摘要訊息 : 隨機變數, 期望以及一些它們的特徵和性質.

0. 前言

在本文中, 我們將要討論具有有限個結局試驗的機率模型可能產生的一些數字特徵.

更新紀錄 :

• 2022 年 6 月 10 日進行第一次更新和修正.

1. 隨機變數

設 $(\Omega, \mathscr {A}, \mathbf {P})$ 是某個具有有限個結局試驗的機率模型. 其中, $\mathscr {A} = \left \{ A : A \subseteq \Omega \right \}$. 之前所討論的關於事件 $A \in \mathscr {A}$, 其機率的計算以及基本事件空間 $\Omega$ 的自然本性並不重要, 重要的是某種數字特徵, 它依賴於基本事件. 我們想要知道的是, 在一系列 $n$ 次試驗中, 出現 $k$ 次 ($k \leq n$) 成功的機率如何等這一類問題.

定義 1. 稱任意定義在有限基本事件空間 $\Omega$ 上的數值函數 $\xi = \xi(\omega)$ 為隨機變數 (random variable).

例題 1. 對於接連兩次投擲硬幣模型, 其基本事件空間為 $\displaystyle {\Omega = \left \{ \mathrm {HH}, \mathrm {HT}, \mathrm {TH}, \mathrm {TT} \right \}}.$ 我們使用下列表格定義隨機變數 $\xi = \xi(\omega)$ :

其中, $\xi(\omega)$ 是事件 $\omega$ 中, 投擲硬幣得到正面的次數.

通過例題 1, 我們不難得到關於隨機變數 $\xi$ 的另外一個簡單例子, 即集合的特徵函數 $\displaystyle {\xi = \mu(\omega)}.$

當實驗者遇到描繪某些記載或者讀數的隨機變數時, 他關心的基本問題是隨機變數取各個數值的機率如何? 從這種觀點出發, 實驗者關心的並非機率 $\mathop {\mathbf {P}}$ 在 $(\Omega, \mathscr {A})$ 上的分佈, 而是機率在隨機變數之可能值的集合上的分佈, 即隨機變數 $\xi$ 取到集合 $C = \left \{ c_{1}, c_{2}, ..., c_{n} \right \}$ 中某個值的機率分佈. 由於所研究的情形, $\Omega$ 由有限個點所構成, 則隨機變數 $\xi$ 的值域 $R_{\xi}$ 也是有限的. 設 $R_{\xi} = \left \{ r_{1}, r_{2}, ..., r_{m} \right \}$, 其中, $r_{1}, r_{2}, ..., r_{m}$ 是 $\xi$ 的全部可能值.

記 $\mathscr {R}$ 是值域 $R_{\xi}$ 上的一切子集的全體, 並且設 $B \in \mathscr {R}$. 當 $R_{\xi}$ 是隨機變數 $\xi$ 的值域時, 集合 $B$ 也可以視為某個事件.

在 $(R_{\xi}, \mathscr {R})$ 上考慮由隨機變數 $\xi$ 按 $\displaystyle {P_{\xi}(B) = \mathop {\mathbf {P}} \left \{ \omega : \xi(\omega) \in B \right \}, B \in \mathscr {R}}$ 產生的機率 $P_{\xi}(\cdot)$. 顯然, 這些機率的值完全取決於 $\displaystyle {P_{\xi}(r_{i}) = \mathop {\mathbf {P}} \left \{ \omega : \xi(\omega) = r_{i} \right \}, r_{i} \in R_{\xi}}.$ 機率組 $\left \{ P_{\xi}(r_{1}), P_{\xi}(r_{2}), ..., P_{\xi}(r_{m}) \right \}$ 稱作隨機變數 $\xi$ 的機率分佈 (probability distribution), 簡稱為分佈 (distribution).

例題 2-1. 設隨機變數 $\xi$ 分別以機率 $p$ 和 $q$ 取到 $1$ 和 $0$ 兩個值. 其中, $p$ 稱作成功的機率, $q$ 稱作失敗的機率. 則稱 $\xi$ 為 Bernoulli 隨機變數, 也稱隨機變數 $\xi$ 服從 Bernoulli 分佈. 顯然, 對於隨機變數 $\xi$ 有 $\displaystyle {P_{\xi}(x) = p^{x}q^{1 - x}, x \in \left \{ 0, 1 \right \}}.$

例題 2-2. 設隨機變數 $\xi$ 以機率 $P_{\xi}(x) = \binom {x}{n}p^{x}q^{n - x}$. 其中, $x = 0, 1, 2, …, n$. 取 $0, 1, ..., n$ 等 $n + 1$ 個值, 則稱隨機變數 $\xi$ 為二項隨機變數 (binomial random variable), 或者說隨機變數 $\xi$ 服從二項分佈.

我們可以發現, 在例題 2 中, 我們已經不關心基本機率空間 $(\Omega, \mathscr {A}, \mathbf {P})$ 的構造了, 而只是關心隨機變數的值及其機率分佈.

2. 分佈函數

定義 2. 設 $x \in R$, 函數 $\displaystyle {F_{\xi}(x) = \mathop {\mathbf {P}} \left \{ \omega : \xi(\omega) \leq x \right \}}$ 稱作隨機變數 $\xi$ 的分佈函數 (distribution function).

根據隨機變數分佈函數的定義, 我們容易得到 $\displaystyle {F_{\xi}(x) = \sum \limits_{\left \{ i : x_{i} \leq x \right \}}P_{\xi}(x_{i})}.$ 除此之外, $\displaystyle {P_{\xi}(x_{i}) = F_{\xi}(x_{i}) - F_{\xi}(x_{i}^{-})}.$ 其中, $F_{\xi}(x_{i}^{-}) = \lim \limits_{x \to x_{i}^{-}}F_{\xi}(x)$. 特別地, 若有 $x_{1} < x_{2} < ... < x_{m}$ 且 $F_{\xi}(x_{0}) = 0$, 則有 $\displaystyle {P_{\xi}(x_{i}) = F_{\xi}(x_{i}) - F_{\xi}(x_{i - 1}), i = 1, 2, ..., n}.$

直接由定義 2, 我們可以得到分佈函數的三條性質 :

1. $\lim \limits_{x \to -\infty}F_{\xi}(x) = 0$;
2. $\lim \limits_{x \to +\infty}F_{\xi}(x) = 1$;
3. $F_{\xi}(x)$ 具有右連續性, 即 $\lim \limits_{x \to x_{0}^{+}}F_{\xi}(x_{0}) = x_{0}$, 且 $F_{\xi}(x)$ 為階梯函數.

除了隨機變數之外, 我們通常還研究隨機向量 (random vector) $\xi = (\xi_{1}, \xi_{2}, ..., \xi_{r})$, 其各分量都是隨機變數. 對於多項分佈來說, 對象為隨機向量 $\upsilon = (\upsilon_{1}, \upsilon_{2}, ..., \upsilon_{r})$. 其中, $\upsilon_{i} = \upsilon_{i}(\omega)$ 是序列 $\omega = (a_{1}, a_{2}, ..., a_{n})$ 中等於 $b_{i}$ ($i = 1, 2, …, r$) 的分量的數量.

對於 $x_{i} \in X_{i}$ ($X_{i}$ 是 $\xi_{i}$ 一切可能值的集合, $i = 1, 2, ..., r$), 機率 $\displaystyle {P_{\xi}(x_{1}, x_{2}, ..., x_{r}) = \mathop {\mathbf {P}} \left \{ \omega : \xi(\omega) = x_{1}, \xi_{2}(\omega) = x_{2}, ..., \xi_{r}(\omega) = x_{r} \right \}}$ 的全體稱為隨機向量 $\xi = (\xi_{1}, \xi_{2}, ..., \xi_{r})$ 的機率分佈, 而函數 $\displaystyle {F_{\xi}(x_{1}, x_{2}, ..., x_{r}) = \mathop {\mathbf {P}} \left \{ \omega : \xi_{1}(\omega) \leq x_{1}, \xi_{2}(\omega) \leq x_{2}, ..., \xi_{r}(\omega) \leq x_{r} \right \}}$ 稱為隨機向量 $\xi = (\xi_{1}, \xi_{2}, ..., \xi_{r})$ 的分佈函數. 其中, $x_{i} \in R\ (i = 1, 2, ..., r)$. 那麼對於多項分佈的向量 $\upsilon = (\upsilon_{1}, \upsilon_{2}, ..., \upsilon_{r})$, 有 $\displaystyle {P_{\upsilon}(n_{1}, n_{2}, ..., n_{r}) = \dbinom {n_{1}, n_{2}, ..., n_{r}}{n}p_{1}^{n_{1}}p_{2}^{n_{2}}...p_{r}^{n_{r}}}.$

定義 3. 設 $\xi_{1}, \xi_{2}, ..., \xi_{r}$ 是一組在 $R$ 中的有限集合 $X$ 上取值的隨機變數. 記 $\mathscr {X}$ 是 $X$ 中所有子集的代數. 若對於任意 $x_{1}, x_{2}, ..., x_{r} \in X$, 有 $\displaystyle {\mathop {\mathbf {P}} \left \{ \xi_{1} = x_{1}, \xi_{2} = x_{2}, ..., \xi_{r} = x_{r} \right \} = \mathop {\mathbf {P}} \left \{ \xi_{1} = x_{1} \right \}\mathop {\mathbf {P}} \left \{ \xi_{2} = x_{2} \right \}...\mathop {\mathbf {P}} \left \{ \xi_{r} = x_{r} \right \}};$ 或等價地, 對於任意 $B_{1}, B_{2}, ..., B_{r} \in \mathscr {X}$, 有 $\displaystyle {\mathop {\mathbf {P}} \left \{ \xi \in B_{1}, \xi_{2} \in B_{2}, ..., \xi_{r} \in B_{r} \right \} = \mathop {\mathbf {P}} \left \{ \xi_{1} \in B_{1} \right \}\mathop {\mathbf {P}} \left \{\xi_{2} \in B_{2} \right \}...\mathop {\mathbf {P}} \left \{ \xi_{r} \in B_{r} \right \}},$ 則稱隨機變數 $\xi_{1}, \xi_{2}, ..., \xi_{r}$ 為全體獨立 (mutually independent) 的.

我們之前所討論的 Bernoulli 概型, 就是獨立隨機變數的一個例子. 具體地, 設 $\displaystyle {\begin {aligned} &\Omega = \left \{ \omega : \omega = (a_{1}, a_{2}, ..., a_{n}), a_{i} = 0, 1 \ (i = 1, 2, ..., n) \right \}, \\ &\mathscr {A} = \left \{ A : A \subseteq \Omega \right \}, \mathop {\mathbf {P}}(\left \{ \omega \right \}) = p(\omega) = p^{\sum \limits_{i}a_{i}}(1 - p)^{n - \sum \limits_{i}a_{i}}, \end {aligned}}$ 並且對於 $\omega = (a_{1}, a_{2}, ..., a_{n})$, $\xi_{i}(\omega) = a_{i}\ (i = 1, 2, ..., n)$, 由於事件 $\displaystyle {A_{1} = \left \{ \omega : a_{1} = 1 \right \}, A_{2} = \left \{ \omega : a_{2} = 1 \right \}, ..., A_{n} = \left \{ \omega : a_{n} = 1 \right \}}$ 獨立, 從而隨機變數 $\xi_{1}, \xi_{2}, ..., \xi_{n}$ 獨立.

如果隨機變數 $\xi$ 的值域 $R_{\xi} = \left \{ x_{1}, x_{2}, ..., x_{k} \right \}$, 隨機變數 $\eta$ 的值域為 $R_{\eta} = \left \{ y_{1}, y_{2}, ..., y_{l} \right \}$, 則隨機變數 $\zeta = \xi + \eta$ 的值域為 $\displaystyle {R_{\zeta} = \left \{ z : z = x_{i} + y_{j}, i = 1, 2, ..., k, j = 1, 2, ..., l \right \}}.$ 那麼顯然, $\displaystyle {P_{\zeta}(z) = \mathop {\mathbf {P}} \left \{ \zeta = z \right \} = \mathop {\mathbf {P}} \left \{ \xi + \eta = z \right \} = \sum \limits_{\left \{ (i, j) : x_{i} + y_{j} = z \right \}}\mathop {\mathbf {P}} \left \{ \xi = x_{i}, \eta = y_{j} \right \}}.$

例題 3. 若隨機變數 $\xi_{1}, \xi_{2}, ..., \xi_{n}$ 是獨立的 Bernoulli 隨機變數, 且 $\displaystyle {\mathop {\mathbf {P}} \left \{ \xi_{i} = 1 \right \} = p, \mathop {\mathbf {P}} \left \{ \xi_{i} = 0 \right \} = q, i = 1, 2, ..., n}.$ 證明 : 隨機變數 $\zeta = \xi_{1} + \xi_{2} + ... \xi_{n}$ 服從二項分佈.

$證明$ :

我們使用歸納法進行證明.

當 $n = 2$ 時, $R_{\zeta} = \left \{ 0, 1, 2 \right \}$. 那麼有 $\displaystyle {\begin {aligned} &P_{\zeta}(0) = P_{\xi_{1}}(0)P_{\xi_{2}}(0) = q^{2} = \binom {0}{2}p^{0}q^{2}, \\ &P_{\zeta}(1) = P_{\xi_{1}}(0)P_{\xi_{2}}(1) + P_{\xi_{1}}(1)P_{\xi_{2}}(0) = 2pq = \binom {1}{2}p^{1}q^{1}, \\ &P_{\zeta}(2) = P_{\xi_{1}}(1)P_{\xi_{2}}(1) = p^{2} = \binom {2}{2}p^{2}q^{0}. \end {aligned}}$ 故當 $n = 2$ 時, 隨機變數 $\zeta = \xi_{1} + \xi_{2}$ 服從二項分佈.

不妨假設當 $n < l$ 時, 隨機變數 $\zeta = \xi_{1} + \xi_{2} + ... + \xi_{n}$ 服從二項分佈.

當 $n = l$ 時, $R_{\zeta} = \left \{ 0, 1, 2, ..., l \right \}$. 那麼有 $\displaystyle {\begin {aligned} &P_{\zeta}(0) = P_{\xi_{1}}(0)P_{\xi_{2}}(0)...P_{\xi_{l}}(0) = q^{l} = \binom {0}{l}p^{0}q^{l}, \\ &\begin {aligned} P_{\zeta}(1) &= P_{\xi_{1}}(1)P_{\xi_{2}}(0)...P_{\xi_{l}}(0) + P_{\xi_{1}}(0)P_{\xi_{2}}(1)P_{\xi_{3}}(0)...P_{\xi_{l}}(0) + ... + \\ &\ \ \ \ P_{\xi_{1}}(0)P_{\xi_{2}}(0)...P_{\xi_{l - 1}}(0)P_{\xi_{l}}(1) \\ &= qp^{l - 1} = \binom {1}{l}p^{1}q^{l - 1}, \end {aligned} \\ &\vdots \\ &P_{\zeta}(l) = P_{\xi_{1}}(1)P_{\xi_{2}}(1)...P_{\xi_{l}}(1) = p^{l} = \binom {l}{l}p^{l}q^{0}. \end {aligned}}$ 故當 $n = l$ 時, 隨機變數 $\zeta = \xi_{1} + \xi_{2} + ... + \xi_{n}$ 仍然服從二項分佈.

綜上所述, 隨機變數 $\zeta = \xi_{1} + \xi_{2} + ... \xi_{n}$ 服從二項分佈.

$\blacksquare$

3. 數學期望

設 $(\Omega, \mathscr {A}, \mathbf {P})$ 是有限機率空間, 而 $\xi = \xi(\omega)$ 是某一隨機變數, 其值域為 $\displaystyle {R_{\xi} = \left \{ x_{1}, x_{2}, ..., x_{k} \right \}}.$ 如果設 $A_{i} = \left \{ \omega : \xi(\omega) = x_{i} \right \}$, 則顯然 $\xi$ 可以表示為 $\displaystyle {\xi = \xi(\omega) = \sum \limits_{i = 1}^{k}x_{i}\mu_{A_{i}}(\omega)}.$ 其中, $A_{1}, A_{2}, ..., A_{k}$ 構成集合 $\Omega$ 的分割, $i = 1, 2, ..., k$, $\mu_{A_{i}}(\omega)$ 是集合 $A_{i}$ 的特徵函數.

$說明$:

我來說明一下為何 $\xi(\omega) = \sum \limits_{i = 1}^{k}x_{i}\mu_{A_{i}}(\omega)$ 成立. 首先, 能夠使得 $\xi(\omega) = x_{i}$ 成立的事件 $\omega$, 必有 $\omega \in A_{i}$. 那麼, $\mu_{A_{i}}(\omega) = 1$. 對於其它 $A_{j}\ (j \neq i)$, 就有 $\omega \notin A_{j}$, 即這些 $\omega$ 會使得 $\mathop {\mu_{A_{j}}} \limits_{\left \{ j \neq i \right \}}(\omega) = 0$. 展開 $\xi(\omega)$, 可得 $\displaystyle {\begin {aligned} \xi(\omega) &= x_{1}\mu_{A_{1}}(\omega) + x_{2}\mu_{A_{2}}(\omega) + ... + \\ &\ \ \ \ x_{i - 1}\mu_{A_{i - 1}}(\omega) + x_{i}\mu_{A_{i}}(\omega) + x_{i + 1}\mu_{A_{i + 1}}(\omega) + ... + \\ &\ \ \ \ x_{k}\mu_{A_{k}}(\omega). \end {aligned}}$ 根據說明, 我們可以得到 $\mu_{A_{1}}(\omega), \mu_{A_{2}}(\omega), ..., \mu_{A_{i - 1}}(\omega), \mu_{A_{i + 1}}(\omega), ..., \mu_{A_{k}}(\omega)$ 都為零. 因此, 不論 $x_{1}, x_{2}, ..., x_{i - 1}, x_{i + 1}, ..., x_{k}$ 為何值, $\displaystyle {x_{1}\mu_{A_{1}}(\omega), x_{2}\mu_{A_{2}}(\omega), ..., x_{i - 1}\mu_{A_{i - 1}}(\omega), x_{i + 1}\mu_{A_{i + 1}}(\omega), ..., x_{k}\mu_{A_{k}}(\omega)}$ 都為零. 最終有 $\displaystyle {\xi(\omega) = x_{i}\mu_{A_{i}}(\omega) = x_{i} = \sum \limits_{i = 1}^{k}x_{i}\mu_{A_{i}}(\omega)}.$

$\blacksquare$

記 $p_{i} = \mathop {\mathbf {P}} \left \{ \xi = x_{i} \right \}, i = 1, 2, ..., k$. 直觀上, 如果在 $n$ 次獨立重複試驗中觀測隨機變數 $\xi$ 的取值, 則取 $x_{i}$ 的值大致上應該出現 $np_{i}$ 次. 考慮投擲 $n$ 次均勻硬幣, 則 $p_{0} = p_{1} = \frac {1}{2}$. 其中, $p_{0}$ 代表反面向上的機率, 而 $p_{1}$ 表示正面向上的機率. 直觀上, 投擲 $n$ 次硬幣, 正面應該大致出現 $\frac {n}{2}$ 次, 即 $np_{1}$ 次. 因此, 根據 $n$ 次試驗的結果, 計算該隨機變數的平均值大致為 $\displaystyle {\frac {1}{n}(np_{1}x_{1} + np_{2}x_{2} + ... + np_{k}x_{k}) = \sum \limits_{i = 1}^{k}p_{i}x_{i}}.$

定義 4. 實數 $\displaystyle {\mathop {\mathrm {E}}(\xi) = \sum \limits_{i = 1}^{k}x_{i}\mathop {\mathbf {P}}(A_{i})}$ 稱作隨機變數 $\xi = \sum \limits_{i = 1}^{k}x_{i}\mu_{A_{i}}(\omega)$ 的數學期望 (mathematical expectation) 或者平均值 (mean value), 簡稱期望 (expectation) 或者均值. 其中, $A_{i} = \left \{ \omega : \xi(\omega) = x_{i} \right \}$ ($i = 1, 2, …, k$).

我們注意到, $P_{\xi}(x_{i}) = \mathop {\mathbf {P}}(A_{i})$. 根據定義 4, 又有 $\displaystyle {\mathop {\mathrm {E}}(\xi) = \sum \limits_{i = 1}^{k}x_{i}P_{\xi}(x_{i})}.$ 另外, 我們記 $\Delta F_{\xi}(x) = F_{\xi}(x) - F_{\xi}(x^{-})$, 根據定義 3, 我們可得 $P_{\xi}(x_{i}) = \Delta F_{\xi}(x_{i})$, 從而有 $\displaystyle {\mathop {\mathrm {E}}(\xi) = \sum \limits_{i = 1}^{k}x_{i}\Delta F_{\xi}(x_{i})}.$

有時, 隨機變數 $\xi$ 的值域中可能出現相同的值, 即 $x_{i} = x_{j}$ 但 $i \neq j$ 的情形. 此時, 我們可以將隨機變數表示為 $\displaystyle {\xi(\omega) = \sum \limits_{j = 1}^{l}x_{j}'\mu_{B_{j}}(\omega)}.$ 其中, $B_{1} + B_{2} + ... B_{l} = \Omega$, $x_{j}$ 之中可能出現相同的值, $j = 1, 2, ..., l$. 但之前, 我們所討論的隨機變數的值域是一個集合, 裡面不能存在相同的元素. 這時, 我們仍然可以按照上式計算期望, 而並不需要將其變換為所有的 $x_{i}$ 值兩兩不等的 $\displaystyle {\xi(\omega) = \sum \limits_{i = 1}^{k}x_{i}\mu_{A_{i}}(\omega)}.$ 其中, $i = 1, 2, …, l$.

事實上, 根據 $\displaystyle {\sum \limits_{\left \{ j : x_{j}' = x_{i} \right \}}x_{j}'\mathop {\mathbf {P}}(B_{j}) = x_{i}\sum \limits_{\left \{ j : x_{j}' = x_{i} \right \}}\mathop {\mathbf {P}}(B_{j}) = x_{i}\mathop {\mathbf {P}}(A_{i})},$ 於是有 $\displaystyle {\sum \limits_{j = 1}^{l}x_{j}\mathop {\mathbf {P}}(B_{j}) = \sum \limits_{i = 1}^{k}x_{i}\mathop {\mathbf {P}}(A_{i})}.$

設 $\xi$ 和 $\eta$ 為隨機變數, 則期望有以下基本性質 :

1. 若 $\xi \geq 0$, 則 $\mathop {\mathrm {E}}(\xi) \geq 0$;
2. $\mathop {\mathrm {E}}(a\xi + b\eta) = a\mathop {\mathrm {E}}(\xi) + b\mathop {\mathrm {E}}(\eta)$. 其中, $a$ 和 $b$ 是常數;
3. 若 $\xi \geq \eta$, 則 $\mathop {\mathrm {E}}(\xi) \geq \mathop {\mathrm {E}}(\eta)$;
4. $\left | \mathop {\mathrm {E}}(\xi) \right | \leq \mathop {\mathrm {E}}(\left | \xi \right |)$;
5. 若 $\xi$ 和 $\eta$ 獨立, 則 $\mathop {\mathrm {E}}(\xi\eta) = \mathop {\mathrm {E}}(\xi)\mathop {\mathrm {E}}(\eta)$;
6. $\mathop {\mathrm {E}}^{2}(\left | \xi\eta \right |) \leq \mathop {\mathrm {E}}(\xi^{2})\mathop {\mathrm {E}}(\eta^{2})$, 這是 Cauchy-Schwarz 不等式的機率形式;
7. 若 $\xi = \mu_{A}(\omega)$, 則 $\mathop {\mathrm {E}}(\xi) = \mathop {\mathbf {P}}(A)$;
8. 若 $c$ 為常數, 則 $\mathop {\mathrm {E}}(c) = c$.

$證明$ :

根據期望的定義, $\mathop {\mathrm {E}}(\xi) = \sum \limits_{i = 1}^{k}x_{i}\mathop {\mathbf {P}}(A_{i})$, 由於 $0 \leq \mathop {\mathbf {P}}(A_{i}) \leq 1$, 故當 $\xi \geq 0$ 時, 必有 $x_{i} \geq 0$. 其中, $i = 1, 2, ..., k$. 於是 $\mathop {\mathrm {E}}(\xi) \geq 0$.

(1) $\square$

設 $\xi = \sum \limits_{i}x_{i}\mu_{A_{i}}(\omega)$ 且 $\eta = \sum \limits_{j}y_{j}\mu_{A_{j}}(\omega)$, 則 $\displaystyle {a\xi + b\eta = a\sum \limits_{i}x_{i}\mu_{A_{i}}(\omega) + b\sum \limits_{j}y_{j}\mu_{A_{j}}(\omega)}.$ 由於 $A_{i} = A_{i} \cup \Omega = A_{i} \bigcup \limits_{j} B_{j}$ 且 $B_{j} = B_{j} \bigcup \limits_{i} A_{i}$, 故有 $\displaystyle {\begin {aligned} a\xi + b\eta &= a\sum \limits_{i, j}x_{i}\mu_{A_{i}B_{j}}(\omega) + b\sum \limits_{i, j}y_{j}\mu_{A_{i}B_{j}}(\omega) \\ &= \sum \limits_{i, j}ax_{i}\mu_{A_{i}B_{j}}(\omega) + \sum \limits_{i, j}by_{j}\mu_{A_{i}B_{j}}(\omega) \\ &= \sum \limits_{i, j}(ax_{i} + by_{j})\mu_{A_{i}B_{j}}(\omega). \end {aligned}}$ 因此, 根據定義 4 有 $\displaystyle {\begin {aligned} \mathop {\mathrm {E}}(a\xi + b\eta) &= \sum \limits_{i, j}(ax_{i} + by_{j})\mathop {\mathbf {P}}(A_{i}B_{j}) \\ &= \sum \limits_{i}ax_{i}\mathop {\mathbf {P}}(A_{i}) + \sum \limits_{j}by_{j}\mathop {\mathbf {P}}(B_{j}) \\ &= a\sum \limits_{i}x_{i}\mathop {\mathbf {P}}(A_{i}) + b\sum \limits_{j}y_{j}\mathop {\mathbf {P}}(B_{j}) \\ &= a\mathop {\mathrm {E}}(\xi) + b\mathop {\mathrm {E}}(\eta). \end {aligned}}$

(2) $\square$

由性質 2 可知, $\displaystyle {\mathop {\mathrm {E}}(\xi) - \mathop {\mathrm {E}}(\eta) = \mathop {\mathrm {E}}(\xi - \eta)}.$ 由於 $\xi \geq \eta$, 結合性質 1, 於是有 $\displaystyle {\mathop {\mathrm {E}}(\xi) \geq \mathop {\mathrm {E}}(\eta)}.$

(3) $\square$

根據絕對值不等式 (《【數學分析】實數——實數的四則運算》定理 3), 我們可以得到 $\displaystyle {\left | \mathop {\mathrm {E}}(\xi) \right | = \left |\sum \limits_{i}x_{i}\mathop {\mathbf {P}}(A_{i}) \right | \leq \sum \limits_{i}\left | x_{i} \right |\mathop {\mathbf {P}}(A_{i}) = \mathop {\mathrm {E}}(\left | \xi \right |)}.$

(4) $\square$

設 $\xi = \sum \limits_{i}x_{i}\mu_{A_{i}}(\omega)$ 且 $\eta = \sum \limits_{j}y_{j}\mu_{A_{j}}(\omega)$, 則 $\displaystyle {\begin {aligned} \mathop {\mathrm {E}}(\xi\eta) &= \mathop {\mathrm {E}} \left (\sum \limits_{i}x_{i}\mu_{A_{i}}(\omega)\sum \limits_{j}y_{j}\mu_{A_{j}}(\omega) \right ) \\ &= \mathop {\mathrm {E}} \left (\sum \limits_{i}\sum \limits_{j}x_{i}y_{j}\mu_{A_{i}B_{j}}(\omega) \right ) \\ &= \sum \limits_{i, j}x_{i}y_{j}\mathop {\mathbf {P}}(A_{i}B_{j}). \end {aligned}}$ 由於隨機變數 $\xi$ 和 $\eta$ 獨立, 故事件 $A_{i} = \left \{ \omega : \xi(\omega) = x_{i} \right \}$ 和 $B_{j} = \left \{ \omega : \eta(\omega) = y_{j} \right \}$ 相互獨立, 於是 $\displaystyle {\begin {aligned} \mathop {\mathrm {E}}(\xi\eta) &= \sum \limits_{i, j}x_{i}y_{j}\mathop {\mathbf {P}}(A_{i})\mathop {\mathbf {P}}(B_{j}) \\ &= \sum \limits_{i}x_{i}\mathop {\mathbf {P}}(A_{i})\sum \limits_{j}y_{j}\mathop {\mathbf {P}}(B_{j}) \\ &= \mathop {\mathrm {E}}(\xi)\mathop {\mathrm {E}}(\eta). \end {aligned}}$

(5) $\square$

對於 $\xi = \sum \limits_{i}x_{i}\mu_{A_{i}}(\omega), \eta = \sum \limits_{j}y_{j}\mu_{A_{j}}(\omega)$, 我們注意到 $\displaystyle {\xi^{2} = \sum \limits_{i}x_{i}^{2}\mu_{A_{i}}(\omega), \eta^{2} = \sum \limits_{j}y_{j}\mu_{B_{j}}(\omega)}.$ 因此根據期望的定義, 有 $\displaystyle {\mathop {\mathrm {E}}(\xi^{2}) = \sum \limits_{i}x_{i}^{2}\mathop {\mathbf {P}}(A_{i}), \mathop {\mathrm {E}}(\eta^{2}) = \sum \limits_{j}y_{j}^{2}\mathop {\mathbf {P}}(B_{j})}.$ 設 $\mathop {\mathrm {E}}(\xi^{2}) > 0$ 且 $\mathop {\mathrm {E}}(\eta^{2}) > 0$, 記 $\displaystyle {\widetilde {\xi} = \frac {\xi}{\sqrt {\mathop {\mathrm {E}}(\xi^{2})}}, \widetilde {\eta} = \frac {\eta}{\sqrt {\mathop {\mathrm {E}}(\eta^{2})}}},$ 於是, $\displaystyle {\mathop {\mathrm {E}} \left ({\widetilde {\xi}}^{2} + {\widetilde {\eta}}^{2} \right ) = \frac {\xi^{2}}{\mathop {\mathrm {E}}(\xi^{2})} + \frac {\eta^{2}}{\mathop {\mathrm {E}}(\eta^{2})}, 2 \left |\widetilde {\xi}\widetilde {\eta} \right | = \frac {2|\xi\eta|}{\sqrt {\mathop {\mathrm {E}}(\xi^{2})\mathop {\mathrm {E}}(\eta^{2})}}}.$ 由均值不等式可知, $2|\widetilde {\xi}\widetilde {\eta}| \leq {\widetilde {\xi}}^{2} + {\widetilde {\eta}}^{2}$, 可見 $\displaystyle {2\mathop {\mathrm {E}} \left ( \left | \widetilde {\xi}\widetilde {\eta} \right | \right ) \leq \mathop {\mathrm {E}}(\widetilde {\xi}^{2}) + \mathop {\mathrm {E}}(\widetilde {\eta}^{2}) = 2}.$ 因此, $\mathop {\mathrm {E}} \left ( \left |\widetilde {\xi}\widetilde {\eta} \right | \right ) \leq 1$ 且 $\mathop {\mathrm {E}}^{2}(\xi\eta) \leq \mathop {\mathrm {E}}(\xi^{2})\mathop {\mathrm {E}}(\eta^{2})$. 若 $\mathop {\mathrm {E}}(\xi^{2}) = 0$, 則 $\displaystyle {\sum \limits_{i}x_{i}^{2}\mathop {\mathbf {P}} \left \{ A_{i} \right \} = 0}.$ 從而 $0$ 是隨機變數 $\xi$ 的可能取值, 並且 $\displaystyle {\mathop {\mathbf {P}} \left \{ \omega : \xi(\omega) = 0 \right \} = 1}.$ 於是, 不論 $\mathop {\mathrm {E}}(\xi^{2}) = 0$ 或者 $\mathop {\mathrm {E}}(\eta^{2}) = 0$, 顯然有 $\mathop {\mathrm {E}}(\left | \xi\eta \right |) = 0$. 此時有 $\displaystyle {\mathop {\mathrm {E}}^{2}(\xi\eta) \leq \mathop {\mathrm {E}}(\xi^{2})\mathop {\mathrm {E}}(\eta^{2})}$ 成立.

綜上所述, $\mathop {\mathrm {E}}^{2}(\xi\eta) = \mathop {\mathrm {E}}(\xi^{2})\mathop {\mathrm {E}}(\eta^{2})$.

(6) $\square$

根據隨機變數 $\xi$ 的性質 : $\displaystyle {\xi = \sum \limits_{i}x_{i}\mu_{A_{i}}(\omega) = \mu_{A_{i}}(\omega)},$ 即 $x_{i} = 1$, 於是 $\displaystyle {\mathop {\mathrm {E}}(\xi) = \sum \limits_{i}x_{i}\mathop {\mathbf {P}}(A_{i}) = \sum \limits_{i}\mathop {\mathbf {P}}(A_{i}) = \mathop {\mathbf {P}}(A)}$

(7) $\square$

由於 $c$ 有且唯有一種取值, 故 $\mathop {\mathbf {P}}(c) = 1$. 根據期望的定義, 有 $\displaystyle {\mathop {\mathrm {E}}(c) = \sum \limits_{i}x_{i}\mathop {\mathbf {P}}(A_{i}) = c \cdot \mathop {\mathbf {P}}(c) = c}.$

(8) $\square$

$\blacksquare$

推論 1. 若隨機變數 $\xi_{1}, \xi_{2}, ..., \xi_{r}$ 獨立, 則有 $\displaystyle {\mathop {\mathrm {E}}(\xi_{1}\xi_{2}...\xi_{r}) = \mathop {\mathrm {E}}(\xi_{1})\mathop {\mathrm {E}}(\xi_{2})...\mathop {\mathrm {E}}(\xi_{r})}.$

$證明$ :

我們使用歸納法進行證明.

當 $r = 1$ 時, 顯然有 $\mathop {\mathrm {E}}(\xi_{1}) = \mathop {\mathrm {E}}(\xi_{1})$; 當 $r = 2$ 時, 由期望的性質可知, $\mathop {\mathrm {E}}(\xi_{1}\xi_{2}) = \mathop {\mathrm {E}}(\xi_{1})\mathop {\mathrm {E}}(\xi_{2})$.

不妨假設當 $r < k$ 時, 有 $\mathop {\mathrm {E}}(\xi_{1}\xi_{2}...\xi_{r}) = \mathop {\mathrm {E}}(\xi_{1})\mathop {\mathrm {E}}(\xi_{2})...\mathop {\mathrm {E}}(\xi_{r})$ 成立.

當 $r = k$ 時, $\displaystyle {\begin {aligned} \mathop {\mathrm {E}}(\xi_{1}\xi_{2}...\xi_{k - 1}\xi_{k}) &= \mathop {\mathrm {E}}((\xi_{1}\xi_{2}...\xi_{k - 1})\xi_{k}) \\ &= \mathop {\mathrm {E}}(\xi_{1}\xi_{2}...\xi_{k - 1})\mathop {\mathrm {E}}(\xi_{k}) \\ &= \mathop {\mathrm {E}}(\xi_{1})\mathop {\mathrm {E}}(\xi_{2})...\mathop {\mathrm {E}}(\xi_{k - 1})\mathop {\mathrm {E}}(\xi_{k}). \end {aligned}}$ 因此, 當 $r = k$ 時, $\mathop {\mathrm {E}}(\xi_{1}\xi_{2}...\xi_{r}) = \mathop {\mathrm {E}}(\xi_{1})\mathop {\mathrm {E}}(\xi_{2})...\mathop {\mathrm {E}}(\xi_{r})$ 仍然成立.

綜上所述, 若隨機變數 $\xi_{1}, \xi_{2}, ..., \xi_{r}$ 獨立, 則有 $\displaystyle {\mathop {\mathrm {E}}(\xi_{1}\xi_{2}...\xi_{r}) = \mathop {\mathrm {E}}(\xi_{1})\mathop {\mathrm {E}}(\xi_{2})...\mathop {\mathrm {E}}(\xi_{r})}.$

$\blacksquare$

例題 4. 設 $\xi$ 是 Bernoulli 隨機變數, 以機率 $p$ 和 $q$ 取 $0$ 和 $1$, 則 $\displaystyle {\mathop {\mathrm {E}}(\xi) = 1 \times \mathop {\mathbf {P}} \left \{ \xi = 1 \right \} + 0 \times \mathop {\mathbf {P}} \left \{ \xi = 0 \right \} = \mathop {\mathbf {P}} \left \{ \xi = 1 \right \} = p}.$

例題 5. 設 $\xi_{1}, \xi_{2}, ..., \xi_{n}$ 是 $n$ 個 Bernoulli 隨機變數, 以機率 $\mathop {\mathbf {P}} \left \{ \xi_{i} = 1 \right \} = p$ 和 $\mathop {\mathbf {P}} \left \{ \xi_{i} = 0 \right \} = q$ 及 $p + q = 1$ 取 $1$ 和 $0$ 為值. 其中, $i = 1, 2, ..., n$. 那麼對於 $S_{n} = \xi_{1} + \xi_{2} + ... + \xi_{n}$ 的期望為 $\displaystyle {\begin {aligned} \mathop {\mathrm {E}}(S_{n}) &= \mathop {\mathrm {E}}(\xi_{1} + \xi_{2} + ... \xi_{n}) = \mathop {\mathrm {E}}(\xi_{1}) + \mathop {\mathrm {E}}(\xi_{2}) + ... + \mathop {\mathrm {E}}(\xi_{n}) \\ &= \underbrace {(1 \times p + 0 \times q) + (1 \times p + 0 \times q) + ... + (1 \times p + 0 \times q)}_{n \text { 個}} \\ &= np. \end {aligned}}$

例題 5'. 設 $\xi_{1}, \xi_{2}, ..., \xi_{n}$ 是 $n$ 個 Bernoulli 隨機變數, 以機率 $\mathop {\mathbf {P}} \left \{ \xi_{i} = 1 \right \} = p$ 和 $\mathop {\mathbf {P}} \left \{ \xi_{i} = 0 \right \} = q$ 及 $p + q = 1$ 取 $1$ 和 $0$ 為值. 其中, $i = 1, 2, ..., n$. 求 $S_{n} = \xi_{1} + \xi_{2} + ... + \xi_{n}$.

$解$ :

若 $\xi_{1}, \xi_{2}, ..., \xi_{n}$ 是 $n$ 個獨立的 Bernoulli 隨機變數, 則 $\mathop {\mathrm {E}}(S_{n})$ 仍保持不變. 此時, 由二項分佈可知 $\displaystyle {\mathop {\mathbf {P}} \left \{ S_{n} = k \right \} = \binom {k}{n}p^{k}q^{n - k}}.$ 其中, $k = 0, 1, 2, …, n$. 於是, $\displaystyle {\begin {aligned} \mathop {\mathrm {E}}(S_{n}) &= \sum \limits_{k = 0}^{n}k\mathop {\mathbf {P}} \left \{ S_{n} = k \right \} \\ &= \sum \limits_{k = 0}^{n}k\binom {k}{n}p^{k}q^{n - k} \\ &= \sum \limits_{k = 0}^{n}k \cdot \frac {n!}{k!(n - k)!}p^{k}q^{n - k} \\ &= \frac {0 \times n!}{0!(n - 0)!}p^{0}q^{n} + \sum \limits_{k = 1}^{n}k \frac {n!}{k!(n - k)!}p^{k}q^{n - k} \\ &= \sum \limits_{k = 1}^{n}k \frac {n!}{k!(n - k)!}p^{k}q^{n - k} \\ &= \sum \limits_{k = 1}^{n}k\frac {n(n - 1)!}{k!(n - k + 1 - 1)!}p \cdot p^{k - 1}q^{n - k + 1 -1} \\ &= \sum \limits_{k = 1}^{n}\frac {n(n - 1)!}{(k - 1)!((n - 1) - (k - 1))!}p \cdot p^{k - 1}q^{(n - 1) - (k - 1)} \\ &= np\sum \limits_{k = 1}^{n}\frac {(n - 1)!}{(k - 1)!((n - 1) - (k - 1))!}p^{k - 1}q^{(n - 1) - (k - 1)} \\ &= np\sum \limits_{l = 1}^{n - 1}\frac {(n - 1)!}{l!((n - 1) - l)!}p^{l}q^{(n - 1) - l} \ (\text {令 } l = k - 1) \\ &= np. \end {aligned}}$

$\blacksquare$

設 $\xi = \sum \limits_{i}x_{i}\mu_{A_{i}}(\omega)$, 其中, $A_{i} = \left \{ \omega : \xi(\omega) = x_{i} \right \}$, 從而 $\varphi = \varphi(\xi(\omega))$ 是 $\xi(\omega)$ 的某一函數. 如果 $B_{j} = \left \{ \omega : \varphi(\xi(\omega)) = y_{j} \right \}$, 則 $\displaystyle {\varphi(\xi(\omega)) = \sum \limits_{j}y_{j}\mu_{B_{j}}(\omega)}.$ 從而有 $\displaystyle {\mathop {\mathrm {E}}(\varphi) = \sum \limits_{j}y_{j}\mathop {\mathbf {P}}(B_{j}) = \sum \limits_{j}y_{j}P_{\varphi}(y_{j})}.$ 顯然, $\displaystyle {\varphi(\xi(\omega)) = \sum \limits_{i}\varphi(x_{i})\mu_{A_{i}}(\omega)}.$ 於是, 為了求 $\varphi = \varphi(\xi(\omega))$ 的期望值, 既可以使用 $\mathop {\mathrm {E}}(\varphi) = \sum \limits_{j}y_{j}\mathop {\mathbf {P}}(B_{j}) = \sum \limits_{j}y_{j}P_{\varphi}(y_{j})$, 也可以使用 $\displaystyle {\mathop {\mathrm {E}}(\varphi(\xi(\omega))) = \sum \limits_{i}\varphi(x_{i})P_{\xi}(x_{i})}.$ 我們稱 $\mathop {\mathrm {E}}(\varphi(\xi(\omega)))$ 為隨機變數函數的數學期望.

定義 5. 稱 $\displaystyle {\mathop {\mathrm {Var}}(\xi) = \mathop {\mathrm {E}} \left ((\xi - \mathop {\mathrm {E}}(\xi))^{2} \right )}$ 為隨機變數 $\xi$ 的方差 (variance).

定義 6. 稱 $\displaystyle {\sigma = \sqrt {\mathop {\mathrm {Var}}(\xi)}}$ 為隨機變數 $\xi$ 的標準差 (standard deviation).

隨機變數 $\xi$ 的方差和標準差表徵了 $\xi$ 的取值的散佈程度.

由於 $\displaystyle {\begin {aligned} \mathop {\mathrm {E}} \left ((\xi - \mathop {\mathrm {E}}(\xi))^{2} \right) &= \mathop {\mathrm {E}} \left (\xi^{2} - 2\xi \mathop {\mathrm {E}}(\xi) + \mathop {\mathrm {E}}^{2}(\xi) \right ) \\ &= \mathop {\mathrm {E}}(\xi^{2}) - \mathop {\mathrm {E}}(2\xi \mathop {\mathrm {E}}(\xi)) + \mathop {\mathrm {E}} \left (\mathop {\mathrm {E}}^{2}(\xi) \right ) \\ &= \mathop {\mathrm {E}}(\xi^{2}) - 2\mathop {\mathrm {E}}(\xi) \cdot \mathop {\mathrm {E}}(\xi) + \mathop {\mathrm {E}}^{2}(\xi) \\ &= \mathop {\mathrm {E}}(\xi^{2}) - 2\mathop {\mathrm {E}}^{2}(\xi) + \mathop {\mathrm {E}}^{2}(\xi) \\ &= \mathop {\mathrm {E}}(\xi^{2}) - \mathop {\mathrm {E}}^{2}(\xi) \end {aligned}}$ 可見, $\mathop {\mathrm {Var}}(\xi) = \mathop {\mathrm {E}}(\xi^{2}) - \mathop {\mathrm {E}}^{2}(\xi)$. 上面推導過程中, 由於 $\mathop {\mathrm {E}}(\xi)$ 是常數, 因此根據期望的性質 8 有 $\mathop {\mathrm {E}}(\mathop {\mathrm {E}}(\xi)) = \mathop {\mathrm {E}}(\xi)$.

設 $\xi$ 為隨機變數, $a$ 與 $b$ 均為常數, 可導出方差如下性質 :

1. $\mathop {\mathrm {Var}}(\xi) \geq 0$;
2. $\mathop {\mathrm {Var}}(a + b\xi) = b^{2}\mathop {\mathrm {Var}}(\xi)$;
3. $\mathop {\mathrm {Var}}(a) = 0$;
4. $\mathop {\mathrm {Var}}(b\xi) = b^{2}\mathop {\mathrm {Var}}(\xi)$.

$證明$ :

由定義 5 可知, $\mathop {\mathrm {Var}}(\xi) = \mathop {\mathrm {E}} \left ((\xi - \mathop {\mathrm {E}}(\xi))^{2} \right )$. 而 $(\xi - \mathop {\mathrm {E}}(\xi))^{2} \geq 0$, 由期望的性質可知, $\mathop {\mathrm {Var}}(\xi) \geq 0$.

(1) $\square$

根據定義 5 , 我們將其展開 : $\displaystyle {\begin {aligned} \mathop {\mathrm {Var}}(a + b\xi) &= \mathop {\mathrm {E}} \left ((a + b\xi)^{2} \right ) - \mathop {\mathrm {E}}^{2}(a + b\xi) \\ &= \mathop {\mathrm {E}}(a^{2} + 2ab\xi + b^{2}\xi^{2}) - \mathop {\mathrm {E}}(a + b\xi)\mathop {\mathrm {E}}(a + b\xi) \\ &= \mathop {\mathrm {E}}(a^{2}) + \mathop {\mathrm {E}}(2ab\xi) + \mathop {\mathrm {E}}(b^{2}\xi^{2}) - \left ( \mathop {\mathrm {E}}^{2}(a) + \mathop {\mathrm {E}}^{2}(b\xi) + 2\mathop {\mathrm {E}}(a)\mathop {\mathrm {E}}(b\xi) \right ) \\ &= a^{2} + 2ab\mathop {\mathrm {E}}(\xi) + b^{2}\mathop {\mathrm {E}}(\xi^{2}) - a^{2} - b^{2}\mathop {\mathrm {E}}^{2}(\xi) - 2ab\mathop {\mathrm {E}}(\xi) \\ &= b^{2}\mathop {\mathrm {E}}(\xi^{2}) - b^{2}\mathop {\mathrm {E}}^{2}(\xi) \\ &= b^{2}\left ( \mathop {\mathrm {E}}(\xi^{2}) - \mathop {\mathrm {E}}^{2}(\xi) \right ) \\ &= b^{2}\mathop {\mathrm {Var}}(\xi). \end {aligned}}$

(2) $\square$

根據定義 5 , 我們將其展開 : $\displaystyle {\mathop {\mathrm {Var}}(a) = \mathop {\mathrm {E}}(a^{2}) - \mathop {\mathrm {E}}^{2}(a) = a^{2} - a^{2} = 0}.$

(3) $\square$

根據定義 5 , 我們將其展開 : $\displaystyle {\begin {aligned} \mathop {\mathrm {Var}}(b\xi) &= \mathop {\mathrm {E}}(b^{2}\xi^{2}) - \mathop {\mathrm {E}}^{2}(b\xi) \\ &= b^{2}\mathop {\mathrm {E}}(\xi^{2}) - b^{2}\mathop {\mathrm {E}}^{2}(\xi) \\ &= b^{2} \left ( \mathop {\mathrm {E}}(\xi^{2}) - \mathop {\mathrm {E}}^{2}(\xi) \right ) \\ &= b^{2}\mathop {\mathrm {Var}}(\xi). \end {aligned}}$

(3) $\square$

$\blacksquare$

對於兩個隨機變數 $\xi$ 和 $\eta$, 有 $\displaystyle {\begin {aligned} \mathop {\mathrm {Var}}(\xi + \eta) &= \mathop {\mathrm {E}} \left ((\xi + \eta)^{2} \right ) + \mathop {\mathrm {E}}^{2}(\xi + \eta) \\ &= \mathop {\mathrm {E}}(\xi^{2} + 2\xi\eta + \eta^{2}) - (\mathop {\mathrm {E}}(\xi) - \mathop {\mathrm {E}}(\eta))^{2} \\ &= \mathop {\mathrm {E}}(\xi^{2}) + 2\mathop {\mathrm {E}}(\xi + \eta) + \mathop {\mathrm {E}}(\eta^{2}) - \mathop {\mathrm {E}}^{2}(\xi) - 2\mathop {\mathrm {E}}(\xi)\mathop {\mathrm {E}}(\eta) + \mathop {\mathrm {E}}^{2}(\eta) \\ &= \mathop {\mathrm {Var}}(\xi) + \mathop {\mathrm {Var}}(\eta) + 2\mathop {\mathrm {E}}(\xi\eta) - 2\mathop {\mathrm {E}}(\xi)\mathop {\mathrm {E}}(\eta). \end {aligned}}$ 而 $\displaystyle {\begin {aligned} \mathop {\mathrm {E}} \big ((\xi - \mathop {\mathrm {E}}(\xi)) + (\eta - \mathop {\mathrm {E}}(\eta)) \big ) &= \mathop {\mathrm {E}} \big ( \xi^{2} + \mathop {\mathrm {E}}^{2}(\xi) + \eta^{2}+ \mathop {\mathrm {E}}^{2}(\eta) - 2\xi \mathop {\mathrm {E}}(\xi) + 2\xi\eta - \\ &\ \ \ \ \ 2\xi \mathop {\mathrm {E}}(\eta) - 2\eta \mathop {\mathrm {E}}(\xi) + 2\mathop {\mathrm {E}}(\eta)\mathop {\mathrm {E}}(\eta) - 2\eta \mathop {\mathrm {E}}(\eta) \big ) \\ &= \mathop {\mathrm {E}}(\xi^{2}) + \mathop {\mathrm {E}}^{2}(\xi) + \mathop {\mathrm {E}}(\eta^{2}) + \mathop {\mathrm {E}}^{2}(\eta) - 2\mathop {\mathrm {E}}^{2}(\xi) + 2\mathop {\mathrm {E}}(\xi\eta) - \\ &\ \ \ \ \ 2\mathop {\mathrm {E}}(\eta)\mathop {\mathrm {E}}(\xi) - 2\mathop {\mathrm {E}}(\eta)\mathop {\mathrm {E}}(\xi) + 2\mathop {\mathrm {E}}(\xi)\mathop {\mathrm {E}}(\eta) - 2\mathop {\mathrm {E}}^{2}(\eta) \\ &= \mathop {\mathrm {Var}}(\xi) + \mathop {\mathrm {Var}}(\eta) + 2\mathop {\mathrm {E}}(\xi\eta) - 2\mathop {\mathrm {E}}(\eta)\mathop {\mathrm {E}}(\xi), \end {aligned}}$ 因此, $\mathop {\mathrm {Var}}(\xi + \eta) = \mathop {\mathrm {E}} \left ( \big ( (\xi - \mathop {\mathrm {E}}(\xi)) + (\eta - \mathop {\mathrm {E}}(\eta)) \big )^{2} \right )$. 除此之外, $\displaystyle {\begin {aligned} 2\mathop {\mathrm {E}} \big ( (\xi - \mathop {\mathrm {E}}(\xi))(\eta - \mathop {\mathrm {E}}(\eta)) \big ) &= 2\mathop {\mathrm {E}} \big ( \xi\eta - \xi \mathop {\mathrm {E}}(\eta) - \eta \mathop {\mathrm {E}}(\xi) + \mathop {\mathrm {E}}(\xi)\mathop {\mathrm {E}}(\eta) \big ) \\ &= 2\big ( \mathop {\mathrm {E}}(\xi\eta) - \mathop {\mathrm {E}}(\xi)\mathop {\mathrm {E}}(\eta) - \mathop {\mathrm {E}}(\eta)\mathop {\mathrm {E}}(\xi) + \mathop {\mathrm {E}}(\xi)\mathop {\mathrm {E}}(\eta) \big ) \\ &= 2\mathop {\mathrm {E}}(\xi\eta) - 2\mathop {\mathrm {E}}(\xi)\mathop {\mathrm {E}}(\eta), \end {aligned}}$ 故有 $\displaystyle {\begin {aligned} \mathop {\mathrm {Var}}(\xi + \eta) &= E\big ( (\xi - \mathop {\mathrm {E}}(\xi)) + (\eta - \mathop {\mathrm {E}}(\eta)) \big ) \\ &= \mathop {\mathrm {Var}}(\xi) + \mathop {\mathrm {Var}}(\eta) - 2\mathop {\mathrm {E}} \big ( (\xi - \mathop {\mathrm {E}}(\xi))(\eta - \mathop {\mathrm {E}}(\eta)) \big ). \end {aligned}}$

定義 7. 我們稱 $\displaystyle {\mathop {\mathrm {Cov}}(\xi, \eta) = \mathop {\mathrm {E}} \big ( (\xi - \mathop {\mathrm {E}}(\xi))(\eta - \mathop {\mathrm {E}}(\eta)) \big )}$ 為隨機變數 $\xi$ 和 $\eta$ 的協方差 (covariance, 也稱共變異數).

定義 8. 若 $\mathop {\mathrm {Var}}(\xi) \geq 0, \mathop {\mathrm {Var}}(\eta) \geq 0$, 我們稱 $\displaystyle {\rho(\xi, \eta) = \frac {\mathop {\mathrm {Cov}}(\xi, \eta)}{\sqrt {\mathop {\mathrm {Var}}(\xi)\mathop {\mathrm {Var}}(\eta)}}}$ 為隨機變數 $\xi$ 和 $\eta$ 的相關係數 (correlation coefficient).

定理 1. 若 $\rho(\xi, \eta) = \pm 1$, 則隨機變數 $\xi$ 和 $\eta$ 線性相關, 即 $\displaystyle {\eta = a\xi + b}.$ 其中, 當 $\rho(\xi, \eta) = 1$ 時, $a > 0$; 當 $\rho(\xi, \eta) = -1$ 時, $a < 0$, $a$ 和 $b$ 都為常數.

$證明$ :

當 $\rho(\xi, \eta) = 1$ 時, 有 $\displaystyle {\frac {\mathop {\mathrm {Cov}}(\xi, \eta)}{\sqrt {\mathop {\mathrm {Var}}(\xi)\mathop {\mathrm {Var}}(\eta)}} = \frac {\mathop {\mathrm {E}} \big ( (\xi - \mathop {\mathrm {E}}(\xi))(\eta - \mathop {\mathrm {E}}(\eta)) \big )}{\sqrt {\mathop {\mathrm {E}} \big ( (\xi - \mathop {\mathrm {E}}(\xi))^{2} \big )\mathop {\mathrm {E}} \big ( (\eta - \mathop {\mathrm {E}}(\eta))^{2} \big )}} = 1}.$ 記隨機變數 $\varphi = \xi - \mathop {\mathrm {E}}(\xi)$ 和 $\psi = \eta - \mathop {\mathrm {E}}(\eta)$, 由上式可知 $\displaystyle {\mathop {\mathrm {E}}(\varphi\psi) = \sqrt {\mathop {\mathrm {E}}(\varphi^{2})\mathop {\mathrm {E}}(\psi^{2})}}.$ 兩邊平方可得 $\displaystyle {\mathop {\mathrm {E}}^{2}(\varphi\psi) = \mathop {\mathrm {E}}(\varphi^{2})\mathop {\mathrm {E}}(\psi^{2})}.$ 根據期望的性質 6, 有 $\displaystyle {\mathop {\mathrm {E}}^{2}(\varphi\psi) \leq \mathop {\mathrm {E}}(\varphi^{2})\mathop {\mathrm {E}}(\psi^{2})}.$ 顯然, 要使得等號成立, 若且唯若隨機變數 $\varphi$ 和 $\psi$ 線性相關, 即存在不為零的 $k_{1}$ 和 $k_{2}$ 使得 $\displaystyle {k_{1}\varphi + k_{2}\psi = 0}.$ 記 $k = -\frac {k_{1}}{k_{2}}$, 於是有 $\varphi = k\psi$. 那麼, 我們可以得到 $\displaystyle {\mathop {\mathrm {E}}^{2}(\varphi\psi) = \mathop {\mathrm {E}}^{2}(\varphi \cdot k\varphi) = k^{2}\mathop {\mathrm {E}}^{2}(\varphi^{2})}$ 和 $\displaystyle {\mathop {\mathrm {E}}(\varphi^{2})\mathop {\mathrm {E}}(\psi^{2}) = \mathop {\mathrm {E}}(\varphi^{2})\mathop {\mathrm {E}}(k^{2}\varphi^{2}) = k^{2}\mathop {\mathrm {E}}^{2}(\varphi^{2})}.$ 因此, $\mathop {\mathrm {E}}^{2}(\varphi\psi) = \mathop {\mathrm {E}}(\varphi)\mathop {\mathrm {E}}(\psi)$. 此時, $\displaystyle {\varphi = \xi - \mathop {\mathrm {E}}(\xi) = k\psi = k(\eta - \mathop {\mathrm {E}}(\eta))}.$ 變換可得 $\displaystyle {\eta = \frac {1}{k}\xi + \frac {k\mathop {\mathrm {E}}(\eta) + \mathop {\mathrm {E}}(\xi)}{k}}.$ 記 $a = \frac {1}{k}, b = \frac {k\mathop {\mathrm {E}}(\eta) + \mathop {\mathrm {E}}(\xi)}{k}$, 則 $\displaystyle {\eta = a\xi + b}.$ 當 $\rho(\xi, \eta) = -1$ 時, 同樣有 $\eta = a\xi + b$.

接下來證明當 $\rho(\xi, \eta) = 1$ 時, $a > 0$; 當 $\rho(\xi, \eta) = -1$ 時, $a < 0$. 當 $\rho(\xi, \eta) = 1$ 時, 有 $\eta = a\xi + b$. 則 $\displaystyle {\begin {aligned} \rho(\xi, \eta) &= \rho(\xi, a\xi + b) = \frac {\mathop {\mathrm {Cov}}(\xi, a\xi + b)}{\sqrt {\mathop {\mathrm {Var}}(\xi)\mathop {\mathrm {Var}}(a\xi + b)}} \\ &= \frac {\mathop {\mathrm {E}}(\xi - \mathop {\mathrm {E}}(\xi))\mathop {\mathrm {E}}(a\xi + b - \mathop {\mathrm {E}}(a\xi + b))}{|a|\mathop {\mathrm {Var}}(\xi)} \\ &= \frac {a}{|a|} = 1. \end {aligned}}$ 故當 $a > 0$ 時, $\rho(\xi, \eta) = 1$. 對於 $\frac {a}{|a|} = -1$, 此時 $a < 0$. 因此當 $a < 0$ 時, $\rho(\xi, \eta) = -1$.

綜上所述, 若 $\rho(\xi, \eta) = \pm 1$, 則隨機變數 $\xi$ 和 $\eta$ 線性相關, 即 $\displaystyle {\eta = a\xi + b}.$ 其中, 當 $\rho(\xi, \eta) = 1$ 時, $a > 0$; 當 $\rho(\xi, \eta) = -1$ 時, $a < 0$, $a$ 和 $b$ 都為常數.

$\blacksquare$

顯然我們可以立即指出 : 若隨機變數 $\xi$ 和 $\eta$ 相互獨立, 則隨機變數 $\xi - \mathop {\mathrm {E}}(\xi)$ 和 $\eta - \mathop {\mathrm {E}}(\eta)$ 獨立, 於是 $\displaystyle {\mathop {\mathrm {Cov}}(\xi, \eta) = \mathop {\mathrm {E}} \big ( (\xi - \mathop {\mathrm {E}}(\xi))(\eta - \mathop {\mathrm {E}}(\eta)) \big ) = \mathop {\mathrm {E}}(\xi\eta) - \mathop {\mathrm {E}}(\xi)\mathop {\mathrm {E}}(\eta) = 0}.$ 另外, 由定義 7 可知 $\displaystyle {\mathop {\mathrm {Var}}(\xi + \eta) = \mathop {\mathrm {Var}}(\xi) + \mathop {\mathrm {Var}}(\eta) + 2\mathop {\mathrm {Cov}}(\xi, \eta) = \mathop {\mathrm {Var}}(\xi) + \mathop {\mathrm {Var}}(\eta)}.$ 相比於 "隨機變數 $\xi$ 和 $\eta$ 相互獨立" 這個條件, "隨機變數 $\xi$ 和 $\eta$ 不相關" 這個條件要稍弱一些. 因此, 一般隨機變數 $\xi$ 和 $\eta$ 不相關並不能推導出 $\xi$ 和 $\eta$ 獨立.

4. 估計量

例題 6. 假設隨機變數 $\alpha$ 以 $\frac {1}{3}$ 的機率分別取 $0, \frac {\pi}{2}, \pi$ 三個值. 證明 : 隨機變數 $\xi = \sin {\alpha}$ 與 $\eta = \cos {\alpha}$ 不相關但是 $\xi$ 和 $\eta$ 關於機率 $\mathop {\mathbf {P}}$ 不獨立, 並求 $\xi$ 和 $\eta$ 之間的函數關係.

$證明$ :

顯然, 隨機變數 $\xi$ 和 $\eta$ 的值域為 $R_{\xi} = \left \{ 0, 1 \right \}, R_{\eta} = \left \{ 0, 1, -1 \right \}$, 那麼有 $\displaystyle {\begin {aligned} &\mathop {\mathbf {P}} \left \{ \xi = 0 \right \} = \frac {2}{3}, \mathop {\mathbf {P}} \left \{ \xi = 1 \right \} = \frac {1}{3}, \\ &\mathop {\mathbf {P}} \left \{ \eta = 0 \right \} = \mathop {\mathbf {P}} \left \{ \eta = 1 \right \} = \mathop {\mathbf {P}} \left \{ \eta = -1 \right \} = \frac {1}{3}, \\ &\mathop {\mathbf {P}} \left \{ \xi = 0, \eta = 0 \right \} = 0, \mathop {\mathbf {P}} \left \{ \xi = 1, \eta = 0 \right \} = \frac {1}{3}, \\ &\mathop {\mathbf {P}} \left \{ \xi = 0, \eta = 1 \right \} = \frac {1}{3}, \mathop {\mathbf {P}} \left \{ \xi = 1, \eta = 1 \right \} = 0, \\ &\mathop {\mathbf {P}} \left \{ \xi = 0, \eta = -1 \right \} = \frac {1}{3}, \mathop {\mathbf {P}} \left \{ \xi = 1, \eta = 1 \right \} = 0. \end {aligned}}$ 因此, 我們有 $\displaystyle {\mathop {\mathrm {E}}(\xi\eta) = 0 \times 0 + 0 \times \frac {1}{3} + \frac {1}{3} \times 0 + 0 \times 1 + \frac {1}{3} \times 0 + 0 \times (-1) = 0}$ 和 $\displaystyle {\mathop {\mathrm {E}}(\xi) = \frac {2}{3} \times 0 + \frac {1}{3} \times 1 = \frac {1}{3}, \mathop {\mathrm {E}}(\eta) = \frac {1}{3} \times 0 + \frac {1}{3} \times (-1) + \frac {1}{3} \times 1 = 0}.$ 於是, $\displaystyle {\mathop {\mathrm {Cov}}(\xi, \eta) = \mathop {\mathrm {E}}(\xi\eta) - \mathop {\mathrm {E}}(\xi)\mathop {\mathrm {E}}(\eta) = 0}.$ 故隨機變數 $\xi$ 和 $\eta$ 不相關. 然而, 對於任意 $i \in R_{\xi}, j \in R_{\eta}$, 都有 $\displaystyle {\mathop {\mathbf {P}} \left \{ \xi = i, \eta = j \right \} \neq \mathop {\mathbf {P}} \left \{ \xi = i \right \}\mathop {\mathbf {P}} \left \{ \eta = j \right \}},$ 且由 $\sin^{2} {\alpha} + \cos^{2} {\alpha} = 1$ 可得 : 隨機變數 $\xi$ 和 $\eta$ 存在函數關係 $\displaystyle {\xi^{2} + \eta^{2} = 1}.$

$\blacksquare$

特別地, 若隨機變數 $\xi_{1}, \xi_{2}, ..., \xi_{n}$ 兩兩不相關 (不一定兩兩獨立), 則有 $\displaystyle {\mathop {\mathrm {Var}} \left (\sum \limits_{i = 1}^{n}\xi_{i} \right ) = \sum \limits_{i = 1}^{n}\mathop {\mathrm {Var}}(\xi_{i})}.$ 考慮兩個隨機變數 $\xi$ 和 $\eta$, 假設只對隨機變數 $\xi$ 進行觀測. 如果隨機變數 $\xi$ 和 $\eta$ 相關, 則可以預期 : 已知 $\xi$ 的值可以對未觀測的隨機變數 $\eta$ 的值作出某種判斷.

我們把隨機變數 $\xi$ 的任何一個函數 $f = f(\xi)$ 稱作隨機變數 $\eta$ 的一個估計量. 若有 $\displaystyle {\mathop {\mathrm {E}} \left ( (\eta - f^{*}(\xi))^{2} \right ) = \inf \limits_{f}{\mathop {\mathrm {E}} \left ( (\eta - f(\xi))^{2} \right )}},$ 則稱 $f^{*} = f^{*}(\xi)$ 為在均方意義下的最佳估計量 (the optimal estimator).

對於線性估計 $\lambda(\xi) = a + b\xi$ 類中, 我們接下來討論如何求得線性估計類中的最佳線性估計. 考慮函數 $\displaystyle {g(a, b) = \mathop {\mathrm {E}} \left ( (\eta - f(\xi))^{2} \right )},$ 對多變數函數 $g(a, b)$ 針對 $a$ 和 $b$ 求偏導數, 得 $\displaystyle {\frac {\partial g}{\partial a} = -2\mathop {\mathrm {E}} \big ( \eta - (a + b\xi) \big ) \text { 和 } \frac {\partial g}{\partial b} = -2\mathop {\mathrm {E}} \big ( (\eta - (a + b)\xi)\xi \big )}.$ 令 $\frac {\partial g}{\partial a} = 0$ 且 $\frac {\partial g}{\partial b} = 0$, 即可求出 $\lambda$ 均方線性估計 $\lambda^{*} = a^{*} + b^{*}\xi$. 其中, $a^{*} = \mathop {\mathrm {E}}(\eta) - b^{*}\mathop {\mathrm {E}}(\xi)$, $b^{*} = \frac {\mathop {\mathrm {Cov}}(\xi, \eta)}{\mathop {\mathrm {Var}}(\xi)}$. 即 $\displaystyle {\lambda^{*}(\xi) = \mathop {\mathrm {E}}(\eta) + \frac {\mathop {\mathrm {Cov}}(\xi, \eta)}{\mathop {\mathrm {Var}}(\xi)}(\xi - \mathop {\mathrm {E}}(\xi))}.$ 稱 $\Delta^{*} = \mathop {\mathrm {E}} \big ( (\eta - \lambda^{*}(\xi))^{2} \big )$ 為估計值的均方誤差 (the mean-square error of estimation), 則有 $\displaystyle {\Delta^{*} = \mathop {\mathrm {E}} \big ( (\eta - \lambda^{*}(\xi))^{2} \big ) = \mathop {\mathrm {Var}}(\eta) - \frac {\mathop {\mathrm {Cov}} \left (\xi, \eta \right )}{\mathop {\mathrm {Var}}(\xi)} = \mathop {\mathrm {Var}}(\eta) \left ( 1 - \rho^{2}(\xi, \eta) \right )}.$ 可見, $\left | \rho(\xi, \eta) \right |$ 越大, 均方誤差 $\Delta^{*}$ 越小. 特別地, 當 $|\rho(\xi, \eta)| = 1$, 則 $\Delta^{*} = 0$. 如果隨機變數 $\xi$ 和 $\eta$ 不相關, 即 $\rho(\xi, \eta) = 0$, 則 $\lambda^{*}(\xi) = \mathop {\mathrm {E}}(\eta)$. 於是, 在隨機變數 $\xi$ 和 $\eta$ 不相關的情形下, 根據 $\xi$ 對 $\eta$ 的估計為 $\mathop {\mathrm {E}}(\eta)$.

5. 練習題

練習題 1. 向 $n$ 個箱子中獨立地投擲 $k$ 個球, 假設每個球落入各箱子的機率為 $\frac {1}{n}$, 求非空箱子數量的期望.

$解$ :

定義隨機變數 $\xi_{1}, \xi_{2}, ..., \xi_{n}$. 當第 $i$ 個箱子中有球時, $\xi_{i} = 1$; 否則, $\xi_{i} = 0$. 其中, $i = 1, 2, ..., n$. 記 $\eta = \xi_{1} + \xi_{2} + ... + \xi_{n}$, 則隨機變數 $\eta$ 可以表示有球箱子的數量.

將球投入任一箱子中的機率為 $\frac {1}{n}$, 則球不投入任何箱子的機率為 $1 - \frac {1}{n}$. 由於 $n$ 個球的投擲相互獨立, 那麼 $n$ 個球都不投入任何箱子中的機率為 $\displaystyle {\left ( 1 - \frac {1}{n} \right )^{n}}.$ 反之, 至少有一個球投入箱子的機率為 $1 - \left ( 1 - \frac {1}{n} \right )^{n}$. 於是, $\displaystyle {\mathop {\mathbf {P}} \left \{ \xi_{i} = 0 \right \} = \left ( 1 - \frac {1}{n} \right )^{n}, \mathop {\mathbf {P}} \left \{ \xi_{i} = 1 \right \} = 1 - \left ( 1 - \frac {1}{n} \right )^{n}}.$ 根據定義 4, 故有 $\displaystyle {\mathop {\mathrm {E}}(\xi_{i}) = 0 \times \mathop {\mathbf {P}} \left \{ \xi_{i} = 0 \right \} + 1 \times \mathop {\mathbf {P}} \left \{ \xi_{i} = 1 \right \} = 1 - \left ( 1 - \frac {1}{n} \right )^{n}}.$ 那麼, 有 $\displaystyle {\begin {aligned} \mathop {\mathrm {E}}(\eta) &= \mathop {\mathrm {E}}(\xi_{1} + \xi_{2} + ... \xi_{n}) \\ &= \mathop {\mathrm {E}}(\xi_{1}) + \mathop {\mathrm {E}}(\xi_{2}) + ... + \mathop {\mathrm {E}}(\xi_{n}) \\ &= n \left (1 - \left ( 1 - \frac {1}{n} \right )^{n} \right ). \end {aligned}}$

$\blacksquare$

自主習題 1. 設 $\xi_{1}, \xi_{2}, ..., \xi_{n}$ 是獨立的 Bernoulli 隨機變數, 且 $\displaystyle {\mathop {\mathbf {P}} \left \{ \xi_{i} = 0 \right \} = 1 - \lambda_{i}\Delta, \mathop {\mathbf {P}} \left \{ \xi_{i} = 1 \right \} = \lambda_{i}\Delta}.$ 其中, $\Delta > 0, \lambda_{i} > 0, i = 1, 2, ..., n$, 而 $\Delta$ 是較小的數. 記 $o(\cdot)$ 表示 $\cdot$ 的高階無窮小. 證明 :

1. $\mathop {\mathbf {P}} \left \{ \xi_{1} + \xi_{2} + ... + \xi_{n} = 1 \right \} = \Delta \sum \limits_{i = 1}^{n}\lambda_{i} + o(\Delta^{2})$;
2. $\mathop {\mathbf {P}} \left \{ \xi_{1} + \xi_{2} + ... + \xi_{n} > 1 \right \} = o(\Delta^{2})$.

自主習題 2. 證明 : 當 $a = \mathop {\mathrm {E}}(\xi)$ 時, $\mathop {\mathrm {E}} \left ((\xi - a)^{2} \right )$ 達到最大下界 $\displaystyle {\inf \limits_{-\infty < a < +\infty}{\mathop {\mathrm {E}} \left ((\xi - a)^{2} \right )}},$ 即 $\inf \limits_{-\infty < a < +\infty}{\mathop {\mathrm {E}} \left ((\xi - a)^{2} \right )} = \mathop {\mathrm {Var}}(\xi)$.

自主習題 3. 設 $F_{\xi}(x)$ 是隨機變數 $\xi$ 的分佈函數, 而 $m_{e}$ 是 $F_{\xi}(x)$ 的中位數, 即下列條件的點 $\displaystyle {F_{\xi}(m_{e}^{-}) \leq \frac {1}{2} \leq F_{\xi}(m_{e})}.$ 證明 : $\inf \limits_{-\infty < a < +\infty} \mathop {\mathrm {E}}( \left | \xi - a \right |) = \mathop {\mathrm {E}}(\left | \xi - m_{e} \right |)$.

自主習題 4. 設 $P_{\xi}(x) = \mathop {\mathbf {P}} \left \{ \xi = x \right \}, F_{\xi}(x) = \mathop {\mathbf {P}} \left \{ \xi \leq x \right \}$, 證明 :

1. 對於 $a > 0, -\infty < b < +\infty$, 有 $\displaystyle {P_{a\xi + b}(x) = P_{\xi} \left ( \frac {x - a}{a} \right )}$ 和 $\displaystyle {P_{a\xi + b}(x) = F_{\xi} \left ( \frac {x - b}{a} \right )}$ 成立;
2. 如果 $y \geq 0$, 則 $\displaystyle {F_{\xi^{2}}(y) = F_{\xi}(\sqrt {y}) - F_{\xi}(-\sqrt {y}) + P_{\xi}(-\sqrt {y})};$
3. 設 $\xi^{+} = \max \left \{ \xi, 0 \right \}$, 則 $\displaystyle {F_{\xi^{+}}(x) = \begin {cases} 0 & {x < 0} \\ F_{\xi}(x) & {x = 0} \\ F_{\xi}(x) & {x > 0}. \end {cases}}$

自主習題 5. 假設對於隨機變數 $\xi$ 和 $\eta$ 有 $\mathop {\mathrm {E}}(\xi) = \mathop {\mathrm {E}}(\eta) = 0, \mathop {\mathrm {Var}}(\xi) = \mathop {\mathrm {Var}}(\eta) = 1$, 而 $\xi$ 和 $\eta$ 的相關係數為 $\rho(\xi, \eta)$. 證明 : $\displaystyle {\mathop {\mathrm {E}} \left (\max \left \{ \xi^{2}, \eta^{2} \right \} \right ) \leq 1 + \sqrt {1 - \rho^{2}}}.$

自主習題 6. 設 $\xi_{1}, \xi_{2}, ..., \xi_{n}$ 是獨立隨機變數, $\varphi_{1} = \varphi_{1}(\xi_{1}, \xi_{2}, ..., \xi_{k})$ 和 $\varphi_{2} = \varphi_{2}(\xi_{k + 1}, \xi_{k + 2}, ..., \xi_{n})$ 分別是 $(\xi_{1}, \xi_{2}, ..., \xi_{k})$ 和 $(\xi_{k + 1}, \xi_{k + 2}, ..., \xi_{n})$ 的函數. 證明 : $\varphi_{1}$ 和 $\varphi_{2}$ 獨立.

自主習題 7. 證明 : 隨機變數 $\xi_{1}, \xi_{2}, ..., \xi_{n}$ 獨立, 若且唯若對於一切 $x_{1}, x_{2}, ..., x_{n}$, 有 $\displaystyle {F_{\xi_{1}, \xi_{2}, ..., \xi_{n}}(x_{1}, x_{2}, ..., x_{n}) = F_{\xi_{1}}(x_{1})F_{\xi_{2}}(x_{2})...F_{\xi_{n}}(x_{n})}.$ 其中, $F_{\xi_{1}, \xi_{2}, ..., \xi_{n}}(x_{1}, x_{2}, ..., x_{n}) = \mathop {\mathbf {P}} \left \{ \xi_{1} \leq x_{1}, \xi_{2} \leq x_{2}, ..., \xi_{n} \leq x_{n} \right \}$.

自主習題 8. 證明隨機變數 $\xi$ 與自己獨立, 即 $\xi$ 與 $\xi$ 獨立, 若且唯若 $\xi$ 為常數.

自主習題 9. 問 : 隨機變數 $\xi$ 滿足什麼條件時, $\xi$ 與 $\sin {\xi}$ 獨立.

自主習題 10. 設 $\xi$ 和 $\eta$ 時獨立隨機變數, 且 $\eta \neq 0$. 請通過機率 $P_{\xi}(x)$ 和 $P_{\eta}(y)$ 的形式表示機率 $\displaystyle {\mathop {\mathbf {P}} \left \{ \xi\eta \leq z \right \} \text { 和 } \mathop {\mathbf {P}} \left \{ \frac {\xi}{\eta} \leq z \right \}}.$

自主習題 11. 設隨機變數 $\xi, \eta, \zeta$ 滿足 $\left | \xi \right | \leq 1, \left | \eta \right | \leq 1, \left | \zeta \right | \leq 1$. 證明 A. G. Bell 不等式 $\displaystyle {\left | \mathop {\mathrm {E}}(\xi\eta) - \mathop {\mathrm {E}}(\eta\zeta) \right | \leq 1 - \mathop {\mathrm {E}}(\xi\eta)}$ 成立.