随机变量函数中包含Φ(X)的数学期望研究

Study on the Mathematical Expectation of Functions of Random Variables Involving Φ(X)

摘要: 论文研究随机变量函数中包含$\Phi \left(X\right)$ 的数学期望，给出了几个定理的证明，而证明方法上具有较好的借鉴作用。

Abstract: This paper studies the mathematical expectation of functions of random variables that involve Φ(X), provides proofs of several theorems, and the proof methods offer good reference value for related research.

1. 引言

正态分布在概率论与数理统计中起着举足轻重的作用，也是最为重要的教学内容之一，具体可查阅文献[1]-[5]。设随机变量$X~N\left(0,1\right)$ ，其密度函数与分布函数分别记为$\phi \left(x\right)=\frac{1}{\sqrt{2\pi }}{\text{e}}^{-x^2/2}$ ，$\Phi \left(x\right)={\displaystyle {\int }_{-\infty }^x\phi \left(t\right)\text{d}t}$ 。若随机变量$X~N\left(\mu ,{\sigma }^2\right)$ ，其密度函数与分布函数分别记为$\frac{1}{\sigma }\phi \left(\frac{x-\mu }{\sigma }\right),\Phi \left(\frac{x-\mu }{\sigma }\right)$ 。本文研究随机变量函数中包含$\Phi \left(X\right)$ 的数学期望，给出了几个定理的证明，证明方法上具有较好的借鉴作用。

2. 随机变量函数中包含$\Phi \left(X\right)$ 的数学期望

引理1：记$t$ 的函数$g_1\left(t\right)={\displaystyle {\int }_{-\infty }^{+\infty }x\phi \left(x\right)\Phi \left(tx\right)\text{d}x}$ ，$g_2\left(t\right)={\displaystyle {\int }_{-\infty }^{+\infty }{x}^2\phi \left(x\right)\Phi \left(tx\right)\text{d}x}$ ， $g_3\left(t\right)={\displaystyle {\int }_{-\infty }^{+\infty }x\left|x\right|\phi \left(x\right)\Phi \left(tx\right)\text{d}x}$ 以及$g_4\left(t\right)={\displaystyle {\int }_0^{+\infty }\phi \left(x\right)\Phi \left(tx\right)\text{d}x}$ ，则

$g_1\left(t\right)=\frac{1}{\sqrt{2\pi }}\frac{t}{\sqrt{1+t^2}},g_2\left(t\right)=\frac{1}{2},g_3\left(t\right)=\frac{1}{\pi }\left(\frac{t}{1+t^2}+\arctan t\right),g_4\left(t\right)=\frac{1}{4}+\frac{\arctan t}{2\pi }$

证明：(1) 易见 $g_1\left(0\right)=\frac{1}{2}{\displaystyle {\int }_{-\infty }^{+\infty }x\phi \left(x\right)\text{d}x}=0$

$\begin{array}{c}g{\text{'}}_1\left(t\right)={\displaystyle {\int }_{-\infty }^{+\infty }{x}^2\phi \left(x\right)\phi \left(tx\right)\text{d}x}=\frac{1}{2\pi }{\displaystyle {\int }_{-\infty }^{+\infty }{x}^2\exp \left(-\frac{1+t^2}{2}{x}^2\right)\text{d}x}=\frac{1}{\pi }{\displaystyle {\int }_0^{+\infty }{x}^2\exp \left(-\frac{1+t^2}{2}{x}^2\right)\text{d}x}\\ =\frac{\sqrt{2}}{\pi }\frac{1}{\left(1+t^2\right)\sqrt{1+t^2}}{\displaystyle {\int }_0^{+\infty }{y}^{\frac{1}{2}}{\text{e}}^{-y}\text{d}y}=\frac{\sqrt{2}}{\pi }\frac{1}{\left(1+t^2\right)\sqrt{1+t^2}}\Gamma \left(\frac{3}{2}\right)=\frac{1}{\sqrt{2\pi }}\frac{1}{\left(1+t^2\right)\sqrt{1+t^2}}\end{array}$

则 $g_1\left(t\right)={\displaystyle {\int }_0^t{g^{\prime }}_1\left(x\right)\text{d}x}=\frac{1}{\sqrt{2\pi }}{\displaystyle {\int }_0^t\frac{1}{\left(1+x^2\right)\sqrt{1+x^2}}\text{d}x}=\frac{1}{\sqrt{2\pi }}\frac{t}{\sqrt{1+t^2}}$

(2) 易见$g_2\left(t\right)={\displaystyle {\int }_{-\infty }^{+\infty }{x}^2\phi \left(x\right)\Phi \left(tx\right)\text{d}x}={\displaystyle {\int }_0^{+\infty }{x}^2\phi \left(x\right)\Phi \left(tx\right)\text{d}x}+{\displaystyle {\int }_{-\infty }^0{x}^2\phi \left(x\right)\Phi \left(tx\right)\text{d}x}$

$={\displaystyle {\int }_0^{+\infty }{x}^2\phi \left(x\right)\Phi \left(tx\right)\text{d}x}+{\displaystyle {\int }_0^{+\infty }{x}^2\phi \left(x\right)\Phi \left(-tx\right)\text{d}x}={\displaystyle {\int }_0^{+\infty }{x}^2\phi \left(x\right)\text{d}x}=\frac{1}{2}{\displaystyle {\int }_{-\infty }^{+\infty }{x}^2\phi \left(x\right)\text{d}x}=\frac{1}{2}$

(3) 易见 $g_3\left(0\right)=\frac{1}{2}{\displaystyle {\int }_{-\infty }^{+\infty }x\left|x\right|\phi \left(x\right)\text{d}x}=0$

$\begin{array}{c}{g^{\prime }}_3\left(t\right)={\displaystyle {\int }_{-\infty }^{+\infty }{x}^2\left|x\right|\phi \left(x\right)\phi \left(tx\right)\text{d}x}=2{\displaystyle {\int }_0^{+\infty }{x}^3\phi \left(x\right)\phi \left(tx\right)\text{d}x}\\ =\frac{1}{\pi }{\displaystyle {\int }_0^{+\infty }{x}^3\exp \left\{-\frac{\left(1+t^2\right)x^2}{2}\right\}\text{d}x}=\frac{2}{\pi }\frac{1}{{\left(1+t^2\right)}^2}{\displaystyle {\int }_0^{+\infty }t{\text{e}}^{-t}\text{d}t}=\frac{2}{\pi }\frac{1}{{\left(1+t^2\right)}^2}\end{array}$

$g_3\left(t\right)={\displaystyle {\int }_0^t{g^{\prime }}_3\left(z\right)\text{d}z}={\displaystyle {\int }_0^t}\frac{2}{\pi }\frac{1}{{\left(1+z^2\right)}^2}\text{d}z=\frac{2}{\pi }\cdot {\left[\frac{1}{2}\left(\frac{z}{1+z^2}+\arctan z\right)\right]|}_0^t=\frac{1}{\pi }\left(\frac{t}{1+t^2}+\arctan t\right)$

(4) 易见 $g_4\left(0\right)=\frac{1}{2}{\displaystyle {\int }_0^{+\infty }\phi \left(x\right)\text{d}x}=\frac{1}{4}$

${g^{\prime }}_4\left(t\right)={\displaystyle {\int }_0^{+\infty }x\phi \left(x\right)\phi \left(tx\right)\text{d}x}=\frac{1}{2\pi }{\displaystyle {\int }_0^{+\infty }x\exp \left(-\frac{1+t^2}{2}{x}^2\right)\text{d}x}=\frac{1}{2\pi }\frac{1}{1+t^2}$

则 $g_4\left(t\right)=g_4\left(0\right)+\frac{1}{2\pi }{\displaystyle {\int }_0^t\frac{1}{1+x^2}\text{d}x}=\frac{1}{4}+\frac{\arctan t}{2\pi }$

定理1：设随机变量$X~N\left(0,1\right)$ ，则(1) $E\left[\Phi \left(X\right)\right]=\frac{1}{2}$ ；(2) $E\left[X\Phi \left(X\right)\right]=\frac{1}{2\sqrt{\pi }}$ ； (3) $E\left[X^2\Phi \left(X\right)\right]=\frac{1}{2}$ ；(4) $E\left[\left|X\right|\Phi \left(X\right)\right]=\frac{1}{\sqrt{2\pi }}$ ；(5) $E\left[X\left|X\right|\Phi \left(X\right)\right]=\frac{1}{4}+\frac{1}{2\pi }$

证明：(1) 由于$\Phi \left(X\right)~U\left(0,1\right)$ ，则$E\left[\Phi \left(X\right)\right]=\frac{1}{2}$ ，并由引理1可知：

(2) $E\left[X\Phi \left(X\right)\right]={\displaystyle {\int }_{-\infty }^{+\infty }x\phi \left(x\right)\Phi \left(x\right)\text{d}x}=g_1\left(1\right)=\frac{1}{2\sqrt{\pi }}$

(3) $E\left[X^2\Phi \left(X\right)\right]={\displaystyle {\int }_{-\infty }^{+\infty }{x}^2\phi \left(x\right)\Phi \left(x\right)\text{d}x}=g_2\left(1\right)=\frac{1}{2}$

(4) $\begin{array}{c}E\left[\left|X\right|\Phi \left(X\right)\right]={\displaystyle {\int }_{-\infty }^{+\infty }\left|x\right|\phi \left(x\right)\Phi \left(x\right)\text{d}x}={\displaystyle {\int }_0^{+\infty }x\phi \left(x\right)\Phi \left(x\right)\text{d}x}-{\displaystyle {\int }_{-\infty }^{0}x\phi \left(x\right)\Phi \left(x\right)\text{d}x}\\ ={\displaystyle {\int }_0^{+\infty }x\phi \left(x\right)\Phi \left(x\right)\text{d}x}+{\displaystyle {\int }_0^{+\infty }x\phi \left(x\right)\Phi \left(-x\right)\text{d}x}={\displaystyle {\int }_0^{+\infty }x\phi \left(x\right)\text{d}x}=\frac{1}{\sqrt{2\pi }}\end{array}$

(5) $E\left[X\left|X\right|\Phi \left(X\right)\right]={\displaystyle {\int }_{-\infty }^{+\infty }x\left|x\right|\phi \left(x\right)\Phi \left(x\right)\text{d}x}=g_3\left(1\right)=\frac{1}{\pi }\left(\frac{1}{2}+\frac{\pi }{4}\right)=\frac{1}{4}+\frac{1}{2\pi }$

引理2：记$t$ 的函数$I\left(t\right)={\displaystyle {\int }_{-\infty }^{+\infty }{x}^n\phi \left(x\right)\Phi \left(tx\right)\text{d}x}$ ，则

(1) 若$n$ 为偶数，$I\left(t\right)=\frac{1}{\sqrt{\pi }}{2}^{n/2-1}\Gamma \left(\frac{n+1}{2}\right)$ ；

(2) 若$n$ 为奇数，

$\begin{array}{c}I\left(t\right)=\frac{1}{\pi }{2}^{n/2}\text{Γ}\left(\frac{n}{2}+1\right){\displaystyle {\int }_0^t\frac{1}{{\left(1+x^2\right)}^{n/2+1}}\text{d}x}\\ =\frac{1}{\pi }{2}^{n/2}\text{Γ}\left(\frac{n}{2}+1\right)\frac{t}{\sqrt{1+t^2}}{\displaystyle \sum }_{i=0}^{\left(n+1\right)/2-1}{\left(-1\right)}^i\frac{C_{\left(n+1\right)/2-1}^i}{2i+1}{\left(\frac{t^2}{1+t^2}\right)}^i\end{array}$

证明：(1)若$n$ 为偶数，

$\begin{array}{c}I\left(t\right)={\displaystyle {\int }_0^{+\infty }{x}^n\phi \left(x\right)\Phi \left(tx\right)\text{d}x}+{\displaystyle {\int }_{-\infty }^0{x}^n\phi \left(x\right)\Phi \left(tx\right)\text{d}x}\\ ={\displaystyle {\int }_0^{+\infty }{x}^n\phi \left(x\right)\Phi \left(tx\right)\text{d}x}+{\displaystyle {\int }_0^{+\infty }{x}^n\phi \left(x\right)\Phi \left(-tx\right)\text{d}x}\\ ={\displaystyle {\int }_0^{+\infty }{x}^n\phi \left(x\right)\text{d}x}\\ =\frac{1}{\sqrt{2\pi }}{\displaystyle {\int }_0^{+\infty }{x}^n{\text{e}}^{-x^2/2}\text{d}x}=\frac{1}{\sqrt{\pi }}{2}^{n/2-1}\Gamma \left(\frac{n+1}{2}\right)\end{array}$

(2) 若$n$ 为奇数，

$\begin{array}{c}I\left(t\right)={\displaystyle {\int }_0^{+\infty }{x}^n\phi \left(x\right)\Phi \left(tx\right)\text{d}x}+{\displaystyle {\int }_{-\infty }^0{x}^n\phi \left(x\right)\Phi \left(tx\right)\text{d}x}\\ ={\displaystyle {\int }_0^{+\infty }{x}^n\phi \left(x\right)\Phi \left(tx\right)\text{d}x}-{\displaystyle {\int }_0^{+\infty }{x}^n\phi \left(x\right)\Phi \left(-tx\right)\text{d}x}\\ =2{\displaystyle {\int }_0^{+\infty }{x}^n\phi \left(x\right)\Phi \left(tx\right)\text{d}x}-{\displaystyle {\int }_0^{+\infty }{x}^n\phi \left(x\right)\text{d}x}\end{array}$

易见 $I\left(0\right)=0$

$\begin{array}{c}{I}^{\prime }\left(t\right)=2{\displaystyle {\int }_0^{+\infty }{x}^{n+1}\phi \left(x\right)\phi \left(tx\right)\text{d}x}=\frac{1}{\pi }{\displaystyle {\int }_0^{+\infty }{x}^{n+1}\exp \left[-\frac{\left(1+t^2\right)x^2}{2}\right]\text{d}x}\\ =\frac{1}{\pi }\frac{2^{n/2}}{{\left(1+t^2\right)}^{n/2+1}}{\displaystyle {\int }_0^{+\infty }{x}^{n/2}{\text{e}}^{-x}\text{d}x}=\frac{1}{\pi }\frac{2^{n/2}}{{\left(1+t^2\right)}^{n/2+1}}\Gamma \left(\frac{n}{2}+1\right)\end{array}$

$I\left(t\right)={\displaystyle {\int }_0^t{I}^{\prime }\left(x\right)\text{d}x}=\frac{1}{\pi }{2}^{n/2}\Gamma \left(\frac{n}{2}+1\right){\displaystyle {\int }_0^t\frac{1}{{\left(1+x^2\right)}^{n/2+1}}\text{d}x}$

由于$n$ 是奇数，记$n=2k-1,k=1,2,\cdots$ ，考虑不定积分：

$\begin{array}{c}{I}_k={\displaystyle \int \frac{1}{{\left(1+x^2\right)}^{n/2+1}}\text{d}x}={\displaystyle \int \frac{1}{{\left(1+x^2\right)}^{k+1/2}}\text{d}x}={\displaystyle \int \frac{{\sec }^2\theta }{{\sec }^{2k+1}\theta }\text{d}\theta }={\displaystyle \int \frac{1}{{\sec }^{2k-1}\theta }\text{d}\theta }={\displaystyle \int {\cos }^{2k-1}\theta \text{d}\theta }\\ ={\displaystyle \int {\left(1-{\sin }^2\theta \right)}^{k-1}\text{d}\sin \theta }={\displaystyle \int {\left(1-u^2\right)}^{k-1}\text{d}u}={\displaystyle \int {\displaystyle \sum }_{i=0}^{k-1}{\left(-1\right)}^i{C}_{k-1}^i{u}^{2i}\text{d}u}={\displaystyle \sum }_{i=0}^{k-1}{\left(-1\right)}^i\frac{C_{k-1}^i}{2i+1}{u}^{2i+1}+C\\ =\frac{x}{\sqrt{1+x^2}}{\displaystyle \sum }_{i=0}^{k-1}{\left(-1\right)}^i\frac{C_{k-1}^i}{2i+1}{\left(\frac{x^2}{1+x^2}\right)}^i+C\end{array}$

由此 ${\displaystyle {\int }_0^t\frac{1}{{\left(1+x^2\right)}^{n/2+1}}\text{d}x}=\frac{t}{\sqrt{1+t^2}}{\displaystyle \sum }_{i=0}^{\left(n+1\right)/2-1}{\left(-1\right)}^i\frac{C_{\left(n+1\right)/2-1}^i}{2i+1}{\left(\frac{t^2}{1+t^2}\right)}^i$

进而 $I\left(t\right)=\frac{1}{\pi }{2}^{n/2}\Gamma \left(\frac{n}{2}+1\right)\frac{t}{\sqrt{1+t^2}}{\displaystyle \sum }_{i=0}^{\left(n+1\right)/2-1}{\left(-1\right)}^i\frac{C_{\left(n+1\right)/2-1}^i}{2i+1}{\left(\frac{t^2}{1+t^2}\right)}^i$

注：记$t$ 的函数$I\left(t\right)={\displaystyle {\int }_{-\infty }^{+\infty }{\left|x\right|}^n\phi \left(x\right)\Phi \left(tx\right)\text{d}x}$ ，类似于引理的证明有：

$I\left(t\right)=\frac{1}{\sqrt{\pi }}{2}^{n/2-1}\Gamma \left(\frac{n+1}{2}\right)$

由引理2易见如下定理2：

定理2：(1) 若$n$ 为偶数，$E\left[X^n\Phi \left(X\right)\right]=\frac{1}{\sqrt{\pi }}{2}^{n/2-1}\Gamma \left(\frac{n+1}{2}\right)$ ；(2) 若$n$ 为奇数，

$E\left[X^n\Phi \left(X\right)\right]=\frac{1}{\pi }{2}^{n/2}\Gamma \left(\frac{n}{2}+1\right)\frac{1}{\sqrt{2}}{\displaystyle \sum }_{i=0}^{\left(n+1\right)/2-1}{\left(-1\right)}^i\frac{C_{\left(n+1\right)/2-1}^i}{2i+1}{\left(\frac{1}{2}\right)}^i$

(3) $E\left[{\left|X\right|}^n\Phi \left(X\right)\right]=\frac{1}{\sqrt{\pi }}{2}^{n/2-1}\Gamma \left(\frac{n+1}{2}\right)$

特别地，当$n=1$ 时，

$I\left(t\right)=\frac{1}{\pi }{2}^{1/2}\Gamma \left(\frac{1}{2}+1\right){\displaystyle {\int }_0^t\frac{1}{{\left(1+x^2\right)}^{1/2+1}}\text{d}x}=\frac{1}{\sqrt{2\pi }}{\displaystyle {\int }_0^t\frac{1}{{\left(1+x^2\right)}^{1/2+1}}\text{d}x}=\frac{1}{\sqrt{2\pi }}\frac{t}{\sqrt{1+t^2}}$ ，

$E\left[X\Phi \left(X\right)\right]=\frac{1}{2\sqrt{\pi }}$ ，$E\left[\left|X\right|\Phi \left(X\right)\right]=\frac{1}{\sqrt{2\pi }}$

当$n=2$ 时，

$E\left[X^2\Phi \left(X\right)\right]=\frac{1}{\sqrt{\pi }}\Gamma \left(\frac{2+1}{2}\right)=\frac{1}{2}$

当$n=3$ 时，

$I\left(t\right)=\frac{1}{\pi }{2}^{3/2}\Gamma \left(\frac{3}{2}+1\right){\displaystyle {\int }_0^t\frac{1}{{\left(1+x^2\right)}^{3/2+1}}\text{d}x}=\frac{3}{\sqrt{2\pi }}{\displaystyle {\int }_0^t\frac{1}{{\left(1+x^2\right)}^{3/2+1}}\text{d}x}=\frac{1}{\sqrt{2\pi }}\frac{t}{\sqrt{1+t^2}}\frac{3+2t^2}{1+t^2}$ ，

$E\left[X^3\Phi \left(X\right)\right]=\frac{5}{4\sqrt{\pi }}$ ，$E\left[{\left|X\right|}^3\Phi \left(X\right)\right]=\sqrt{\frac{2}{\pi }}$

当$n=4$ 时，

$E\left[X^4\Phi \left(X\right)\right]=\frac{1}{\sqrt{\pi }}2\Gamma \left(\frac{4+1}{2}\right)=\frac{3}{2}$

当$n=5$ 时，

$I\left(t\right)=\frac{1}{\pi }{2}^{5/2}\Gamma \left(\frac{5}{2}+1\right){\displaystyle {\int }_0^t\frac{1}{{\left(1+x^2\right)}^{5/2+1}}\text{d}x}=\frac{15}{\sqrt{2\pi }}{\displaystyle {\int }_0^t\frac{1}{{\left(1+x^2\right)}^{5/2+1}}\text{d}x}=\frac{1}{\sqrt{2\pi }}\frac{t}{\sqrt{1+t^2}}\frac{15+20t^2+8t^4}{{\left(1+t^2\right)}^2}$ ，

$E\left[X^5\Phi \left(X\right)\right]=\frac{43}{8\sqrt{\pi }}$ ，$E\left[{\left|X\right|}^5\Phi \left(X\right)\right]=4\sqrt{\frac{2}{\pi }}$

定理3：设随机变量$X~N\left(\mu ,{\sigma }^2\right)$ ，则$E\left[\Phi \left(X\right)\right]=\Phi \left(\frac{\mu }{\sqrt{{\sigma }^2+1}}\right)$

证明： $E\left[\Phi \left(X\right)\right]={\displaystyle {\int }_{-\infty }^{+\infty }\Phi \left(x\right)\frac{1}{\sqrt{2\pi }\sigma }\exp \left[-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}\right]\text{d}x}$

$={\displaystyle {\int }_{-\infty }^{+\infty }\frac{1}{\sqrt{2\pi }\sigma }\exp \left[-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}\right]\text{d}x}{\displaystyle {\int }_{-\infty }^x\frac{1}{\sqrt{2\pi }}\exp \left(-\frac{y^2}{2}\right)\text{d}y}$

考虑如下问题：若$Z_1~N\left(\mu ,{\sigma }^2\right),Z_2~N\left(0,1\right)$ ，且$Z_1,Z_2$ 相互独立，求$P\left(Z_2-Z_1<0\right)$

由于$Z_2-Z_1~N\left(-\mu ,{\sigma }^2+1\right)$ ，$\frac{Z_2-Z_1+\mu }{\sqrt{{\sigma }^2+1}}~N\left(0,1\right)$ ，则

$P\left(Z_2-Z_1<0\right)=P\left(\frac{Z_2-Z_1+\mu }{\sqrt{{\sigma }^2+1}}<\frac{\mu }{\sqrt{{\sigma }^2+1}}\right)=\Phi \left(\frac{\mu }{\sqrt{{\sigma }^2+1}}\right)$

将$\left(Z_1,Z_2\right)$ 视作二维正态分布，然后求$P\left(Z_2-Z_1<0\right)$

易见 $P\left(Z_2-Z_1<0\right)={\displaystyle {\int }_{-\infty }^{+\infty }{f}_{Z_2}\left(x\right)\text{d}x}{\displaystyle {\int }_{-\infty }^x{f}_{Z_1}\left(y\right)\text{d}y}$

则 $E\left[\Phi \left(X\right)\right]=\Phi \left(\frac{\mu }{\sqrt{{\sigma }^2+1}}\right)$

另证：$E\left[\Phi \left(X\right)\right]={\displaystyle {\int }_{-\infty }^{+\infty }\frac{1}{\sigma }\phi \left(\frac{x-\mu }{\sigma }\right)\Phi \left(x\right)\text{d}x}={\displaystyle {\int }_{-\infty }^{+\infty }\phi \left(x\right)\Phi \left(\sigma x+\mu \right)\text{d}x}={\displaystyle {\int }_{-\infty }^{+\infty }\phi \left(x\right)\text{d}x}{\displaystyle {\int }_{-\infty }^{\sigma x+\mu }\phi \left(y\right)\text{d}y}$

考虑如下问题：若$Z_1~N\left(0,1\right),Z_2~N\left(0,1\right)$ ，且$Z_1,Z_2$ 相互独立

将$\left(Z_1,Z_2\right)$ 视作二维正态分布，则$P\left(Z_2-\sigma Z_1\le \mu \right)={\displaystyle {\int }_{-\infty }^{+\infty }\phi \left(x\right)\text{d}x}{\displaystyle {\int }_{-\infty }^{\sigma x+\mu }\phi \left(y\right)\text{d}y}$

又易见 $Z_2-\sigma Z_1~N\left(0,1+{\sigma }^2\right),\frac{Z_2-\sigma Z_1}{\sqrt{1+{\sigma }^2}}~N\left(0,1\right)$

$P\left(Z_2-\sigma Z_1\le \mu \right)=P\left(\frac{Z_2-\sigma Z_1}{\sqrt{1+{\sigma }^2}}\le \frac{\mu }{\sqrt{1+{\sigma }^2}}\right)=\Phi \left(\frac{\mu }{\sqrt{1+{\sigma }^2}}\right)$

则 $E\left[\Phi \left(X\right)\right]=\Phi \left(\frac{\mu }{\sqrt{{\sigma }^2+1}}\right)$

定理4：设随机变量$X~N\left(0,1\right)$ ，$\sigma >0$ 则$E\left[\Phi \left(\frac{X-\mu }{\sigma }\right)\right]=1-\Phi \left(\frac{\mu }{\sqrt{{\sigma }^2+1}}\right)$

注2：若将定理3改为：若随机变量$X~N\left(0,1\right)$ ，$\sigma >0$ ，则

$E\left[\Phi \left(\frac{X-\mu }{\sigma }\right)\right]={\displaystyle {\int }_{-\infty }^{+\infty }\phi \left(x\right)\Phi \left(\frac{x-\mu }{\sigma }\right)\text{d}x}={\displaystyle {\int }_{-\infty }^{+\infty }\phi \left(x\right)\text{d}x}{\displaystyle {\int }_{-\infty }^{\frac{x-\mu }{\sigma }}\phi \left(y\right)\text{d}y}={\displaystyle {\int }_{-\infty }^{+\infty }\phi \left(y\right)\text{d}y}{\displaystyle {\int }_{\sigma y+\mu }^{+\infty }\phi \left(x\right)\text{d}x}$

考虑如下问题：若$Z_1~N\left(0,1\right),Z_2~N\left(0,1\right)$ ，且$Z_1,Z_2$ 相互独立

将$\left(Z_1,Z_2\right)$ 视作二维正态分布，则$P\left(Z_1-\sigma Z_2\ge \mu \right)={\displaystyle {\int }_{-\infty }^{+\infty }\phi \left(y\right)\text{d}y}{\displaystyle {\int }_{\sigma y+\mu }^{+\infty }\phi \left(x\right)\text{d}x}$

又易见 $Z_1-\sigma Z_2~N\left(0,1+{\sigma }^2\right),\frac{Z_1-\sigma Z_2}{\sqrt{1+{\sigma }^2}}~N\left(0,1\right)$

$P\left(Z_2-\sigma Z_1\ge \mu \right)=P\left(\frac{Z_1-\sigma Z_2}{\sqrt{1+{\sigma }^2}}\ge \frac{\mu }{\sqrt{1+{\sigma }^2}}\right)=1-\Phi \left(\frac{\mu }{\sqrt{1+{\sigma }^2}}\right)$

则 $E\left[\Phi \left(\frac{X-\mu }{\sigma }\right)\right]=1-\Phi \left(\frac{\mu }{\sqrt{{\sigma }^2+1}}\right)$

定理5：设随机变量$X~{\chi }^2\left(1\right)$ ，则$E\left[\Phi \left(\sqrt{X}\right)\right]=\frac{3}{4}$

证明：易见$E\left[\Phi \left(\sqrt{X}\right)\right]={\displaystyle {\int }_0^{+\infty }\frac{1}{\sqrt{2\pi x}}{\text{e}}^{-x/2}\Phi \left(\sqrt{x}\right)\text{d}x}=2{\displaystyle {\int }_0^{+\infty }\phi \left(x\right)\Phi \left(x\right)\text{d}x}=2{\displaystyle {\int }_0^{+\infty }\phi \left(x\right)\text{d}x}{\displaystyle {\int }_{-\infty }^x\phi \left(y\right)\text{d}y}$

考虑如下问题：若$Z_1~N\left(0,1\right),Z_2~N\left(0,1\right)$ ，且$Z_1,Z_2$ 相互独立，则

${\displaystyle {\int }_0^{+\infty }\phi \left(x\right)\text{d}x}{\displaystyle {\int }_{-\infty }^x\phi \left(y\right)\text{d}y}=\frac{1}{8}+\frac{1}{4}=\frac{3}{8}$

则 $E\left[\Phi \left(\sqrt{X}\right)\right]=\frac{3}{4}$

另证：由引理1知：$g_4\left(t\right)={\displaystyle {\int }_0^{+\infty }\phi \left(x\right)\Phi \left(tx\right)\text{d}x}=\frac{1}{4}+\frac{\arctan t}{2\pi }$

$E\left[\Phi \left(\sqrt{X}\right)\right]=2g_4\left(1\right)=2\left(\frac{1}{4}+\frac{1}{8}\right)=\frac{3}{4}$

注：定理3~5中引入随机变量$Z_1,Z_2$ 也是求解这类题目的常用方法。

引理3：设$X~N\left(0,1\right)$ ，$k$ 为正整数，(1) 若$k$ 为奇数，则$E\left(X^k\right)=0$ ；(2) 若$k=2l$ 为偶数，$l$ 为正整数，则$E\left(X^k\right)=\frac{2^{k/2}}{\sqrt{\pi }}\Gamma \left(\frac{k+1}{2}\right)=\frac{2^l}{\sqrt{\pi }}\Gamma \left(l+\frac{1}{2}\right)=\left(2l-1\right)\left(2l-3\right)\cdots 3\cdot 1$

证明：易见，若$k$ 为奇数，$E\left(X^k\right)=0$

若$n$ 为偶数，即$n=2l$ ，则$E\left(X^k\right)={\displaystyle {\int }_{-\infty }^{+\infty }{x}^k\frac{1}{\sqrt{2\pi }}{\text{e}}^{-x^2/2}\text{d}x}=\frac{2}{\sqrt{2\pi }}{\displaystyle {\int }_0^{+\infty }{x}^k{\text{e}}^{-x^2/2}\text{d}x}$

$=\frac{2^{k/2}}{\sqrt{\pi }}{\displaystyle {\int }_0^{+\infty }{x}^{\left(k-1\right)/2}{\text{e}}^{-t}\text{d}t}=\frac{2^{k/2}}{\sqrt{\pi }}\Gamma \left(\frac{k+1}{2}\right)=\frac{2^l}{\sqrt{\pi }}\Gamma \left(l+\frac{1}{2}\right)$

$=\frac{2^l}{\sqrt{\pi }}\left(l+\frac{1}{2}-1\right)\left(l+\frac{1}{2}-2\right)\cdots \frac{3}{2}\frac{1}{2}\Gamma \left(\frac{1}{2}\right)=\left(2l-1\right)\left(2l-3\right)\cdots 3\cdot 1$

若记函数$M\left(k\right)=\{\begin{array}{ll}1,\hfill & k为偶数\hfill \\ 0,\hfill & k为奇数\hfill \end{array}$ ，于是$E\left(X^k\right)=M\left(k\right)\frac{2^{k/2}}{\sqrt{\pi }}\text{Γ}\left(\frac{k+1}{2}\right)$

特别地，若$X~N\left(0,1\right)$ ，则$E\left(X\right)=E\left(X^3\right)=E\left(X^5\right)=E\left(X^7\right)=0$ ，$E\left(X^2\right)=1$ ，$D\left(X\right)=1$

$E\left(X^4\right)=3$ ，$D\left(X^2\right)=2$ ，$E\left(X^6\right)=15$ ，$E\left(X^8\right)=105$

定理6：设随机变量$X~N\left(0,1\right)$ ，$n$ 是正整数，则

$E\left[X^n\Phi \left(aX+b\right)\right]=M\left(n\right)\frac{2^{n/2}}{\sqrt{\pi }}\Gamma \left(\frac{n+1}{2}\right)\Phi \left(b\right){\displaystyle {\int }_0^1\exp \left\{-\frac{b^2}{2\left(a^2{x}^2+1\right)}\right\}\frac{1}{{\left(a^2{x}^2+1\right)}^{n/2+1}}{\left(\frac{abx}{\sqrt{a^2{x}^2+1}}\right)}^{n+1-i}\text{d}x}$

证明：记函数$I\left(t\right)={\displaystyle {\int }_{-\infty }^{+\infty }{x}^n\phi \left(x\right)\Phi \left(atx+b\right)\text{d}x}$

由引理3易见：$I\left(0\right)={\displaystyle {\int }_{-\infty }^{+\infty }{x}^n\phi \left(x\right)\Phi \left(b\right)\text{d}x}=M\left(n\right)\frac{2^{n/2}}{\sqrt{\pi }}\Gamma \left(\frac{n+1}{2}\right)\Phi \left(b\right)$

则 $\begin{array}{c}I\left(t\right)=I\left(0\right)+\frac{a}{\sqrt{2}\pi }{\displaystyle \sum }_{i=0}^{n+1}{\left(-1\right)}^{n+1-i}M\left(i\right)2^{i/2}{C}_{n+1}^i\Gamma \left(\frac{i+1}{2}\right)\\ \cdot {\displaystyle {\int }_0^t\exp \left\{-\frac{b^2}{2\left(a^2{x}^2+1\right)}\right\}\frac{1}{{\left(a^2{x}^2+1\right)}^{n/2+1}}{\left(\frac{abx}{\sqrt{a^2{x}^2+1}}\right)}^{n+1-i}\text{d}x}\end{array}$

进而

$E\left[X^n\Phi \left(aX+b\right)\right]=M\left(n\right)\frac{2^{n/2}}{\sqrt{\pi }}\Gamma \left(\frac{n+1}{2}\right)\Phi \left(b\right){\displaystyle {\int }_0^1\exp \left\{-\frac{b^2}{2\left(a^2{x}^2+1\right)}\right\}\frac{1}{{\left(a^2{x}^2+1\right)}^{n/2+1}}{\left(\frac{abx}{\sqrt{a^2{x}^2+1}}\right)}^{n+1-i}\text{d}x}$

定理7：设随机变量$X~N\left(0,1\right)$ ，$n$ 是正整数，则$E\left[{\Phi }^n\left(X\right)\right]=\frac{1}{n+1}$

证明：$E\left[{\Phi }^k\left(X\right)\right]={\displaystyle {\int }_{-\infty }^{+\infty }{\Phi }^k\left(x\right)\phi \left(x\right)\text{d}x}={{\Phi }^{k+1}\left(x\right)|}_{-\infty }^{+\infty }-k{\displaystyle {\int }_{-\infty }^{+\infty }{\Phi }^k\left(x\right)\phi \left(x\right)\text{d}x}=1-kE\left[{\Phi }^k\left(X\right)\right]$

则 $E\left[{\Phi }^k\left(X\right)\right]=\frac{1}{k+1}$

致 谢

作者非常感谢各位审稿专家提出的宝贵建议！

基金项目

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