[機率論] 淺論 弱大數法則

以下我們討論一些關於 弱大數法則(Weak Law of Large Numbers, WLLN) 的結果，首先介紹 一組隨機變數 數列 的 機率收斂  (Convergence in Probability)

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Definition: Convergence in Probability
令 $Y_n$ 為一組隨機變數 sequence，我們說 $Y_n$ converges to $Y$ in probability 若下列條件成立：對任意 $\varepsilon >0$
$$P(|Y_n - Y| > \varepsilon) \rightarrow 0 \;\; \text{ as  $n \rightarrow \infty$}$$ =============================

Comments:
1. 上述定義等價為
$$P(|Y_n - Y| \leq \varepsilon) \rightarrow 1 \;\; \text{ as  $n \rightarrow \infty$}$$
2. 上述定義中 $Y_n \to^P Y$ 的 $Y$ 可為隨機變數或者為常數。
3. 機率收斂在 機率論與隨機過程，以及 統計理論中 扮演重要角色，比如機率收斂在統計中等價稱為 consistent estimator，在此不做贅述。

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Definition: Uncorrelated Random Variables

接著再回憶我們說一組隨機變數 $X_i, \; i \in \mathbb{N}$ 且 $E [X_i^2] <\infty$ 為 uncorrelated 若下列條件成立：當 $i \neq j$
$$E[X_i X_j] = E[X_i] E[X_j]$$ ============================

現在我們看個 uncorrelated 隨機變數的結果

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Theorem:
令 $X_1, X_2,...X_n$ 為 uncorrelated 且 $E[X_i^2] < \infty$ 則
$$var(X_1 + ... + X_n ) = var(X_1) + ... + var(X_n)$$ 其中 $var(X)$ 為 variance of $X$。
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Proof: omitted.


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Lemma:
若 $p >0$ 且 $E[|X_n|^p] \rightarrow 0$ 則 $X_n \rightarrow 0$ in probability。
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Proof:
要證明 $X_n \rightarrow 0$ in probability，亦即
$$P(|{X_n} - 0| > \varepsilon ) \to 0$$ 首先觀察下列事件等價
$$\begin{array}{l} \left\{ {|{X_n}| > \varepsilon } \right\} = \left\{ {|{X_n}{|^p} > {\varepsilon ^p}} \right\}\\  \Rightarrow P\left\{ {|{X_n}| > \varepsilon } \right\} = P\left\{ {|{X_n}{|^p} > {\varepsilon ^p}} \right\} \end{array}$$ 由 Markov inequality 可知
$$P\left\{ {|{X_n}| > \varepsilon } \right\} = P\left\{ {|{X_n}{|^p} > {\varepsilon ^p}} \right\} \le \frac{{E\left[ {|{X_n}{|^p}} \right]}}{{{\varepsilon ^p}}}$$ 由於 $E[|X_n|^p] \rightarrow 0$ ，且 $\varepsilon^p$ 為定值，故
$$P\left\{ {|{X_n}| > \varepsilon } \right\} \le \frac{{E\left[ {|{X_n}{|^p}} \right]}}{{{\varepsilon ^p}}} \to {\rm{0}}$$


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Theorem: $L^2$ Weak Law
令 $X_1, ..., X_n$ 為 uncorrelated 隨機變數 且 $E[X_i] = \mu$，$var(X_i) \le C < \infty$。現在定義隨機變數的和 $S_n := X_1 + X_2 + ... + X_n$ 則
$$S_n/n \rightarrow \mu \text{  in Probability as  $n \rightarrow \infty$}$$ =============================

Proof
要證明 給定 $\varepsilon>0$ 當 $n \rightarrow \infty$
$$P\left( {\left| {\frac{{{S_n}}}{n} - \mu } \right| > \varepsilon } \right) \to 0 \ \ \ \ (*)$$ 不過如前一個定理所述，若我們可以證明 $E\left[ {{{\left| {\frac{{{S_n}}}{n} - \mu } \right|}^2}} \right] \to 0$ 則 $(*)$ 自動滿足。

首先觀察
$$E\left[ {\frac{{{S_n}}}{n}} \right] = \frac{1}{n}E\left[ {{S_n}} \right] = \frac{1}{n}\sum\limits_{i = 1}^n {E{X_i}}  = \frac{\mu }{n}n = \mu$$ 則
$$\begin{array}{l} E\left[ {{{\left| {\frac{{{S_n}}}{n} - \mu } \right|}^2}} \right] = E\left[ {{{\left| {\frac{{{S_n}}}{n} - E\left[ {\frac{{{S_n}}}{n}} \right]} \right|}^2}} \right]\\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{}&{} \end{array} = var\left( {\frac{{{S_n}}}{n}} \right) = \frac{1}{{{n^2}}}var\left( {{S_n}} \right)\\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{}&{} \end{array} = \frac{1}{{{n^2}}}var\left( {\sum\limits_{i = 1}^n {{X_i}} } \right) \end{array}$$ 由於 $X_1, ..., X_n$ 為 uncorrelated 隨機變數，故我們有 $var\left( {\sum\limits_{i = 1}^n {{X_i}} } \right) = \sum\limits_{i = 1}^n {var\left( {{X_i}} \right)}$ 將此結果帶入上式可得
$$\begin{array}{l} E\left[ {{{\left| {\frac{{{S_n}}}{n} - \mu } \right|}^2}} \right] = \frac{1}{{{n^2}}}var\left( {\sum\limits_{i = 1}^n {{X_i}} } \right) = \frac{1}{{{n^2}}}\sum\limits_{i = 1}^n {var\left( {{X_i}} \right)} \\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{}&{} \end{array} \le \frac{1}{{{n^2}}}Cn = \frac{1}{n}C \to 0 \end{array}$$ 當 $n \rightarrow \infty$ 故 $L^2$-convergence。又 $L^2$ convergence 保證 convergence in Probability，故
$$P\left( {\left| {\frac{{{S_n}}}{n} - \mu } \right| > \varepsilon } \right) \to 0 \ \ \ \ \square$$

現在問題變成若我們想要拓展上述的 Weak law (e.g., 拔除 finite 2nd moment 條件 )，則我們必須引入一些新的定義如下：

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Definition: Tail Equivalence of Two Sequences of Random Variables
我們說兩隨機變數的 sequences $\{X_n\}$ 與 $\{Y_n \}$ 為 Tail Equivalent 若下列條件成立
$$\sum_n P(X_n \neq Y_n) < \infty$$ ===========================

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Definition: Truncation Function
定義以下剪切函數(truncation function)
$$X{1_{\left| X \right| \le M}} := \left\{ \begin{array}{l} X,\begin{array}{*{20}{c}} {}&{} \end{array}\left| X \right| \le M\\ 0,\begin{array}{*{20}{c}} {}&{} \end{array}\left| X \right| > M \end{array} \right.$$ ============================

首先看幾個結果
FACT 1: 若 $Y \ge 0$ 且 $p >0$ 則
$$E[Y^p] = \int_0^\infty p y^{p-1} P(Y>y)dy$$ Proof: 首先觀察積分
$$\begin{array}{l} \int_0^\infty  p {y^{p - 1}}P(Y > y)dy = \int_0^\infty  p {y^{p - 1}}\int_\Omega ^{} {{1_{Y > y}}dP} dy\\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{}&{}&{} \end{array} = \int_0^\infty  {\int_\Omega ^{} {{1_{Y > y}}} p} {y^{p - 1}}dPdy\\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{}&{}&{} \end{array} = \int_\Omega ^{} {\int_0^\infty  {{1_{Y > y}}} p{y^{p - 1}}dydP} \\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{}&{}&{} \end{array} = \int_\Omega ^{} {\int_0^Y {p{y^{p - 1}}} dydP} \end{array}$$ 因為 $\int_0^Y {p{y^{p - 1}}} dy = {Y^p}$ 故代入上式可得
$$\Rightarrow \int_0^\infty  p {y^{p - 1}}P(Y > y)dy = \int_\Omega ^{} {\int_0^Y {p{y^{p - 1}}} dydP = } \int_\Omega ^{} {{Y^p}dP = } E[{Y^p}] \ \ \ \ \square$$


現在我們可以介紹 General Weak Law of Large Number：

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Theorem: Weak Law of Large Number
令 $X_1, X_2, ...$ 為 i.i.d. 隨機變數 且 $S_n := X_1 + X_2 + ... + X_n$。若
(1) $\sum_{j=1}^n P(|X_j > n|) \rightarrow 0$
(2) $\frac{1}{n^2} \sum_{j=1}^n E[ X_j^2 1_{|X_j \le n|}] \rightarrow 0$
則
$$S_n / n - \mu_n \rightarrow 0 \text{  in Probability}$$ 其中 $a_n := \sum_{j=1}^{n}E [X_j 1 _{|X_j| \le n}]$
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Comments
在證明之前有幾點值得注意，上述 Weak Law of Large Number 並無對 $E[X_i^2]$ 有做假設

Proof
首先定義 $X_{nj}' := X_j 1_{|X_j| \le n}$ 且 $S_n' := \sum_{j=1}^n X_{nj}'$ 則觀察 Tail parts
$$\sum\limits_{j = 1}^n {P\left( {{X_{nj}}' \ne {X_j}} \right)}  = \sum\limits_{j = 1}^n {P\left( {\left| {{X_j}} \right| > n} \right) \to 0}$$ 上述收斂成立由 Hypothesis $(1)$。接著我們觀察
$$\begin{array}{l} P\left( {\left| {{S_n} - {S_n}'} \right| > \varepsilon } \right) \le P\left( {{S_n} \ne {S_n}'} \right)\\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{}&{} \end{array} \le P\left( {\bigcup\limits_{j = 1}^n {\left\{ {{X_{nj}}' \ne {X_j}} \right\}} } \right)\\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{}&{} \end{array} \le \sum\limits_{j = 1}^n {P\left( {{X_{nj}}' \ne {X_j}} \right)}  = \sum\limits_{j = 1}^n {P\left( {\left| {{X_j}} \right| > n} \right) \to 0} \end{array}$$ 故可知
$$S_n - S_n' \rightarrow 0 \text{  in Probability }$$
現在觀察
$$\begin{array}{l} P\left( {\frac{{\left| {{S_n}' - E{S_n}'} \right|}}{n} > \varepsilon } \right) \le \frac{{E\left[ {{{\left| {{S_n}' - E{S_n}'} \right|}^2}} \right]}}{{{n^2}{\varepsilon ^2}}} = \frac{{{\mathop{\rm var}} \left( {{S_n}'} \right)}}{{{n^2}{\varepsilon ^2}}}\\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{}&{}&{}&{} \end{array}{\rm{ = }}\frac{{{\mathop{\rm var}} \left( {{S_n}'} \right)}}{{{n^2}{\varepsilon ^2}}}{\rm{ = }}\frac{1}{{{n^2}{\varepsilon ^2}}}\sum\limits_{j = 1}^n {{\mathop{\rm var}} \left( {{X_{nj}}'} \right)} \\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{}&{}&{}&{} \end{array} \le \frac{1}{{{n^2}{\varepsilon ^2}}}\sum\limits_{j = 1}^n {E\left[ {{{\left( {{X_{nj}}'} \right)}^2}} \right]} \\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{}&{}&{}&{} \end{array} = \frac{1}{{{n^2}{\varepsilon ^2}}}\sum\limits_{j = 1}^n {E\left[ {{{\left( {{X_j}{1_{\left\{ {\left| {{X_j}} \right| \le n} \right\}}}} \right)}^2}} \right]} \\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{}&{}&{}&{} \end{array} = \frac{1}{{{n^2}{\varepsilon ^2}}}\sum\limits_{j = 1}^n {E\left[ {{X_j}^2{1_{\left\{ {\left| {{X_j}} \right| \le n} \right\}}}} \right]}  \to 0 \ \ \ \ (**) \end{array}$$ 上式收斂結果來自 Hypothesis (2)。

注意到 $${a_n}: = \sum\limits_{j = 1}^n E [{X_j}{1_{|{X_j}| \le n}}] = E\left[ {\sum\limits_{j = 1}^n {{X_j}{1_{|{X_j}| \le n}}} } \right] = E\left[ {{S_n}'} \right]$$ 故由 $(**)$ 可知
$$\frac{{{S_n}' - {a_n}}}{n}\mathop  \to \limits^P 0$$ 故現在觀察
$$\frac{{{S_n} - {a_n}}}{n} = \frac{{{S_n} - {S_n}' + {S_n}' - {a_n}}}{n} = \frac{{{S_n} - {S_n}'}}{n} + \frac{{{S_n}' - {a_n}}}{n}\mathop  \to \limits^P 0$$