# Let $X_1,X_2,\ldots$ be iid random variables with Cauchy distribution and $S_n=X_1+X_2+\cdots+X_n$, find $P(S_n>an)$, $a>0$

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 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.

• 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 $\lim_n P(S_n > an)$?" Commented May 20, 2023 at 15:06
• $S_n$ 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. Commented May 20, 2023 at 15:21
• @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 Commented May 20, 2023 at 15:30
• It is a small-sample (i.e., accurate) result, see my answer below. Commented May 20, 2023 at 15:31

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.
• 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. Commented May 20, 2023 at 16:23