Noob question about bayes rule denominator estimation A known problem of Bayes rule is the intractability of the estimation of $p(D)$ given a multiparametric problem, since $p(D)$ is found by marginalizing the joint probability $p(D, \theta_{1..n})$ over the whole n-dimensions $\theta$ space.
But I was wondering, since $p(D)$, in theory, does not depend on the actual parameterization of the model (otherwise it would be conditional on $\theta$ and not marginal), wouldn't be sufficient to estimate in a way that doesn't use $\theta$ at all?
I wouldn't know how to do it with no parameters at all, but the first thing that comes to my mind would be to use a simplified model with just one parameter to integrate over and then use the $p(D)$ estimated this way as a denominator of the joint density of the complex model.
What am I missing? If one can compute $p(D)$ analytically (or numerically but fast) from a simpler model, couldn't she plug it into a complex model to avoid using MCMC to estimate the normalized posterior?

There is more than one possible model for the joint distribution. Let's index two of them by $A$ and $B$:
\begin{equation}
p(D,\theta|A) \quad\text{versus} \quad p(D,\theta|B) . 
\end{equation}
It follows that there are two different marginal likelihoods for the data, one for each model:
\begin{equation}
p(D|A) = \int p(D,\theta|A)\,d\theta 
\quad\text{versus}\quad
p(D|B) = \int p(D,\theta|B)\,d\theta. 
\end{equation}
Is one of the models correct? Which one is better?
To strip things down to the bare essentials, consider two models for $D$, neither of which has any parameters:
\begin{equation}
p(D|A) = \textsf{N}(D|0,1)
\quad\text{and}\quad
p(D|B) = \textsf{N}(D|3,1) . 
\end{equation}
Now suppose $D =1$. Which model is more consistent with the observation $D$? I.e., which model has the greater marginal likelihood of the fixed data?
