Almost sure convergence of subsequence

Question

If we have for a sequence of identically independent variables with $\limsup \{|X_n|>n\}$ happens almost surely. Can we conclude that $\limsup \{n^{-1}|\sum X_j|>1\}$ happens a.s.? One of the problem I was working on seem to need this to conclude but I don't see why this is true. Any help?

Answer 1

Yes, it is true. Here is a proof of a generalization:

Claim:

Suppose $\{X_n\}_{n=1}^{\infty}$ are independent and identically distributed (i.i.d.) and there is an $\alpha>0$ such that $$ P[\{|X_n|> \alpha n\} \quad i.o.] = 1$$ where "i.o." stands for "infinitely often." Then for any $\beta>0$ we have: $$P[\{|X_n|>\beta n\} \quad i.o.]=1 \quad (Eq. *)$$ and $$P\left[\left\{\left|\frac{1}{n}\sum_{i=1}^n X_i\right|>\beta \right\} \quad i.o.\right] = 1 \quad (Eq. **)$$

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Fact 1:

Let $Y$ be a nonnegative random variable. Fix $\theta>0$. Then $E[Y]=\infty$ if and only if $$ \sum_{n=1}^{\infty}P[Y>\theta n] = \infty$$

Proof: This can be proven by $E[Y]=\int_0^{\infty} P[Y>t]dt$ and bounding the integral by a sum.

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Proof of Claim:

Suppose $\{X_n\}_{n=1}^{\infty}$ are i.i.d. and there is an $\alpha>0$ that satisfies $P[\{|X_n|> \alpha n\} \quad i.o.]=1$. By Borel-Cantelli we know $$ \sum_{n=1}^{\infty} P[|X_n|> \alpha n] = \infty$$ since if the sum were finite, $\{|X_n|> \alpha n\}$ would happen only finitely often (with prob 1). Since $\{X_n\}$ are i.i.d. we know $P[|X_n|> \alpha n]=P[|X_1|> \alpha n]$ and so $$ \sum_{n=1}^{\infty} P[|X_1|> \alpha n] = \infty$$ It follows from Fact 1 that $E[|X_1|]=\infty$ and so for any $\theta>0$ $$ \sum_{n=1}^{\infty} P[|X_1|>\theta n] = \infty$$ and so $$ \sum_{n=1}^{\infty} P[|X_n|>\theta n] = \infty$$ Since $\{X_n\}$ are mutually independent, by Borel-Cantelli this implies $$ P[\{|X_n|>\theta n\} \quad i.o.] = 1 \quad (Eq. ***)$$ This proves (Eq. *).

Now fix $\beta>0$. Define $L_n = \frac{1}{n}\sum_{i=1}^n X_i$. We have by the Kolmogorov 0-1 law that $$ P[\{|L_n|>\beta \} \quad i.o.] \in \{0,1\}$$ If the probability is 1 then we are done. Now suppose the probability is zero (we reach a contradiction). Then $|L_n|\leq \beta$ for all but finitely many indices $n$ (with prob 1). But if we define $\theta = 3\beta$ and use (Eq. ***), we know $\{|X_n|>3\beta n\}$ happens infinitely often (with prob 1). Thus, with prob 1, the events $\{|X_{n+1}|>3\beta (n+1) \}\cap\{|L_n|\leq \beta \}$ occur infinitely often. Any such time this event occurs it must be that $|L_{n+1}|>\beta$ because: $$ |L_{n+1}| = \left|\frac{nL_n}{n+1} + \frac{X_n}{n+1}\right| \geq \frac{|X_n|-|nL_n|}{n+1}> \frac{3\beta n - \beta n}{n+1} \geq \beta $$ Thus $$P[|L_{n+1}|> \beta \quad i.o. ] = 1$$ which contradicts our assumption that the probability is 0. This proves (Eq. **). $\Box$