佚名

Fourier变换和中心极限定理

随机变量的概率密度函数

  • 随机变量X的概率密度函数记为p_X(x)，x \in \mathbb{R}
  • 如果\lambda \in \mathbb{R}_+，那么

        \begin{equation*}\begin{split} \mathbb{P}(\lambda X \leq a) &= \int_{-\infty}^a p_{\lambda X}(x)dx, \\ \mathbb{P}(X \leq \lambda^{-1}a) &= \int_{-\infty}^{\lambda^{-1}a} p_{X}(x)dx \\ &= \int_{-\infty}^a p_{X}(\lambda^{-1}y)d(\lambda^{-1}y) \\ &= \int_{-\infty}^a \lambda^{-1}p_{X}(\lambda^{-1}y)dy. \end{split}\end{equation*}

    因此，p_{\lambda X}(x) = \lambda^{-1}p_{X}(\lambda^{-1}x)
  • 如果X_1、X_2为独立的随机变量，那么

        \[ \mathbb{P}(X_1 + X_2 \leq a) = \int_{-\infty}^a\int_\mathbb{R} p_{X_1}(x - y)p_{X_2}(y)dydx. \]

    因此，p_{X_1 + X_2} = p_{X_1} * p_{X_2}。利用Fourier变换，

        \[ \widehat{p}_{X_1 + X_2} = \widehat{p}_{X_1}\widehat{p}_{X_2}. \]

中心极限定理

  • 设X_1, \ldots, X_n, \ldots为独立的随机变量，其分布和一个固定的随机变量X相同，均值为0，方差为1
  • 令S_n = X_1 + \cdots + X_n。那么，\widehat{p}_{S_n} = (\widehat{p}_X)^n，从而

        \begin{equation*}\begin{split} \widehat{p}_{\frac{S_n}{\sqrt{n}}} &= (\widehat{p}_{\frac{X}{\sqrt{n}}})^n \\ &= \bigg\{\bigg[\sqrt{n}p_X(\sqrt{n}x)\bigg]^{\wedge}\bigg\}^n \\ &= \bigg[\widehat{p}_X\bigg(\frac{\xi}{\sqrt{n}}\bigg)\bigg]^n. \end{split}\end{equation*}

  • 注意，X的均值为0，方差为1，

        \[ \int_\mathbb{R} p_X(x)dx = 1,\; \int_\mathbb{R} xp_X(x)dx = 0,\; \int_\mathbb{R} x^2p_X(x)dx = 1. \]

    因此，

        \begin{equation*}\begin{split} \widehat{p}_X(0) &= \int_\mathbb{R} p_X(x)dx = 1, \\ \frac{d\widehat{p}_X}{d\xi}(0) &= -2\pi i\int_\mathbb{R} xp_X(x)dx = 0, \\ \frac{d^2\widehat{p}_X}{d\xi^2}(0) &= -4\pi^2\int_\mathbb{R} x^2p_X(x)dx = -4\pi^2. \end{split}\end{equation*}

  • 利用Taylor展开式，

        \[ \bigg[\widehat{p}_X\bigg(\frac{\xi}{\sqrt{n}}\bigg)\bigg]^n = \bigg[1 - 2\pi^2\frac{\xi^2}{n} + o\bigg(\frac{\xi^2}{n}\bigg)\bigg]^n \to e^{-2\pi^2\xi^2} \text{ as } n \to \infty, \]

    故

        \[ p_{\frac{S_n}{\sqrt{n}}}(x) \to (e^{-2\pi^2\xi^2})^{\vee}(x) \text{ as } n \to \infty. \]

    又由于

        \begin{equation*}\begin{split} (e^{-\pi(\sqrt{2\pi}\xi)^2})^{\vee}(x) &= \frac{1}{\sqrt{2\pi}}(e^{-\pi\xi^2})^{\vee}\bigg[\frac{x}{\sqrt{2\pi}}\bigg] \\ &= \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}, \end{split}\end{equation*}

    故

        \[ p_{\frac{S_n}{\sqrt{n}}}(x) \to \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} \text{ as } n \to \infty. \]

  • 一般地，设X_1, \ldots, X_n, \ldots为独立、同分布的随机变量，均值为\mu，方差为\sigma^2。我们可以使用平移、伸缩，将其化归为均值为0，方差为1的情形，

        \[ X_n \mapsto \frac{X_n - \mu}{\sigma},\; S_n \mapsto \frac{S_n - n\mu}{\sigma}. \]

    从而，我们可以得到中心极限定理，

        \[ p_{\frac{S_n - n\mu}{\sqrt{n}\sigma}}(x) \to \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} \text{ as } n \to \infty. \]