# A convergence property for iid sequence of Cauchy random variables

A real random variable $${X}$$ is said to have a standard Cauchy distribution if it has the probability density function $$\displaystyle {x \mapsto \frac{1}{\pi} \frac{1}{1+x^2}}$$. If $${X_1,X_2,\dots}$$ are iid copies of a random variable $${X}$$ with the standard Cauchy distribution, show that $$\displaystyle {\frac{|X_1|+\dots+|X_n|}{n \log n}}$$ converges in probability to $$\displaystyle {\frac{2}{\pi}}$$.

Attempt: We use a truncation argument as posted below. Verifications and suggestions for a different approach are welcomed.

I'll use the following theorem, a proof of this theorem is available in Allan Gut's Probability: A Graduate Course:

Theorem 4.2 Suppose that $$X, X_1, X_2, \ldots$$ are independent, identically distributed random variables with partial sums $$S_n, n \ge 1$$. Further, let,for $$x > 0, b(x) = x^{1/\rho}\ell(x)$$, where $$\ell \in \text{SV}$$ and $$\rho \in (0, 1]$$. Finally, set $$b_n = b(n), n \ge 1$$. Then, $$\dfrac{S_n - n \mathbb{E}[X \cdot 1\{\vert X \vert \le b_n\}]}{b_n} \stackrel{\mathbb{P}}{\rightarrow} 0$$ if and only if $$n\mathbb{P}(\vert X \vert > b_n) \rightarrow 0 \text{ as } n \rightarrow \infty$$

where $$\text{SV}$$ is a collection of slowly varying functions. You can check that, for $$b(x) = x \ln x, x > 0$$, $$\ln x \in \text{SV}$$. Now we need to check that $$n\mathbb{P}(\vert X \vert > n \ln n) \rightarrow 0 \text{ as } n \rightarrow \infty$$ The LHS can be found in its exact form: $$n\mathbb{P}(\vert X \vert > n \ln n) = n\left(1 - \int_{-n \ln n}^{n\ln n} \dfrac{1}{\pi (1 + x^2)}dx\right) = n\left[1 - \dfrac{2}{\pi}\text{arctan}(n \ln n)\right] \rightarrow 0$$ The limit can be computed via L'Hospital. Thus, the theorem above implies that $$\dfrac{S_n}{n \ln n} - \dfrac{\mathbb{E}[\vert X \vert \cdot 1\{\vert X \vert \le n \ln n\}]}{\ln n} = \dfrac{S_n - n \mathbb{E}[\vert X \vert \cdot 1\{\vert X \vert \le n \ln n\}]}{n \ln n} \stackrel{\mathbb{P}}{\rightarrow} 0$$ where $$S_n = \sum_{k = 1}^n \vert X_k \vert$$. But, $$\dfrac{\mathbb{E}[\vert X \vert \cdot 1\{\vert X \vert \le n \ln n\}]}{\ln n} = \dfrac{\ln(1 + n^2 (\ln n)^2)}{\pi \ln n} \xrightarrow{n \rightarrow \infty} \dfrac{2}{\pi}$$ Therefore, $$\dfrac{\vert X_1 \vert + \ldots + \vert X_n \vert}{n \ln n} = \dfrac{S_n}{n \ln n} = \left[\dfrac{S_n}{n \ln n} - \dfrac{\mathbb{E}[\vert X \vert \cdot 1\{\vert X \vert \le n \ln n\}]}{\ln n}\right] + \dfrac{\mathbb{E}[\vert X \vert \cdot 1\{\vert X \vert \le n \ln n\}]}{\ln n} \xrightarrow{\mathbb{P}} \dfrac{2}{\pi}$$

• This is a nice proof as well. On the other hand, is there any issue with the original solution? Commented Jun 21 at 2:08

Let $$S_n := |X_1| + \dots + |X_n|$$. We truncate each $$|X_i|$$ for $$1 \leq i \leq n$$ at $$Cn$$ for some $$C$$ to be chosen later, by writing $$|X_i| = |X_i|_{\leq Cn} + |X_i|_{> Cn}$$, where $$|X_i|_{\leq Cn} := |X_i|1_{|X_i| \leq Cn}$$ and $$|X_i|_{> Cn} := |X_i|1_{|X_i| > Cn}$$, similarly decompose $$S_n = S_{n, \leq} + S_{n, >}$$, where $$S_{n, \leq} := |X_1|_{\leq Cn} + \dots + |X_n|_{\leq Cn}$$ and $$S_{n, >} := |X_1|_{> Cn} + \dots + |X_n|_{> Cn}$$.

The random variable $$|X|1_{|X| \leq Cn}$$ can be computed to have mean

$$\displaystyle {\bf E}(|X|1_{|X| \leq Cn}) = \frac{1}{\pi} \int_{\bf R} \frac{|x|1_{|x| \leq Cn}}{{1+x^2}}\ dx = \frac{1}{\pi} \int_{-Cn}^{Cn} \frac{|x|}{1 + x^2}\ dx = \log(C^2 n^2 + 1) / \pi$$

and we can upper bound the variance by

$$\displaystyle {\bf Var}(|X|1_{|X| \leq Cn}) \leq {\bf E}(|X|1_{|X| \leq Cn})^2 = \frac{1}{\pi} \int_{\bf R} \frac{x^21_{|x| \leq Cn}}{1+x^2}\ dx = \frac{2}{\pi}(Cn - \tan^{-1}(Cn))$$

and hence $$S_{n,\leq}/n$$ has mean $$\log(C^2 n^2 + 1)/\pi$$ and variance at most $$\displaystyle \frac{2/\pi \cdot (Cn - \tan^{-1}(Cn))}{n}$$. By Chebyshev’s inequality, we thus have

$$\displaystyle {\bf P}(|S_{n,\leq}/n - \log(C^2 n^2 + 1)/\pi| \geq \lambda) \leq \frac{2/\pi \cdot (Cn - \tan^{-1}(Cn))}{n{\lambda}^2}$$

for any $$\lambda > 0$$.

We now turn to $$S_{n,>}$$. Observe that the random variable $${|X|1_{|X|>Cn}}$$ is only nonzero with probability

$$\displaystyle 1 - {\bf P}(|X| \leq Cn) = 1 - \frac{1}{\pi}\int_{-Cn}^{Cn} \frac{1}{1 + x^2}\ dx = 1 - \frac{2 \tan^{-1}(Cn)}{\pi}$$.

Thus $$S_{n,>}$$ is nonzero with probability at most $$\displaystyle n(1 - \frac{2 \tan^{-1}(Cn)}{\pi})$$. By the triangle inequality, we conclude that

$$\displaystyle {\bf P}(|S_n/n - \log(C^2 n^2 + 1)/\pi| \geq \lambda) \leq \frac{2/\pi(Cn - \tan^{-1}(Cn))}{n{\lambda}^2} + n(1 - \frac{2 \tan^{-1}(Cn)}{\pi})$$

which is valid for any $$\lambda > 0$$. If we pick $$C = \lambda = \sqrt{\log n}$$, we see that

$$\displaystyle {\bf P}(|S_n/n\log n - \log(n^2 \log n + 1)/\pi \log n| \geq \frac{1}{\sqrt{\log n}}) \leq O(\frac{1}{\log^{1/2} n})$$

which for large $$n$$ implies

$$\displaystyle {\bf P}(|S_n/n\log n - 2/\pi| \geq \frac{1}{\sqrt{\log n}}) \leq O(\frac{1}{\log^{1/2} n})$$, giving the result.