贝叶斯分析的题

战巡

我们的作业之一，拿来给大家玩玩

已知$u_1\sim N(\mu_1,\sigma^2), u_2\sim N(\mu_2,\sigma^2), s^2\sim \frac{\sigma^2}{v}\chi^2(v)$，且$u_1,u_2,s^2$独立
令$\zeta=\frac{\mu_1-\mu_2}{\sqrt{2}\sigma}$为需要考察的变量
令先验分布$\pi(\mu_1,\mu_2,\sigma)=\frac{1}{\sigma}$

（1）、证明关于$\zeta$的后验分布$\pi(\zeta|u_1,u_2,s^2)$表达式中只含有$z=\frac{u_1-u_2}{\sqrt{2}s}$

（2）、求$z$关于$\zeta$的分布$f(z|\zeta)$，并证明$f(z|\zeta)\ne \pi(\zeta|u_1,u_2,s^2)$

（3）、证明当先验分布为$\pi(\mu_1,\mu_2,\sigma)=\frac{1}{\sigma^2}$时$f(z|\zeta)= \pi(\zeta|u_1,u_2,s^2)$

战巡

回复 1# 战巡


又没人玩.....

给点提示吧，第一问：

\[\pi(\mu_1,\mu_2,\sigma|x)=\frac{f(x|\mu_1,\mu_2,\sigma^2\pi(\mu_1,\mu_2,\sigma^2)}{\int f(x|\mu_1,\mu_2,\sigma^2\pi(\mu_1,\mu_2,\sigma^2)d\mu_1d\mu_2d\sigma^2}\]
\[=\frac{N(\mu_1,\sigma^2)N(\mu_2,\sigma^2)\frac{v}{\sigma^2}\chi^2_v(\frac{vs^2}{\sigma^2})\frac{1}{\sigma}}{\int N(\mu_1,\sigma^2)N(\mu_2,\sigma^2)\frac{v}{\sigma^2}\chi^2_v(\frac{vs^2}{\sigma^2})\frac{1}{\sigma}d\mu_1d\mu_2d\sigma^2}\]
\[=\frac{N(\mu_1,\sigma^2)N(\mu_2,\sigma^2)\frac{v}{\sigma^2}\chi^2_v(\frac{vs^2}{\sigma^2})\frac{1}{\sigma}}{\int_0^\infty \frac{v}{\sigma^2}\chi^2_v(\frac{vs^2}{\sigma^2})\frac{1}{\sigma}d\sigma^2}=N(\mu_1,\sigma^2)N(\mu_2,\sigma^2)\frac{v}{\sigma^2}\chi^2_v(\frac{vs^2}{\sigma^2})\frac{2s^2}{\sigma}\]
令$\mu=\mu_1-\mu_2$，可得
\[\pi_{\mu_1,\mu,\zeta}(\mu_1,\mu,\zeta|x)=\pi_{\mu_1,\mu_2,\sigma}(\mu_1,\mu_1-\mu,\frac{\mu}{\sqrt{2}\zeta}|x)|J|\]
其中$|J|$为雅克比式，有：
\[|J|=|\frac{\partial(\mu_1,\mu_2,\sigma^2)}{\partial(\mu_1,\mu,\zeta)}|=|\frac{\mu}{\sqrt{2}\zeta^2}|\]
于是有：
\[\pi_{\mu_1,\mu,\zeta}(\mu_1,\mu,\zeta|x)=N(\mu_1,\frac{\mu^2}{2\zeta^2})N(\mu_1-\mu,\frac{\mu^2}{2\zeta^2})\frac{2v\zeta^2}{\mu^2}\chi^2_v(\frac{2\zeta^2vs^2}{\mu^2})\frac{2\sqrt{2}\zeta s^2}{\mu}|\frac{\mu}{\sqrt{2}\zeta^2}|\]
而
\[\pi(\zeta|x)=\int \pi_{\mu_1,\mu,\zeta}(\mu_1,\mu,\zeta|x)d\mu_1d\mu\]
\[\propto \int_{-\infty}^{+\infty} d\mu \int_{-\infty}^{+\infty}\frac{\zeta}{\mu^2}(\frac{s^2\zeta^2}{\mu^2})^{\frac{v}{2}}\exp (-\frac{s^2v\zeta^2}{\mu^2}-\frac{(\mu_1-u_1)^2\zeta^2}{\mu^2}-\frac{(\mu_1-\mu-u_2)^2\zeta^2}{\mu^2})d\mu_1\]
\[\propto \int_{-\infty}^{+\infty} (\frac{s\zeta}{\mu})^{v-1}\exp(-\frac{s^2\zeta^2}{\mu^2}(v+(\frac{\mu}{\sqrt{2}s}-z)^2))\frac{s\zeta}{\mu^2}d\mu\]
做换元$\frac{s\zeta}{\mu}=t$，就有：
\[\pi(\zeta|x)\propto\int_{-\infty}^{+\infty} t^{v-1}\exp(-t^2(v+(\frac{\zeta}{\sqrt{2}t}-z)^2))dt\]
显然$t$在积分后就没了，整个只剩下$z$而已
所以$\pi(\zeta|x)$里面只含有$z$