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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*}

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5
+50
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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.

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