 Solution

Put $M_0=\alpha +x_0\beta$. As a linear combination of independent normal random variables under the posterior distribution,The shortest 95% credible interval isHere $\Phi$ is the standard normal distribution cumulative distribution function. To justify shortest length, consider all intervals of a fixed length $d$. If $f$ is this normal distribution posterior density, the interval probability ${\int }_a^{a+d}f(t)dt$ has derivative $f(a+d)-f(a)$ with respect to $a$. The normal distribution is symmetric and its density strictly decreases with distance from its mean, so this probability is uniquely maximized when the midpoint is $m_0$. The centered interval can therefore attain 95% posterior probability with the least length. It is both an equal-tailed credible interval and a highest posterior density interval.

The target is the experiment's mean at $x_0$. A posterior predictive distribution for a new noisy observation would instead have variance $v_0+{\sigma }^2$; that extra observation variance does not belong in the requested interval.