Solution
= Solution
Conditionally on $\lambda_i$, the proposed precision gives $\theta_i\mid\lambda_i\sim N(\mu,4\tau^2/\lambda_i)$. Thus, with independent $Z\sim N(0,1)$ and $\lambda_i\sim\chi_4^2$,
$$
\frac{\theta_i-\mu}{\tau}=\frac{Z}{\sqrt{\lambda_i/4}}.
$$
By the <Student t-distribution> construction, $\boxed{(\theta_i-\mu)/\tau\sim t_4}$. Here $\tau$ is a scale rather than the standard deviation: the prior variance of $\theta_i$ is $2\tau^2$.