Q:

Proof of Bayesian Prediction Formula

I was wondering how to proof this formula, commonly used in Bayesian Prediction:
$$ \mathrm{P}(x|\alpha) = \int_\theta \mathrm{P}(x|\theta)\mathrm{P}(\theta|\alpha) \, \mathrm{d}\theta$$
The left hand side can be expressed as the following, through marginalizing:
$$ \mathrm{P}(x|\alpha) = \int_\theta \mathrm{P}(x, \theta | \alpha) \, \mathrm{d}\theta \quad \quad \ldots \text{(1)}$$
Expanding the right hand side,
$$ \int_\theta \mathrm{P}(x|\theta) \mathrm{P}(\theta|\alpha) \, \mathrm{d}\theta = \int_\theta \frac{\mathrm{P}(x,\theta)}{\mathrm{P}(\theta)} \frac{\mathrm{P}(\theta,\alpha)}{\mathrm{P}(\alpha)} \, \mathrm{d} \theta \quad \quad \ldots \text{(2)}$$
Note that in equation (1), there will be a $\mathrm{P}(x,\theta,\alpha)$ term, but in equation (2), I can't see how that term will appear.
Thanks.

A:

We have, in general
$$ \mathrm{P}(x) = \int_\theta \mathrm{P}(x|\theta)\mathrm{P}(\theta) \, \mathrm{d}\theta$$
and conditioning everything on $\alpha$ :
$$ \mathrm{P}(x | \alpha) = \int_\theta \mathrm{P}(x|\theta \alpha) \mathrm{P}(\theta | \alpha) \, \mathrm{d}\theta$$
In the Bayesian setting, $\mathrm{P}(x|\theta \alpha)=\mathrm{P}(x|\theta)$  because, if we are given the parameter $\theta$ we know the density of $x$, and the values of $\alpha$ adds nothing ($\alpha$ only gives us information about $\theta$ - once we know it, they contribute nothing).

