Ytukyg

I have been told that the following fact is true. Let $X_1,X_2,X_3,dots$ be i.i.d. random variables. Then there exists $epsilon$ such that for all $n$,
$$mathbb{P}left(frac{|X_1+dots+X_n|}{sqrt{n}}geq epsilonright)geq delta.$$

Observe that $epsilon$ does not depend on $n$. Both $epsilon$ and $delta$ are $>0$.

However, I am struggling to prove this, or find any reference. The kicker is that the $X_i$'s need not have finite mean or variance. In fact, I am interested in applying this "fact" to a situation where the $X_i$'s must have infinite variance (but possibly zero mean), so elementary things like Markov or Chebyshev's inequality won't help. I am unsure on how to proceed. Any hint would be greatly appreciated!

Update on progress: For the $X_i$ that I am interested in, I have deduced the condition
$$X_1+X_2+dots+X_{2^k} sim2^{k/4}X_i.$$
I also have proved the inequality
$$mathbb{P}(|S_n|>t) geq frac{1}{2}mathbb{P}(max_j|X_j|>t)geqfrac{1}{2}(1-e^{-n(1-F(t)+F(-t))}),$$
where $F$ is the c.d.f. of $X_i$. By $S_n$, I mean the $S_n=X_1+dots+X_n$. Thus it seems like the issue boils down to analyzing the distribution of $X_i$.

Let me give a direct proof using characteristic functions. The setting is as follows:

• 
  $(X_n)$ and $(X'_n)$ are i.i.d.


• 
  $tilde{X}_n = X_n - X'_n$ are symmetrized variables.


• 
  $S_n = X_1 + cdots + X_n$ and $tilde{S}_n = tilde{X}_1 + cdots + tilde{X}_n$.

Under this setting, we want to prove that

Claim. If the law of $X_1$ is not degenerate, then there exists $epsilon > 0$ such that
$$ inf_{ngeq 1} mathbb{P}left( |S_n| geq epsilonsqrt{n} right) > 0. $$

We prove the contraposition. To this end, assume that $inf_n mathbb{P}left(|S_n|geq epsilon sqrt{n}right) = 0$ for any $epsilon > 0$. Then exists $(n_k)$ such that $S_{n_k}/sqrt{n_k} to 0$ in probability. This implies that $tilde{S}_{n_k}/sqrt{n_k} to 0$ in probability as well. So, if $varphi(t) = mathbb{E}[cos(ttilde{X}_1)] $ denotes the characteristic funtion of $tilde{X}_1$, then

$$ varphileft( frac{t}{sqrt{n_k}} right)^{n_k} = mathbb{E}[exp{mathrm{i}t tilde{S}_{n_k}/sqrt{n_k}}] xrightarrow[ktoinfty]{} 1 $$

by the Portmanteau theorem. By taking $log|cdot|$, we have $n_k logleft| varphileft( frac{t}{sqrt{n_k}} right) right| to 0$. But since

• $ varphileft( frac{t}{sqrt{n_k}} right) = 1 - 2 mathbb{E}left[ sin^2left( frac{ttilde{X}_1}{2sqrt{n_k}}right) right] $ by the double-angle identity;




• $mathbb{E}left[ sin^2left( frac{ttilde{X}_1}{2sqrt{n_k}}right) right] to 0$ by the dominated convergence theorem;

it follows that

$$ n_k mathbb{E}left[ sin^2left( frac{ttilde{X}_1}{2sqrt{n_k}}right) right] xrightarrow[ktoinfty]{} 0. $$

Plugging $t = 2$ and applying the monotone convergence theorem and the squeezing lemma,

$$ mathbb{E}[tilde{X}_1^2]
= lim_{ktoinfty} mathbb{E}left[ n_k sin^2left( frac{tilde{X}_1}{sqrt{n_k}}right) mathbf{1}_{{ |tilde{X}_1| leq frac{pi}{2}sqrt{n_k} }} right]
= 0, $$

and therefore $X_1$ is degenerate.

I think what we want to show is that $S_n/sqrt{n}$ does not converge in probability to 0.

If $X_i$ have finite mean and variance > 0, the result follows from the CLT. Thus, we only have to consider the case of infinite variance. In some sense, this should be even easier to prove since it's more likely that the sum $S_n=X_1+...+X_n$ is large. In fact, if $S_n/sqrt{n}$ converges in probability to 0, then $X_i$ must have finite variance. This was an exercise in Durrett's probability book. The idea is to symmetrize by considering the random variables $Y_i = X_i - X'_i$. Assume that $X_i$ have infinite variance. Then we can consider truncated versions of $Y_i$ with arbitrarily large finite variance. This then allows us to get a bound like $mathbb{P}(sum Y_i ge K sqrt{n}) ge 1/5$ for arbitrary $K$. (Essentially, if that probability is too small, you have no chance of obtaining the required large variance.) But the probability was supposed to go to 0 for $K > 0$. Thus, $X_i$ can be assumed to have finite variance and the CLT applies.