n維球坐標變換的Jacobi行列式和卡方分布的概率密度函數(shù)

問題 1. 定義n維球坐標變換x_{1}=r\cos\theta_{1}, x_{i}=r\sin\theta_{1}\cdots\sin\theta_{i-1}\cos\theta_{i}, x_{n}=r\sin\theta_{1}\cdots\sin\theta_{n-1}. Jacobi矩陣行列式J_{n}=\left|\frac{\partial\left(x_{1},\dots,x_{n}\right)}{\partial\left(r,\theta_{1},\dots,\theta_{n-1}\right)}\right|？

按照最后一列的最后兩個元素展開，有J_{n}=r\sin\theta_{1}\cdots\sin\theta_{n-2}\cos\theta_{n-1}\cdot\cos\theta_{n-1}J_{n-1}+r\sin\theta_{1}\cdots\sin\theta_{n-2}\sin\theta_{n-1}\cdot\sin\theta_{n-1}J_{n-1}=r\sin\theta_{1}\cdots\sin\theta_{n-2}\cdot J_{n-1}=J_{n-1}r\underset{j\le n-2}{\prod}\sin\theta_{j}=J_{2}\stackrel[k=1]{n-2}{\prod}\left(r\underset{j\le k}{\prod}\sin\theta_{j}\right)=r\cdot r^{n-2}\stackrel[j=1]{n-2}{\prod}\stackrel[k=j]{n-2}{\prod}\sin\theta_{j}=r^{n-1}\stackrel[j=1]{n-1}{\prod}\left(\sin\theta_{j}\right)^{n-1-j}.

特別注意球坐標變換的空間。球坐標變換可分解為以下步驟，首先令x_{1}=r\cos\theta_{1}, x_{2}^{2}+\dots+x_{n}^{2}=\left(r\sin\theta_{1}\right)^{2}，此時\theta_{1}的范圍是\left[0,\pi\right]，恰好滿足x_{1}的范圍是\mathbb{R}且r\sin\theta_{1}\geq0能作為下一個單位球的半徑。0\leq r<\infty, 0\leq\theta_{1\sim n-2}\leq\pi, 0\leq\theta_{n-1}\leq2\pi.

問題 2. 計算\int_{0}^{2\pi}\left|\sin\left(x\right)\right|^{n}dx

當n\ge2時，I_{n}\overset{def}{=}\int_{0}^{\pi}\sin\left(x\right)^{n}dx=\left.-\sin\left(x\right)^{n-1}\cos\left(x\right)\right|_{0}^{\pi}+\left(n-1\right)\int_{0}^{\pi}\cos\left(x\right)^{2}\sin\left(x\right)^{n-2}dx=\left(n-1\right)\left(I_{n-2}-I_{n}\right)，所以I_{n}=\frac{n-1}{n}I_{n-2}. I_{0}=\pi, I_{1}=2.

I_{2n}=\frac{2n-1}{2n}\cdots\frac{1}{2}\cdot\pi=\frac{\left(2n-1\right)!!}{\left(2n\right)!!}\pi, I_{2n+1}=\frac{2n}{2n+1}\cdots\frac{2}{3}\cdot2=\frac{\left(2n\right)!!}{\left(2n+1\right)!!}2

問題 3. 卡方分布的概率密度函數(shù)

設x_{1\sim n} i.i.d. \sim N\left(0,1\right). 對x_{1\sim n}做球坐標變換，\left(r,\theta_{1\sim n-1}\right)的聯(lián)合概率密度函數(shù)是\left(2\pi\right)^{-\frac{n}{2}}\exp\left(-\frac{r^{2}}{2}\right)\left|J_{n}\right|=\left(2\pi\right)^{-\frac{n}{2}}\exp\left(-\frac{r^{2}}{2}\right)r^{n-1}\stackrel[j=1]{n-1}{\prod}\left|\sin\left(\theta_{j}\right)\right|^{n-1-j}.

r,\theta_{1\sim n-1}相互獨立。\theta_{1\sim n-2}的概率密度函數(shù)是\left(I_{n-1-j}\right)^{-1}\sin\left(\theta_{j}\right)^{n-1-j}I\left(0\leq\theta_{j}\leq\pi\right)，\theta_{n-1}的概率密度函數(shù)是\left(2I_{0}\right)^{-1}\left|\sin\left(\theta_{n-1}\right)\right|^{0}I\left(0\leq\theta_{n-1}\leq2\pi\right).

所以r的概率密度函數(shù)是\left(2\pi\right)^{-\frac{n}{2}}\exp\left(-\frac{r^{2}}{2}\right)r^{n-1}2\stackrel[j=1]{n-1}{\prod}\left(I_{n-1-j}\right)=2\left(2\pi\right)^{-\frac{n}{2}}\exp\left(-\frac{r^{2}}{2}\right)r^{n-1}\stackrel[j=0]{n-2}{\prod}I_{j}.

\stackrel[j=0]{n-2}{\prod}I_{j}=\left\langle n\text{偶數(shù)}\right\rangle \frac{2^{\frac{n-2}{2}}\pi^{\frac{n}{2}}}{\left(n-2\right)!!}+\left\langle n\text{奇數(shù)}\right\rangle \frac{\left(2\pi\right)^{\frac{n-1}{2}}}{\left(n-2\right)!!}

r的概率密度函數(shù)是\left(\left\langle n\text{偶數(shù)}\right\rangle 1+\left\langle n\text{奇數(shù)}\right\rangle \sqrt{\frac{2}{\pi}}\right)\frac{1}{\left(n-2\right)!!}\exp\left(-\frac{r^{2}}{2}\right)r^{n-1}.

r^{2}的概率密度函數(shù)是\left(\left\langle n\text{偶數(shù)}\right\rangle \frac{1}{2}+\left\langle n\text{奇數(shù)}\right\rangle \frac{1}{\sqrt{2\pi}}\right)\frac{1}{\left(n-2\right)!!}\exp\left(-\frac{x}{2}\right)x^{\frac{n}{2}-1}，這也是n個自由度的卡方分布的概率密度函數(shù)。