i=1}^{n}X_i$. Since $$\limsup_{n\to\infty}\frac{|H_n|}{n}\leq\limsup_{n\to\infty}\left|\frac{S_n}{n}-a_n\right|+\limsup_{n\to\infty}\left|\frac{\tilde{S}_n}{n}-a_n\right|$$ So by Kolmogorov $0$-$1$ law, it is sufficient to prove $$\limsup_{n\to\infty}\frac{|H_n|}{n}\geq\infty\text{ (P-a.s.)}$$ Let $A>0$, $$\infty=E\frac{|X_i|}{A}=\int_0^\infty P(|X_i|>\lambda A)d\lambda\leq\sum_{k=0}^\infty P(|X_i|>kA)$$ so $$\sum_{i=1}^\infty P(|X_i|>iA)\geq\infty$$ Since $\{X_i\}$ are independent, by Borel-Catelli lemma $$P(|X_i|>iA\text{ i.o.})=1$$ Since $H_{n+1}-H_n=X_{n+1}$, we have $$P(|H_n|>\frac{nA}{2}\text{ i.o.})=1$$ i.e. $$\limsup_{n\to\infty}\frac{|H_n|}{n}\geq\frac{A}{2}\text{ (P-a.s.)}$$ Take a sequence $0<A_m\uparrow\infty$, we get $$\limsup_{n\to\infty}\frac{|H_n|}{n}\geq\infty\text{ (P-a.s.)}$$ So the problem has been proved.