Cauchy：为何方差必为无穷 & 均值同分布

Is 2X the Same

专题: Probability / 概率
难度: L4
来源: QuantQuestion

题目详情

Let X be a continuous rv with density f, and let X1, X2 be two inde- pendent rvs, both distributed as is X. It is then not usually the case that the rv 2X is distributed as is $\mathrm{X1 + X2}$ . However, the Cauchy density whose standardized form is given by $f(x) = \frac{1}{\pi}\cdot \frac{1}{1 + x^{2}}$ possesses this property: $\mathrm{X1 + X2}$ has the same distribution as the rv 2X. It is illustrated and compared to the standard normal distribution in Figure 1.1.

a. Based on the property described earlier, argue without any explicit calculation that the variance of the Cauchy distribution is necessarily infinite.

b. Give an inductive argument for the rather unintuitive feature that for the Cauchy distribution the arithmetic mean from a sample of $n = 2^{k}$ independent realizations of X has exactly the same distribution as each contributing summand itself.

图5.4: The standard Cauchy and the standard normal density look basically similar: they are both unimodal and symmetric around $x = 0$ . However, the Cauchy has thicker tails, whereas the normal density is more concentrated around its mean.

解析

(a) 已知对 Cauchy 分布：若 $X_1,X_2$ 独立同分布于 $X$ ，则

X_1+X_2\stackrel{d}=2X.

若方差存在且为 $\sigma^2$ ，则

\mathrm{Var}(X_1+X_2)=2\sigma^2,\qquad \mathrm{Var}(2X)=4\sigma^2,

同分布要求方差相等，推出 $2\sigma^2=4\sigma^2$ ，只能 $\sigma^2=0$ 或 $\sigma^2=\infty$ 。

Cauchy 非退化，故 $\boxed{\mathrm{Var}(X)=\infty}$ 。

(b) 由 (a) 得

\frac{X_1+X_2}{2}\stackrel{d}=X.

若已知 $\frac{1}{2^k}\sum_{i=1}^{2^k}X_i\stackrel{d}=X$ ，则把 $2^{k+1}$ 个样本分成两组各 $2^k$ 个：

\frac{1}{2^{k+1}}\sum_{i=1}^{2^{k+1}}X_i =\frac12\left(\frac{1}{2^k}\sum_{i=1}^{2^k}X_i+\frac{1}{2^k}\sum_{i=2^k+1}^{2^{k+1}}X_i\right).

两项独立且各自同分布于 $X$ ，因此右侧在分布上等于 $\frac{X_1+X_2}{2}\stackrel{d}=X$ 。

归纳得对任意 $n=2^k$ ，样本均值与单个样本同分布。