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Question: Let $(X_k)_{k \geq 1}$ be a sequence of independent random variables with uniform distribution on $[-1,1]$ Compute $\lim_{n \to \infty} \mathbb{P}(|\sum_{k=1}^{n} X_{k}^{-1}|> \pi n/2)$

I tried to solve it using characteristic function, but the calculation seems very complicated However, I got the characteristic function to be $1$ as $n$ approaches infinity, which doesn't really give any assumption about the distribution function.

Gary
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Tas
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  • Maybe it's easier to consider as $1-\lim_{n\to\infty}\mathbb{P}\left(\left|\frac{1}{n}\sum_{k=1}^nX^{-1}_k\right|<\frac{\pi}{2}\right)$ instead? It would be a lot easier if $\mathbb{E}[X^{-1}_k]$ existed, but unfortunately it doesn't. – Varun Vejalla Jun 02 '23 at 03:51
  • @VarunVejalla also, we could work out if the random variables are identically distributed as well. But it's not – Tas Jun 02 '23 at 03:56
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    If they are all independent and have a uniform distribution on $[-1,1]$, wouldn't that mean they're identically distributed? Then with the characteristic function of $X^{-1}k$, you could take the products (or really just the power to $k$ since they're identically distributed) to get the characteristic function of $\sum{k=1}^n X^{-1}_k$. – Varun Vejalla Jun 02 '23 at 03:56
  • @VarunVejalla I wasn't sure that they are identically distributed . Then I will be able to solve. Thanks very much for the help – Tas Jun 02 '23 at 04:08
  • $\frac{1}{n}\sum\limits_{k=1}^n X_k^{-1}$ has a stable distribution with index $1$ as limiting distribution, with proper normalization. Based on the formula on page 11 of this document (https://escholarship.org/content/qt8940b4k8/qt8940b4k8.pdf?t=lnqpml), it seems the limit of your probability is one. I believe you would have to replace $\frac{\pi}{2}$ with something close to $\log(n)$ in order to have a limit probability not either $1$ or $0$ – charmd Jun 02 '23 at 08:30

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Let $X$ be uniformly distributed in $[-1,1]$, then $$ \mathbb{P}(X^{-1}\leqslant x)= \begin{cases} -\frac1{2x}, & \hfill x\leqslant-1\hfill \\ \frac12, & -1\leqslant x\leqslant 1 \\ 1-\frac1{2x}, & \hfill x\geqslant1\hfill \end{cases} $$ and we easily find \begin{align} \varphi(t)&:=\mathbb{E}e^{itX^{-1}} =\int_1^\infty\frac{\cos tx}{x^2}\,dx \\&=\cos t-t\int_1^\infty\frac{\sin tx}{x}\,dx \\&=\cos t-\frac\pi2|t|+t\int_0^1\frac{\sin tx}{x}\,dx, \end{align} so that $\varphi(t)=1-\pi|t|/2+O(t^2)$ as $t\to 0$.

Now, if $Y_n=\frac1n\sum_{k=1}^n X_k^{-1}$, then $\mathbb{E}e^{itY_n}=\varphi^n(t/n)$, and inversion gives$$P_n:=\mathbb{P}\left(|Y_n|\leqslant\frac\pi2\right)=\frac1\pi\int_{-\infty}^\infty\frac{\sin(\pi t/2)}{t}\,\varphi^n\left(\frac tn\right)dt.$$

To take $n\to\infty$ here, we see that the above implies $\varphi^n(t/n)\to e^{-\pi|t|/2}$ (and this is a dominated convergence). Using $\int_{-\infty}^\infty=2\int_0^\infty$ and substituting $x=\pi t/2$, we then get $$ \lim_{n\to\infty}P_n=\frac2\pi\int_0^\infty\frac{\sin x}x\,e^{-x}\,dx=\frac12 $$ (the integral can be computed in a number of ways).

metamorphy
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