Marginal density over parameter This is from my textbook, I was wondering why the last equation hold?

I got
$$\begin{aligned}
p(y|M_i)&amp; =\int p(y,\theta_i|M_i)d\theta_i\\
&amp; =\int \frac{p(y,\theta_i,M_i)}{p(M_i)}d\theta_i\\
&amp; =\int \frac{p(y,\theta_i,M_i)}{p(\theta_i,M_i)}\frac{p(\theta_i,M_i)}{p(M_i)}d\theta_i\\
&amp;=\int p(y|\theta_i,M_i)p(\theta_i|M_i)d\theta_i
\end{aligned}$$

Many thanks~
The correct series of equations can be obtained in a more direct way.

\begin{equation}
\begin{aligned}
p(y \mid M_i)
&amp; = \int p(y, \theta_i \mid M_i) d\theta \text{ by marginalizing} \\
&amp; = \int p(p, \mid M_i, \theta_i) p(\theta_i \mid M_i) d\theta_i \text{ by factoring pdf} \\
\end{aligned}
\end{equation}

The last step is justified in general because $ p(x, y \mid z) = p(x \mid y, z)p(y \mid z) $, where $ x, y, z $ are allowed to be vectors (and in particular, $ z $ can be empty).
The correct series of equations can be obtained in a more direct way.

\begin{equation}
\begin{aligned}
p(y \mid M_i)
&amp; = \int p(y, \theta_i \mid M_i) d\theta \text{ by marginalizing} \\
&amp; = \int p(p, \mid M_i, \theta_i) p(\theta_i \mid M_i) d\theta_i \text{ by factoring pdf} \\
\end{aligned}
\end{equation}

The last step is justified in general because $ p(x, y \mid z) = p(x \mid y, z)p(y \mid z) $, where $ x, y, z $ are allowed to be vectors (and in particular, $ z $ can be empty).