Section02_卡方、t、F分布

$\chi^{2}$分布

1. 构造 设 $X_{1},X_{2},\cdots,X_{n}$ 独立同 $\mathcal{N}(0,1)$ 分布，则 $$\chi^{2} = \sum_{i=1}^{n}X_{i}^{2}\sim \chi^{2}(n)$$
2. 性质
   1. $X\sim \mathcal{N}(0,1)$，则 $X^{2}\sim \chi^{2}(1), EX^{2} = 1, DX^{2} = 2$

      Proof $$\begin{array}{ll} & EX^{2} = DX + (EX)^{2} = 1 \\ & DX^{2} = EX^{4} - (EX^{2})^{2} = EX^{4} - 1 \\ & \displaystyle = \int_{-\infty}^{+\infty}x^{4}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^{2}}{2}} \cdot dx - 1 \\ & = 3 - 1 = 2 \end{array}$$

   2. $\chi^{2}\sim \chi^{2}(n)$，则 $E\chi^{2} = n, D\chi^{2} = 2n$
   3. 可加性 设 $\chi^{2}_{1}\sim \chi^{2}(n_{1});\ \chi_{2}^{2}\sim \chi(n_{2})$，且 $\chi_{1}^{2}$ 同 $\chi_{2}^{2}$ 独立，则 $$\chi^{2}_{1} + \chi^{2}_{2} \sim \chi^{2}(n_{1}+ n_{2})$$
3. 上侧 $\alpha$ 分位点
   1. 设 $X\sim \mathcal{N}(0,1), Y = X^{2}\sim \chi^{2}(1)$，求 $f_{Y}(y)$ $$\begin{array}{ll} & F_{Y}(y) = \mathbb{P}\{Y\le y\} = \mathbb{P}\{X^{2}\le y\} \\ & = \begin{cases} 0, & y<0 \\ \displaystyle \int_{-\sqrt{y}}^{\sqrt{y}}f_{X}(x)\cdot dx,& y \ge 0 \end{cases} \\ \because & \displaystyle f_{X}(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{x^{2}}{2}} \\ \therefore & \text{当} y\ge 0 \text{时, }\displaystyle F_{Y}(y) = \int_{-\sqrt{y}}^{\sqrt{y}}f_{X}(x)\cdot dx \\ & \displaystyle = \int_{-\sqrt{y}}^{\sqrt{y}}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^{2}}{2}} \cdot dx \\ \therefore & \displaystyle f_{y}(y) = \frac{dF_{Y}(y)}{dy} = \frac{1}{\sqrt{2\pi}}e^{-\frac{y}{2}} \bigg(\frac{1}{2 \sqrt{y}}+ \frac{1}{2 \sqrt{y}}\bigg) \\ & = \frac{1}{\sqrt{2\pi y}} e^{-\frac{y}{2}} \\ \therefore & f_{Y}(y) = \begin{cases} 0, & y < 0 \\ \frac{1}{\sqrt{2\pi y}}e^{-\frac{y}{2}}, & y \ge 0 \end{cases} \\ \end{array}$$
   2. $X\sim \mathcal{N}(0,1), Y\sim \mathcal{N}(0,1)$，且$X,Y$独立，则$Z=X^{2}+Y^{2}\sim \chi^{2}(2)$ $$f_{Z}(z) = \begin{cases} \frac{1}{2}e^{\frac{-z}{2}}, & z \ge 0 \\ 0, & z< 0 \end{cases}$$
      • 注 $Z\sim \chi^{2}(2)$ 相当于 $Z\sim E(\frac{1}{2})$
   3. $n\ge 3, \chi^{2}(n)$ 的概率密度草图
      • $\mathbb{P}\{\chi^{2}>\chi^{2}_{\alpha})(n)\} = \alpha$ 称 $\chi^{2}_{\alpha}(n)$ 为自由度为 $n$ 的 $\chi^{2}$ 上侧 $\alpha$ 分位点
      • 注 $\chi^{2}_{\alpha}(n)$ 从几何上看，指该点右侧面积为 $\alpha$

例题

1. 设 $X_{1},X_{2},X_{3},X_{4}$ 是来自正态总体 $\mathcal{N}(0,2^{2})$ 的简单随机样本，$X= a(X_{1}- 2X_{2})^{2} + b(3X_{3} -4X_{4})^{2}, a\ne 0, b\ne 0$，当$a=$        ，$b=$        时，$X$服从$\chi^{2}$分布，其自由度为         $$\begin{array}{ll} \because & X_{i}\sim \mathcal{N}(0,2^{2}),\quad i = 1,2,3,4 \\ \therefore & X_{1}- 2X_{2} \sim \mathcal{N}(0, 2^{2} + 2^{2}\times 2^{2} = 20) \\ & 3X_{3}- 4X_{4} \sim \mathcal{N}(0, 3^{2}\times 2^{2} + 4^{2}\times 2^{2} = 100) \\ \because & X = a(X_{1}- 2X_{2})^{2} + b(3X_{3} - 4X_{4})^{2}\sim \chi^{2} \\ \therefore & \sqrt{a}(X_{1}-2X_{2}) \sim \mathcal{N}(0,1) \\ & \sqrt{b}(3X_{3}-4X_{4}) \sim \mathcal{N}(0,1) \\ \therefore & a = \frac{1}{D(X_{1} - 2X_{2})} = \frac{1}{20} \\ & b = \frac{1}{D(3X_{3} - 4X_{4})} = \frac{1}{100} \\ \therefore & X \sim \chi^{2}(2) \\\\ \therefore & \begin{cases} a = \frac{1}{20} \\ b = \frac{1}{100} \\ n = 2 \end{cases} \\ \end{array}$$

$t$ 分布

1. 构造 $X\sim \mathcal{N}(0,1), Y\sim \chi^{2}(n)$，且 $X,Y$ 独立，则 $$T = \frac{X}{\sqrt{Y/n}} \sim t(n)$$

2. 上侧$\alpha$分位点

   

   • $\mathbb{P}\{T>t_{\alpha}(n)\} = \alpha$，称 $t_{\alpha}(n)$ 为 $t(n)$ 的上侧 $\alpha$ 分位点

例题

1. 设随机变量 $X$ 和 $Y$ 相互独立，且都服从正态分布 $\mathcal{N}(0,3^{2})$，而 $X_{1}, X_{2},X_{3},\cdots,X_{9}$ 和 $Y_{1}, Y_{2},\cdots ,Y_{9}$ 分别是来自总体 $X$ 和 $Y$ 的简单随机样本，则统计量 $\displaystyle U = \frac{X_{1}+X_{2}+\cdots+X_{9}}{\sqrt{Y_{1}^{2} + Y_{2}^{2}+ \cdots Y_{9}^{2}}}$ 服从        分布，自由度为         $$\begin{array}{ll} & X_{1}+X_{2}+\cdots + X_{9}\sim \mathcal{N}(0, 9^{2}) \\ \therefore & \frac{1}{9}(X_{1} + X_{2} + \cdots + X_{9}) \sim \mathcal{N}(0,1) \\ \because & \frac{Y}{3}\sim \mathcal{N}(0,1) \\ \therefore & \frac{1}{9}(Y_{1}^{2} + Y_{2}^{2}+\cdots+Y_{9}^{2})\sim\chi^{2}(9) \\ \therefore & U = \frac{\frac{1}{9}(X_{1}+X_{2}+\cdots+X_{9})}{\sqrt{\frac{1}{9\times 9}(Y_{1}^{2} + Y_{2}^{2}+\cdots+Y_{9}^{2})}} \sim t(9) \\\\ \therefore & U \text{服从} t \text{分布，自由度为}9 \\ \end{array}$$

$F$ 分布

1. 构造 设 $X\sim \chi^{2}(n_{1}), Y\sim \chi^{2}(n_{2})$且 $X,Y$ 独立，则 $$F = \frac{X/n_{1}}{Y/n_{2}}\sim F(n_{1},n_{2})$$
2. 性质
   1. $F\sim F(n_{1},n_{2})$，则 $\frac{1}{F}\sim F(n_{2},n_{1})$
   2. $T\sim t(n)$，则 $T^{2} = (\frac{X}{\sqrt{Y/n}})^{2} = \frac{X^{2}/1}{Y/n}\sim F(1,n)\Rightarrow \frac{1}{T^{2}}\sim F(n,1)$

3. 上侧 $\alpha$ 分位点

   

   • $\mathbb{P}\{F>F_{\alpha}(n_{1},n_{2})\} = \alpha$

例题

1. $(X_{1},X_{2},X_{3},X_{4})$ 为来自总体 $X\sim \mathcal{N}(0,1)$ 的一个样本，指出下列随机变量服从的分布
   1. $X_{1}^{2}$
   2. $X_{1}^{2} + X_{2}^{2}$
   3. $\frac{(X_{2}+X_{3})^{2}}{2}$
   4. $\frac{\sqrt{2}X_{1}}{\sqrt{X_{2}^{2}+X_{3}^{2}}}$
   5. $\frac{\sqrt{2}X_{1}}{\vert X_{2}+ X_{3} \vert}$
   6. $\frac{2X_{1}^{2}}{(X_{2}+X_{3})^{2}}$
   7. $\frac{X_{1}^{2}+X_{2}^{2}}{X_{3}^{2}+X_{4}^{2}}$
   8. $\frac{2X_{1}^{2}}{X_{2}^{2} + X_{3}^{2}}$ $$\begin{array}{ll} & X_{1}^{2}\sim \chi^{2}(1) \\\\ & X_{1}^{2} + X_{2}^{2} \sim \chi^{2}(2) \\\\ \because & \frac{X_{2}+X_{3}}{\sqrt{2}} \sim \mathcal{N}(0,1) \\ \therefore & \frac{(X_{2}+X_{3})^{2}}{2} \sim \chi^{2}(1) \\\\ & \frac{\sqrt{2}X_{1}}{\sqrt{X_{2}^{2}+X_{3}^{2}}} = \frac{X_{1}}{\sqrt{(X_{2}^{2}+X_{3}^{2})/2}} \sim t(2) \\\\ & \frac{\sqrt{2}X_{1}}{\vert X_{2} + X_{3} \vert} = \frac{\sqrt{2}X_{1}}{\sqrt{(X_{2}+X_{3})^{2}}} = \frac{X_{1}}{\sqrt{(X_{2}+X_{3})^{2}/2}} \sim t(1) \\\\ & \frac{2X_{1}^{2}}{(X_{2}+X_{3})^{2}} = \frac{X_{1}^{2}}{(X_{2}+X_{3})^{2}/2} \sim F(1,1) \\\\ & \frac{X_{1}^{2} + X_{2}^{2}}{X_{3}^{2} + X_{4}^{2}} = \frac{(X_{1}^{2}+X_{2}^{2})/2}{(X_{3}^{2}+X_{4}^{2})/2}\sim F(2,2) \\\\ & \frac{2X_{1}^{2}}{X_{2}^{2}+X_{3}^{2}} = \frac{X_{1}^{2}}{(X_{2}^{2}+X_{3}^{2})/2} \sim F(1,2) \end{array}$$