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