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Let $X_1,X_2,....$ be iid random variables with Cauchy distribution and $S_n=X_1+X_2+\cdots+X_n$, find $P(S_n>an)$, $a>0$. This is exercise 8.44 of the intro book of Grimmet and Welsh. We cannot use the Large deviation theorem, because all the $X_i$ have no defined mean. The only thing that came in my mind was that $S_n$ must also be a random variable with Cauchy distribution, hence : \begin{align} & P(S_n>an)=1-P(S_n<an) \\[6pt] = {} & 1-\int_{-\infty}^{an} \frac{1}{\pi(1+x^2)}\, dx =\frac{3}{2}-\arctan(an)\pi^{-1}.\end{align} The correct answer is $\frac{1}{2}-\arctan(a)\pi^{-1}$. My method is clearly wrong. I also could not use the central limit theorem, because the mean is not defined.

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  • $\begingroup$ Can you double-check if the result is asymptotic or small-sample? Looks like the former because the "correct answer" does not contain $n$. If so, the question should be modified to "what is $\lim_n P(S_n > an)$?" $\endgroup$
    – Zhanxiong
    Commented May 20, 2023 at 15:06
  • $\begingroup$ $S_n$ is the sum of the first $X_i$, I do not know if it's small-sample or asymptotic, but in the section they let n to infinity for another example given in the book. $\endgroup$ Commented May 20, 2023 at 15:21
  • $\begingroup$ @Zhanxiong $S_n$ is the sum of the first $X_i$, I do not know if it's small-sample or asymptotic, but in the section they let n to infinity for another example given in the book $\endgroup$ Commented May 20, 2023 at 15:30
  • $\begingroup$ It is a small-sample (i.e., accurate) result, see my answer below. $\endgroup$
    – Zhanxiong
    Commented May 20, 2023 at 15:31

1 Answer 1

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If $X_1, \ldots, X_n$ are i.i.d. standard Cauchy distribution $C(0, 1)$, then since Cauchy distribution is closed under independent summation (see the second item of this link), the distribution of $S_n = X_1 + \cdots + X_n$ is still Cauchy, with location parameter $0$, and scale parameter $n$ (i.e., $C(0, n)$ with density function $f(x) = \frac{1}{\pi n(1 + x^2/n^2)}$), it then follows that \begin{align} P(S_n > an) = \int_{an}^\infty\frac{1}{\pi n(1 + x^2/n^2)}dx = \int_a^\infty \frac{1}{\pi (1 + u^2)}du = \frac{1}{\pi} \left(\frac{\pi}{2} - \arctan a\right), \end{align} matching the answer key.

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  • $\begingroup$ thanks man I understand it now. Could there be an alternative way, with the CLT or Large deviation theorem? $\endgroup$ Commented May 20, 2023 at 15:43
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    $\begingroup$ I don't think so. First, as you said, both CLT and LD requires the existence of mean/variance of $S_n$, which is not the case for Cauchy. Second, if an exact small-sample result holds, there is no need to seek for large-sample result that is CLT designed for. $\endgroup$
    – Zhanxiong
    Commented May 20, 2023 at 16:23

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