概率论 (H) 习题

1引言: 随机现象与概率论
2从古典概率模型、几何概率模型到概率论的公理
3经典条件概率与事件独立性
4Lebesgue 积分理论简介
5随机变量及其分布律 (I)
6数学期望与分布律
7条件数学期望与条件分布律
8随机变量及其分布律 (II)
9随机变量及其分布律 (III)
10随机变量列的收敛与大数律

1引言: 随机现象与概率论

2从古典概率模型、几何概率模型到概率论的公理

2.2

(1) $A_{n} ↗ A ⟺ A_{n} \subset A_{n + 1} and n = 1 ⋃ \infty A_{n} = A ⟺ A_{n} \subset A_{n + 1} and \forall ω \in A, \exists N st ω \in A_{n}, \forall n \geq N ⟺ 1_{A_{n}} ↗ and n \to \infty lim 1_{A_{n}} = 1_{A} ⟺ 1_{A_{n}} ↗ 1_{A}$ ,
$A_{n} ↘ A ⟺ A_{n + 1} \subset A_{n} and n = 1 ⋂ \infty A_{n} = A ⟺ A_{n + 1} \subset A_{n} and \forall ω \in A, \exists N st ω \in A_{n}, \forall n \geq N ⟺ 1_{A_{n}} ↘ and n \to \infty lim 1_{A_{n}} = 1_{A} ⟺ 1_{A_{n}} ↗ 1_{A}$ .

(2) $ω \in n \to \infty \overline{lim} A_{n} ⟺ \forall N, \exists n \geq N st ω \in A_{n} ⟺ \forall N, \exists 无数个 n \geq N st ω \in A_{n} ⟺ n = 1 \sum \infty 1_{A_{n}} (ω) = \infty$ ;
$ω \in n \to \infty \overline{lim} A_{n} ⟺ \forall N, \exists n \geq N st ω \in A_{n} ⟺ \forall N, \exists n \geq N st 1_{A_{n}} (ω) = 1 ⟺ n \to \infty \overline{lim} 1_{A_{n}} (ω) = 1$ ,
$ω \in / n \to \infty \overline{lim} A_{n} ⟺ \exists N st \forall n \geq N, ω \in / A_{n} ⟺ \exists N st \forall n \geq N, 1_{A_{n}} (ω) = 0 ⟺ n \to \infty \overline{lim} 1_{A_{n}} (ω) = 0$ .

(3) $ω \in n \to \infty \overline{lim} A_{n} ⟺ \exists N st \forall n \geq N, ω \in / A_{n}^{c} ⟺ 只有有限个 n st ω \in A_{n}^{c} ⟺ n = 1 \sum \infty 1_{A_{n}^{c}} (ω) < \infty$ ;
$n \to \infty \underline{lim} 1_{A_{n}} = n \to \infty \underline{lim} (1 - 1_{A_{n}^{c}}) = 1 - n \to \infty \overline{lim} 1_{A_{n}^{c}} = 1 - 1_{n \to \infty \overline{lim} A_{n}^{c}} = 1_{(n \to \infty \overline{lim} A_{n}^{c})^{c}} = 1_{n \to \infty \underline{lim} A_{n}}$ .

(4) $A_{n} \to A ⟺ n \to \infty \overline{lim} A_{n} = n \to \infty \underline{lim} A_{n} = A ⟺ n \to \infty \overline{lim} 1_{A_{n}} = n \to \infty \underline{lim} 1_{A_{n}} = 1_{A} ⟺ n \to \infty lim 1_{A_{n}} = 1_{A}$ .

(5) $ω \in n \to \infty \overline{lim} (A_{n} Δ A_{n + 1}) ⟺ ω \in 无数个 A_{n} Δ A_{n + 1} ⟺ ω \in 无数个 A_{n} \cup A_{n + 1} 不过不能 \forall n \geq 某个 N, ω \in A_{n} \cap A_{n + 1} ⟺ ω \in 无数个 A_{n} 不过不能 \forall n \geq 某个 N, ω \in A_{n} ⟺ ω \in (n \to \infty \overline{lim} A_{n}) \ (n \to \infty \underline{lim} A_{n})$ .

(6) $ω \in n \to \infty \overline{lim} {f_{n} \geq C} ⟺ \exists N st \forall n \geq N, f_{n} (ω) \geq C ⟹ \forall N, \exists n \geq N st f_{n} (ω) \geq C ⟺ ω \in {n \to \infty \overline{lim} f_{n} \geq C}$ , 后面三个一样.

2.3

(1) 对于集合 $Σ_{n + 1}$ , 首先给元素 $n + 1$ 找同伴, 它们是确定的 $n - k$ 个数 ( $0 \leq k \leq n$ ) , 有 $(n - k n) = (k n)$ 个取法, 接下来剩下的 $k$ 个数作分划, 于是 $T_{n + 1} = k = 0 \sum n (k n) T_{k} .$

(2) $T_{0} = B_{0}$ , 设 $T_{m} = B_{m}$ 对 $m \leq n$ 成立, 注意 $B_{n + 1} = e^{- 1} ℓ = 1 \sum \infty \frac{ℓ ^{n + 1}}{ℓ !} = e^{- 1} ℓ = 0 \sum \infty \frac{( ℓ + 1 ) ^{n}}{ℓ !} = e^{- 1} k = 0 \sum n + 1 ℓ = 1 \sum \infty (k n) \frac{ℓ ^{k}}{ℓ !} = e^{- 1} k = 0 \sum n + 1 (k n) B_{k},$ 故 $T_{n + 1} = B_{n + 1}$ . 并且, $B = B_{0} + n = 1 \sum \infty \frac{B _{n}}{n !} x^{n} = 1 + e^{- 1} n = 1 \sum \infty k = 1 \sum \infty \frac{( k x ) ^{n}}{n ! k !} = e^{- 1} (k = 1 \sum \infty \frac{1}{k !} + k = 1 \sum \infty \frac{e ^{k x} - 1}{k !}) = e^{e^{x} - 1} .$

2.4

$∣ A ∣ = m \geq k \sum (m n) .$

2.6

$n^{2}$ 个人标上编号, 有 $(n^{2})!$ 种方法; $n^{2}$ 个人分成不带编号的 $n$ 组, 有 $C (n)$ 种方法, 再给每组标上编号, 每组 $n$ 人标上编号, 有 $n! (n!)^{n}$ 种方法. 所以 $(n^{2})! = C (n) (n!)^{n + 1}$ .

数论的方法是, 要说明对任何素数 $p$ , $0 < α \leq ⌊ l o g_{p} n ⌋ \sum ⌊ \frac{n ^{2}}{p ^{α}} ⌋ = 0 < α \leq ⌊ l o g_{p} n ⌋ \sum ⌊ \frac{n ^{2}}{p ^{α}} ⌋ + ⌊ l o g_{p} n ⌋ < α \leq ⌊ l o g_{p} n^{2} ⌋ \sum ⌊ \frac{n ^{2}}{p ^{α}} ⌋ \geq (n + 1) 0 < α \leq ⌊ l o g_{p} n ⌋ \sum ⌊ \frac{n}{p ^{α}} ⌋,$ 这是因为 $⌊ \frac{n ^{2}}{p ^{α}} ⌋ \geq n ⌊ \frac{n}{p ^{α}} ⌋, ⌊ lo g_{p} n^{2} ⌋ - ⌊ lo g_{p} n ⌋ \geq ⌊ lo g_{p} n ⌋, \frac{n ^{2}}{p ^{α + ⌊ l o g_{p} n ⌋}} \geq \frac{n}{p ^{⌊ l o g_{p} n ⌋}} .$

2.9

一方面, 集合 $F$ 可由 $E$ 通过可数并与取补集生成; 另一方面, $F$ 是一个 $σ$ -代数.

2.14

不是. 令 $A_{i} = (2^{2 i - 2}, 2^{2 i - 1}] \cap Z, i \in Z_{+}$ , 渐进密度均为 $0$ 因而皆属于 $E$ ; 而 $A = i = 1 ⋃ \infty A_{i}$ 不属于 $E$ , 因为 $n \to \infty lim \frac{∣ A \cap ( 0 , 2 ^{2 n - 1} ] ∣}{2 ^{2 n - 1}} n \to \infty lim \frac{∣ A \cap ( 0 , 2 ^{2 n} ] ∣}{2 ^{2 n}} = n \to \infty lim \frac{2 ^{0} + 2 ^{2} + \dots + 2 ^{2 n - 2}}{2 ^{2 n - 1}} = \frac{2}{3}, = n \to \infty lim \frac{2 ^{0} + 2 ^{2} + \dots + 2 ^{2 n - 2}}{2 ^{2 n}} = \frac{1}{3} .$

2.15

记 $E = {A_{i_{1}} \cap \dots \cap A_{i_{k}} \cap A_{i_{k + 1}}^{c} \cap \dots \cap A_{i_{n}}^{c} : i_{1}, \dots, i_{n} 是 1, \dots, n 的排列}$ , 有 $F = {e \in E ⋃ e : E \subseteq E - {\emptyset}},$ 故 $∣ F ∣ \leq 2^{∣ E ∣ - 1} \leq 2^{2^{n} - 1}$ .

2.18

(i)	$A_{1} \cap A_{2}^{c} \cap A_{3}^{c}$ ;
(ii)	$A_{1} \cap A_{2} \cap A_{3}^{c}$ ;
(iii)	$A_{1} \cap A_{2} \cap A_{3}$ ;
(iv)	$A_{1} \cup A_{2} \cup A_{3}$ ;
(v)	$(A_{1} \cap A_{2}) \cup (A_{1} \cap A_{3}) \cup (A_{2} \cap A_{3})$ ;
(vi)	$(A_{1} \cap A_{2}^{c} \cap A_{3}^{c}) \cup (A_{1}^{c} \cap A_{2} \cap A_{3}^{c}) \cup (A_{1}^{c} \cap A_{2}^{c} \cap A_{3})$ ;
(vii)	$(A_{1}^{c} \cap A_{2} \cap A_{3}) \cup (A_{1} \cap A_{2}^{c} \cap A_{3}) \cup (A_{1} \cap A_{2} \cap A_{3}^{c})$ ;
(viii)	$(A_{1} \cap A_{2} \cap A_{3})^{c}$ .

2.21

简写 $P = P (A \cap B)$ , 有 $P (A) + P (B) \leq 1 + P$ , 得 $- \frac{1}{4} \leq \frac{P ( 2 - P ) - 1}{4} = P - (\frac{1 + P}{2})^{2} \leq P (A \cap B) - P (A) P (B) \leq P - P^{2} \leq \frac{1}{4} .$

2.23

即 $\int ∣ 1_{A} - 1_{B} ∣ \leq \int ∣ 1_{A} - 1_{C} ∣ + \int ∣ 1_{C} - 1_{B} ∣$ .

2.28

性质 (1), (2) 是说 $A_{i} \cap A_{j} \cap A_{k} = \emptyset, A_{i} = j \neq = i ⨄ (A_{i} \cap A_{j}) .$ 用 Jordan 公式 $P (i ⋃ A_{i}) = i \sum P (A_{i}) - i < j \sum P (A_{i} \cap A_{j}) + \dots + (- 1)^{n - 1} P (i ⋂ A_{i}),$ 得到 $2 P (i ⋃ A_{i}) = 2 i \sum P (A_{i}) - i \neq = j \sum P (A_{i} \cap A_{j}) = 2 i \sum P (A_{i}) - i \sum j \neq = given i \sum P (A_{i} \cap A_{j}) = 2 i \sum P (A_{i}) - i \sum P (A_{i}) = i \sum P (A_{i}) .$

2.32

能组成的四位数有 $9 \cdot 9 \cdot 8 \cdot 7$ 个, 按照末尾是 $0$ 或 $2, 4, 6, 8$ , 得里面偶数有 $9 \cdot 8 \cdot 7 \cdot 1 + 8 \cdot 8 \cdot 7 \cdot 4$ 个, 概率为 $\frac{9 \cdot 1 + 8 \cdot 4}{9 \cdot 9} = \frac{41}{81}$ .

2.35

$1 - \frac{11 \cdot 10 \cdot 9}{12 \cdot 12 \cdot 12} = \frac{41}{96}$ . (反直觉地大)

2.42

(1) 前者大, $1 - (1 - \frac{1}{6})^{4} > 1 - (1 - \frac{1}{6 ^{2}})^{6 \cdot 4}$ .

$1 - \frac{1}{6} < (1 - \frac{1}{6 ^{2}})^{6}$ 是因为, 对 $n$ 归纳易有, 对于 $a_{1}, \dots, a_{n} > 0$ , $1 - a_{1} - \dots - a_{n} < (1 - a_{1}) \dots (1 - a_{n}) .$

(2) 前者大. 扔 $6 n$ 个骰子得到的 $6$ 的数量写为 $X_{n}$ (这是个随机变量) , 那么 $X_{n + 1} = X_{n} + X_{1}$ , 且 $P (X_{n + 1} \geq n + 1) = i = 0 \sum 6 P (X_{n} \geq n + 1 - i) P (X_{1} = i) = P (X_{n} \geq n) + i = 0 \sum 6 [P (X_{n} \geq n + 1 - i) - P (X_{n} \geq n)] P (X_{1} = i) .$ 观察到 $P (X_{n} = n + 1 - 6) < \dots < P (X_{n} = n + 1)$ , 从而 $i > 1$ 时 $P (X_{n} \geq n + 1 - i) - P (X_{n} \geq n) < (i - 1) P (X_{n} = n) .$ 由此, 最后一项小于 $P (X_{n} = n) i = 0 \sum 6 (i - 1) P (X_{1} = i) = P (X_{n} = n) (E (X_{1}) - 1) = 0.$

2.55

有人找零困难等价于存在某时间排到的人里面 $50$ 元的比 $100$ 元的少 $1$ 人. 沿票数差为 $- 1$ 的直线反射这个时间后的图像, 最终票数差为 $n - m - 2$ (本来是 $m - n$ ) , 相当于 $n - 1$ 人 $50$ 元, $m + 1$ 人 $100$ 元. 所以不发生找零困难的概率是 $1 - \frac{( m + 1 m + n )}{( m m + n )} = 1 - \frac{n}{m + 1}$ .

2.57

相当于随机选三条边不是同一条边的概率, 是 $1 - (\frac{1}{4})^{2} = \frac{15}{16}$ .

2.62

本题说明, 无法说: 对任何正整数 $a$ , 一个正整数是 $a$ 的倍数的概率是 $\frac{1}{a}$ (尽管能说渐进密度是 $\frac{1}{a}$ ) .

假如概率测度 $P$ 存在, 不妨令 $P$ 是完备的. 对 $a_{1}, \dots, a_{n} \in N$ , 有 $P (a_{1} N \cup \dots \cup a_{n} N) = i \sum P (a_{i} N) - i < j \sum P (a_{i} N \cap a_{j} N) + \dots + (- 1)^{n - 1} P (a_{1} N \cap \dots \cap a_{n} N) = i \sum \frac{1}{a _{i}} - i < j \sum \frac{1}{[ a _{i} , a _{j} ]} + \dots + (- 1)^{n - 1} \frac{1}{[ a _{1} , \dots , a _{n} ]} .$ 当 $a_{1}, \dots, a_{n}$ 互素时最小公倍数就是乘积, 于是 $P (a_{1} N \cup \dots \cup a_{n} N) = 1 - i = 1 \prod n (1 - \frac{1}{a _{i}}) .$ 对任意的 $m \in N$ , 取不是 $m$ 的因数的不同素数 $p_{1}, \dots, p_{n}$ , 当 $n \to \infty$ 时 $P (p_{1} N \cup \dots \cup p_{n} N) = 1 - i = 1 \prod n (1 - \frac{1}{p _{i}}) \to 1,$ 于是 $P ({m}) = 0$ , 不成立. 这里用了 $p prime \prod (1 - \frac{1}{p}) \leq s \to 1^{+} lim p prime \prod (1 - \frac{1}{p ^{s}}) = s \to 1^{+} lim \frac{1}{ζ ( s )} = 0.$

2.63

对圆周上的点 $P$ , 以之为顶点顺时针作内接正方形, 四个顶点中无白点的 $P$ 的范围的长度不小于 $1 - 4 q > 0$ .

3经典条件概率与事件独立性

3.1

由 $⎩ ⎨ ⎧ p (A \cup B \cup C) p (A \cap B \cap C) p (A ∖ (B \cup C)) = 3 p - 3 p^{2} \leq 1, = 0 \neq = p^{3}, = p - 2 p^{2} \geq 0,$ 解得 $p \in (0, \frac{1}{2}]$ .

3.2

设事件 $Γ$ 使 $P (Γ) = 0$ 且 ${A_{n} ∖ Γ}$ 互不相交, 有 $B = n = 1 ⨄ N ((A_{n} ∖ Γ) \cap B) ⊎ (Γ \cap B) ⊎ (B ∖ (n = 1 ⋃ N A_{n} \cup Γ)),$ 于是 $P (B) = n = 1 \sum N P (A_{n} \cap B)$ 同样成立, 其余推理不变.

3.3

$\frac{2}{3}$ .

3.4

3.1 就是.

3.6

(1.a)	如果有无数个 $x_{n} > 0.5$ , 结论成立; 如果 $x_{n} > 0.5$ 的个数有限, 由 (1.b) 结论成立.
(1.b)	$0$ 附近 $- x \sim ln (1 - x)$ , 得 $[0, 1 - δ]$ 上 $- x, ln (1 - x)$ 相互控制, 再用 $ln (x)$ 的连续性, 有 $n = 1 \sum \infty x_{n} = \infty \Leftrightarrow n = 1 \sum \infty ln (1 - x_{n}) = - \infty \Leftrightarrow n = 1 \prod \infty (1 - x_{n}) = 0.$
(2.a)	取 $x_{n} = P (E_{n}^{c} ∣ ∣ k = 1 ⋂ n - 1 E_{k})$ , 成立 $n = 1 \prod \infty P (E_{n} ∣ ∣ k = 1 ⋂ n - 1 E_{k}) = P (k = 1 ⋂ \infty E_{k})$ .
(2.b)	同上.

3.7

$A$ 与 $A^{c}$ 独立, 推出 $0 = P (A \cap A^{c}) = P (A) P (A^{c})$ .

3.8

$\Leftrightarrow \Leftrightarrow \Leftrightarrow P (B ∣ ∣ A) = P (B ∣ ∣ A^{c}) P (A \cap B) / P (A) = P (A^{c} \cap B) / P (A^{c}) P (A \cap B) (1 - P (A)) = (P (B) - P (A \cap B)) P (A) P (A \cap B) = P (A) P (B) .$

3.9

由于 $P (A \cap B) = P (A) P (B) \in [0, \frac{1}{4}],$ 有 $P (A Δ B ∣ ∣ A \cup B) = \frac{P ( A Δ B )}{P ( A \cup B )} = \frac{P ( A ) + P ( B ) - 2 P ( A \cap B )}{P ( A ) + P ( B ) - P ( A \cap B )} = \frac{1 - 2 P ( A \cap B )}{1 - P ( A \cap B )} \in [\frac{2}{3}, 1] .$

3.10

$\frac{P ( A \cap B )}{P ( A \cup B )} \leq \frac{P ( A \cap B )}{P ( A )} \leq \frac{P ( A \cap B ) + P ( B ∖ A )}{P ( A ) + P ( B ∖ A )} = \frac{P ( B )}{P ( A \cup B )} .$

3.12

$σ ({A_{n} : 1 \leq n \leq N})$ 的元素形如 $n \in I ⨄ A_{n}$ 与 $Ω ∖ n \in I ⨄ A_{n}$ , 其中 $I$ 是 ${1, \dots, N}$ 的子集. 对任意的 $I$ , $P (n \in I ⨄ A_{n}) P (B) P (Ω ∖ n \in I ⨄ A_{n}) P (B) = n \in I \sum P (A_{n}) P (B) = n \in I \sum P (A_{n} \cap B) = P (n \in I ⨄ A_{n} \cap B), = (P (Ω) - P (n \in I ⨄ A_{n})) P (B) = P (B) - P (n \in I ⨄ A_{n} \cap B) = P ((Ω ∖ n \in I ⨄ A_{n}) \cap B) .$

3.15

设概率为 $P (a, b)$ . 按第一局的输赢, $P (a, b) = (P (a + 1, b - 1) + P (a - 1, b + 1)) /2,$ 即 $P (a + \cdot, b - \cdot)$ 是等差数列. 由于 $P (a + b, 0) = 1, P (0, a + b) = 0$ , 有 $P (a, b) = \frac{a}{a + b}$ .

3.17

$\frac{P ( H ∣ ∣ E )}{P ( G ∣ ∣ E )} = \frac{P ( H ∣ ∣ E ) P ( E )}{P ( G ∣ ∣ E ) P ( E )} = \frac{P ( H \cap E )}{P ( G \cap E )} = \frac{P ( H ) P ( E ∣ ∣ H )}{P ( G ) P ( E ∣ ∣ G )}$ .

$\frac{P ( H ∣ ∣ E )}{P ( G ∣ ∣ E )} = 3 \cdot \frac{1}{2}$ 表明 $E$ 出现时 $H$ 更可能成立.

3.18

$P (B ∣ ∣ 看成 B) = \frac{P ( B \cap 看成 B )}{P ( 看成 B )} = \frac{15% \cdot 80%}{85% \cdot 10% + 15% \cdot 80%} = \frac{24}{41} .$

3.21

$n = 1 \sum \infty (1 - P (A_{1} ⊎ A_{2}))^{n - 1} P (A_{1}) = P (A_{1}) / P (A_{1} ⊎ A_{2}) .$

3.24

(1)	$P (患病 ∣ ∣ 检出患病) = \frac{P ( 患病 \cap 检出患病 )}{P ( 检出患病 )} = \frac{1 0 ^{- 4}}{1 0 ^{- 4} + ( 1 - 1 0 ^{- 4} ) 5%} = \frac{20}{10019} .$
(2)	$P (患病 ∣ ∣ 两次检出患病) = \frac{P ( 患病 \cap 两次检出患病 )}{P ( 两次检出患病 )} = \frac{1 0 ^{- 4}}{1 0 ^{- 4} + ( 1 - 1 0 ^{- 4} ) ( 5% ) ^{2}} = \frac{400}{10399} .$

3.51

设 ${A_{ij}}_{i, j = 1}^{n}$ 是一些相互独立的事件矩阵, 满足 $P (A_{ij}) = p_{ij}$ .

设事件 $M$ 为 “每列至少有一个事件不发生” 的事件, 事件 $N$ 为 “每行至少有一个事件发生” 的事件,

则 $P (M) = j = 1 \prod n (1 - i = 1 \prod n p_{ij})$ , $P (N) = i = 1 \prod n (1 - j = 1 \prod n (1 - p_{ij}))$ .

假设 $M$ , $N$ 都不发生, 则意味着 $A$ 中至少有一列全发生, 且至少有一行全不发生, 这是不可能的.

因此 $P (M^{c} \cap N^{c}) = 0$ .

从而 $P (M) + P (N) \geq 1$ , 即 $j = 1 \prod n (1 - i = 1 \prod n p_{ij}) + i = 1 \prod n (1 - j = 1 \prod n (1 - p_{ij})) \geq 1.$

4Lebesgue 积分理论简介

5随机变量及其分布律 (I)

5.1	$1_{A}$ 是随机变量 $⟺ \forall x \in R, {1_{A} \leq x} \in F ⟺ \emptyset, A^{c}, Ω \in F ⟺ A \in F$ .
5.2	${n \geq 1 sup X_{n} \leq x} = n \geq 1 ⋂ {X_{n} \leq x}$ , ${n \geq 1 in f X_{n} < x} = n \geq 1 ⋃ {X_{n} < x}$ , $n \to \infty \overline{lim} X_{n} = N \geq 1 in f n \geq N sup X_{n}$ , $n \to \infty \underline{lim} X_{n} = N \geq 1 sup n \geq N in f X_{n}$ .
5.4	$\forall δ > 0$ , 取 $N$ 大使得 $F (- N) < δ, 1 - F (N) < δ$ , 并且 $F$ 在 $[- N, N]$ 一致连续.
5.5	事实上 $P (ξ = a) = P (ξ \leq a) - P (n \geq 1 ⋃ {ξ \leq a - \frac{1}{n}}) = P (ξ \leq a) - n \to \infty lim P (ξ \leq a - \frac{1}{n}) = F (a) - F (a^{-})$ , 于是, $F$ 在 $x = a$ 处连续 $⟺ F (a^{-}) = F (a) ⟺ P (ξ = a) = 0$ .
5.8	$\forall x < y$ , $P (X \leq x < Y \leq y) = P (X \leq x, Y \leq y) - P (X \leq x, Y \leq x) = F (x) - F (x) = 0$ , 故 $P (X < Y) = P (x, y \in Q ⋃ X \leq x < Y \leq y) = 0$ , 同理 $P (X > Y) = 0$ .
5.9	注意 $S_{F}^{c} = {x : 存在 n \geq 1 使 P (∥ X - x ∥ \leq \frac{1}{n}) = 0}$ 是开集, 毕竟对 $x \in S_{F}^{c}$ , $B_{\frac{1}{n ( x )}} (x) \subset S_{F}^{c}$ , 其中 $n (x)$ 为 $x$ 对应的 $n$ . 第二可数空间 $S_{F}^{c}$ 有开覆盖 $x \in S_{F}^{c} ⋃ B_{\frac{1}{n ( x )}} (x)$ , 从而存在可数子覆盖, 每个开集都零测, 得 $P (X \in S_{F}^{c}) = 0$ .
5.12	$F (x) = ⎩ ⎨ ⎧ 0, 1 - p, 1, x < 0 0 \leq x < 1 x \geq 1 .$ $Y \sim P (F (X) \leq x) = ⎩ ⎨ ⎧ 0, P (X = 0), 1, x < F (0) F (0) \leq x < F (1) x \geq F (1) = ⎩ ⎨ ⎧ 0, 1 - p, 1, x < 1 - p 1 - p \leq x < 1 x \geq 1 .$
5.14	$X_{I} \sim P (X_{I} \leq \cdot) = P (I = 0) P (X_{0} \leq \cdot) + P (I = 1) P (X_{1} \leq \cdot) = (1 - p) F_{0} + p F_{1}$ .
5.21	$P (X = n + 1∣ X > n) = \frac{2}{3}, \forall n \geq 1,$ $X$ 满足无记忆性, 因此 $X \sim Geo (\frac{2}{3})$ .
5.23	$Y_{k} \sim ((- 1)^{2} (\frac{1}{2})^{2} (- 1) \cdot 1 (\frac{1}{2})^{2} 1 \cdot (- 1) (\frac{1}{2})^{2} 1^{2} (\frac{1}{2})^{2}) = (- 1 \frac{1}{2} 1 \frac{1}{2})$ . $P (Y_{1} = ε_{1}, Y_{2} = ε_{2}) = P (X_{2} = ε_{1} / X_{3}, X_{1} = ε_{2} / X_{3}) = P (X_{2} = ε_{1} / X_{3}) P (X_{1} = ε_{2} / X_{3}) = P (Y_{1} = ε_{1}) P (Y_{2} = ε_{2})$ , 这里 $ε_{1}, ε_{2} = \pm 1$ . $P (Y_{1} = Y_{2} = Y_{3} = - 1) = 0 \neq = (\frac{1}{2})^{3} = P (Y_{1} = - 1) P (Y_{2} = - 1) P (Y_{3} = - 1)$ .
5.26	设 ${X_{N}^{(k)}}_{N = 1}^{\infty}$ 是相互独立的概率为 $p_{k}$ 的事件列, $P (τ = ℓ) = P (1 \leq k \leq n min τ_{k} = ℓ) = P (1 \leq k \leq n min in f {N : X_{N}^{(k)} = 1} = ℓ) = P (in f {N : 1 \leq k \leq n ⋃ X_{N}^{(k)} = 1} = ℓ)$ , 且 $P (1 \leq k \leq n ⋃ X_{N}^{(k)}) = 1 - q_{1} \dots q_{n}$ .
$习题$	已知 $f : (Ω, F, P) \to (Ω_{1}, F_{1}, P_{1}) \otimes (Ω_{2}, F_{2}, P_{2})$ 是概率空间之间的同构. 记 $\tilde{ω} = (ω_{1}, ω_{2}) \in Ω_{1} \times Ω_{2}, π_{1} (\tilde{ω}) := ω_{1}, π_{2} (\tilde{ω}) := ω_{2}$ . 又设 $ξ_{k}$ 为 $(Ω_{k}, F_{k}, P_{k})$ 上的随机变量 ( $k = 1, 2$ ). 令 $\hat{ξ}_{k} = ξ_{k} \circ π_{k} \circ f$ , 求证: (1) $ξ_{k} d \hat{ξ}_{k}$ , $k = 1, 2$ . (2) $\hat{ξ}_{1}$ 与 $\hat{ξ}_{2}$ 相互独立. (1) $f$ 是同构是说 $P (A) = P_{1} \otimes P_{2} (f (A)), \forall A \in F$ . $P (\hat{ξ}_{1} \leq x) = P (ξ_{1} \circ π_{1} \circ f \leq x) = P (f^{- 1} ({ξ_{1} \leq x} \times Ω_{2})) = P_{1} \otimes P_{2} ({ξ_{1} \leq x} \times Ω_{2}) = P_{1} (ξ_{1} \leq x) P_{2} (Ω_{2}) = P_{1} (ξ_{1} \leq x) .$ (2) $P (\hat{ξ}_{1} \leq x) P (\hat{ξ}_{2} \leq y) = P_{1} (ξ_{1} \leq x) P_{2} (ξ_{2} \leq y) = P_{1} \otimes P_{2} (ξ_{1} \leq x, ξ_{2} \leq y) = P (f^{- 1} (ξ_{1} \leq x, ξ_{2} \leq y)) = P (ξ_{1} \circ π_{1} \circ f \leq x, ξ_{2} \circ π_{2} \circ f \leq y) = P (\hat{ξ}_{1} \leq x, \hat{ξ}_{2} \leq y) .$
$习题$	设 $X, Y$ 是一维随机变量, $f$ 是二元可测函数. 求证: $f (X, Y)$ 是随机变量. 只需证明对任意二维可测集 $A$ , ${(X, Y) \in A}$ 还是可测集, 要使用单调类定理: 对 $A =$ 开集 $O_{1} \times$ 开集 $O_{2}$ , 由 $X, Y$ 是可测函数, $X^{- 1} (O_{1}) \times Y^{- 1} (O_{2})$ 可测; 这类 $A$ 生成的 $σ$ -代数是二维可测集, 二维可测集又是单调系.
5.27	(1) 取事件 $X_{k} = {1, 0, 第 k 次 A 出现第 k 次 A 未出现 \sim B (1, p_{A})$ , 则 $τ_{A} \sim Geo (p_{A})$ . (2) 见 5.26, 当 $A, B$ 互斥时 $p_{A} p_{B} = p_{A \cap B} = 0$ , $P (A ⊎ B) = p_{A} + p_{B}$ . (3) $τ_{A}, τ_{B}$ 相互独立 $⟺ P (τ_{A} > ℓ_{A}) P (τ_{B} > ℓ_{B}) = P (τ_{A} > ℓ_{A}, τ_{B} > ℓ_{B})$ $⟺ q_{A}^{ℓ_{A}} q_{B}^{ℓ_{B}} = q_{A \cup B}^{ℓ_{A}} q_{B}^{ℓ_{B} - ℓ_{A}}$ 当 $ℓ_{A} \leq ℓ_{B}$ , $q_{A}^{ℓ_{A}} q_{B}^{ℓ_{B}} = q_{A \cup B}^{ℓ_{B}} q_{B}^{ℓ_{A} - ℓ_{B}}$ 当 $ℓ_{A} \geq ℓ_{B}$ $⟺ q_{A} q_{B} = q_{A \cup B} ⟺ p_{A} p_{B} = p_{A \cap B} ⟺ A, B 相互独立$ . (5) $P (τ_{[2]} = ℓ) = P (τ_{A} - τ_{B} = ℓ) + P (τ_{B} - τ_{A} = ℓ) = P (B 先出现) P (再过 ℓ 次 A 才出现) + P (A 先出现) P (再过 ℓ 次 B 才出现) = \frac{p _{B}}{p _{A} + p _{B}} Geo (p_{A}, ℓ) + \frac{p _{A}}{p _{A} + p _{B}} Geo (p_{B}, ℓ),$ 即 $τ_{[2]} \sim \frac{p _{B}}{p _{A} + p _{B}} Geo (p_{A}) + \frac{p _{A}}{p _{A} + p _{B}} Geo (p_{B})$ . 特别地, $p_{A} = p_{B} : = p$ 时, $τ_{[2]} \sim Geo (p)$ . $τ_{[1]}, τ_{[2]}$ 相互独立是因为 $P (τ_{[1]} = ℓ_{1}, τ_{[2]} = ℓ_{2}) = P (τ_{A} = ℓ_{1} + ℓ_{2}, τ_{B} = ℓ_{1}) + P (τ_{A} = ℓ_{1}, τ_{B} = ℓ_{1} + ℓ_{2}) = p_{A ⊎ B}^{ℓ_{1} - 1} p_{B} Geo (p_{A}, ℓ_{2}) + p_{A ⊎ B}^{ℓ_{1} - 1} p_{A} Geo (p_{B}, ℓ_{2}) = \frac{Geo ( p _{A ⊎ B} , ℓ _{1} )}{p _{A} + p _{B}} p_{B} Geo (p_{A}, ℓ_{2}) + \frac{Geo ( p _{A ⊎ B} , ℓ _{1} )}{p _{A} + p _{B}} p_{A} Geo (p_{B}, ℓ_{2}) = Geo (p_{A ⊎ B}, ℓ_{1}) (\frac{p _{B}}{p _{A} + p _{B}} Geo (p_{A}, ℓ_{2}) + \frac{p _{A}}{p _{A} + p _{B}} Geo (p_{B}, ℓ_{2})) = P (τ_{[1]} = ℓ_{1}) P (τ_{[2]} = ℓ_{2}) .$ (6) $P (τ_{[k]} = ℓ) = i_{1}, \dots, i_{k - 1} \sum P (已出现的是 A_{i_{1}}, \dots, A_{i_{k - 1}}) P (τ_{A_{k} ⊎ \dots ⊎ A_{n}} = ℓ) = Geo ((n - k + 1) p, ℓ) .$ $P (τ_{[1]} = ℓ_{1}, \dots, τ_{[n]} = ℓ_{n}) = i_{1}, \dots, i_{n} \sum P (τ_{i_{1}} = ℓ_{1}, \dots, τ_{i_{n}} = ℓ_{1} + \dots + ℓ_{n}) = n! (1 - n p)^{ℓ_{1} - 1} p \dots (1 - p)^{ℓ_{n} - 1} p = Geo (n p, ℓ_{1}) \dots Geo (p, ℓ_{n}) = P (τ_{[1]} = ℓ_{1}) \dots P (τ_{[n]} = ℓ_{n}) .$
5.30	对所有 $1 \leq k \leq n - 1$ , 各 $P (X = k, X + Y = n) = P (X = k, Y = n - k) = P (X = k) P (Y = n - k) = p q^{k - 1} p q^{n - k - 1} = p^{2} q^{n - 2}$ 相同, 又 $k = 1 \sum n - 1 P (X = k, X + Y = n) = P (X + Y = n),$ 故 $P (X = k ∣ ∣ X + Y = n) = \frac{P ( X = k , X + Y = n )}{P ( X + Y = n )} = \frac{1}{n - 1} .$
5.31	注: 熵 (entropy) 是反映分布为 $p_{1}, \dots, p_{N}$ 的随机变量的不确定性的值. 例如, 分布为 $1, 0, \dots, 0$ 的随机变量不具有任何不确定性, 熵为 $0$ ; 分布为 $\frac{1}{N}, \dots, \frac{1}{N}$ 的随机变量具有最大的不确定性, 毕竟都不能说哪个事件的可能性更大些, 熵最大. (1) $(x ln x)^{''} = (ln x + 1)^{'} > 0$ 得 $x ln x$ 是凸函数, 由 Jensen 不等式, $H (X) \leq - N_{X} \cdot \frac{1}{N _{X}} ln \frac{1}{N _{X}} = ln N_{X}$ ; (2) 用 $ln x \leq x - 1$ , 等号当且仅当 $x = 1$ . 由 $H (X, Y) = - x, y \sum P (X = x, Y = y) ln P (X = x, Y = y),$ $H (X) + H (Y) = - x \sum P (X = x) ln P (X = x) - y \sum P (Y = y) ln P (Y = y) = - x, y \sum P (X = x, Y = y) ln P (X = x) - x, y \sum P (X = x, Y = y) ln P (Y = y),$ 有 $H (X) + H (Y) - H (X, Y) = - x, y \sum P (X = x, Y = y) ln \frac{P ( X = x ) P ( Y = y )}{P ( X = x , Y = y )} \geq - x, y \sum P (X = x, Y = y) (\frac{P ( X = x ) P ( Y = y )}{P ( X = x , Y = y )} - 1) = - x, y \sum P (X = x) P (Y = y) - P (X = x, Y = y) = 0,$ 等号成立当且仅当 $P (X = x) P (Y = y) / P (X = x, Y = y) = 1$ 对任意的 $x, y$ , 即 $X, Y$ 相互独立; (3) $H (X, Y) = - x, y \sum P (X = x, Y = y) ln P (X = x, Y = y) = - x, y \sum P (X = x, Y = y) (ln P (X = x) + ln P (Y = y ∣ ∣ X = x)) = H (X) + H (Y ∣ ∣ X) .$
5.34	$L (p) = P ⎝ ⎛ ⎝ ⎛ X_{1} ⋮ X_{n} ⎠ ⎞ = ⎝ ⎛ x_{1} ⋮ x_{n} ⎠ ⎞ ⎠ ⎞ = p^{n} (1 - p)^{x_{1} + \dots + x_{n} - n},$ 令 $\frac{\partial ln L ( p )}{\partial p} = \frac{n}{p} + \frac{x _{1} + \dots + x _{n} - n}{p - 1} = 0$ , 解得 $\overset{p}{^} = \frac{n}{x _{1} + \dots + x _{n}}$ .

6数学期望与分布律

6.5	$E [τ] = k = 1 \sum \infty k p q^{k - 1} = p k = 1 \sum \infty (x^{k})^{'} ∣ ∣_{x = q} = p (\frac{1}{1 - x})^{'} ∣ ∣_{x = q} = p \frac{1}{( 1 - q ) ^{2}} = \frac{1}{p}$ , $E [τ (τ - 1)] = k = 1 \sum \infty k (k - 1) p q^{k - 1} = pq k = 1 \sum \infty (x^{k})^{''} ∣ ∣_{x = q} = pq (\frac{1}{1 - x})^{''} ∣ ∣_{x = q} = pq \frac{2}{( 1 - q ) ^{3}} = \frac{2 q}{p ^{2}}$ , $V ar (τ) = E [τ^{2}] - (E τ)^{2} = \frac{2 q}{p ^{2}} + \frac{1}{p} - \frac{1}{p ^{2}} = \frac{q}{p ^{2}}$ .
6.8	$\int F (x + a) - F (x) d x = \int \int_{(x, x + a]} d P d x = \int \int_{[x, x + a)} d x d P = a$ .
6.9	注意 $[F (x + a) - F (x)]^{2023} \leq F (x + a) - F (x)$ , 等号成立当且仅当 $P (x, x + a] = F (x + a) - F (x) = 0 或 1$ . 不能对所有的 $x$ 都 $P (x - a, x] = 0$ , 即存在 $\overset{x}{ˉ}$ 使 $P (\overset{x}{ˉ} - a, \overset{x}{ˉ}] = 1$ . 令 $b = in f {x ∣ P (x - a, x] = 1} \leq b_{0}$ , 则 $\forall b^{-} < b, P (b^{-} - a, b^{-}] = 0$ , 不然 $P (b^{-} - a, b^{-}] = 1$ , 与 $b$ 的最小性矛盾; 且 $\forall b^{+} > b, P (b^{+}, \infty) = 0$ , 不然对任何 $b^{'} \in [b, b^{+}), P (b^{'} - a, b^{'}] < 1$ 故 $= 0$ , 同样与 $b$ 的定义矛盾. 于是 $P (X = b) = P (\overset{x}{ˉ} - a, \overset{x}{ˉ}] - P (\overset{x}{ˉ} - a, b) - P (b, \overset{x}{ˉ}] = 1$ . 反过来显然.
6.10	分部积分有 $a F (a) - \int_{0}^{F (a)} x d F (x) = - a (1 - F (a)) - \int_{F (a)}^{1} x d (1 - F (x)),$ 得 $a = \int_{0}^{1} x d F (x) = E X .$
6.12	由于 $E X^{+} + E X^{-} = \int_{0}^{\infty} P (X^{+} > x) d x + \int_{0}^{\infty} P (X^{-} > x) d x = \int_{0}^{\infty} P (∣ X ∣ > x) d x,$ 且 $n = 1 \sum \infty P (∣ X ∣ \geq n) \leq \int_{0}^{\infty} P (∣ X ∣ > x) d x \leq n = 0 \sum \infty P (∣ X ∣ \geq n) .$
6.19	即要证 $(k = 1 \sum n E (X_{k} - E X_{k}) (Y_{k} - E Y_{k}))^{2} \leq k = 1 \sum n E (X_{k} - E X_{k})^{2} k = 1 \sum n E (Y_{k} - E Y_{k})^{2},$ 交换 $k = 1 \sum n$ 和 $E$ 并用 Cauchy 不等式即可.
6.34	对 ${1, \dots, n}$ 的各种排列 ${i_{1}, \dots, i_{n}}$ , $(X_{i_{1}}, \dots, X_{i_{n}})$ 相互独立且同分布, 由推论 5.1.1, $\frac{X _{i_{1}} + \dots + X _{i_{m}}}{X _{i_{1}} + \dots + X _{i_{n}}}$ 同分布故同期望, 其平均值是 (从而平均期望也是) $\frac{m}{n}$ , 故期望都是 $\frac{m}{n}$ .
6.35	即 $k = 0 \sum \infty \frac{λ ^{k}}{k !} e^{- λ} k g (k) = λ k = 0 \sum \infty \frac{λ ^{k}}{k !} e^{- λ} g (k + 1),$ 左右皆是 $k = 1 \sum \infty \frac{λ ^{k}}{( k - 1 )!} e^{- λ} g (k)$ .
6.38	$f (X) - f (X^{'}), g (X) - g (X^{'})$ 均与 $X - X^{'}$ 同号, 所以 $2 (E [f (X) g (X)] - E [f (X)] E [g (X)]) = E [(f (X) - f (X^{'})) (g (X) - g (X^{'}))] \geq 0.$

7条件数学期望与条件分布律

7.1

当 $E [X^{2}] < \infty$ , $E [(X - Y)^{2}] = 2 (E [X^{2}] - E [X Y]) \leq 4 E [X^{2}] < \infty$ .

当 $E [(X - Y)^{2}] < \infty$ , 即 $\int E [(X - y)^{2}] d F (y) < \infty$ , 那么对某个 $y$ 有 $E [(X - y)^{2}] < \infty$ , 于是 $E [X^{2}] \leq E [2 ((X - y)^{2} + y^{2})] < \infty$ .

7.3

由定理 6.3.4, 对于任意的 $y$ , $E ∣ X - y ∣ = \int_{0}^{+ \infty} P (∣ X - y ∣ > x) d x = \int_{0}^{+ \infty} (P (X > y + x) + P (X < y - x)) d x = \int_{0}^{+ \infty} (1 - F (y + x)) d x + \int_{0}^{+ \infty} F (y - x) d x = \int_{y}^{+ \infty} (1 - F (x)) d x + \int_{- \infty}^{y} F (x) d x = \int_{- \infty}^{+ \infty} ((1 - F (x)) 1_{{y \leq x}} + F (x) 1_{{y \geq x}}) d x,$ 然后用定理 7.3.3, 以及 Fubini 定理, $E ∣ X - Y ∣ = E [E ∣ X - y ∣ ∣ ∣_{y = Y}] = E [\int_{- \infty}^{+ \infty} ((1 - F (x)) 1_{{Y \leq x}} + F (x) 1_{{Y \geq x}}) d x] = \int_{- \infty}^{+ \infty} ((1 - F (x)) E [1_{{Y \leq x}}] + F (x) E [1_{{Y \geq x}}]) d x = \int_{- \infty}^{+ \infty} [(1 - F (x)) F (x) + F (x) (1 - F (x))] d x = 2 \int_{- \infty}^{+ \infty} F (x) (1 - F (x)) d x .$

$E [∣ X - m ∣] = - \int_{{x - m < 0}} x - m d F (x) - \int_{{x - m \geq 0}} x - m d (1 - F (x)) = \int_{(- \infty, 0)} F (x) d (x - m) + \int_{[0, \infty)} 1 - F (x) d (x - m) \leq \int_{(- \infty, 0)} 2 [1 - F (x)] F (x) d (x - m) + \int_{[0, \infty)} 2 F (x) [1 - F (x)] d (x - m) = E [∣ X - Y ∣] .$

7.5

与习题 7.3 相同, 对于任意的 $y$ , 有 $E ∣ X - y ∣ = \int_{y}^{+ \infty} (1 - F (x)) d x + \int_{- \infty}^{y} F (x) d x,$ 从而 $E ∣ X - m_{X} ∣ = \int_{m_{X}}^{+ \infty} (1 - F (x)) d x + \int_{- \infty}^{m_{X}} F (x) d x .$ 若 $y \geq m_{X}$ , $E ∣ X - y ∣ - E ∣ X - m_{X} ∣ = - \int_{m_{X}}^{y} (1 - F (x)) d x + \int_{m_{X}}^{y} F (x) d x = \int_{m_{X}}^{y} (2 F (x) - 1) d x \geq 0,$ 若 $y < m_{X}$ , $E ∣ X - y ∣ - E ∣ X - m_{X} ∣ = \int_{y}^{m_{X}} (1 - F (x)) d x - \int_{y}^{m_{X}} F (x) d x = \int_{y}^{m_{X}} (1 - 2 F (x)) d x \geq 0.$ 综上, 有 $E ∣ X - m_{X} ∣ \leq E ∣ X - y ∣$ .

由定理 7.3.3, $E ∣ X - Y ∣ = E [E ∣ X - y ∣ ∣ ∣_{y = Y}] \geq E [E ∣ X - m_{X} ∣ ∣ ∣_{y = Y}] = E ∣ X - m_{X} ∣.$

7.12

(1)	$E S_{τ} = \int E (S_{t}) d F (t) = \int t E X_{1} d F (t) = (E τ) \cdot E X_{1} .$
(2)	$E [S_{τ}^{2}] = \int E (S_{t}^{2}) d F (t) = \int t E [X_{1}^{2}] + (t^{2} - t) E (X_{1} X_{2}) d F (t) = (E τ) \cdot E [X_{1}^{2}] + (E [τ^{2}] - E τ) (E X_{1})^{2},$ 于是 $V ar (S_{τ}) = E [S_{τ}^{2}] - (E S_{τ})^{2} = (E τ) \cdot (E [X_{1}^{2}] - (E X_{1})^{2}) + (E X_{1})^{2} (E [τ^{2}] - (E τ)^{2}) .$

7.23

知道! $φ (E [ξ ∣ G]) = φ (\int_{B} ξ d P ∣_{G}) = (a, b) \in S sup a \int_{B} ξ d P ∣_{G} + b = (a, b) \in S sup \int_{B} a ξ + b d P ∣_{G} \leq \int_{B} φ (ξ) d P ∣_{G} = E [φ (ξ) ∣ G] .$

8随机变量及其分布律 (II)

例 8.19	基于 $X_{1} = \frac{T _{1} / m}{T _{2} / n}, X_{2} = T_{2}$ 得到 $\iint ρ_{1, 2} (x_{1}, x_{2}) d x_{1} d x_{2} = \iint \frac{( \frac{1}{2} ) ^{\frac{m}{2}} ( \frac{m}{n} x _{1} x _{2} ) ^{\frac{m}{2} - 1}}{Γ ( \frac{m}{2} )} e^{- \frac{m}{n} \cdot \frac{x _{1} x _{2}}{2}} 1_{(0, \infty)} (\frac{m}{n} x_{1} x_{2}) \frac{( \frac{1}{2} ) ^{\frac{n}{2}} x _{2}^{\frac{n}{2} - 1}}{Γ ( \frac{n}{2} )} e^{- \frac{x _{2}}{2}} 1_{(0, \infty)} (x_{2}) d (\frac{m}{n} x_{1} x_{2}) d x_{2},$ 从而 $ρ_{1, 2} (x_{1}, x_{2}) = \frac{( \frac{1}{2} ) ^{\frac{m}{2}} ( \frac{m}{n} x _{1} x _{2} ) ^{\frac{m}{2} - 1}}{Γ ( \frac{m}{2} )} e^{- \frac{m}{n} \frac{x _{1} x _{2}}{2}} 1_{(0, \infty)} (x_{1}) \frac{( \frac{1}{2} ) ^{\frac{n}{2}} x _{2}^{\frac{n}{2} - 1}}{Γ ( \frac{n}{2} )} e^{- \frac{x _{2}}{2}} 1_{(0, \infty)} (x_{2}) \frac{m}{n} x_{2} .$ 进而 $f_{m, n} (x) = \int ρ_{1, 2} (x, x_{2}) d x_{2} = \int_{0}^{\infty} \frac{( \frac{m}{n} \frac{x x _{2}}{2} ) ^{\frac{m}{2} - 1}}{Γ ( \frac{m}{2} )} e^{- \frac{m}{n} \frac{x x _{2}}{2}} 1_{(0, \infty)} (x) \frac{( \frac{x _{2}}{2} ) ^{\frac{n}{2} - 1}}{Γ ( \frac{n}{2} )} e^{- \frac{x _{2}}{2}} x_{2} d (\frac{m}{n} \frac{x _{2}}{2}) = \frac{1}{Γ ( \frac{m}{2} ) Γ ( \frac{n}{2} )} (\frac{m}{n})^{\frac{m}{2}} x^{\frac{m}{2} - 1} \int_{0}^{\infty} (\frac{x _{2}}{2})^{\frac{m + n}{2} - 1} e^{- (1 + \frac{m}{n} x) \frac{x _{2}}{2}} x_{2} d (\frac{x _{2}}{2}) 1_{(0, \infty)} (x) = \frac{Γ ( \frac{m + n}{2} )}{Γ ( \frac{m}{2} ) Γ ( \frac{n}{2} )} (\frac{m}{n})^{\frac{m}{2}} x^{\frac{m}{2} - 1} (1 + \frac{m}{n} x)^{- \frac{m + n}{2}} 1_{(0, \infty)} (x) .$
$思考题$	设 $X_{1}, X_{2}$ 独立同分布于 $N (0, 1)$ . 证明: $(X_{1}, X_{2}) = d (X_{1} sgn (X_{2}), X_{2})$ . $P (X_{1} sgn (X_{2}) \leq x, X_{2} \leq y) = P (X_{1} \leq x, 0 < X_{2} \leq y) + P (0 \leq x, X_{2} = 0) + P (- X_{1} \leq x, X_{2} < 0) = P (X_{1} \leq x) P (0 < X_{2} \leq y) + 0 + P (- X_{1} \leq x) P (X_{2} < 0) = P (X_{1} \leq x) (P (0 < X_{2} \leq y) + 0 + P (X_{2} < 0)) = P (X_{1} \leq x, X_{2} \leq y) .$
$一个引理$	设 $X, Y$ 为连续型变量, 且其联合分布的概率密度函数 $ρ_{X, Y}$ 可以分离变量,i.e. $ρ_{X, Y} (x, y) = ρ_{1} (x) \cdot ρ_{2} (y)$ , 并且 $\int_{R} ρ_{1} = \int_{R} ρ_{2} = 1$ , 那么 $X, Y$ 独立, 且 $ρ_{1}, ρ_{2}$ 分别为 $X, Y$ 的密度函数. 证明. 由于 $X, Y$ 是连续型变量, 故由其边缘分布性质可得: $ρ_{X} (x) = \int_{- \infty}^{+ \infty} ρ_{X, Y} (x, y) d y = ρ_{1} (x) \int_{- \infty}^{+ \infty} ρ_{2} (y) d y = ρ_{1} (x)$ . 同理 $ρ_{2} = ρ_{Y}$ , 因此由 $ρ_{X, Y} = ρ_{X} \cdot ρ_{Y}$ 知 $X, Y$ 独立. $□$
8.1	(1) 由 $Γ$ 的半生成性直接得到 $Y_{1} \sim Γ (α_{1} + α_{2}, λ)$ , 考虑变换 $ϕ : (X_{1}, X_{2}) \to (Y_{1}, Y_{2}), (x_{1}, x_{2}) \mapsto (x_{1} + x_{2}, \frac{x _{1}}{x _{1} + x _{2}})$ , 则 $∣ det (D ϕ^{- 1}) ∣ = ∣ - (x_{1} + x_{2}) ∣$ , 因此: $ρ_{Y_{1}, Y_{2}} (y_{1}, y_{2}) = ρ_{X_{1}, X_{2}} (ϕ^{- 1} (y_{1}, y_{2})) \cdot ∣ det (D ϕ^{- 1}) ∣ = ρ_{X_{1}, X_{2}} (y_{1} y_{2}, y_{1} (1 - y_{2})) \cdot ∣ y_{1} ∣ = ρ_{X_{1}} (y_{1} y_{2}) \cdot ρ_{X_{2}} (y_{1} (1 - y_{2})) \cdot ∣ y_{1} ∣ = \frac{λ ^{α_{1} + α_{2}}}{Γ ( α _{1} ) Γ ( α _{2} )} y_{1}^{α_{1} + α_{2} - 1} \cdot e^{- λ y_{1}} \cdot y_{2}^{α_{1} - 1} \cdot (1 - y_{2})^{α_{2} - 1} = (\frac{λ ^{α_{1} + α_{2}}}{Γ ( α _{1} + α _{2} )} y_{1}^{α_{1} + α_{2} - 1} \cdot e^{- λ y_{1}}) \cdot (\frac{y _{2}^{α_{1} - 1} ( 1 - y _{2} ) ^{α_{2} - 1}}{B ( α _{1} , α _{2} )}) .$ 并且容易注意到 $\int_{R} \frac{λ ^{α_{1} + α_{2}}}{Γ ( α _{1} + α _{2} )} y_{1}^{α_{1} + α_{2} - 1} \cdot e^{- λ y_{1}} d y_{1} = 1, \int_{R} \frac{y _{2}^{α_{1} - 1} ( 1 - y _{2} ) ^{α_{2} - 1}}{B ( α _{1} , α _{2} )} d y_{2} = 1$ . 因此由引理可知 $Y_{1}, Y_{2}$ 独立且 $Y_{2} \sim Beta (α_{1}, α_{2})$ . (2) 考虑变换 $ϕ : (X_{1}, X_{2}) \to (Y_{1}, Z_{2}), (x_{1}, x_{2}) \mapsto (x_{1} + x_{2}, \frac{x _{1}}{x _{2}})$ , 则 $D ϕ = [1 \frac{1}{x _{2}} 1 - \frac{x _{1}}{x _{2}^{2}}]$ , 因此 $∣ det (D ϕ^{- 1}) ∣ = ∣ - \frac{x _{1} + x _{2}}{x _{2}^{2}} ∣$ , 因此: $ρ_{Y_{1}, Z_{2}} (y_{1}, z_{2}) = ρ_{X_{1}, X_{2}} (ϕ^{- 1} (y_{1}, z_{2})) \cdot ∣ det (D ϕ^{- 1}) ∣ = ρ_{X_{1}, X_{2}} (\frac{z _{2} y _{1}}{z _{2} + 1}, \frac{y _{1}}{z _{2} + 1}) \cdot \frac{y _{1}}{( z _{2} + 1 ) ^{2}} = ρ_{X_{1}} (\frac{z _{2} y _{1}}{z _{2} + 1}) \cdot ρ_{X_{2}} (\frac{y _{1}}{z _{2} + 1}) \cdot \frac{y _{1}}{( z _{2} + 1 ) ^{2}} = \frac{λ ^{α_{1}} ( \frac{z _{2} y _{1}}{z _{2} + 1} ) ^{α_{1} - 1}}{Γ α _{1}} \cdot \frac{λ ^{α_{2}} ( \frac{y _{1}}{z _{2} + 1} ) ^{α_{2} - 1}}{Γ ( α _{2} )} \cdot e^{λ y_{1}} \cdot \frac{y _{1}}{( z _{2} + 1 ) ^{2}} = (\frac{λ ^{α_{1} + α_{2}} \cdot y _{1}^{α_{1} + α_{2} - 1} \cdot e ^{λ y_{1}}}{Γ ( α _{1} ) Γ ( α _{2} )}) \cdot (\frac{z _{2}^{α_{1} - 1}}{B ( α _{1} , α _{2} ) \cdot ( 1 + z _{2} ) ^{α_{1} + α_{2}}}) .$ 并且容易注意到: $\int_{R} \frac{λ ^{α_{1} + α_{2}} \cdot y _{1}^{α_{1} + α_{2} - 1} \cdot e ^{λ y_{1}}}{Γ ( α _{1} ) Γ ( α _{2} )} d y_{1} = 1, \int_{R} \frac{z _{2}^{α_{1} - 1}}{B ( α _{1} , α _{2} ) \cdot ( 1 + z _{2} ) ^{α_{1} + α_{2}}} d z_{2} = 1$ . 因此由引理可知 $Y_{1}, Z_{2}$ 独立, 并且 $Z_{2} \sim Beta (α_{1}, α_{2}) .$
8.2	(1) 利用定义容易得到: $ρ_{X^{2}} (t) = ρ_{X} (t) \cdot \frac{1}{2 t} \cdot 2 = \frac{e ^{\frac{t}{2}}}{2 π t} = \frac{\frac{1}{2} \cdot e ^{- \frac{t}{2}}}{Γ \frac{1}{2} \cdot t} .$ 因此 $X^{2} \sim Γ (\frac{1}{2}, \frac{1}{2})$ , 利用 $Γ$ 的半生成性得到: $χ^{2} (n) \sim Γ (\frac{n}{2}, \frac{1}{2})$ . (2) 此时 $ρ_{X^{2}} (t) = \frac{2 ρ ( t ; μ _{1} , 1 )}{2 t} = \frac{- e ^{\frac{( t - μ _{1} ) ^{2}}{2}}}{2 π t} .$

9随机变量及其分布律 (III)

9.4	下证明: 对任意 $M > 0$ , 有 $P (x_{1} = x_{2} \in [- M, M]) = 0$ . (在此基础上, 利用 ${x_{1} = x_{2}} = \cup_{M \in Z} {x_{1} = x_{2} \in [- M, M]}$ 即可得证!) 由于 $F_{1}$ 连续, 得知 $F_{1}$ 在 $[- M, M]$ 中一致连续, 因此 $\forall ϵ > 0,$ 存在一个对 $[- M, M]$ 的剖分: $- M = m_{0} < m_{1} < \dots < m_{N} = M$ 使得 $F_{1} (m_{i}) - F_{1} (m_{i - 1}) < ϵ, \forall i = 1, 2, \dots, N$ , 那么: $P (x_{1} = x_{2} \in [- M, M]) = P (x_{1} = x_{2} \in i = 1 ⋃ N [m_{i - 1}, m_{i})) \leq i = 1 \sum N P (x_{1} \in [m_{i - 1}, m_{i}) 且 x_{2} \in [m_{i - 1}, m_{i})) = i = 1 \sum N P (x_{1} \in [m_{i - 1}, m_{i})) \cdot P (x_{2} \in [m_{i - 1}, m_{i})) \leq i = 1 \sum N ϵ \cdot P (x_{2} \in [m_{i - 1}, m_{i})) = ϵ \cdot P (x_{2} \in [- M, M]) .$ 然后令 $ϵ \to 0$ , 得到 $P (x_{1} = x_{2} \in [- M, M]) = 0$ . 因而得证!
9.7	即说明对 $(0, 1)$ 的所有 Borel 子集 $A$ 有 $P (n = 1 \sum \infty \frac{X _{n}}{2 ^{n}} \in A) = ∣ A ∣.$ 此式对二进制区间 $A = [a, a + \frac{1}{2 ^{m}})$ 成立, 其中 $a = n = 1 \sum m \frac{a _{n}}{2 ^{n}}$ , 这是因为 $P (n = 1 \sum \infty \frac{X _{n}}{2 ^{n}} \in [a, a + \frac{1}{2 ^{m}})) = P (X_{1} = a_{1}, \dots, X_{m} = a_{m}) = (\frac{1}{2})^{m} = ∣ ∣ [a, a + \frac{1}{2 ^{m}}) ∣ ∣ .$ 对一般的 $A$ , 对任何 $ε > 0$ , 存在二进制区间的有限组合 $A^{+}, A^{-}$ , 使得 $A^{-} \subset A \subset A^{+}$ 且 $∣ A^{+} ∖ A ∣, ∣ A ∖ A^{-} ∣ < ε$ , 用下式完成逼近 $∣ A ∣ - ε < ∣ A^{-} ∣ < P (n = 1 \sum \infty \frac{X _{n}}{2 ^{n}} \in A) < ∣ A^{+} ∣ < ∣ A ∣ + ε .$
9.11	由习题 8.12, 可以设 ${X_{1} = R cos Θ X_{2} = R sin Θ,$ 那么有 $R, Θ$ 互相独立. 记 $Y = \frac{X _{1} X _{2}}{X _{1}^{2} + X _{2}^{2}} = R cos Θ sin Θ = \frac{R s i n 2Θ}{2}$ , 由于 $R$ 与 $sin Θ$ 独立, 那么 $R$ 与 $sin 2Θ$ 也独立, 并且 $sin Θ$ 和 $sin 2Θ$ 同分布 (由于 $[0, 1)$ 的均匀分布乘 $2$ 再取小数部分还是 $[0, 1)$ 均匀分布), 故: $\frac{R sin 2Θ}{2} \sim \frac{R sin Θ}{2} \sim \frac{1}{2} X_{1} \sim N (0, \frac{1}{4}) .$ $□$
定理 9.2.3	证明如下. $P (\dots, U_{(k)} \in (x_{k}, x_{k} + d x_{k}], \frac{U _{(k + 1)} - U _{(k)}}{U _{(ℓ)} - U _{(k)}} \in (t_{k + 1}, t_{k + 1} + d t_{k + 1}], \dots, \frac{U _{(ℓ - 1)} - U _{(k)}}{U _{(ℓ)} - U _{(k)}} \in (t_{ℓ - 1}, t_{ℓ - 1} + d t_{ℓ - 1}], U_{(ℓ)} \in (x_{ℓ}, x_{ℓ} + d x_{ℓ}], \dots) = P (\dots, U_{(k)} \in (x_{k}, x_{k} + d x_{k}], U_{(k + 1)} \in (t_{k + 1} U_{(ℓ)} + (1 - t_{k + 1}) U_{(k)}, (t_{k + 1} + d t_{k + 1}) U_{(ℓ)} + (1 - t_{k + 1} - d t_{k + 1}) U_{(k)}], \dots, U_{(ℓ - 1)} \in (t_{ℓ - 1} U_{(ℓ)} + (1 - t_{ℓ - 1}) U_{(k)}, (t_{ℓ - 1} + d t_{ℓ - 1}) U_{(ℓ)} + (1 - t_{ℓ - 1} - d t_{ℓ - 1}) U_{(k)}], U_{(ℓ)} \in (x_{ℓ}, x_{ℓ} + d x_{ℓ}], \dots) = (ρ_{(1, \dots, n)} (x_{1}, \dots, x_{k}, t_{k + 1} x_{ℓ} + (1 - t_{k + 1}) x_{k}, \dots, t_{ℓ - 1} x_{ℓ} + (1 - t_{ℓ - 1}) x_{k}, x_{ℓ}, \dots, x_{n}) + o (1)) d x_{1} \dots d x_{k} (x_{ℓ} - x_{k}) d x_{k + 1} \dots (x_{ℓ} - x_{k}) d x_{ℓ - 1} d x_{ℓ} \dots d x_{n} = n! (x_{ℓ} - x_{k})^{ℓ - k - 1} 1_{{0 < x_{1} < \dots < x_{k} < t_{k + 1} x_{ℓ} + (1 - t_{k + 1}) x_{k} < \dots < t_{ℓ - 1} x_{ℓ} + (1 - t_{ℓ - 1}) x_{k} < x_{ℓ} < \dots < x_{n} < 1}} d x_{1} \dots d x_{n} = \frac{n !}{( ℓ - k - 1 )!} (ℓ - k - 1)! (x_{ℓ} - x_{k})^{ℓ - k - 1} 1_{{0 < x_{1} < \dots < x_{k} < x_{ℓ} < \dots < x_{n} < 1}} 1_{{0 < t_{k + 1} < \dots < t_{ℓ - 1} < 1}} d x_{1} \dots d x_{n} = ρ_{(1, \dots, k, ℓ, \dots, n)} (x_{1}, \dots, x_{k}, x_{ℓ}, \dots, x_{n}) ρ_{(1, \dots, ℓ - k - 1)} (t_{k + 1}, \dots, t_{ℓ - 1}) d x_{1} \dots d x_{n} .$

10随机变量列的收敛与大数律

10.2	设 $X_{n} ⟶ P 0$ , 则 $E [\frac{∣ X _{n} ∣}{1 + ∣ X _{n} ∣}] = \int_{∣ X_{n} ∣ \geq ε} \frac{∣ X _{n} ∣}{1 + ∣ X _{n} ∣} d P + \int_{∣ X_{n} ∣ < ε} \frac{∣ X _{n} ∣}{1 + ∣ X _{n} ∣} d P \leq P (∣ X_{n} ∣ \geq ε) + \frac{ε}{1 + ε},$ 令 $ε$ 小再令 $n$ 大, 上式就任意小. 反过来, 设 $X_{n} ⟶ P 0$ 不成立, 即存在 $ε, δ$ 使得存在任意大的 $n$ 使 $P (∣ X_{n} ∣ \geq ε) \geq δ$ , 则 $E [\frac{∣ X _{n} ∣}{1 + ∣ X _{n} ∣}] \geq δ \frac{ε}{1 + ε} .$
10.3	$x E [\frac{1}{X} \cdot 1_{{X > x}}] = x \int_{{X > x}} \frac{d P ( X )}{X} \leq \int_{{X > x}} d P (X) = P (X > x) \to 0.$
10.20	(i) 对 $ξ_{n} ⟶ P 0$ , 对任意 $ε > 0$ , 取 $N$ 使 $\forall n > N, P (∣ ξ_{n} ∣ \geq 1) < ε$ , 对 $k = 1, \dots, N$ , 取 $M_{k}$ 使 $P (∣ ξ_{k} ∣ \geq M_{k}) < ε$ , 令 $M = max (1, M_{1}, \dots, M_{N})$ . (ii) $P (∣ ξ_{n} + η_{n} ∣ \geq ε) \leq P (∣ ξ_{n} ∣ \geq \frac{ε}{2}) + P (∣ η_{n} ∣ \geq \frac{ε}{2}) \to 0$ . $sup_{n} P (∣ ξ_{n} + η_{n} ∣ \geq 2 M) \leq sup_{n} P (∣ ξ_{n} ∣ \geq M) + sup_{n} P (∣ η_{n} ∣ \geq M)$ 充分小. (iii) $P (∣ ξ_{n} η_{n} ∣ \geq ε) \leq P (∣ ξ_{n} ∣ \geq \frac{ε}{M}) + sup_{n} P (∣ η_{n} ∣ \geq M)$ , 令 $M$ 大再令 $n$ 大. $sup_{n} P (∣ ξ_{n} η_{n} ∣ \geq M^{2}) \leq sup_{n} P (∣ ξ_{n} ∣ \geq M) + sup_{n} P (∣ η_{n} ∣ \geq M)$ 充分小.
10.28	先证明一个引理: $N$ 个人至少有一个人拿对了信封的概率是 $k = 1 \sum N \frac{( - 1 ) ^{(k - 1)}}{k !} .$ 记 $A_{i}$ 表示第 $i$ 个人拿到对的信封, $P (A_{i}) = \frac{1}{N}, P (A_{i} A_{j}) = \frac{1}{N ( N - 1 )}$ , 其余同理, 则引理所求事件的概率为 $P (⋃_{i = 1}^{N} A_{i}) = \sum_{i = 1}^{N} \frac{( - 1 ) ^{(i - 1)}}{i !}$ (用容斥原理). 接下来考虑 $N_{n} = k, n$ 封信中挑出 $k$ 个有 $(k n)$ 种取法, 对于指定的 $k$ 个人拿对信封的概率为 $\frac{1}{n ( n - 1 ) \dots ( n - k + 1 )}$ , 对于剩下的 $n - k$ 个人, 利用引理, 可知他们都没拿对的概率为 $1 - \sum_{i = 1}^{n - k} \frac{( - 1 ) ^{(i - 1)}}{i !} = \sum_{i = 0}^{n - k} \frac{( - 1 ) ^{i}}{i !}$ ．于是 $P (N_{n} = k) = (k n) \frac{1}{n ( n - 1 ) \dots ( n - k + 1 )} \sum_{i = 0}^{n - k} \frac{( - 1 ) ^{i}}{i !}$ 令 $n \to \infty, P (N_{n} = k) \to \frac{e ^{- 1}}{k !} ．$ , 从而得证!
10.31	考虑 $A_{n, t} = {X_{n} \geq t a_{n}}$ , 其中 $a_{n} = - lo g_{q} n$ ． $\sum_{n = 1}^{\infty} P (A_{n, t}) = q^{[t a_{n}]}$ , 将 $a_{n}$ 写成以 $e$ 为底的形式, 可以发现当 $t > 1$ 时级数收敛; 当 $t \leq 1$ 时, 级数发散．根据 Borel-Cantelli 引理, 再结合 10.30 的结论可得到关于上极限的结论．接下来, 因为 $max {X_{1}, X_{2} \dots X_{n}} \geq X_{n}$ , 因此同除 $a_{n}$ , 再令 $n \to \infty$ , 得到下极限对应的结论．
10.40	先注意到一个事实: $a, b \in [0, 1]$ , 则 $(1 - a - b)^{n} \leq (1 - a)^{n} (1 - b)^{n}$ . 设离散型分布的值域为 $S$ (为一个至多可列集合) , 记 $p_{x} = P (X_{1} = x)$ , 则 $R_{n} = \sum_{x \in S} 1_{x \in {X_{1}, \dots X_{n}}}$ , $E R_{n} = \sum_{x \in S} (1 - (1 - p_{x})^{n})$ , $Var (R_{n}) = x \in S, y \in S \sum Cov (1_{x \in {X_{1} \dots X_{n}}}, 1_{y \in {X_{1} \dots X_{n}}}) = x \in S \sum Var (1_{x \in {X_{1} \dots X_{n}}}) + x \neq = y \sum Cov (1_{x \in {X_{1} \dots X_{n}}}, 1_{y \in {X_{1} \dots X_{n}}}) .$ 记 $A_{x} = {x \in / {X_{1} \dots X_{n}}}$ , 下面利用注意到的事实和 $n = 2$ 的容斥原理, 易证交叉项非正 (不严谨的说, 交叉项的两个事件是负相关的) 由 $E R_{n}$ 的表达式, 易验证满足定理 10．3．2 条件.

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