Solution
= Solution
Let $q_0=n_0$, $q_1=n_0/c^2$ and $y=\overline y$. These are the two prior <precision parameters>. Completing the square in $n(\beta-y)^2+q_i\beta^2$ gives the <Normal-normal conjugacy> update
$$
\boxed{\beta\mid y,H_i\sim N\left(\frac{n}{n+q_i}y,\frac1{n+q_i}\right),\qquad i=0,1.}
$$
Thus the <posterior mean> is a precision-weighted average of the observed mean and the prior center zero. The <posterior variance> is the reciprocal of the total precision.