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Let $\left(X_n\right)_{n\geq 1}$ be a sequence of i.i.d. real random variables, with $\mathbb E(X_1)=0$, $\operatorname{var}(X_1)=1$. Let $S_n=X_1+\cdots+X_n$.

Prove that for any $A>0$, $$\displaystyle \left\{\limsup_{n\rightarrow \infty}\frac{S_n}{\sqrt{n}}>A\right\}\in\cap_{n\geq 1}\sigma(X_i,i\geq n)$$

And deduce that $$\displaystyle \mathbb P\left(\limsup_{n\rightarrow \infty}\frac{S_n}{\sqrt{n}}=+\infty\right)=1$$

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We observe that for any $m \in \mathbb N$,

$$\limsup_{n\rightarrow\infty} \frac{S_n}{\sqrt n} = \limsup_{n\rightarrow\infty} (\frac{1}{\sqrt n}\sum_{i=m}^n X_i + \frac{1}{\sqrt n}\sum_{i=1}^{m-1} X_i ).$$

The second term in parentheses goes to 0 as $n\rightarrow\infty$. Thus we find

$$\limsup_{n\rightarrow\infty} \frac{S_n}{\sqrt n} = \limsup_{n\rightarrow\infty} \frac{1}{\sqrt n}\sum_{i=m}^n X_i .$$

and so

$$\{\limsup_{n\rightarrow\infty} \frac{S_n}{\sqrt n} > A\} = \{\limsup_{n\rightarrow\infty} \frac{1}{\sqrt n}\sum_{i=m}^n X_i > A\} .$$

The limsup of a sequence of random variables which is measurable with respect to a certain $\sigma$-algebra is also measurable with respect to that $\sigma$-algebra. Thus the event on the right side above is in $\bigcap_{n=m}^\infty \sigma(X_i : i\geq n)$. This proves the first statement.

For the second, we apply the Kolmogorov zero-one law to deduce that $P\{\limsup_{n\rightarrow\infty} \frac{S_n}{\sqrt n} > A\} = 0$ or $1$. So, to prove the second claim it suffices to show that this event has positive probability. By the central limit theorem, $\lim_{n\rightarrow\infty} P\{\frac{S_n}{\sqrt n} > A\} > 0$. So, with positive probability there exist arbitrarily large $n$ with $\frac{S_n}{\sqrt n} > A$. Thus $P\{\limsup_{n\rightarrow\infty} \frac{S_n}{\sqrt n} > A\} > 0$, which proves the result.

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