Bayes factor and Posterior odds Consider the following posterior odds
\begin{equation*}
\frac{P(H|D_1,D_2)}{P(\overline{H}|D_1,D_2)}=\frac{P(D_2|H,D_1)\times P(D_1|H)P(H)}{P(D_2|\overline{H},D_1)\times P(D_1|\overline{H})P(\overline{H})}
\end{equation*}
Under what circumstances can we write
\begin{equation*}
\frac{P(H|D_1,D_2)}{P(\overline{H}|D_1,D_2)}=\frac{P(D_2|H,D_1)}{P(D_2|\overline{H},D_1)}\times \frac{P(H|D_1)}{P(\overline{H}|D_1)}
\end{equation*}
 A: The equivalence you mention always holds, no need for particular circumstances.
By the definition of a conditional probability
\begin{align} P(D_1|H) & := \frac{P(D_1,H)}{P(H)}\\
              P(D_1|H)P(H) & = P(D_1,H) \\
                           & = P(H,D_1) \\
                           & = P(H|D_1) P(D_1) \end{align}
Similarily 
\begin{align} P(D_1|\overline{H}) & := \frac{P(D_1,\overline{H})}{P(\overline{H})}\\
              P(D_1|\overline{H})P(\overline{H}) & = P(D_1,\overline{H}) \\
                           & = P(\overline{H},D_1) \\
                           & = P(\overline{H}|D_1) P(D_1) \end{align}
Replacing the $P(D_1|\bar{H})P(\bar{H})$ and $P(D_1|H)P(H)$ in your original equation we get
\begin{equation*}
\frac{P(H|D_1,D_2)}{P(\overline{H}|D_1,D_2)}=\frac{P(D_2|H,D_1)\times P(H|D_1) P(D_1)}{P(D_2|\overline{H},D_1)\times P(\overline{H}|D_1) P(D_1)} = \frac{P(D_2|H,D_1)}{P(D_2|\overline{H},D_1)}\times \frac{P(H|D_1)}{P(\overline{H}|D_1)}
\end{equation*}
the desired result.
