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For a Cauchy distribution with density $f(x) = \frac{1}{\pi(1 + x^2)}$ , it doesn't have well-defined moments. Therefore both the law of large numbers and the central limit theorem can't apply to the distribution. For example, Given iid sample points $X_1, \dots, X_n$ of the distribution, as $n\to \infty$, $\bar{X}$ will not converge to a constant in probability.

However a note says under the above Cauchy distribution, "asymptotic variance of $\sqrt{n} \bar{X}$ is $\infty$", which seems like a version of CLT?

The definition of the asymptotic variance of a statistic is defined in Casella's Statistical Infernce:

Definition 10.1.9 For an estimator $T_n$, suppose that $k_n(T_n - \tau(\theta)) \to n(0, \sigma^2)$ in distribution. The parameter $\sigma^2$ is called the asymptotic variance or variance of the limit distribution of $T_n$.

So I wonder if the above definition applies to the case here, for example, with $\tau(\theta) =0$ and $k_n = \sqrt{n}$? If yes, the limit distribution of $\sqrt{n} \bar{X}$ is $N(0, \infty)$, the asymptotic mean of $\sqrt{n} \bar{X}$ is $0$ and its asymptotic variance is $\infty$? If no, how shall we understand "the asymptotic variance of $\sqrt{n} \bar{X}$ is $\infty$"?

Btw, for each $n$, $\sqrt{n} \bar{X}$ doesn't admit mean and variance, right?

Thanks and regards!

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For $X_i \sim \operatorname{Cauchy}(0,1)$, their linear combination is also Cauchy distributed, specifically $$ Z = \sum_{k=1}^n a_k X_k \sim \operatorname{Cauchy}\left(0, \sum_{k=1}^n |a_k| \right) $$ which is easily proven by considering a characteristic function: $$ \varphi_Z(t) = \prod_{k=1}^n \phi_X\left( a_k t \right) = \exp\left( - |t| \sum_{k=1}^n |a_k| \right) $$ which is the characteristic function of the claimed distribution.

With this say, $\bar{X} \sim \operatorname{Cauchy}\left(0, 1\right) $ for all $n$, and neither $\mathbb{E}(\bar{X})$ nor $\mathbb{Var}(\bar{X})$ is defined.

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Thanks, Sasha! Why is "the asymptotic variance of n√X¯ is ∞"? – Tim Mar 27 '13 at 20:41
I am not certain what is meant by this. $\bar{X}$ does not admit variance, and neither does $\sqrt{n} \bar{X}$, thus I would say $\mathbb{Var}(\sqrt{n} \bar{X}) = \infty$, but I see no need to bring up "asymptotic" here. Asymptotic would be appropriate if $\sqrt{n} \bar{X}$ had finite variance, which would grown boundlessly as $n$ were increasing. – Sasha Mar 27 '13 at 20:57
Thanks! (1) the variance of $\bar{X}$ is actually not well defined, which can have different values ($0$ or $\infty$ or ...) when the improper integral is interpreted in different ways. If $\sqrt{n} \bar{X}$ is also such case, why would its variance is $\infty$ instead of $0$? – Tim Mar 27 '13 at 21:09
(2) "Asymptotic would be appropriate if $\sqrt{n} \bar{X}$ had finite variance, which would grown boundlessly as n were increasing." Casella's book says, which I quoted in my post, the asymptotic variance is the variance of the limit distribution of $\sqrt{n} \bar{X}$. So the asymptotic variance is not the same as the limit of the variance of $\sqrt{n} \bar{X}$. – Tim Mar 27 '13 at 21:09
By $\bar{X}$ is equal in distribution to the standard Cauchy, thus $\sqrt{n} \bar{X}$ is distributed as $\operatorname{Cauchy}(0,\sqrt{n})$. The scale parameter of it grows unboundedly, but the scale parameter has nothing to do with variance. – Sasha Mar 27 '13 at 21:31

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