背就完事了

$$\Gamma(\alpha) = \begin{cases}\int_0^{+\infty}x^{\alpha-1}e^{-x}dx \\ \\ 2\int_0^{+\infty}x^{2\alpha-1}e^{-x^2}dx \end{cases}\\ \Gamma(\frac{1}{2}) = \sqrt{\pi},\quad \Gamma(1) = 1,\quad \Gamma(n+1) = n\Gamma(n)$$

推导过程

$$\int_{-\infty}^{+\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}dx = 2\int_{0}^{+\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}dx = 1\quad \Rightarrow\quad \Gamma(\frac{1}{2}) = 2\int_{0}^{+\infty} e^{-x^2}dx = \sqrt{\pi} \\ \int_0^{+\infty} e ^ {-x} dx = 1 = \Gamma(1)$$

补充一个二维正态分布的函数

$$f(x,y) = \frac{1}{2\pi \sigma_1 \sigma_2 \sqrt{1-\rho^2}} e ^ {-\frac{1}{2(1-\rho ^ 2)} [(\frac{x-\mu_1 }{\sigma_1})^2 - 2\rho(\frac{x-\mu_1}{\sigma_1})(\frac{y-\mu_2}{\sigma_2}) + (\frac{y-\mu_2}{\sigma_2})^2]}$$

• 联合是正太 推出 边缘是正太
• 边缘是正太 （独立的话）推出 联合是正太
• X,Y是正太，线性组合仍然是正太