柯西分布和正态分布的关系

柯西分布和正态分布的关系

X_{1}\sim N\left(0,1\right), X_{2}\sim N\left(0,1\right)相互独立。

问题 1. 随机变量\left(r,\theta\right)由如下关系定义：X_{1}=r\cos\theta, X_{2}=r\sin\theta, \theta\in\left[0,2\pi\right). 求其概率密度函数。

\left(X_{1},X_{2}\right)的概率密度函数p\left(x_{1},x_{2}\right)=\left(2\pi\right)^{-1}\exp\left(-\frac{x_{1}^{2}+x_{2}^{2}}{2}\right). 定义映射F\left(r,\theta\right)\overset{def}{=}\left(r\cos\theta,r\sin\theta\right)'. 计算Jacobi矩阵的行列式\left|\frac{\partial\left(X_{1},X_{2}\right)}{\partial\left(r,\theta\right)}\right|=r. P\left(\left(r,\theta\right)\in V\right)=P\left(\left(X_{1},X_{2}\right)\in F^{-1}\left(V\right)\right)=\int_{F^{-1}\left(V\right)}p\left(x_{1},x_{2}\right)dx_{1}dx_{2}=\int_{V}p\left(r\cos\theta,r\sin\theta\right)rdrd\theta，所以\left(r,\theta\right)的PDF是p\left(r\cos\theta,r\sin\theta\right)r=\left(2\pi\right)^{-1}\exp\left(-\frac{r^{2}}{2}\right)r，所以r、θ相互独立，r的PDF是\exp\left(-\frac{r^{2}}{2}\right)rI\left(r\geq0\right)，θ的PDF是\frac{1}{2\pi}I\left(0\leq\theta<2\pi\right).

推论2. \sqrt{\chi^{2}\left(2\right)}的PDF是\exp\left(-\frac{r^{2}}{2}\right)rI\left(r\geq0\right)，\chi^{2}\left(2\right)的PDF是指数分布\frac{1}{2}\exp\left(-\frac{x}{2}\right)I\left(x\geq0\right)

问题 3. Y\overset{def}{=}\frac{X_{2}}{X_{1}}=\tan\theta的PDF

容易证明\tan\theta\leq t在0\leq\theta<2\pi内的区间长度是2\arctan t+\pi，所以P\left(\tan\theta\leq t\right)=\frac{1}{2\pi}\left(2\arctan t+\pi\right)=\frac{1}{\pi}\arctan t+\frac{1}{2}，所以\tan\theta的PDF是\frac{1}{\pi\left(1+t^{2}\right)}. 即Y服从Cauchy分布。