= Solution
For a fixed observed colour $\widetilde c_s$, <Bayes theorem> gives
$$
p(E_s\mid\widetilde c_s)
\propto
\exp\!\left[-\frac{(\widetilde c_s-E_s-c_0)^2}{2\sigma_c^2}-\frac{E_s}{\tau}\right]
\mathbf1_{\{E_s\geq0\}}.
$$
Completing the square shows that this is a <truncated normal distribution> with variance $\sigma_c^2$, lower endpoint zero, and untruncated location
$$
e_s=\widetilde c_s-c_0-\frac{\sigma_c^2}{\tau}.
$$
Hence
$$
\overline E_s
\equiv\mathbb E[E_s\mid\widetilde c_s]
=e_s+\sigma_c\frac{\phi(e_s/\sigma_c)}{\Phi(e_s/\sigma_c)}.
$$
Conditioning first on $E_s$ and using the <law of total expectation> yields the dusty colour-magnitude relation
$$
\mathbb E[\widetilde M_s\mid\widetilde c_s]
=M_0+\beta\widetilde c_s+(R-\beta)\overline E_s.
$$
As $\widetilde c_s\to-\infty$, the <Inverse Mills ratio> implies $\overline E_s\to0$ with vanishing derivative, so the asymptotic slope is $\beta$. As $\widetilde c_s\to+\infty$, $\overline E_s\sim\widetilde c_s-c_0-\sigma_c^2/\tau$, so the asymptotic slope is $R$.
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