Beta函数证明

数学 机器学习

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$Beta$函数证明：

$$Beta(m,n) = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)} = \int_0^1 t^{m-1}(1-t)^{n-1}\mathrm{d}t, \qquad \Gamma(x) = \int_0^{+\infty}e^{-t}t^{x-1}\mathrm{d}t$$

$$\begin{align*} \Gamma(x) &= \int_0^{+\infty}e^{-t}t^{x-1}\mathrm{d}t \\ &= -\int_0^{+\infty}t^{x-1}\mathrm{d}e^{-t} \\ &= -\left( \left. e^{-t}t^{x-1} \right|_0^{+\infty} - \int_0^{+\infty}(x-1)t^{x-2}e^{-t}\mathrm{d}t \right) \\ &= (x-1)\cdot\int_0^{+\infty}e^{-t}t^{x-2}\mathrm{d}t \\ &= (x-1)\cdot\Gamma(x-1) \\ &= (x-1)! \end{align*}$$

$$\begin{align*} \Gamma(x)\Gamma(y) & = \int_0^{+\infty}e^{-t}t^{x-1}\mathrm{d}t \cdot \int_0^{+\infty}e^{-s}s^{y-1}\mathrm{d}s \\ & = \int_0^{+\infty}\int_0^{+\infty}e^{-(t+s)}t^{x-1}s^{y-1}\mathrm{d}t\mathrm{d}s \\ & = \int_0^{+\infty}\int_0^{+\infty}e^{-(u^2+v^2)}u^{2x-2}s^{2y-2}2u2v\mathrm{d}u\mathrm{d}v \quad //t=u^2,s=v^2 \\ & = 4\cdot\int_0^{+\infty}\int_0^{+\infty}e^{-(u^2+v^2)}u^{2x-1}v^{2y-1}\mathrm{d}u\mathrm{d}v \\ & = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-(u^2+v^2)}|u|^{2x-1}|v|^{2y-1}\mathrm{d}u\mathrm{d}v \\ & = \int_0^{+\infty}\int_{0}^{2\pi}e^{-r^2}|r\cos\theta|^{2x-1}|r\sin\theta|^{2y-1}r\mathrm{d}r\mathrm{d}\theta \\ & = \int_0^{+\infty}e^{-r^2}r^{2(x+y-1)}r\mathrm{d}r \cdot \int_0^{2\pi}|\cos\theta|^{2x-1}|\sin\theta|^{2y-1}\mathrm{d}\theta \\ & = \frac{1}{2}\int_0^{+\infty}e^{-r^2}r^{2(x+y-1)}\mathrm{d}r^2 \cdot 4 \int_0^{\frac{\pi}{2}}\cos^{2x-1}\theta\sin^{2y-1}\theta\mathrm{d}\theta \\ & = \Gamma(x+y) \cdot 2\int_1^0 t^{x-\frac{1}{2}}(1-t)^{y-\frac{1}{2}}\mathrm{d}\arccos\sqrt{t} \quad //t=\cos^2\theta \\ & = \Gamma(x+y)\cdot 2\int_1^0t^{x-\frac{1}{2}}(1-t)^{y-\frac{1}{2}}(-(1-t)^{-\frac{1}{2}}\frac{1}{2}t^{-\frac{1}{2}})\mathrm{d}t \\ & = \Gamma(x+y) \cdot \int_0^1 t^{x-1}(1-t)^{y-1}\mathrm{d}t \\ & = \Gamma(x+y) \cdot Beta(x,y) \\ Beta(x,y) & = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} \end{align*}$$

Reference

• 暴力推导 Beta 函数与 Gamma 函数关系式