几道题

写在前面

一些注意事项.

期望的线性性
$E (\sum_{i = 1}^{n} X_{i}) = \sum_{i = 1}^{n} E (X_{i})$
$X_i$ 之间的关联附加任何限制关系，常用；
随机向量函数的期望
$\begin{matrix} E (g (X)) = \int_{R} g (x) f (x) d x, \\ E (g (X, Y)) = \int_{R^{2}} g (x, y) f (x, y) d x d y \end{matrix}$
$f$ 为分布函数.
$E(g(X,Y))$ 的情形，理论上我们先通过随机向量函数的分布
$F_{Z} (z) = P (Z ⩽ z) = P (g (X, Y) ⩽ z) = \iint_{g (x, y) ⩽ z} f (x, y) d x d y$
$Z=g(X,Y)$ 的分布，然后再通过期望的定义
$E (Z) = \int_{R} z d F_{Z} (z)$
$Z$ 的期望，但这里的两次积分往往都比较复杂，尤其当分布比较难算的时候这个方法就不适用了. 这里就可以直接利用随机向量函数的期望直接求解.
$0\leqslant x\leqslant y$ ，则
$\int_{0}^{+ \infty} d x \int_{x}^{+ \infty} x y f (x, y) d y = \int_{0}^{+ \infty} d y \int_{0}^{y} x y f (x, y) d x$
（2022 春二.（3））.
$\Gamma$ 函数的一些公式：
- $\int_{0}^{+ \infty} x^{α} e^{- β x^{2}} d x = \frac{Γ ((α + 1) / 2)}{2 β^{(α + 1) / 2}};$
- $\Gamma(x+1)=x\Gamma(x),x>0$ . 特别地，
  $\begin{matrix} Γ (1) = 1, Γ (\frac{1}{2}) = \sqrt{π}; \\ Γ (n) = (n - 1)!, Γ (n + \frac{1}{2}) = \frac{(2 n - 1)!!}{2^{n}} \sqrt{π}, n > 0. \end{matrix}$
$X_1,\cdots,X_n$ $c_1,\cdots,c_n$ $n$ 个常数，则
$Var (\sum_{i = 1}^{n} c_{i} X_{i}) = \sum_{i = 1}^{n} c_{i}^{2} Var (X_{i}) .$
$X,Y$ $\mathrm{Var}(X-Y)=\mathrm{Var}(X)+\mathrm{Var}(Y)$ ！
$\chi^2_n$ ：
- $X\sim \chi^2_n$ $E(X)=n,\mathrm{Var}(X)=2n$ ；
- $X\sim \chi^2_m,Y\sim\chi^2_n$ $X,Y$ 独立 $Z=X+Y\sim \chi^2_{m+n}$ .
$t_n$ ：
- $T\sim t_n$ $E(T)=0,n\geqslant 2$ $\mathrm{Var}(T)=n/(n-2),n\geqslant 3$ ；
- $lim_{n \to \infty} f_{n} (t) = φ (t) .$
$F_{m,n}$ ：
- $Z\sim F_{m,n},1/Z\sim F_{n,m}$ ；
- $F_{m,n}(1-\alpha)=1/F_{n,m}(\alpha)$ .
抽样分布的其他结论：
- 独立同分布的随机正态变量的线性组合服从正态分布，
  $T = \sum_{i = 1}^{n} c_{i} X_{i} \sim N (μ \sum_{i = 1}^{n} c_{i}, σ^{2} \sum_{i = 1}^{n} c_{i}^{2}) .$
  $T=\overline X$ 为样本均值时，有
  $\overset{―}{X} \sim N (μ, \frac{σ^{2}}{n});$
- $\frac{(n - 1) S^{2}}{σ^{2}} \sim χ_{n - 1}^{2};$
- $\overline X$ $S^2$ 相互独立；
- $\frac{\sqrt{n} (\overset{―}{X} - μ)}{S} \sim t_{n - 1} .$
- （补充，与上一条对比）林德伯格-莱维中心极限定理，
  $\frac{\sqrt{n} (\overset{―}{X} - μ)}{σ} \overset{L}{\to} N (0, 1) .$

题目解答

2022 秋

$A=\set{两个孩子中有一个在春季出生的男孩}$ $B=\set{两个孩子都是男孩}$ ，则

\begin{matrix} P (A) = \frac{(\binom{2}{1}) 7 + 1}{64} = \frac{15}{64}, \\ P (A B) = \frac{(\binom{2}{1}) 3 + 1}{64} = \frac{7}{64}, \end{matrix}

所以

P (B | A) = \frac{P (A B)}{P (A)} = \frac{7}{15} .

A：
$\int_{- \infty}^{+ \infty} (f_{1} (x) + f_{2} (x)) d x = \int_{- \infty}^{+ \infty} f_{1} (x) d x + \int_{- \infty}^{+ \infty} f_{2} (x) d x = 2 \neq 1;$
$f_1(x)=f_2(x)=\varphi(x)=\dfrac{1}{\sqrt{2\pi}}\mathrm e^{-x^2/2}$ ，则
$\begin{matrix} g (x) = f_{1} (x) f_{2} (x) = \frac{1}{2 π} e^{- x^{2}}, \\ \int_{- \infty}^{+ \infty} g (x) d x = \frac{1}{2 π} \int_{- \infty}^{+ \infty} e^{- x^{2}} d x = \frac{1}{2 π} \cdot \sqrt{π} = \frac{1}{2 \sqrt{π}} \neq 1; \end{matrix}$
$f_1(x)+f_2(x)$ 为概率密度函数，但由 A 它不是；
$G(x)=F_1(x)F_2(x)$ $g(x)=G'(x)=f_1(x)F_2(x)+F_1(x)f_2(x)$ .
$\begin{aligned} \int_{R} g (x) d x & = \int_{R} (f_{1} (x) F_{2} (x) + F_{1} (x) f_{2} (x)) d x \\ = \int_{R} f_{1} F_{2} d x + \int_{R} F_{1} f_{2} d x \\ = \int_{R} F_{2} d F_{1} + \int_{R} F_{1} d F_{2} \\ = {- \int_{R} F_{1} d F_{2} + 2 F_{1} F_{2} |}_{- \infty}^{+ \infty} - \int_{R} F_{2} d F_{1} \\ = 2 - (\int_{R} F_{2} d F_{1} + \int_{R} F_{1} d F_{2}) \\ = 2 - \int_{R} g (x) d x, \end{aligned}$
从而
$\int_{R} g (x) d x = 1.$
$\binom{10}{3}=120$ 个三角形.
$X_i=\{第\ i\ 个三角形存在\},i=1,2,\cdots,120$ $X_i\sim B(1,1/8)$ .
$X = \sum_{i = 1}^{120} X_{i} \sim B (120, 1 / 8)$
$E(X)=120\times1/8=15$ .
上面的方法或许会有争议，其实这里可以直接用期望的可加性
$E (X) = E (\sum_{i = 1}^{n} X_{i}) = \sum_{i = 1}^{n} E (X_{i})$
求解.

$X,Y$ $f(x,y)=f_1(x)f_2(y)=\varphi(x)\varphi(y)$ .
$(X,Y)\sim N(0,0,1,1,0)$ .
B：
C：
D：
$Z_1=X+Y,Z_2=X-Y$ .
$\begin{matrix} u = g_{1} (x, y) = x + y, v = g_{2} (x, y) = x - y . \\ x = φ_{1} (u, v) = \frac{u + v}{2}, y = φ_{2} (u, v) = \frac{u - v}{2}, \\ | \frac{\partial (u, v)}{\partial (x, y)} | = 2, | \frac{\partial (x, y)}{\partial (u, v)} | = \frac{1}{2} . \end{matrix}$ $\begin{aligned} f_{Z} (z_{1}, z_{2}) & = f (φ_{1} (z_{1}, z_{2}), φ_{2} (z_{1}, z_{2})) {| \frac{\partial (φ_{1} (u, v), φ_{2} (u, v))}{\partial (u, v)} |}_{(u, v) = (z_{1}, z_{2})} \\ = \frac{1}{2} φ (φ_{1} (z_{1}, z_{2})) φ (φ_{2} (z_{1}, z_{2})) \\ = \frac{1}{2} φ (\frac{z_{1} + z_{2}}{2}) φ (\frac{z_{1} - z_{2}}{2}) \\ = \frac{1}{2} \cdot \frac{1}{\sqrt{2 π}} e^{- ((z_{1} + z_{2}) / 2)^{2} / 2} \cdot \frac{1}{\sqrt{2 π}} e^{- ((z_{1} - z_{2}) / 2)^{2} / 2} \\ = \frac{1}{4 π} e^{- (z_{1}^{2} + z_{2}^{2}) / 4} \\ = \frac{1}{4 π} e^{- z_{1}^{2} / 4} \cdot e^{- z_{2}^{2} / 4}, \end{aligned}$
$f_\boldsymbol Z(z_1,z_2)$ $Z_1$ $Z_2$ 相互独立.

根据熵的定义

H (X) = - \int_{R} f (x) \ln f (x) d x

$X$ $\lambda$ 的指数分布，即

f (x) = λ e^{- λ x} I_{(0, + \infty)} (x),

得

\begin{aligned} H (X) & = - \int_{0}^{+ \infty} λ e^{- λ x} \ln (λ e^{- λ x}) d x \\ = - \int_{0}^{+ \infty} λ e^{- λ x} (\ln λ - λ x) d x \\ = - λ \ln λ \int_{0}^{+ \infty} e^{- λ x} d x + λ^{2} \int_{0}^{+ \infty} x e^{- λ x} d x \\ = - λ \ln λ \cdot {\frac{e^{- λ x}}{- λ} |}_{0}^{+ \infty} + λ^{2} \cdot \frac{1}{- λ} (x e^{- λ x} |_{0}^{+ \infty} - \int_{0}^{\infty} e^{- λ x} d x) \\ = - λ \ln λ (0 - \frac{1}{- λ}) - λ (0 - 0 - (0 - \frac{1}{- λ})) \\ = - \ln λ + 1. \end{aligned}

$X\sim P(\lambda)$ $E(X)=\lambda=3$ $\mathrm{Var}(X)=\lambda=3$ .

$Z=X-Y$ ，则

\begin{matrix} E (Z) = E (X - Y) = E (X) - E (Y) = 0, \\ Var (Z) = Var (X - Y) = Var (X) + Var (Y) = 4, \end{matrix}

$X,Y$ 相互独立这一条件. 由切比雪夫不等式，

\begin{matrix} P (| Z - E (Z) | ⩾ ε) = \frac{Var (Z)}{ε^{2}}, \\ P (| X - Y | ⩾ ε) = \frac{4}{ε^{2}} . \end{matrix}

$\varepsilon=3$ ，解得

\begin{matrix} P (| X - Y | ⩾ 3) = \frac{4}{9}, \\ P (| X - Y | < 3) = \frac{5}{9}, \\ P (X - 3 < Y < X + 3) = \frac{5}{9} . \end{matrix}

$(X_i-\mu)\sim N(0,1)$ ，所以
$\sum_{i = 1}^{n} (X_{i} - μ)^{2} \sim χ_{n}^{2} .$
$(X_n-X_1)\sim N(0,2),\dfrac{X_n-X_1}{\sqrt 2}\sim N(0,1)$ $\left(\dfrac{X_n-X_1}{\sqrt 2}\right)^2=\dfrac12(X_n-X_1)^2\sim\chi_1^2$ .
C：
$\sum_{i = 1}^{n} {(X_{i} - \overset{―}{X})}^{2} = (n - 1) S^{2} \sim χ_{n - 1}^{2} .$
D：
$\begin{aligned} n {(\overset{―}{X} - μ)}^{2} & = n {(\frac{1}{n} \sum_{i = 1}^{n} X_{i} - μ)}^{2} \\ = n {(\frac{1}{n} (\sum_{i = 1}^{n} X_{i} - n μ))}^{2} \\ = n {(\frac{1}{n} \sum_{i = 1}^{n} (X_{i} - μ))}^{2} \\ = {(\frac{1}{\sqrt{n}} \sum_{i = 1}^{n} (X_{i} - μ))}^{2} . \end{aligned}$
其中
$\begin{matrix} (X_{i} - μ) \sim N (0, 1), \\ \sum_{i = 1}^{n} (X_{i} - μ) \sim N (0, n), \\ \frac{1}{\sqrt{n}} \sum_{i = 1}^{n} (X_{i} - μ) \sim N (0, 1), \end{matrix}$
$n\left(\overline X-\mu\right)^2\sim \chi_1^2$ .

$p$ $\alpha$ $p$ $H_0$ 有关，C 错误；

D 还不确定.

这是单个正态总体均值在方差已知情形下的第三个问题的检验.

当

| Z | = \frac{\sqrt{n} | \overset{―}{X} - μ_{0} |}{σ} = 2 | \overset{―}{X} - μ_{0} | > u_{α / 2}

$c=u_{\alpha/2}/2=u_{0.025}/2=1.96/2=0.98$ .

$r=1$ $p$ $k=3$ $k-r-1=1$ .

(1)

f (x, y) = 2 I_{G} (x, y),

$G=\set{(x,y)|0<x<y<1}$ .

(2)

$U\sim U(0,1)$ ，则其概率密度函数为

g_{U} (u) = \frac{1}{1 - 0} I_{(0, 1)} (u) = I_{(0, 1)} (u) .

分布函数为

\begin{matrix} G_{U} (u) = {\begin{cases} 0, & u ⩽ 0, \\ u, & 0 < u < 1, \\ 1, & u ⩾ 1. \end{cases} \end{matrix}

同理，

\begin{matrix} g_{V} (v) = I_{(0, 1)} (v), \\ G_{V} (v) = {\begin{cases} 0, & v ⩽ 0, \\ v, & 0 < v < 1, \\ 1, & v ⩾ 1. \end{cases} \end{matrix}

$W_1=\max\set{U,V}$ 的分布函数为

F_{max} (w) = G_{U} (w) G_{V} (w) = G_{U}^{2} (w),

从而其概率密度函数为

f_{max} (w) = 2 G_{U} (w) g_{U} (w) = 2 w I_{(0, 1)} (w) .

$Y$ 的概率密度函数为

f_{2} (y) = \int_{R} f (x, y) d x = \int_{0}^{y} 2 d x = 2 y, 0 < y < 1,

即

f_{2} (y) = 2 y I_{(0, 1)} (y) .

$W_2=\min\set{U,V}$ 的分布函数为

F_{min} (w) = 1 - (1 - G_{U} (w)) (1 - G_{V} (w)) = 1 - (1 - G_{U} (w))^{2},

从而其概率密度函数为

f_{min} (w) = 2 (1 - G_{U} (w)) g_{U} (w) = 2 (1 - w) I_{(0, 1)} (w) .

$X$ 的概率密度函数为

f_{1} (x) = \int_{R} f (x, y) d y = \int_{x}^{1} 2 d y = 2 (1 - x), 0 < x < 1,

即

f_{1} (x) = 2 (1 - x) I_{(0, 1)} (x) .

证毕.

(3)

f_{Y | X} (y | x) = \frac{f (x, y)}{f_{1} (x)} = \frac{1}{1 - x} I_{(x, 1)} (y),

$Y$ $X=x$ $(x,1)$ 上的均匀分布.

(4)

由

\begin{matrix} f_{1} (x) = 2 (1 - x) I_{(0, 1)} (x), \\ f_{2} (y) = 2 y I_{(0, 1)} (y) \end{matrix}

计算得

\begin{matrix} E (X) = \int_{0}^{1} x f_{1} (x) d x = \int_{0}^{1} 2 x (1 - x) d x = \frac{1}{3}, \\ E (Y) = \int_{0}^{1} y f_{2} (y) d y = \int_{0}^{1} 2 y^{2} d y = \frac{2}{3} . \end{matrix}

\begin{matrix} E (X^{2}) = \int_{0}^{1} x^{2} f_{1} (x) d x = \int_{0}^{1} 2 x^{2} (1 - x) d x = \frac{1}{6}, \\ E (Y^{2}) = \int_{0}^{1} y f_{2} (y) d y = \int_{0}^{1} 2 y^{3} d y = \frac{1}{2} . \end{matrix}

\begin{matrix} Var (X) = E (X^{2}) - (E (X))^{2} = \frac{1}{18}, \\ Var (Y) = E (Y^{2}) - (E (Y))^{2} = \frac{1}{18} . \end{matrix}

$Z=XY$ ，得

F_{Z} (z) = \iint_{x y ⩽ z} f (x, y) d x d y = \iint_{y ⩽ z / x} f (x, y) d x d y = 2 | D |,

$D$ $y=x$ $x=0$ $y=1$ $y=z/x$ $0<z<1$ . 计算得

\begin{matrix} F_{Z} (z) = z (1 - \ln z), 0 < z < 1, \\ f_{Z} (z) = - \ln z, 0 < z < 1. \end{matrix}

期望

E (Z) = \int_{0}^{1} z d F_{Z} (z) = z F_{z} (z) |_{0}^{1} - \int_{0}^{1} F_{z} (z) d z = \frac{1}{2} - \int_{0}^{1} z d F_{Z} (z),

E (Z) = \int_{0}^{1} z d F_{Z} (z) = \frac{1}{4} .

补充：这里应当直接考虑随机向量函数的期望进行计算：

E (X Y) = \iint_{G} x y f (x, y) d x d y = 2 \int_{0}^{1} d y \int_{0}^{y} x y d x = \frac{1}{4} .

所以，

ρ_{X Y} = \frac{Cov (X, Y)}{\sqrt{Var (X)} \sqrt{Var (Y)}} = \frac{E (X Y) - E (X) E (Y)}{\sqrt{Var (X)} \sqrt{Var (Y)}} = \frac{1 / 4 - 1 / 3 \times 2 / 3}{1 / 18} = \frac{1}{2} .

(1)

F_{U_{i}} (x) = x, 0 ⩽ x ⩽ 1.

P (M ⩽ x | Y = n) = (F_{U_{i}} (x))^{n} = x^{n} .

(2)

$\forall x\in[0,1]$ ，

\begin{aligned} P (M ⩽ x) & = \sum_{n = 1}^{\infty} P (M ⩽ x | Y = n) P (Y = n) \\ = \sum_{n = 1}^{\infty} \frac{x^{n}}{(e - 1) n!} \\ = \frac{1}{e - 1} (\sum_{n = 0}^{\infty} \frac{x^{n}}{n!} - 1) \\ = \frac{e^{x} - 1}{e - 1} . \end{aligned}

求导得

f_{M} (x) = \frac{e^{x}}{e - 1} I_{(0, 1)} (x) .

(3)

不会.

(1)

\begin{matrix} f_{1} (x) = \frac{1}{\sqrt{2 π} σ} e^{- \frac{(x - μ)^{2}}{2 σ^{2}}}, \\ f_{2} (y) = \frac{1}{2 \sqrt{π} σ} e^{- \frac{(x - μ)^{2}}{4 σ^{2}}} . \end{matrix}

似然函数

\begin{aligned} L & = \prod_{i = 1}^{m} f_{1} (x_{i}) \prod_{j = 1}^{n} f_{2} (y_{j}) \\ = \frac{1}{\sqrt{(2 π)^{m} (4 π)^{n}} σ^{m + n}} \exp {- \frac{1}{2 σ^{2}} \sum_{i = 1}^{m} (x_{i} - μ)^{2} - \frac{1}{4 σ^{2}} \sum_{j = 1}^{n} (y_{j} - μ)^{2}} \\ = \frac{1}{2^{m / 2 + n} π^{(m + n) / 2} σ^{m + n}} \exp {- \frac{1}{4 σ^{2}} (2 \sum_{i = 1}^{m} (x_{i} - μ)^{2} + \sum_{j = 1}^{n} (y_{j} - μ)^{2})}, \end{aligned}

\begin{aligned} ℓ = \ln L & = \ln σ^{- (m + n)} - \frac{1}{4 σ^{2}} (2 \sum_{i = 1}^{m} (x_{i} - μ)^{2} + \sum_{j = 1}^{n} (y_{j} - μ)^{2}) + \dots \\ = - \frac{m + n}{2} \ln σ^{2} - \frac{1}{4 σ^{2}} (2 \sum_{i = 1}^{m} (x_{i} - μ)^{2} + \sum_{j = 1}^{n} (y_{j} - μ)^{2}) + \dots . \end{aligned}

\frac{\partial ℓ}{\partial σ^{2}} = - \frac{m + n}{2 σ^{2}} + \frac{1}{4 (σ^{2})^{2}} (2 \sum_{i = 1}^{m} (x_{i} - μ)^{2} + \sum_{j = 1}^{n} (y_{j} - μ)^{2}) = 0,

解得

{\hat{σ}}^{2} = \frac{1}{2 (m + n)} (2 \sum_{i = 1}^{m} (X_{i} - μ)^{2} + \sum_{j = 1}^{n} (Y_{j} - μ)^{2}) .

(2)

{\hat{σ}}^{2} (X_{1}, \dots, X_{m}, Y_{1}, \dots, Y_{n}) = \frac{1}{m + n} \sum_{i = 1}^{m} (X_{i} - μ)^{2} + \frac{1}{2 (m + n)} \sum_{j = 1}^{n} (Y_{j} - μ)^{2},

\begin{aligned} E ({\hat{σ}}^{2}) & = \frac{1}{m + n} \cdot σ^{2} \cdot E (\sum_{i = 1}^{m} {(\frac{X_{i} - μ}{σ})}^{2}) + \frac{1}{2 (m + n)} \cdot 2 σ^{2} \cdot E (\sum_{j = 1}^{n} {(\frac{Y_{j} - μ}{\sqrt{2} σ})}^{2}) \\ = \frac{σ^{2}}{m + n} \cdot m + \frac{σ^{2}}{(m + n)} \cdot n \\ = σ^{2}, \end{aligned}

从而是无偏估计.

(3)

T = \frac{(m + n) {\hat{σ}}^{2}}{σ^{2}} = \sum_{i = 1}^{m} {(\frac{X_{i} - μ}{σ})}^{2} + \sum_{j = 1}^{n} {(\frac{Y_{j} - μ}{\sqrt{2} σ})}^{2} \sim χ_{m + n}^{2} .

解不等式

χ_{m + n}^{2} (1 - α / 2) ⩽ \frac{(m + n) {\hat{σ}}^{2}}{σ^{2}} ⩽ χ_{m + n}^{2} (α / 2)

得

σ^{2} \in [\frac{(m + n) {\hat{σ}}^{2}}{χ_{m + n}^{2} (α / 2)}, \frac{(m + n) {\hat{σ}}^{2}}{χ_{m + n}^{2} (1 - α / 2)}] = [\frac{(m + n) {\hat{σ}}^{2}}{χ_{m + n}^{2} (0.025)}, \frac{(m + n) {\hat{σ}}^{2}}{χ_{m + n}^{2} (0.975)}] .

$\boldsymbol X=(X_1,X_2,\cdots,X_m)$ $\boldsymbol Y=(Y_1,Y_2,\cdots,Y_n)$ $m=13,n=10$ $\overline x=14.95,s_X^2=6.85^2;\overline y=22.29,s_Y^2=5.32^2$ ，则

s_{T}^{2} = \frac{(m - 1) s_{X}^{2} + (n - 1) s_{Y}^{2}}{m + n - 2} = \frac{12 \times {6.85}^{2} + 9 \times {5.32}^{2}}{21} = {6.24}^{2} .

(1)

这是两个正态总体方差比的检验，检验问题为

H_{0} : σ_{X}^{2} / σ_{Y}^{2} = 1 \leftrightarrow H_{1} : σ_{X}^{2} / σ_{Y}^{2} \neq 1.

检验统计量为

F = \frac{S_{X}^{2}}{S_{Y}^{2}},

$f=s_X^2/s_Y^2=6.85^2/5.32^2=1.65$ .

$F_{m-1.n-1}(1-\alpha/2)=F_{12,9}(0.975)=(F_{9,12}(0.025))^{-1}=1/3.87=0.26$ $F_{m-1,n-1}(\alpha/2)=F_{12,9}(0.025)=3.44$ .

$0.26\leqslant 1.65\leqslant 3.44$ $H_0$ ，即可以认为体脂率的方差没有性别差异.

(2)

这是两个正态总体均值差的成组比较，检验问题为

H_{0} : μ_{X} - μ_{Y} ⩾ 0 \leftrightarrow H_{1} : μ_{X} - μ_{Y} < 0.

由于方差未知，取检验统计量

T = \sqrt{\frac{m n}{m + n}} \frac{\overset{―}{X} - \overset{―}{Y}}{S_{T}},

其观测值为

t = \sqrt{\frac{m n}{m + n}} \frac{\overset{―}{x} - \overset{―}{y}}{s_{T}} = \sqrt{\frac{13 \times 10}{13 + 10}} \frac{14.95 - 22.29}{6.24} = - 2.83,

$-t_{m+n-2}(\alpha)=-t_{21}(0.05)=-1.72$ .

$-2.83< -1.72$ $H_0$ ，即可以认为成年女性体脂率显著大于成年男性体脂率.

2022 春

$A=\{第\ 1\ 次抛掷结果为正面\}$ $B=\{抛掷\ n\ 次硬币出现\ k\ 次正面\}$ ，则

\begin{matrix} P (A B) = \frac{1}{2} \cdot (\binom{n - 1}{k - 1}) {(\frac{1}{2})}^{k - 1} {(\frac{1}{2})}^{n - k} = (\binom{n - 1}{k - 1}) 2^{- n}, \\ P (B) = (\binom{n}{k}) {(\frac{1}{2})}^{k} {(\frac{1}{2})}^{n - k} = (\binom{n}{k}) 2^{- n}, \end{matrix}

P (A | B) = \frac{P (A B)}{P (B)} = \frac{(\binom{n - 1}{k - 1})}{(\binom{n}{k})} = \frac{k}{n} .

$n$ $k$ $n-k$ $n$ $k/n$ . 这个结果应当是与抛掷硬币出现正面的概率无关的.

$Z=\sqrt{X^2+Y^2}$ ，则

\begin{aligned} F_{Z} (z) & = P (Z ⩽ z) = P (X^{2} + Y^{2} ⩽ z^{2}) \\ = \iint_{x^{2} + y^{2} ⩽ z^{2}} f (x, y) d x d y \\ = \iint_{x^{2} + y^{2} ⩽ z^{2}} φ (x) φ (y) d x d y \\ = \frac{1}{2 π} \iint_{x^{2} + y^{2} ⩽ z^{2}} e^{- (x^{2} + y^{2}) / 2} d x d y \\ = \frac{1}{2 π} \underset{0 ⩽ θ ⩽ 2 π}{\iint_{0 ⩽ r ⩽ z}} e^{- r^{2} / 2} r d r d θ \\ = - \frac{1}{2 π} \int_{0}^{2 π} d θ \int_{0}^{z} d (e^{- r^{2} / 2}) \\ = - \frac{1}{2 π} \cdot 2 π \cdot (e^{- z^{2} / 2} - 1) \\ = 1 - e^{- z^{2} / 2}, 0 ⩽ z < + \infty . \end{aligned}

期望为

\begin{aligned} E (Z) & = \int_{0}^{+ \infty} z d F_{Z} (z) \\ = \int_{0}^{+ \infty} z^{2} e^{- z^{2} / 2} d z \\ = - \int_{0}^{+ \infty} z d (e^{- z^{2} / 2}) \\ = - ({z e^{- z^{2} / 2} |}_{0}^{+ \infty} - \int_{0}^{+ \infty} e^{- z^{2} / 2} d z) \\ = \int_{0}^{+ \infty} e^{- z^{2} / 2} d z \\ = \frac{1}{2} \int_{- \infty}^{+ \infty} e^{- z^{2} / 2} d z \\ = \frac{1}{2} \cdot \sqrt{2 π} \int_{- \infty}^{+ \infty} \frac{1}{\sqrt{2 π}} e^{- z^{2} / 2} d z \\ = \frac{\sqrt{2 π}}{2} . \end{aligned}

利用随机向量函数的期望直接求解：

\begin{aligned} E (\sqrt{X^{2} + Y^{2}}) & = \iint_{R^{2}} \sqrt{x^{2} + y^{2}} f (x, y) d x d y \\ = \iint_{R^{2}} \sqrt{x^{2} + y^{2}} φ (x) φ (y) d x d y \\ = \frac{1}{2 π} \iint_{R^{2}} \sqrt{x^{2} + y^{2}} e^{- (x^{2} + y^{2}) / 2} d x d y \\ = \frac{1}{2 π} \int_{0}^{2 π} d θ \int_{0}^{+ \infty} r e^{- r^{2} / 2} r d r \\ = \frac{1}{2 π} \cdot 2 π \int_{0}^{+ \infty} r^{2} e^{- r^{2} / 2} d r \\ = \frac{Γ ((2 + 1) / 2)}{2 \cdot (1 / 2)^{(2 + 1) / 2}} \\ = \sqrt{2} Γ (1 + 1 / 2) \\ = \sqrt{2} \cdot \frac{1}{2} \sqrt{π} \\ = \frac{\sqrt{2 π}}{2} . \end{aligned}

注：这里用到了

$\int_{0}^{+ \infty} x^{α} e^{- β x^{2}} d x = \frac{Γ ((α + 1) / 2)}{2 β^{(α + 1) / 2}};$
$Γ (n + \frac{1}{2}) = \frac{(2 n - 1)!!}{2^{n}} \sqrt{π} .$

由全期望公式（平滑公式）得

E (E (X | Y)) = E (g (Y)) = E (Y^{2}) = E (X) .

$E(Y)=2,\mathrm{Var}(Y)=2$ ，得

E (X) = E (Y^{2}) = Var (Y) + (E (Y))^{2} = 6.

$BQ=x,0\leqslant x\leqslant BC$ $Y=\set{直线\ PQ\ 与边\ AB\ 相交}$ ，则

P (Y | X = x) = \frac{S_{A B Q}}{S_{A B C}} = \frac{x}{B C},

$X$ $(0,BC)$ 上的均匀分布，即

P (X = x) = \frac{1}{B C} .

所以由全概率公式，

P (Y) = \sum_{x} P (Y | X = x) P (X = x) = \sum_{x} \frac{x}{B C^{2}} = \int_{0}^{B C} \frac{x}{B C^{2}} d x = \frac{1}{2} .

$X+Y,X-Y\sim N(0,2)$ ，从而

\frac{X + Y}{\sqrt{2}}, \frac{X - Y}{\sqrt{2}} \sim N (0, 1),

从而

\frac{{(\frac{X + Y}{\sqrt{2}})}^{2} / 1}{{(\frac{X - Y}{\sqrt{2}})}^{2} / 1} = \frac{(X + Y)^{2}}{(X - Y)^{2}} \sim F_{1, 1} .

$X_i$ $N(0,1)$ ；
$X_i/\sigma \sim N(0,1)$ $(X_i/\sigma)^2\sim \chi^2_1$ ；另一方面，
$\frac{(n - 1) S^{2}}{σ^{2}} \sim χ_{n - 1}^{2} .$
$\chi^2_n$ ；
C：一眼错.
$X_i/\sigma\sim N(0,1)$ ，记
$Y = \sum_{i = 1}^{n} \frac{X_{i}}{σ} \sim N (0, n) .$
则
$\frac{1}{n} Y^{2} = {(\frac{Y}{\sqrt{n}})}^{2} \sim χ_{1}^{2} .$
$\overline X$ $S^2$ 相互独立，
$\frac{1}{n} Y^{2} = \frac{{\overset{―}{X}}^{2}}{n σ^{2}}$
和
$Z = \frac{(n - 1) S^{2}}{σ^{2}} \sim χ_{n - 1}^{2}$
$\chi^2$ $n$ $1+(n-1)=n$ $\chi^2$ 分布.
$\begin{matrix} E ({\hat{μ}}_{1}) = E ({\hat{μ}}_{2}) = μ, \\ Var ({\hat{μ}}_{1}) = 1, Var ({\hat{μ}}_{2}) = 2, \\ \Rightarrow E ({\hat{μ}}_{1}^{2}) = Var ({\hat{μ}}_{1}) + (E ({\hat{μ}}_{1}))^{2} = μ^{2} + 1, \\ E ({\hat{μ}}_{2}^{2}) = Var ({\hat{μ}}_{2}) + (E ({\hat{μ}}_{2}))^{2} = μ^{2} + 2. \end{matrix}$ $\begin{aligned} ρ_{12} & = \frac{Cov ({\hat{μ}}_{1}, {\hat{μ}}_{2})}{\sqrt{Var ({\hat{μ}}_{1})} \sqrt{Var ({\hat{μ}}_{2})}} \\ = \frac{E ({\hat{μ}}_{1} {\hat{μ}}_{2}) - E ({\hat{μ}}_{1}) E ({\hat{μ}}_{2})}{\sqrt{2}} \\ = \frac{E ({\hat{μ}}_{1} {\hat{μ}}_{2}) - μ^{2}}{\sqrt{2}} \\ = \frac{1}{2 \sqrt{2}} \\ \Rightarrow E ({\hat{μ}}_{1} {\hat{μ}}_{2}) & = μ^{2} + \frac{1}{2} . \end{aligned}$ $\begin{matrix} E (α {\hat{μ}}_{1} + β {\hat{μ}}_{2}) = α E ({\hat{μ}}_{1}) + β E ({\hat{μ}}_{2}) = (α + β) μ = μ \\ \Rightarrow α + β = 1. \end{matrix}$ $\begin{aligned} Var (α {\hat{μ}}_{1} + β {\hat{μ}}_{2}) & = E ((α {\hat{μ}}_{1} + β {\hat{μ}}_{2})^{2}) - (E (α {\hat{μ}}_{1} + β {\hat{μ}}_{2}))^{2} \\ = E (α^{2} {\hat{μ}}_{1}^{2} + 2 α β {\hat{μ}}_{1} {\hat{μ}}_{2} + β^{2} {\hat{μ}}_{2}^{2}) - μ^{2} \\ = α^{2} E ({\hat{μ}}_{1}^{2}) + 2 α β E ({\hat{μ}}_{1} {\hat{μ}}_{2}) + β^{2} E ({\hat{μ}}_{2}^{2}) - μ^{2} \\ = α^{2} (μ^{2} + 1) + 2 α β (μ^{2} + \frac{1}{2}) + β^{2} (μ^{2} + 2) - μ^{2} \\ = (α^{2} + 2 α β + β^{2} - 1) μ^{2} + (α^{2} + α β + 2 β^{2}) \\ = ((α + β)^{2} - 1) μ^{2} + (α^{2} + α β + 2 β^{2}) \\ = α^{2} + α β + 2 β^{2} \\ = (1 - β)^{2} + (1 - β) β + β^{2} \\ = 2 β^{2} - β + 1 \\ = 2 {(β - \frac{1}{4})}^{2} + \frac{7}{8} \\ ⩾ \frac{7}{8}, \end{aligned}$
$\beta=1/4,\alpha=1-\beta=3/4$ $\alpha\beta=3/16$ .
$\begin{matrix} {\hat{μ}}_{1} = \overset{―}{x} - \frac{s}{\sqrt{n}} t_{n - 1} (α / 2), \\ {\hat{μ}}_{2} = \overset{―}{x} + \frac{s}{\sqrt{n}} t_{n - 1} (α / 2), \\ {\hat{μ}}_{1}^{*} = \overset{―}{x} - \frac{s}{\sqrt{n}} t_{n - 1} (α), \\ {\hat{μ}}_{2}^{*} = \overset{―}{x} + \frac{s}{\sqrt{n}} t_{n - 1} (α) . \end{matrix}$
$t_{n-1}(\alpha/2)>t_{n-1}(\alpha)$ $\hat\mu_1<\hat\mu_1^*,\hat\mu_2>\hat\mu_2^*$ ，A、B 正确，这从直观上也是容易看出的；
$\begin{matrix} {\hat{μ}}_{2} - {\hat{μ}}_{1} = \frac{2 s}{\sqrt{n}} t_{n - 1} (α / 2), \\ {\hat{μ}}_{2}^{*} - {\hat{μ}}_{1}^{*} = \frac{2 s}{\sqrt{n}} t_{n - 1} (α), \end{matrix}$
显然两者不等，C 错误；
${\hat{μ}}_{2} - {\hat{μ}}_{2}^{*} = \frac{s}{\sqrt{n}} t_{n - 1} (α / 2) - \frac{s}{\sqrt{n}} t_{n - 1} (α) = {\hat{μ}}_{1}^{*} - {\hat{μ}}_{1},$
D 正确.
显著性水平越小，原假设被保护得越好，从而更不容易被拒绝，所以 A 正确.
(1)
$f_{2} (y) = \int_{0}^{y} f (x, y) d x = \frac{c}{6} y^{3} e^{- y}, 0 < y < + \infty .$ $\int_{0}^{+ \infty} f_{2} (y) d y = \frac{c}{6} \int_{0}^{+ \infty} y^{3} e^{- y} d y = \frac{c}{6} Γ (4) = c = 1 \Rightarrow c = 1.$
(2)
$f_{2} (y) = \frac{1}{6} y^{3} e^{- y} I_{(0, + \infty)} (y),$ $\begin{aligned} f_{1} (x) & = \int_{x}^{+ \infty} f (x, y) d y \\ = x \int_{x}^{+ \infty} (y - x) e^{- y} d y \\ = x e^{- x} \int_{x}^{+ \infty} (y - x) e^{- (y - x)} d y \\ = x e^{- x} \int_{0}^{+ \infty} u e^{- u} d u (u = y - x) \\ = x e^{- x} Γ (2) \\ = x e^{- x} I_{(0, + \infty)} (x) . \end{aligned}$ $\begin{matrix} f_{Y | X} (y | x) = \frac{f (x, y)}{f_{1} (x)} = (y - x) e^{- (y - x)}, 0 < x ⩽ y < + \infty; \\ f_{X | Y} (x | y) = \frac{f (x, y)}{f_{2} (y)} = \frac{6 x (y - x)}{y^{3}}, 0 < x ⩽ y < + \infty . \end{matrix}$
(3)
$\begin{aligned} E (X Y) & = \int_{0}^{+ \infty} d x \int_{x}^{+ \infty} x y f (x, y) d y \\ = \int_{0}^{+ \infty} d y \int_{0}^{y} x y f (x, y) d x \\ = \int_{0}^{+ \infty} d y \int_{0}^{y} x y \cdot x (y - x) e^{- y} d x \\ = \int_{0}^{+ \infty} y e^{- y} d y \int_{0}^{y} x^{2} (y - x) d x \\ = \int_{0}^{+ \infty} y e^{- y} d y (y \int_{0}^{y} x^{2} d x - \int_{0}^{y} x^{3} d x) \\ = \int_{0}^{+ \infty} y e^{- y} d y \cdot \frac{y^{4}}{12} \\ = \frac{1}{12} \int_{0}^{+ \infty} y^{5} e^{- y} d y \\ = \frac{Γ (6)}{12} \\ = 10. \end{aligned}$
这里在一开始用到了积分换序的技巧.
(4)
$\begin{matrix} E (X) = \int_{0}^{+ \infty} x f_{1} (x) d x = \int_{0}^{+ \infty} x^{2} e^{- x} d x = Γ (3) = 2, \\ E (X^{2}) = \int_{0}^{+ \infty} x^{2} f_{1} (x) d x = \int_{0}^{+ \infty} x^{3} e^{- x} d x = Γ (4) = 6, \\ E (Y) = \int_{0}^{+ \infty} y f_{2} (y) d y = \frac{1}{6} \int_{0}^{+ \infty} y^{4} e^{- y} d y = \frac{Γ (5)}{6} = 4, \\ E (Y^{2}) = \int_{0}^{+ \infty} y^{2} f_{2} (y) d y = \frac{1}{6} \int_{0}^{+ \infty} y^{5} e^{- y} d y = \frac{Γ (6)}{6} = 20, \end{matrix}$ $\begin{matrix} Var (X) = E (X^{2}) - (E (X))^{2} = 2, \\ Var (Y) = E (Y^{2}) - (E (Y))^{2} = 4. \end{matrix}$
综上，
$\begin{aligned} Corr (X, Y) & = \frac{Cov (X, Y)}{\sqrt{Var (X)} \sqrt{Var (Y)}} \\ = \frac{E (X Y) - E (X) E (Y)}{\sqrt{Var (X)} \sqrt{Var (Y)}} \\ = \frac{10 - 2 \times 4}{\sqrt{2} \sqrt{4}} \\ = \frac{1}{\sqrt{2}} . \end{aligned}$

(1)

\begin{matrix} f_{1} (x) = \frac{1}{1 - 0} I_{(0, 1)} (x) = I_{(0, 1)} (x); \end{matrix}

$Y\sim Exp(\lambda),E(Y)=\lambda^{-1}=2$ $\lambda=1/2$ .

f_{2} (y) = λ e^{- λ y} I_{(0, + \infty)} (y) = e^{- y / 2} / 2 \cdot I_{(0, + \infty)} (y) .

f (x, y) = f_{1} (x) f_{2} (y) = \frac{1}{2} e^{- y / 2} I_{G} (x, y),

$G=\set{(x,y)|0<x<1,y>0}$ .

(2)

$Z=g(X,Y)=X+Y$ .

$z\geqslant 1$ $Z$ 的分布函数为

\begin{aligned} F_{Z} (z) & = P (Z ⩽ z) = P (X + Y ⩽ z) \\ = \iint_{x + y ⩽ z} f (x, y) d x d y \\ = \int_{0}^{1} d x \int_{0}^{z - x} \frac{1}{2} e^{- y / 2} d y \\ = \frac{1}{2} \int_{0}^{1} 2 (1 - e^{- (z - x) / 2}) d x \\ = \int_{0}^{1} (1 - e^{- (z - x) / 2}) d x \\ = 1 - 2 (e^{- (z - 1) / 2} - e^{- z / 2}), 1 ⩽ z < + \infty . \end{aligned}

$Z$ 的概率密度函数为

p (z) = \frac{d}{d z} F_{z} (z) = (\sqrt{e} - 1) e^{- z / 2} I_{[1, + \infty)} (z) .

$0<z<1$ $Z$ 的分布函数为

\begin{aligned} F_{Z} (z) & = P (Z ⩽ z) = P (X + Y ⩽ z) \\ = \iint_{x + y ⩽ z} f (x, y) d x d y \\ = \int_{0}^{z} d x \int_{0}^{z - x} \frac{1}{2} e^{- y / 2} d y \\ = \frac{1}{2} \int_{0}^{z} 2 (1 - e^{- (z - x) / 2}) d x \\ = \int_{0}^{z} (1 - e^{- (z - x) / 2}) d x \\ = z + 2 e^{- z / 2} - 2, 0 < z < 1. \end{aligned}

$Z$ 的概率密度函数为

p (z) = \frac{d}{d z} F_{z} (z) = (1 - e^{- z / 2}) I_{(0, 1)} (z) .

$Z$ 的概率密度函数为

p (z) = (1 - e^{- z / 2}) I_{(0, 1)} (z) + (\sqrt{e} - 1) e^{- z / 2} I_{[1, + \infty)} (z) .

(3)

$\Delta=(2X)^2-4Y=4(X^2-Y)\geqslant 0\rArr Y\leqslant X^2$ $P(Y\leqslant X^2)$ .

\begin{aligned} P (Y ⩽ X^{2}) & = \iint_{y ⩽ x^{2}} f (x, y) d x d y \\ = \frac{1}{2} \int_{0}^{1} d x \int_{0}^{x^{2}} \frac{1}{2} e^{- y / 2} d y \\ = 1 - \int_{0}^{1} e^{- x^{2} / 2} d x \\ = 1 - \sqrt{2 π} \int_{0}^{1} φ (x) d x \\ = 1 - \sqrt{2 π} (Φ (1) - Φ (0)) \\ \approx 1 - 2.5066 (0.8413 - 0.5) \\ \approx 0.144 . \end{aligned}

(1)

\begin{aligned} α_{1} & = E (X) \\ = \int_{c}^{+ \infty} x f (x) d x \\ = \int_{c}^{+ \infty} \frac{α c^{α}}{x^{α}} d x \\ = α c^{α} \int_{c}^{+ \infty} \frac{d x}{x^{α}} \\ = α c^{α} \cdot - {(α - 1)^{- 1} \frac{1}{x^{α - 1}} |}_{c}^{+ \infty} \\ = α c / (α - 1) . \end{aligned}

\begin{aligned} α_{2} & = E (X^{2}) \\ = \int_{c}^{+ \infty} x^{2} f (x) d x \\ = \int_{c}^{+ \infty} \frac{α c^{α}}{x^{α - 1}} d x \\ = α c^{α} \int_{c}^{+ \infty} \frac{d x}{x^{α - 1}} \\ = α c^{α} \cdot - {(α - 2) \frac{1}{x^{α - 2}} |}_{c}^{+ \infty} \\ = α c^{2} / (α - 2) . \end{aligned}

μ_{2} = Var (X) = E (X^{2}) - (E (X))^{2} = α_{2} - α_{1}^{2} = \frac{α c^{2}}{(α - 2) (α - 1)^{2}} .

$\overline X$ $\alpha_1$ $S^2$ $\mu_2$ ，得

\begin{matrix} {\hat{α}}_{1} = 1 + \sqrt{{\overset{―}{X}}^{2} / S^{2} + 1}, \\ {\hat{c}}_{1} = \frac{\hat{α} - 1}{\hat{α}} \overset{―}{X} = \frac{\sqrt{{\overset{―}{X}}^{2} / S^{2} + 1}}{1 + \sqrt{{\overset{―}{X}}^{2} / S^{2} + 1}} \overset{―}{X} . \end{matrix}

(2)

似然函数

L = \prod_{i = 1}^{n} f (x_{i}) = \frac{α^{n} c^{α n}}{\prod_{i = 1}^{n} x_{i}^{α + 1}},

$c\leqslant x_1,x_2,\cdots,x_n$ $c\leqslant \min\set{x_1,x_2,\cdots,x_i}$ $c$ $\hat c_2=\min\set{X_1,X_2,\cdots,X_n}=X_{(1)}$ .

对数似然函数

\begin{aligned} ℓ (α, c) & = \ln L = \ln α^{n} + \ln c^{α n} - \sum_{i = 1}^{n} \ln x_{i}^{α + 1} \\ = n (\ln α + α \ln c) - (α + 1) \sum_{i = 1}^{n} \ln x_{i}, \\ \frac{\partial ℓ}{\partial α} & = n (α^{- 1} + \ln c) - \sum_{i = 1}^{n} \ln x_{i} = 0 \\ \Rightarrow {\hat{α}}_{2} & = \frac{1}{\frac{1}{n} \sum_{i = 1}^{n} \ln X_{i} - \ln {\hat{c}}_{2}} = \frac{n}{\sum_{i = 1}^{n} \ln (\frac{X_{i}}{X_{(1)}})} . \end{aligned}

(3)

$X$ 的分布函数为

F (x) = \int_{c}^{x} f (t) d t = 1 - {(\frac{c}{x})}^{α}, x ⩾ c .

$\hat c_2=X_{(1)}=\min\set{X_1,X_2,\cdots,X_n}$ 的分布函数为

F_{min} (z) = 1 - (1 - F (z))^{n} = 1 - {(\frac{c}{z})}^{n α}, z ⩾ c;

期望为

E ({\hat{c}}_{2}) = \int_{c}^{+ \infty} z d F_{min} (z) = n α c^{n α} \int_{c}^{+ \infty} \frac{d z}{z^{n α}} = c n α / (n α - 1) \neq c,

修正值为

{\hat{c}}_{2}^{*} = \frac{(n α - 1)}{n α} X_{(1)} .

两个正态总体均值差的成对比较.

$Z=X-Y\sim N(\mu,\sigma^2)$ .

序号	1	2	3	4	5	6	7	8
$(x_i)$	172	168	180	181	160	163	165	177
$(y_i)$	172	167	177	179	159	161	166	175
$(z_i=x_i-y_i)$	0	1	3	2	1	2	-1	2

$n=8,\overline z=1.25,s_Z^2=1.28^2$ .

$H_0:\mu\leqslant 0\lrarr H_1:\mu>0$ .

检验统计量

T = \frac{\sqrt{n} (\overset{―}{Z} - 0)}{S_{Z}},

$\alpha$ $T>t_{n-1}(\alpha)$ $H_0$ $H_0$ .

$T$ $t=\sqrt n\overline z/s_Z=\sqrt 8\times1.25/1.28=2.76$ $t_{n-1}(\alpha)=t_7(0.05)=1.895$ $2.76=t>t_{n-1}(\alpha)=1.895$ $H_0$ ，即可以认为成人早晨的身高显著高于晚上的身高.

其他

\begin{aligned} f_{Y} (y) = & \int_{R} f (x, y) d x \\ = & \frac{1}{2 π} \int_{R} e^{- x^{2} + x y - y^{2} / 2} d x \\ = & \frac{1}{2 π} \int_{R} e^{- (x - y / 2)^{2} - y^{2} / 4} d x \\ = & \frac{1}{2 π} e^{- y^{2} / 4} \int_{R} e^{- (x - y / 2)^{2}} d x \\ \overset{u = x - y / 2}{=} & \frac{1}{2 π} e^{- y^{2} / 4} \int_{- \infty}^{+ \infty} e^{- u^{2}} d u \\ = & \frac{1}{2 π} e^{- y^{2} / 4} \cdot \sqrt{π} \\ = & \frac{1}{2 \sqrt{π}} e^{- y^{2} / 4} . \end{aligned}

$A=\{随机抽取\ 2\ 台均为一等品\}$ $B=\{售出一台一等品\}$ $\overline B=\{售出一台二等品\}$ ，则

\begin{matrix} P (B) = \frac{7}{10}, P (\overset{―}{B}) = \frac{3}{10}, \\ P (A | B) = \frac{(\binom{6}{2})}{(\binom{9}{2})} = \frac{5}{12}, P (A | \overset{―}{B}) = \frac{(\binom{7}{2})}{(\binom{9}{2})} = \frac{7}{12} . \end{matrix}

由贝叶斯公式，

P (\overset{―}{B} | A) = \frac{P (A \overset{―}{B})}{P (A)} = \frac{P (A | \overset{―}{B}) P (\overset{―}{B})}{P (A | B) P (B) + P (A | \overset{―}{B}) P (\overset{―}{B})} = \frac{3}{8} .

X_{1} + X_{n} - 2 \overset{―}{X} = σ (\frac{X_{1} - μ}{σ} + \frac{X_{n} - μ}{σ} - 2 \frac{\overset{―}{X} - μ}{σ}) .

令

Y_{i} = \frac{X_{i} - μ}{σ} \sim N (0, 1), i = 1, 2, \dots, n,

则

\overset{―}{Y} = \frac{1}{n} \sum_{i = 1}^{n} Y_{i} = \frac{\overset{―}{X} - μ}{σ} \sim N (0, \frac{1}{n}) .

从而，

X_{1} + X_{n} - 2 \overset{―}{X} = σ (Y_{1} + Y_{n} - 2 \overset{―}{Y}),

\begin{aligned} {(X_{1} + X_{n} - 2 \overset{―}{X})}^{2} & = σ^{2} {(Y_{1} + Y_{n} - 2 \overset{―}{Y})}^{2} \\ = σ^{2} (Y_{1}^{2} + Y_{n}^{2} + 4 {\overset{―}{Y}}^{2} + 2 Y_{1} Y_{n} - 4 Y_{1} \overset{―}{Y} - 4 Y_{n} \overset{―}{Y}), \\ E {(X_{1} + X_{n} - 2 \overset{―}{X})}^{2} & = σ^{2} E (Y_{1}^{2} + Y_{n}^{2} + 4 {\overset{―}{Y}}^{2} + 2 Y_{1} Y_{n} - 4 Y_{1} \overset{―}{Y} - 4 Y_{n} \overset{―}{Y}) \\ = σ^{2} (E (Y_{1}^{2} + Y_{n}^{2}) + 4 E ({\overset{―}{Y}}^{2}) + 2 E (Y_{1} Y_{n}) - 4 E (Y_{1} \overset{―}{Y}) - 4 E (Y_{n} \overset{―}{Y})) . \end{aligned}

$Y_1,Y_n\sim N(0,1)$ $Y_1^2+Y_n^2\sim \chi^2_2$ $E(Y_1^2+Y_n^2)=2$ ；
$E\left(\overline Y\right)=0,\mathrm{Var}\left(\overline Y\right)=\dfrac1n$ ，则
$E ({\overset{―}{Y}}^{2}) = Var (\overset{―}{Y}) + {(E (\overset{―}{Y}))}^{2} = \frac{1}{n};$
$E(Y_1Y_n)=E(Y_1)E(Y_n)=0$ ；
因为
$Y_{1} \overset{―}{Y} = \frac{1}{n} Y_{1} \sum_{i = 1}^{n} Y_{i} = \frac{1}{n} (Y_{1}^{2} + \sum_{i = 2}^{n} Y_{1} Y_{i}),$
所以
$E (Y_{1} \overset{―}{Y}) = \frac{1}{n} E (Y_{1}^{2} + \sum_{i = 2}^{n} Y_{1} Y_{i}) = \frac{1}{n} (E (Y_{1}^{2}) + \sum_{i = 2}^{n} E (Y_{1} Y_{i})) .$
$Y_1^2\sim\chi_1^2$ $E(Y_1^2)=1$ $E(Y_1Y_i)=E(Y_1)E(Y_i)=0,i=2,3,\cdots,n$ ，所以有
$E (Y_{1} \overset{―}{Y}) = \frac{1}{n} .$
同理，也有
$E (Y_{n} \overset{―}{Y}) = \frac{1}{n} .$

综上，有

E {(X_{1} + X_{n} - 2 \overset{―}{X})}^{2} = σ^{2} (2 + \frac{4}{n} - \frac{4}{n} - \frac{4}{n}) = \frac{2 (n - 2)}{n} σ^{2},

E (T) = c E {(X_{1} + X_{n} - 2 \overset{―}{X})}^{2} = c \cdot \frac{2 (n - 2)}{n} σ^{2} = σ^{2} ⟹ c = \frac{n}{2 (n - 2)} .

从形式上看，随机变量

Z = \frac{\sqrt{2} (Y_{1} - Y_{2})}{S} = \frac{Y_{1} - Y_{2}}{\sqrt{S^{2} / 2}}

$t$ $Y_1-Y_2$ $S^2$ 的分布.

$X_i\mathrm{\ i.i.d.}\sim N(\mu,\sigma^2)$ $Y_1\sim N(\mu,\sigma^2/6)$ $Y_2\sim N(\mu,\sigma^2/3)$ ，从而

\frac{Y_{1} - μ}{σ / \sqrt{6}} = \frac{\sqrt{6} (Y_{1} - μ)}{σ} \sim N (0, 1),

\begin{matrix} \frac{Y_{2} - μ}{σ / \sqrt{3}} = \frac{\sqrt{3} (Y_{2} - μ)}{σ} \sim N (0, 1), \end{matrix}

\begin{matrix} \frac{\sqrt{6} (Y_{1} - μ)}{σ} - \sqrt{2} \cdot \frac{\sqrt{3} (Y_{2} - μ)}{σ} = \frac{\sqrt{6} (Y_{1} - Y_{2})}{σ} \sim N (0 \cdot (1 - \sqrt{2}), 1 \cdot (1^{2} + (\sqrt{2})^{2})) = N (0, 3), \\ Y_{1} - Y_{2} \sim N (0, σ^{2} / 2), \\ \frac{Y_{1} - Y_{2}}{σ / \sqrt{2}} = \frac{\sqrt{2} (Y_{1} - Y_{2})}{σ} \sim N (0, 1) . \end{matrix}

这里用到了：
独立同分布的随机正态变量的线性组合服从正态分布，
$T = \sum_{i = 1}^{n} c_{i} X_{i} \sim N (μ \sum_{i = 1}^{n} c_{i}, σ^{2} \sum_{i = 1}^{n} c_{i}^{2}) .$
及样本均值
$\overset{―}{X} \sim N (μ, σ^{2} / n) .$

原式

Z = \frac{Y_{1} - Y_{2}}{\sqrt{S^{2} / 2}} = \frac{\sqrt{2} (Y_{1} - Y_{2}) / σ}{\sqrt{S^{2} / σ^{2}}},

又因为

\frac{2 S^{2}}{σ^{2}} \sim χ_{2}^{2},

这是因为
$\frac{(n - 1) S^{2}}{σ^{2}} \sim χ_{n - 1}^{2} .$

$Y_2$ $S^2$ 是相互独立的，

$\overline X$ $S^2$ 是相互独立的.

$\sqrt 2(Y_1-Y_2)/\sigma$ $2S^2/\sigma^2$ $Z$ $2$ $t$ 分布，即

Z = \frac{\sqrt{2} (Y_{1} - Y_{2}) / σ}{\sqrt{S^{2} / σ^{2}}} = \frac{\sqrt{2} (Y_{1} - Y_{2}) / σ}{\sqrt{(2 S^{2} / σ^{2}) / 2}} \sim t_{2} .

$X_1,X_2,\cdots,X_n\ \mathrm{i.i.d.}\sim N(\mu,\sigma^2)$ ，得

\begin{matrix} \overset{―}{X} \sim N (μ, \frac{σ^{2}}{n}), \\ \frac{\overset{―}{X} - μ}{σ / \sqrt{n}} = \frac{\sqrt{n} (\overset{―}{X} - μ)}{σ} \sim N (0, 1) . \end{matrix}

$X_{n+1}\sim N(\mu,\sigma^2)$ ，得

\frac{X_{n + 1} - μ}{σ} \sim N (0, 1) .

所以

\begin{matrix} \sqrt{n} \cdot \frac{X_{n + 1} - μ}{σ} - \frac{\sqrt{n} (\overset{―}{X} - μ)}{σ} = \frac{\sqrt{n} (X_{n + 1} - \overset{―}{X})}{σ} \sim N (0, n + 1), \\ \frac{\sqrt{n} (X_{n + 1} - \overset{―}{X})}{σ \cdot \sqrt{n + 1}} \sim N (0, 1) . \end{matrix}

另一方面，

m_{2} = \frac{n - 1}{n} S^{2},

而

\frac{(n - 1) S^{2}}{σ^{2}} \sim χ_{n - 1}^{2},

故

\frac{\frac{\sqrt{n} (X_{n + 1} - \overset{―}{X})}{σ \cdot \sqrt{n + 1}}}{\sqrt{((n - 1) S^{2} / σ^{2}) / (n - 1)}} \sim t_{n - 1} .

\frac{\frac{\sqrt{n} (X_{n + 1} - \overset{―}{X})}{σ \cdot \sqrt{n + 1}}}{\sqrt{((n - 1) S^{2} / σ^{2}) / (n - 1)}} = \sqrt{\frac{n}{n + 1}} \frac{X_{n + 1} - \overset{―}{X}}{\sqrt{S^{2}}} = \sqrt{\frac{n}{n + 1}} \frac{X_{n + 1} - \overset{―}{X}}{\sqrt{n \cdot m_{2} / (n - 1)}} = \sqrt{\frac{n - 1}{n + 1}} \frac{X_{n + 1} - \overset{―}{X}}{\sqrt{m_{2}}} .