Integrals in Probability Theory

概率论常用积分 | Integrals in Probability Theory

指数分布 | Exponential Distribution

指数分布的概率密度函数 | Probability Density Function of Exponential Distribution

$$f_X(x)=\lambda e^{-\lambda x}\text{for}x\ge 0$$

指数分布的期望值 | Expectation of Exponential Distribution

$$\mathbb{E}(X)={\int }_0^{\infty }x\lambda e^{-\lambda x}\mathrm{d}x$$

积分过程：

$$\begin{array}{rl}\mathbb{E}(X)& ={\int }_0^{\infty }x\lambda e^{-\lambda x}\mathrm{d}x\\ \text{令}u& =\lambda x,\mathrm{d}u=\lambda \mathrm{d}x\\ & ={\int }_0^{\infty }\frac{u}{\lambda }\lambda e^{-u}\frac{\mathrm{d}u}{\lambda }\\ & ={\int }_0^{\infty }u{e}^{-u}\mathrm{d}u\\ \text{使用分部积分法}\left(\int u\mathrm{d}v=uv-\int v\mathrm{d}u\right)& \\ u& =u,\mathrm{d}v=e^{-u}\mathrm{d}u\\ v& =-e^{-u},\mathrm{d}u=\mathrm{d}u\\ \int u{e}^{-u}\mathrm{d}u& =-u{e}^{-u}{|}_0^{\infty }+\int e^{-u}\mathrm{d}u\\ & =-u{e}^{-u}{|}_0^{\infty }-e^{-u}{|}_0^{\infty }\\ & =0+1=1\\ \mathbb{E}(X)& =1/\lambda \end{array}$$

指数分布的方差 | Variance of Exponential Distribution

$$\mathrm{Var}(X)={\int }_0^{\infty }{x}^2\lambda e^{-\lambda x}\mathrm{d}x-{\left(\frac{1}{\lambda }\right)}^2=\frac{1}{{\lambda }^2}$$

积分过程：

$$\begin{array}{rl}\mathrm{Var}(X)& ={\int }_0^{\infty }{x}^2\lambda e^{-\lambda x}\mathrm{d}x-{\left(\frac{1}{\lambda }\right)}^2\\ \text{令}u& =\lambda x,\mathrm{d}u=\lambda \mathrm{d}x\\ & ={\int }_0^{\infty }{\left(\frac{u}{\lambda }\right)}^2\lambda e^{-u}\frac{\mathrm{d}u}{\lambda }\\ & ={\int }_0^{\infty }\frac{u^2}{{\lambda }^2}\lambda e^{-u}\frac{\mathrm{d}u}{\lambda }\\ & ={\int }_0^{\infty }{u}^2{e}^{-u}\mathrm{d}u\\ \text{使用Gamma函数}\Gamma (n)={\int }_0^{\infty }{x}^{n-1}{e}^{-x}\mathrm{d}x& \\ \text{对于}\Gamma (3)& ={\int }_0^{\infty }{u}^2{e}^{-u}\mathrm{d}u=2!=2\\ \mathrm{Var}(X)& =\frac{2}{{\lambda }^2}-{\left(\frac{1}{\lambda }\right)}^2=\frac{2}{{\lambda }^2}-\frac{1}{{\lambda }^2}=\frac{1}{{\lambda }^2}\end{array}$$

正态分布 | Normal Distribution

正态分布的概率密度函数 | Probability Density Function of Normal Distribution

$$f_X(x)=\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}}$$

标准正态分布的积分 | Integral of Standard Normal Distribution

$${\int }_{-\infty }^{\infty }{e}^{-\frac{x^2}{2}}\mathrm{d}x=\sqrt{2\pi }$$

高斯积分 (Gaussian Integral) 积分过程：

$$\begin{array}{rl}I& ={\int }_{-\infty }^{\infty }{e}^{-\frac{x^2}{2}}\mathrm{d}x\\ I^2& =\left({\int }_{-\infty }^{\infty }{e}^{-\frac{x^2}{2}}\mathrm{d}x\right)\left({\int }_{-\infty }^{\infty }{e}^{-\frac{y^2}{2}}\mathrm{d}y\right)\\ & ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{e}^{-\frac{x^2+y^2}{2}}\mathrm{d}x\mathrm{d}y\\ \text{转换为极坐标}x=r\cos \theta ,y=r\sin \theta ,\mathrm{d}x\mathrm{d}y=r\mathrm{d}r\mathrm{d}\theta & \\ I^2& ={\int }_0^{2\pi }{\int }_0^{\infty }{e}^{-\frac{r^2}{2}}r\mathrm{d}r\mathrm{d}\theta \\ & ={\int }_0^{2\pi }\mathrm{d}\theta {\int }_0^{\infty }r{e}^{-\frac{r^2}{2}}\mathrm{d}r\\ \text{令}u& =\frac{r^2}{2},\mathrm{d}u=r\mathrm{d}r\\ I^2& ={\int }_0^{2\pi }\mathrm{d}\theta {\int }_0^{\infty }{e}^{-u}\mathrm{d}u\\ & =2\pi {\left[-e^{-u}\right]}_0^{\infty }\\ & =2\pi \left[0-(-1)\right]\\ & =2\pi \\ I& =\sqrt{2\pi }\end{array}$$

正态分布的期望值 | Expectation of Normal Distribution

$$\mathbb{E}(X)={\int }_{-\infty }^{\infty }x\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}}\mathrm{d}x=\mu$$

积分过程：

$$\begin{array}{rl}\mathbb{E}(X)& ={\int }_{-\infty }^{\infty }x\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}}\mathrm{d}x\\ \text{令}u& =\frac{x-\mu }{\sigma },x=\sigma u+\mu ,\mathrm{d}x=\sigma \mathrm{d}u\\ & ={\int }_{-\infty }^{\infty }(\sigma u+\mu )\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{u^2}{2}}\sigma \mathrm{d}u\\ & =\mu {\int }_{-\infty }^{\infty }\frac{1}{\sqrt{2\pi }}{e}^{-\frac{u^2}{2}}\mathrm{d}u+\sigma {\int }_{-\infty }^{\infty }u\frac{1}{\sqrt{2\pi }}{e}^{-\frac{u^2}{2}}\mathrm{d}u\\ & =\mu \left[{\int }_{-\infty }^{\infty }\frac{1}{\sqrt{2\pi }}{e}^{-\frac{u^2}{2}}\mathrm{d}u\right]+0\text{(奇函数在对称区间内积分为零)}\\ & =\mu \cdot 1\\ & =\mu \end{array}$$

正态分布的方差 | Variance of Normal Distribution

$$\mathrm{Var}(X)={\int }_{-\infty }^{\infty }(x-\mu {)}^2\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}}\mathrm{d}x={\sigma }^2$$

积分过程：

$$\begin{array}{rl}\mathrm{Var}(X)& ={\int }_{-\infty }^{\infty }(x-\mu {)}^2\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}}\mathrm{d}x\\ \text{令}u& =\frac{x-\mu }{\sigma },x=\sigma u+\mu ,\mathrm{d}x=\sigma \mathrm{d}u\\ & ={\int }_{-\infty }^{\infty }(\sigma u{)}^2\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{u^2}{2}}\sigma \mathrm{d}u\\ & ={\sigma }^2{\int }_{-\infty }^{\infty }{u}^2\frac{1}{\sqrt{2\pi }}{e}^{-\frac{u^2}{2}}\mathrm{d}u\\ \text{使用分部积分法}\left(\int u^2{e}^{-\frac{u^2}{2}}\mathrm{d}u=\sqrt{2\pi }\right)& \\ \mathrm{Var}(X)& ={\sigma }^2\cdot 1={\sigma }^2\end{array}$$

均匀分布 | Uniform Distribution

均匀分布的概率密度函数 | Probability Density Function of Uniform Distribution

$$f_X(x)=\{\begin{array}{ll}\frac{1}{b-a}& a\le x\le b\\ 0& \text{otherwise}\end{array}$$

均匀分布的期望值 | Expectation of Uniform Distribution

$$\mathbb{E}(X)={\int }_a^{b}x\frac{1}{b-a}\mathrm{d}x=\frac{a+b}{2}$$

积分过程：

$$\begin{array}{rl}\mathbb{E}(X)& ={\int }_a^{b}x\frac{1}{b-a}\mathrm{d}x\\ & =\frac{1}{b-a}{\int }_a^{b}x\mathrm{d}x\\ & =\frac{1}{b-a}{\left[\frac{x^2}{2}\right]}_a^b\\ & =\frac{1}{b-a}\left(\frac{b^2}{2}-\frac{a^2}{2}\right)\\ & =\frac{b^2-a^2}{2(b-a)}\\ & =\frac{(b-a)(b+a)}{2(b-a)}\\ & =\frac{a+b}{2}\end{array}$$

均匀分布的方差 | Variance of Uniform Distribution

$$\mathrm{Var}(X)={\int }_a^b{x}^2\frac{1}{b-a}\mathrm{d}x-{\left(\frac{a+b}{2}\right)}^2=\frac{(b-a{)}^2}{12}$$

积分过程：

$$\begin{array}{rl}\mathrm{Var}(X)& ={\int }_a^b{x}^2\frac{1}{b-a}\mathrm{d}x-{\left(\frac{a+b}{2}\right)}^2\\ & =\frac{1}{b-a}{\int }_a^b{x}^2\mathrm{d}x-{\left(\frac{a+b}{2}\right)}^2\\ & =\frac{1}{b-a}{\left[\frac{x^3}{3}\right]}_a^b-\frac{(a+b{)}^2}{4}\\ & =\frac{1}{b-a}\left(\frac{b^3}{3}-\frac{a^3}{3}\right)-\frac{(a^2+2ab+b^2)}{4}\\ & =\frac{b^3-a^3}{3(b-a)}-\frac{a^2+2ab+b^2}{4}\\ & =\frac{(b-a)(b^2+ba+a^2)}{3(b-a)}-\frac{a^2+2ab+b^2}{4}\\ & =\frac{b^2+ba+a^2}{3}-\frac{a^2+2ab+b^2}{4}\\ & =\frac{4(b^2+ba+a^2)-3(a^2+2ab+b^2)}{12}\\ & =\frac{4b^2+4ba+4a^2-3a^2-6ab-3b^2}{12}\\ & =\frac{b^2-2ab+a^2}{12}\\ & =\frac{(b-a{)}^2}{12}\end{array}$$

伽马分布 | Gamma Distribution

伽马分布的概率密度函数 | Probability Density Function of Gamma Distribution

$$f_X(x)=\frac{{\beta }^{\alpha }}{\Gamma (\alpha )}{x}^{\alpha -1}{e}^{-\beta x}\text{for}x\ge 0$$

伽马分布的期望值 | Expectation of Gamma Distribution

$$\mathbb{E}(X)={\int }_0^{\infty }x\frac{{\beta }^{\alpha }}{\Gamma (\alpha )}{x}^{\alpha -1}{e}^{-\beta x}\mathrm{d}x=\frac{\alpha }{\beta }$$

积分过程：

$$\begin{array}{rl}\mathbb{E}(X)& ={\int }_0^{\infty }x\frac{{\beta }^{\alpha }}{\Gamma (\alpha )}{x}^{\alpha -1}{e}^{-\beta x}\mathrm{d}x\\ & =\frac{{\beta }^{\alpha }}{\Gamma (\alpha )}{\int }_0^{\infty }{x}^{\alpha }{e}^{-\beta x}\mathrm{d}x\\ \text{令}u& =\beta x,\mathrm{d}u=\beta \mathrm{d}x\\ & =\frac{{\beta }^{\alpha }}{\Gamma (\alpha )}{\int }_0^{\infty }{\left(\frac{u}{\beta }\right)}^{\alpha }{e}^{-u}\frac{\mathrm{d}u}{\beta }\\ & =\frac{{\beta }^{\alpha }}{\Gamma (\alpha )}{\int }_0^{\infty }\frac{u^{\alpha }}{{\beta }^{\alpha }\beta }{e}^{-u}\mathrm{d}u\\ & =\frac{{\beta }^{\alpha }}{\Gamma (\alpha )}\cdot \frac{1}{{\beta }^{\alpha +1}}{\int }_0^{\infty }{u}^{\alpha }{e}^{-u}\mathrm{d}u\\ & =\frac{1}{\beta \Gamma (\alpha )}\Gamma (\alpha +1)\\ & =\frac{\Gamma (\alpha +1)}{\beta \Gamma (\alpha )}\\ \text{由于}\Gamma (\alpha +1)& =\alpha \Gamma (\alpha )\\ \mathbb{E}(X)& =\frac{\alpha \Gamma (\alpha )}{\beta \Gamma (\alpha )}\\ & =\frac{\alpha }{\beta }\end{array}$$

伽马分布的方差 | Variance of Gamma Distribution

$$\mathrm{Var}(X)={\int }_0^{\infty }{x}^2\frac{{\beta }^{\alpha }}{\Gamma (\alpha )}{x}^{\alpha -1}{e}^{-\beta x}\mathrm{d}x-{\left(\frac{\alpha }{\beta }\right)}^2=\frac{\alpha }{{\beta }^2}$$

积分过程：

$$\begin{array}{rl}\mathrm{Var}(X)& ={\int }_0^{\infty }{x}^2\frac{{\beta }^{\alpha }}{\Gamma (\alpha )}{x}^{\alpha -1}{e}^{-\beta x}\mathrm{d}x-{\left(\frac{\alpha }{\beta }\right)}^2\\ & =\frac{{\beta }^{\alpha }}{\Gamma (\alpha )}{\int }_0^{\infty }{x}^{\alpha +1}{e}^{-\beta x}\mathrm{d}x\\ \text{令}u& =\beta x,\mathrm{d}u=\beta \mathrm{d}x\\ & =\frac{{\beta }^{\alpha }}{\Gamma (\alpha )}{\int }_0^{\infty }{\left(\frac{u}{\beta }\right)}^{\alpha +1}{e}^{-u}\frac{\mathrm{d}u}{\beta }\\ & =\frac{{\beta }^{\alpha }}{\Gamma (\alpha )}{\int }_0^{\infty }\frac{u^{\alpha +1}}{{\beta }^{\alpha +1}\beta }{e}^{-u}\mathrm{d}u\\ & =\frac{{\beta }^{\alpha }}{\Gamma (\alpha )}\cdot \frac{1}{{\beta }^{\alpha +2}}{\int }_0^{\infty }{u}^{\alpha +1}{e}^{-u}\mathrm{d}u\\ & =\frac{1}{{\beta }^2\Gamma (\alpha )}\Gamma (\alpha +2)\\ & =\frac{\Gamma (\alpha +2)}{{\beta }^2\Gamma (\alpha )}\\ \text{由于}\Gamma (\alpha +2)& =(\alpha +1)\alpha \Gamma (\alpha )\\ \mathrm{Var}(X)& =\frac{(\alpha +1)\alpha \Gamma (\alpha )}{{\beta }^2\Gamma (\alpha )}-{\left(\frac{\alpha }{\beta }\right)}^2\\ & =\frac{(\alpha +1)\alpha }{{\beta }^2}-\frac{{\alpha }^2}{{\beta }^2}\\ & =\frac{{\alpha }^2+\alpha -{\alpha }^2}{{\beta }^2}\\ & =\frac{\alpha }{{\beta }^2}\end{array}$$

贝塔分布 | Beta Distribution

贝塔分布的概率密度函数 | Probability Density Function of Beta Distribution

$$f_X(x)=\frac{x^{\alpha -1}(1-x{)}^{\beta -1}}{\mathrm{B}(\alpha ,\beta )}\text{for}0\le x\le 1$$

其中，$\mathrm{B}(\alpha ,\beta )$ 是 Beta 函数

where $\mathrm{B}(\alpha ,\beta )$ is the Beta function

贝塔分布的期望值 | Expectation of Beta Distribution

$$\mathbb{E}(X)={\int }_0^{1}x\frac{x^{\alpha -1}(1-x{)}^{\beta -1}}{\mathrm{B}(\alpha ,\beta )}\mathrm{d}x=\frac{\alpha }{\alpha +\beta }$$

积分过程：

$$\begin{array}{rl}\mathbb{E}(X)& ={\int }_0^{1}x\frac{x^{\alpha -1}(1-x{)}^{\beta -1}}{\mathrm{B}(\alpha ,\beta )}\mathrm{d}x\\ & =\frac{1}{\mathrm{B}(\alpha ,\beta )}{\int }_0^1{x}^{\alpha }(1-x{)}^{\beta -1}\mathrm{d}x\\ \text{由于}\mathrm{B}(\alpha ,\beta )& ={\int }_0^1{x}^{\alpha -1}(1-x{)}^{\beta -1}\mathrm{d}x\\ \mathbb{E}(X)& =\frac{\mathrm{B}(\alpha +1,\beta )}{\mathrm{B}(\alpha ,\beta )}\\ \text{使用Beta函数性质}\mathrm{B}(\alpha ,\beta )& =\frac{\Gamma (\alpha )\Gamma (\beta )}{\Gamma (\alpha +\beta )}\\ \mathrm{B}(\alpha +1,\beta )& =\frac{\Gamma (\alpha +1)\Gamma (\beta )}{\Gamma (\alpha +1+\beta )}\\ & =\frac{\alpha \Gamma (\alpha )\Gamma (\beta )}{(\alpha +\beta )\Gamma (\alpha +\beta )}\\ \mathbb{E}(X)& =\frac{\alpha \Gamma (\alpha )\Gamma (\beta )}{(\alpha +\beta )\Gamma (\alpha +\beta )}\cdot \frac{\Gamma (\alpha +\beta )}{\Gamma (\alpha )\Gamma (\beta )}\\ & =\frac{\alpha }{\alpha +\beta }\end{array}$$

贝塔分布的方差 | Variance of Beta Distribution

$$\mathrm{Var}(X)={\int }_0^1{x}^2\frac{x^{\alpha -1}(1-x{)}^{\beta -1}}{\mathrm{B}(\alpha ,\beta )}\mathrm{d}x-{\left(\frac{\alpha }{\alpha +\beta }\right)}^2=\frac{\alpha \beta }{(\alpha +\beta {)}^2(\alpha +\beta +1)}$$

积分过程：

$$\begin{array}{rl}\mathrm{Var}(X)& ={\int }_0^1{x}^2\frac{x^{\alpha -1}(1-x{)}^{\beta -1}}{\mathrm{B}(\alpha ,\beta )}\mathrm{d}x-{\left(\frac{\alpha }{\alpha +\beta }\right)}^2\\ & =\frac{1}{\mathrm{B}(\alpha ,\beta )}{\int }_0^1{x}^{\alpha +1}(1-x{)}^{\beta -1}\mathrm{d}x\\ & =\frac{\mathrm{B}(\alpha +2,\beta )}{\mathrm{B}(\alpha ,\beta )}-{\left(\frac{\alpha }{\alpha +\beta }\right)}^2\\ & =\frac{\frac{\Gamma (\alpha +2)\Gamma (\beta )}{\Gamma (\alpha +2+\beta )}}{\frac{\Gamma (\alpha )\Gamma (\beta )}{\Gamma (\alpha +\beta )}}-{\left(\frac{\alpha }{\alpha +\beta }\right)}^2\\ & =\frac{\Gamma (\alpha +2)\Gamma (\beta )\Gamma (\alpha +\beta )}{\Gamma (\alpha +2+\beta )\Gamma (\alpha )\Gamma (\beta )}-{\left(\frac{\alpha }{\alpha +\beta }\right)}^2\\ & =\frac{(\alpha +1)\alpha \Gamma (\alpha )\Gamma (\beta )\Gamma (\alpha +\beta )}{(\alpha +\beta +1)(\alpha +\beta )\Gamma (\alpha +\beta )\Gamma (\alpha )\Gamma (\beta )}-{\left(\frac{\alpha }{\alpha +\beta }\right)}^2\\ & =\frac{(\alpha +1)\alpha }{(\alpha +\beta +1)(\alpha +\beta )}-{\left(\frac{\alpha }{\alpha +\beta }\right)}^2\\ & =\frac{\alpha (\alpha +1)}{(\alpha +\beta {)}^2(\alpha +\beta +1)}-\frac{{\alpha }^2}{(\alpha +\beta {)}^2}\\ & =\frac{\alpha (\alpha +1)-{\alpha }^2}{(\alpha +\beta {)}^2(\alpha +\beta +1)}\\ & =\frac{\alpha (\alpha +1-\alpha )}{(\alpha +\beta {)}^2(\alpha +\beta +1)}\\ & =\frac{\alpha \beta }{(\alpha +\beta {)}^2(\alpha +\beta +1)}\end{array}$$