= Solution
{c}
Put $M_0=\alpha+x_0\beta$. As a <linear combination of independent normal random variables> under the <posterior distribution>,
$$
M_0\mid\boldsymbol y\sim N(m_0,v_0),\qquad
m_0=m_\alpha+x_0m_\beta,\qquad
v_0=v_\alpha+x_0^2v_\beta.
$$
The shortest 95% <credible interval> is
$$
\boxed{\left[m_\alpha+x_0m_\beta
-\Phi^{-1}(0.975)\sqrt{v_\alpha+x_0^2v_\beta},\quad
m_\alpha+x_0m_\beta
+\Phi^{-1}(0.975)\sqrt{v_\alpha+x_0^2v_\beta}\right]}.
$$
Here $\Phi$ 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.
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