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I have a sum of $n$ i.i.d random variables $X_i$ such that $E[X_i] = 0$,$\mathrm{E}[|X_i|^{1 + \delta}]$ exists for some $0 < \delta < 1$ but $\mathrm{E}[|X_i|^{1 + \delta+ \epsilon}]$ does not exists for any $\epsilon > 0$.

I would like to ask if it is possible to provide a lower bound for $P(|\sum_{i=1}^n X_i| > A\sqrt{n})$, showing that this term converges to $1$ when $n$ approaches infinity.

I am aware that when $X_i$ obeys the tail balance condition, we would have $\frac{X_1 + ... + X_n}{a_n}$ converges to a $\alpha-$stable distribution for proper scaling factor $a_n$. So I am mostly interested in the case where the tail balance condition does not hold.

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Certainly. All you need is $EX^2=+\infty$. Then the characteristic function $f_X(t)$ satisfies $\lim_{t\to 0}\frac{1-|f(t)|}{t^2}=+\infty$, so for every finite interval $I\subset \mathbb R$, we have $\lim_{n\to\infty}\int_I|f(t/\sqrt n)|^n\,dt=0$ and the usual Fourier trickery finishes the story. Of course, the efficient lower bound would require some quantitative information about the speed of divergence of the second moment: if you have essentially the uniform distribution on $[-1,1]$ combined with some large value dust of extremely low probability (which can still blow up any moment you like), then you'll have to wait quite long time before the spreading effect kicks in.

Edit: Just to explain the "Fourier trickery". The story is that if you have some *smooth non-negative function $g$ with compact support that is identically $1$ on $[-A,A]$, then $P(|X|\le A)\le Eg(X)=c\int_{\mathbb R}\widehat g(t)f_X(-t)$, so if you know that $\int|f_X|$ is small over a long interval where $\widehat g$ is still noticeable, then you can conclude that this probability is small and, therefore, $P(|X|>A)$ is close to $1$.

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  • $\begingroup$ Thank you! I think this is very helpful but I am not quite aware of the Fourier transformation techniques you mention. Can you please give me some reference on this? $\endgroup$
    – Kaiyue Wen
    Commented Jan 16, 2023 at 10:21
  • $\begingroup$ After some second thought, I realize this is exactly utilizing the inversion formula of characteristic function! This is truly a great answer! Thank you! $\endgroup$
    – Kaiyue Wen
    Commented Jan 16, 2023 at 11:06
  • $\begingroup$ Sorry, I found an error in my previous deduction. For now, if the c.d.f of $\sum_i X_i/\sqrt{n}$ is $F_n$, I am only able to prove for any $C > 0$, $|\int_{- \infty}^{\infty} \frac{sin(Cx)}{Cx} dF_n(x)| \to_n 0$. I am not sure whether this is enough to show the lower bound. $\endgroup$
    – Kaiyue Wen
    Commented Jan 16, 2023 at 15:58
  • $\begingroup$ @KaiyueWen I added more details. Is it clear now? $\endgroup$
    – fedja
    Commented Jan 16, 2023 at 19:07
  • $\begingroup$ Could you please elaborate a little bit how does $EX^2=+\infty$ imply $\lim_{t\to 0}\frac{1-|f(t)|}{t^2}=+\infty$? It looks like it should be a simple Taylor expansion but I cannot make it work (not really sure how to deal with the absolute value $|f(t)|$). $\endgroup$ Commented Jan 16, 2023 at 23:50

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