令 $Y $ 為 任意非負 連續隨機變數 配備機率密度 $f_Y$,則我們有以下非常簡潔的結果來描述 $Y$ 的期望值 $E[Y]$。
============
Lemma:
\[
E[Y] = \int_0^\infty P(Y>y) dy
\]============
Proof:
首先觀察等式右方,由於 $P\left( {Y > y} \right) = \int_y^\infty {{f_Y}\left( x \right)dx} $ 故
\[\begin{array}{l}
\int_0^\infty {P\left( {Y > y} \right)dy} = \int_0^\infty {\left( {\int_y^\infty {{f_Y}\left( x \right)dx} } \right)dy} \\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}
\end{array} = \int_0^\infty {\left( {\int_0^\infty {{f_Y}\left( x \right){1_{\left\{ {x \ge y} \right\}}}\left( x \right)dx} } \right)dy} \\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}
\end{array} = \int_0^\infty {\left( {\int_0^\infty {{f_Y}\left( x \right){1_{\left\{ {y \le x} \right\}}}\left( y \right)dx} } \right)dy}
\end{array}\]由於 integrand 非負,由 Fubini Theorem 我們可互換積分順序並得到如下結果
\[\begin{array}{l}
\int_0^\infty {P\left( {Y > y} \right)dy} = \int_0^\infty {\int_0^\infty {{f_Y}\left( x \right){1_{\left\{ {y \le x} \right\}}}\left( y \right)dydx} } \\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}
\end{array} = \int_0^\infty {{f_Y}\left( x \right)\left( {\int_0^\infty {{1_{\left\{ {y \le x} \right\}}}\left( y \right)dy} } \right)} dx\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}
\end{array} = \int_0^\infty {{f_Y}\left( x \right)\left( {\int_0^x {1dy} } \right)} dx\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}
\end{array} = \int_0^\infty {{f_Y}\left( x \right)x} dx\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}
\end{array} = \int_0^\infty {x{f_Y}\left( x \right)} dx = E[Y] \;\;\;\;\;\;\; \square
\end{array}\]
Comments:
1. 前述假設 非負隨機變數是指 $Y \ge 0$ almost surely, 亦即 $ P(Y \ge 0) = 1$
2. 上述證明中採用的符號 $1_{A} (x)$ 表示 指示函數 (indicator function),我們給出定義如下:令 $X$ 為任意集合 則我們可定義對其上的任意子集 $A \subset X$ 所對應的指示函數( indicator function of a subset $A$ of a set $X$ ) 為 $1_A: X \to \{0,1\}$ 滿足 \[{1_A}\left( x \right): = \left\{ \begin{array}{l}
1,\begin{array}{*{20}{c}}
{}&{x \in A}
\end{array}\\
0,\begin{array}{*{20}{c}}
{}&{x \notin A}
\end{array}
\end{array} \right.\]
3. 上述結果可用 distribution function 改寫,令 $F_Y(y) := P(Y \leq y)$則
\[
E[Y] = \int_0^\infty P(Y>y) dy = \int_0^\infty (1 - F_Y(y)) dy
\]這個結果可以使得我們在計算期望值的同時,不用再困擾需要先求出 pdf ,只要有分配函數 即可計算期望值。
============
Lemma:
\[
E[Y] = \int_0^\infty P(Y>y) dy
\]============
Proof:
首先觀察等式右方,由於 $P\left( {Y > y} \right) = \int_y^\infty {{f_Y}\left( x \right)dx} $ 故
\[\begin{array}{l}
\int_0^\infty {P\left( {Y > y} \right)dy} = \int_0^\infty {\left( {\int_y^\infty {{f_Y}\left( x \right)dx} } \right)dy} \\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}
\end{array} = \int_0^\infty {\left( {\int_0^\infty {{f_Y}\left( x \right){1_{\left\{ {x \ge y} \right\}}}\left( x \right)dx} } \right)dy} \\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}
\end{array} = \int_0^\infty {\left( {\int_0^\infty {{f_Y}\left( x \right){1_{\left\{ {y \le x} \right\}}}\left( y \right)dx} } \right)dy}
\end{array}\]由於 integrand 非負,由 Fubini Theorem 我們可互換積分順序並得到如下結果
\[\begin{array}{l}
\int_0^\infty {P\left( {Y > y} \right)dy} = \int_0^\infty {\int_0^\infty {{f_Y}\left( x \right){1_{\left\{ {y \le x} \right\}}}\left( y \right)dydx} } \\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}
\end{array} = \int_0^\infty {{f_Y}\left( x \right)\left( {\int_0^\infty {{1_{\left\{ {y \le x} \right\}}}\left( y \right)dy} } \right)} dx\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}
\end{array} = \int_0^\infty {{f_Y}\left( x \right)\left( {\int_0^x {1dy} } \right)} dx\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}
\end{array} = \int_0^\infty {{f_Y}\left( x \right)x} dx\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}
\end{array} = \int_0^\infty {x{f_Y}\left( x \right)} dx = E[Y] \;\;\;\;\;\;\; \square
\end{array}\]
Comments:
1. 前述假設 非負隨機變數是指 $Y \ge 0$ almost surely, 亦即 $ P(Y \ge 0) = 1$
2. 上述證明中採用的符號 $1_{A} (x)$ 表示 指示函數 (indicator function),我們給出定義如下:令 $X$ 為任意集合 則我們可定義對其上的任意子集 $A \subset X$ 所對應的指示函數( indicator function of a subset $A$ of a set $X$ ) 為 $1_A: X \to \{0,1\}$ 滿足 \[{1_A}\left( x \right): = \left\{ \begin{array}{l}
1,\begin{array}{*{20}{c}}
{}&{x \in A}
\end{array}\\
0,\begin{array}{*{20}{c}}
{}&{x \notin A}
\end{array}
\end{array} \right.\]
3. 上述結果可用 distribution function 改寫,令 $F_Y(y) := P(Y \leq y)$則
\[
E[Y] = \int_0^\infty P(Y>y) dy = \int_0^\infty (1 - F_Y(y)) dy
\]這個結果可以使得我們在計算期望值的同時,不用再困擾需要先求出 pdf ,只要有分配函數 即可計算期望值。
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