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Choose a fixed $H_{\mathrm{ref}}>0$, and write $h(H_0)=-5{\log }_{10}(H_0/H_{\mathrm{ref}})$ and $d_s(w,{\Omega }_M)=\mu (z_s;H_{\mathrm{ref}},w,{\Omega }_M)$. The predicted magnitude is $M_0+h(H_0)+d_s$. Thus the likelihood depends on $M_0,H_0$ only through $\beta =M_0+h(H_0)$, the absolute magnitude–Hubble constant degeneracy. Since $d{M}_0=d\beta$ at fixed $H_0$, integrating with a flat density over the entire real line removes $H_0$.

Explicitly, let $a_s=(v+r_s{)}^{-1}$, $S={\sum }_s{a}_s$, $x_s={\hat{m}}_s-d_s$, $\bar{x}=S^{-1}{\sum }_s{a}_s{x}_s$ and $Q={\sum }_s{a}_s(x_s-\bar{x}{)}^2$. Completing the square gives ${\sum }_s{a}_s(x_s-\beta {)}^2=Q+S(\beta -\bar{x}{)}^2$, so flat-prior elimination of a Gaussian common mean yieldsindependently of $H_0$. The proportionality includes the arbitrary constant of the flat prior, which cancels within this posterior analysis.

If the remaining posterior integral is finite, prior independence gives $p(H_0\mid \mathcal{D})\propto p(H_0)$ and hence $\boxed{p(H_0\mid \mathcal{D})=p(H_0)}$: these data supply no marginal update of the Hubble constant. The quoted Gaussian is consequently retained with its stated $a,b$. Strictly, the model uses $H_0>0$ inside a logarithm, so the Gaussian prior must be restricted and normalized on that domain:where $\varphi ,\Phi$ are the standard normal density and distribution function. Ignoring its negligible negative tail gives the stated untruncated Gaussian approximation.