# A probability inequality: probability that the normalized sum of i.i.d. random variables is bounded below, is bounded below

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 $$n$$, $$\mathbb{P}\left(\frac{|X_1+\dots+X_n|}{\sqrt{n}}\geq \epsilon\right)\geq \delta.$$

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 $$\mathbb{P}(|S_n|>t) \geq \frac{1}{2}\mathbb{P}(\max_j|X_j|>t)\geq\frac{1}{2}(1-e^{-n(1-F(t)+F(-t))}),$$ 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 $$X_i$$.

• This obviously fails if $\mathbb{P}(X_i = 0) = 1$. Do you have any other assumptions on $X_i$? You're trying to prove an instance of something that resembles "small-ball probabilities", if that helps you look for relevant references. – VHarisop Dec 4 '18 at 23:41
• We have that $X_i$ is nontrivial, should've added that. – Phil Dec 4 '18 at 23:56

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 $$\inf_{n\geq 1} \mathbb{P}\left( |S_n| \geq \epsilon\sqrt{n} \right) > 0.$$

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 $$\varphi(t) = \mathbb{E}[\cos(t\tilde{X}_1)]$$ denotes the characteristic funtion of $$\tilde{X}_1$$, then

$$\varphi\left( \frac{t}{\sqrt{n_k}} \right)^{n_k} = \mathbb{E}[\exp\{\mathrm{i}t \tilde{S}_{n_k}/\sqrt{n_k}\}] \xrightarrow[k\to\infty]{} 1$$

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

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

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

it follows that

$$n_k \mathbb{E}\left[ \sin^2\left( \frac{t\tilde{X}_1}{2\sqrt{n_k}}\right) \right] \xrightarrow[k\to\infty]{} 0.$$

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

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

and therefore $$X_1$$ is degenerate.

• Nice! I was looking for a direct characteristic function approach, but got stuck since we can't use Taylor expansion (no moments given). I'll have to remember this double angle trick. – zoidberg Dec 5 '18 at 16:08
• Cheeky trick, Thank you! – Phil Dec 5 '18 at 19:00

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.