统计数据科学简介|  Introduction to Statistical data Science代写STAT0032代考

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\begin{aligned} h\left(\theta_{1} \mid x_{2}\right) &=\frac{f\left(x_{2} \mid \theta_{1}\right) \pi\left(\theta_{1}\right)}{\sum_{i=1}^{2} f\left(x_{2} \mid \theta_{i}\right) \pi\left(\theta_{i}\right)} \ &=\frac{.3 \times .8}{.3 \times .8+.2 \times .2} \ &=.86 \end{aligned}
Hence,
$$h\left(\theta_{2} \mid x_{2}\right)=.14$$

We next calculate the posterior risk (PR) for $a_{1}$ and $a_{2}$ :
\begin{aligned} \operatorname{PR}\left(a_{1}\right) &=l\left(\theta_{1}, a_{1}\right) h\left(\theta_{1} \mid x_{2}\right)+l\left(\theta_{2}, a_{1}\right) h\left(\theta_{2} \mid x_{2}\right) \ &=0+400 \times .14 \ &=56 \end{aligned}
and
\begin{aligned} \operatorname{PR}\left(a_{2}\right) &=l\left(\theta_{1}, a_{2}\right) h\left(\theta_{1} \mid x_{2}\right)+l\left(\theta_{2}, a_{2}\right) h\left(\theta_{2} \mid x_{2}\right) \ &=100 \times .86+0 \ &=86 \end{aligned}

PHAS00032 COURSE NOTES ：

Proof Let $1 / c=\pi /(1-\pi)$; that is,
$$\pi=\frac{1}{c+1}$$
Then $d^{*}$ accepts if
$$\frac{\pi f\left(\mathbf{x} \mid \theta_{1}\right)}{(1-\pi) f\left(\mathbf{x} \mid \theta_{2}\right)}>1$$