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卡方分布和伽马函数(Chi-Square Distribution)

2022-07-28 13:38:00 【Turbo-shengsong】

伽马函数

要定义卡方分布，我们需要首先定义伽马函数：
Γ ( p ) = ∫ 0 + ∞ x p − 1 e − x d x , p > 0 (1) \Gamma(p) = \int_{0}^{+ \infty} x^{p-1} e^{-x} dx, p >0 \tag{1} Γ(p)=∫0+∞​xp−1e−xdx,p>0(1)

如果我们使用分部积分，
Γ ( p ) = ∫ 0 + ∞ x p − 1 e − x d x = ∫ 0 + ∞ − x p − 1 d e − x = − x p − 1 e − x ∣ 0 + ∞ − ∫ 0 + ∞ [ − e − x ( p − 1 ) x p − 2 ] d x = ( p − 1 ) Γ ( p − 1 ) (2) \begin{aligned} \Gamma(p) &= \int_{0}^{+ \infty} x^{p-1} e^{-x} dx \\ & = \int_{0}^{+ \infty} -x^{p-1} d e^{-x} \\ & = -x^{p-1} e^{-x} |_{0}^{+ \infty} - \int_{0}^{+ \infty} \left [ - e^{-x} (p-1)x^{p-2} \right] dx \\ &= (p-1) \Gamma(p-1) \tag{2} \end{aligned} Γ(p)​=∫0+∞​xp−1e−xdx=∫0+∞​−xp−1de−x=−xp−1e−x∣0+∞​−∫0+∞​[−e−x(p−1)xp−2]dx=(p−1)Γ(p−1)​(2)

通过这种方式，我们可以证明伽玛函数服从一个有趣的递归关系。
Γ ( p ) = ( p − 1 ) Γ ( p − 1 ) = ( p − 1 ) ( p − 2 ) Γ ( p − 2 ) = ( p − 1 ) ( p − 2 ) ⋯ Γ ( 1 ) (3) \begin{aligned} \Gamma(p) &= (p-1) \Gamma(p-1) \\ &= (p-1) (p-2) \Gamma(p-2) & = (p-1) (p-2) \cdots \Gamma(1) \end{aligned} \tag{3} Γ(p)​=(p−1)Γ(p−1)=(p−1)(p−2)Γ(p−2)​=(p−1)(p−2)⋯Γ(1)​(3)

容易计算
Γ ( 1 ) = 1 (4) \Gamma(1) = 1 \tag{4} Γ(1)=1(4)

因此，我们可以得到
Γ ( p ) = ( p − 1 ) ! (5) \Gamma(p) = (p-1)! \tag{5} Γ(p)=(p−1)!(5)

另外，我们可以计算
Γ ( 1 2 ) = ∫ 0 + ∞ x − 1 2 e − x d x = 2 ∫ 0 + ∞ e − x d x 1 2 = 2 ∫ 0 + ∞ e − u 2 d u = 2 ⋅ 2 π 1 2 ⋅ 1 2 π 1 2 ∫ 0 + ∞ e − u 2 d u = 2 π ⋅ 1 2 = π (6) \begin{aligned} \Gamma(\frac{1}{2}) &= \int_{0}^{+ \infty} x^{-\frac{1}{2}} e^{-x} dx \\ &= 2 \int_{0}^{+ \infty} e^{-x} d x^{\frac{1}{2}} \\ & = 2 \int_{0}^{+ \infty} e^{-u^2} d u \\ & = 2 \cdot \sqrt{2 \pi \frac{1}{2}} \cdot \frac{1}{\sqrt{2 \pi \frac{1}{2}}} \int_{0}^{+ \infty} e^{-u^2} d u \\ & = 2 \sqrt \pi \cdot \frac{1}{2} \\ & = \sqrt \pi \tag{6} \end{aligned} Γ(21​)​=∫0+∞​x−21​e−xdx=2∫0+∞​e−xdx21​=2∫0+∞​e−u2du=2⋅2π21​​⋅2π21​​1​∫0+∞​e−u2du=2π​⋅21​=π​​(6)

卡方分布

如果 x \boldsymbol x x由 n n n个独立同分布(i.i.d.)的随机变量构成， x i ∼ N ( 0 , 1 ) , i = 0 , 1 , ⋯   , n x_i \sim \mathcal N(0, 1), i=0,1,\cdots, n xi​∼N(0,1),i=0,1,⋯,n，那么
y = ∑ i = 1 n x i 2 ∼ χ n 2 (7) y = \sum_{i=1}^{n} x^2_i \sim \chi^2_n \tag{7} y=i=1∑n​xi2​∼χn2​(7)

where χ n 2 \chi^2_n χn2​ denotes a χ 2 \chi^2 χ2 random variable with n n n degree of freedom. PDF:
p ( y ) = { 1 2 n 2 Γ ( n 2 ) y n 2 − 1 exp ⁡ − ( 1 2 y )   for  y ≥ 0 0                                       for  y < 0 (8) p(y) = \begin{cases} \frac{1}{ 2^{\frac{n}{2}} \Gamma(\frac{n}{2}) } y^{\frac{n}{2}-1} \exp -(\frac{1}{2} y) \ \ \text{for} \ y \geq 0 \\ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{for} \ y < 0 \end{cases} \tag{8} p(y)={ 22n​Γ(2n​)1​y2n​−1exp−(21​y)  for y≥00                                      for y<0​(8)

其中 Γ ( u ) \Gamma(u) Γ(u)是伽马函数。随机变量 y y y的均值和方差分别为：
E [ y ] = n var [ y ] = 2 n (9) \begin{aligned} \mathbb E[y] &= n \\ \text{var}[y] &= 2n \end{aligned} \tag{9} E[y]var[y]​=n=2n​(9)

补充
如果 x i ∼ N ( 0 , σ 2 ) , i = 0 , 1 , ⋯   , n x_i \sim \mathcal N(0, \sigma^2), i=0,1,\cdots, n xi​∼N(0,σ2),i=0,1,⋯,n，那么
p ( y ) = { 1 σ n 2 n 2 Γ ( n 2 ) y n 2 − 1 exp ⁡ − ( 1 2 σ 2 y )   for  y ≥ 0 0                                       for  y < 0 (10) p(y) = \begin{cases} \frac{1}{ \sigma^n 2^{ \frac{n}{2} } \Gamma(\frac{n}{2}) } y^{\frac{n}{2}-1} \exp -(\frac{1}{2 \sigma^2} y) \ \ \text{for} \ y \geq 0 \\ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{for} \ y < 0 \end{cases} \tag{10} p(y)={ σn22n​Γ(2n​)1​y2n​−1exp−(2σ21​y)  for y≥00                                      for y<0​(10)

这时，随机变量 y y y的均值和方差分别为：
E [ y ] = n σ 2 var [ y ] = 2 n σ 4 (11) \begin{aligned} \mathbb E[y] &= n \sigma^2 \\ \text{var}[y] &= 2n \sigma^4 \end{aligned} \tag{11} E[y]var[y]​=nσ2=2nσ4​(11)

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版权声明
本文为[Turbo-shengsong]所创，转载请带上原文链接，感谢
https://blog.csdn.net/weixin_43413559/article/details/126010570

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