# 对换钱悖论的贝叶斯分析

Posted on Tue 04 April 2017 in ProbStat

## 贝叶斯分析¶

\begin{align} p(x)&=\int p(x|y)\rho(y)dy\\ &=\frac{1}{2}\int [\delta(x-2y)+\delta(x-y)]\rho(y)dy\\ &=\frac{\rho(x)}{2}+\frac{\rho(x/2)}{4} \end{align}

\begin{align} p(y|x)&=\frac{p(x|y)\rho(y)}{p(x)}\\ &=\frac{\delta(x-2y)+\delta(x-y)}{2p(x)}\rho(y)\\ &=\frac{\rho(y)\delta(y-x/2)/2+\rho(y)\delta(x-y)}{2p(x)}\\ &=\frac{\rho(x/2)}{4p(x)}\delta(y-x/2)+\frac{\rho(x)}{2p(x)}\delta(y-x) \end{align}

\begin{align} p(y=x|x)&=\frac{\rho(x)}{2p(x)}\\ p(y=x/2|x)&=\frac{\rho(x/2)}{4p(x)} \end{align}

\begin{align} \mathrm{E}[\Delta x]&=xp(y=x|x)-\frac{x}{2}p(y=x/2|x)\\ &=\frac{x\rho(x)}{2p(x)}-\frac{x\rho(x/2)}{8p(x)} \end{align}

## 无脑交换的收益期望值¶

$\exists A$使得对于$\forall x>A$，满足$\rho(x)\equiv 0$

\begin{align}\overline{\mathrm{E}[\Delta x]}&=\int_0^{2A} p(x)\mathrm{E}[\Delta x]dx\\ &=\int_0^{2A} \frac{x\rho(x)}{2}-\frac{x\rho(x/2)}{8}dx\\ &=\int_0^{2A} \frac{x\rho(x)}{2}-\int_0^A\frac{t\rho(t)}{2}dt,\quad t\stackrel{def}{=}x/2\\ &=0 \end{align} 所以无脑交换在合理假设下无平均收益。