例 3.5.6

hard一级题目发布者: ai-batch

题干

设随机变量 $X \sim N(\mu,\sigma_1^2)$ ，在 $X=x$ 下 $Y$ 的条件分布为 $N(x,\sigma_2^2)$ 。试求 $Y$ 的（无条件）密度函数 $p_Y(y)$ 。

答案点击展开后可查看解析

由题意知

p_X(x) = \frac{1}{\sqrt{2\pi}\,\sigma_1}\exp\left\{-\frac{(x-\mu)^2}{2\sigma_1^2}\right\},

p(y \mid x) = \frac{1}{\sqrt{2\pi}\,\sigma_2}\exp\left\{-\frac{(y-x)^2}{2\sigma_2^2}\right\},

所以由 (3.5.11) 式得

p_Y(y) = \int_{-\infty}^{\infty} p_X(x)p(y \mid x)\mathrm{d}x

= \frac{1}{2\pi\sigma_1\sigma_2}\int_{-\infty}^{\infty}\exp\left\{-\frac{(x-\mu)^2}{2\sigma_1^2}-\frac{(y-x)^2}{2\sigma_2^2}\right\}\mathrm{d}x

\propto \int_{-\infty}^{\infty}\exp\left\{-\frac{1}{2}\left(\frac{1}{\sigma_1^2}+\frac{1}{\sigma_2^2}\right)x^2+\left(\frac{y}{\sigma_2^2}+\frac{\mu}{\sigma_1^2}\right)x\right\}\mathrm{d}x\exp\left\{-\frac{1}{2}\frac{y^2}{\sigma_2^2}\right\},

被积函数可以看作一种特殊的正态分布的核，这里取

a = \frac{1}{2}\left(\frac{1}{\sigma_1^2}+\frac{1}{\sigma_2^2}\right), \quad b = \frac{y}{\sigma_2^2}+\frac{\mu}{\sigma_1^2},

则上式化成

p_Y(y) \propto \int_{-\infty}^{\infty}\exp\left\{-a\left(x-\frac{b}{2a}\right)^2\right\}\mathrm{d}x\exp\left\{\frac{b^2}{4a}-\frac{1}{2}\frac{y^2}{\sigma_2^2}\right\}

\propto \exp\left\{\frac{\sigma_1^2\sigma_2^2}{2(\sigma_1^2+\sigma_2^2)}\left(\frac{y}{\sigma_2^2}+\frac{\mu}{\sigma_1^2}\right)^2-\frac{1}{2}\frac{y^2}{\sigma_2^2}\right\}

= \exp\left\{-\frac{1}{2(\sigma_1^2+\sigma_2^2)}y^2+\frac{\mu}{\sigma_1^2+\sigma_2^2}y\right\},

根据 (2.5.23) 式，该分布是正态分布，即 $N(\mu,\sigma_1^2+\sigma_2^2)$ 。