I am currently reading an introduction to Bayesian Modelling. Im trying to understand why the following is true:
$$p(w|x,t,\alpha,\beta)\propto p(t|x,w,\beta)p(w|\alpha)$$ where:
$w$ is are the parameters
$t$ is the target we are trying to estimate. with mean = $y(x,w)$ and precision $beta$
$\alpha$ is the hyper parameter for the prior distribution of $w$
$\beta$ is the precision parameter for the distribution of $t$.
I am familiar with the product rule, but I can't wrap my head around this product of 2 conditional probabilities. $p(t|x,w,\beta)p(w|\alpha)$
Am I right to think that $P(A|B,C) *P(C|D) = P(A,C|B,D) $?
If that is true, am I then right in thinking that $p(t|x,w,\beta)p(w|\alpha) = p(t,w|x,\alpha,\beta)$?
That formula follows from Bayes' theorem. I will for a moment disregard the hyperparameters (because they are irrelevant for the probabilistic reasoning). From Bayes' theorem, we have that in general $$p(w|x,t) = \frac{p(t|x,w)p(w|x)}{p(t|x)}.$$
This is a general result; be careful with conditioning on $x$! However, now we introduce the information from the model that says that $w$ is independent of $x$ to say that $p(w|x) = p(w)$: this is the prior distribution.
Now, we notice that what we are interested in is a distribution of $w$. So, we can keep in the p.d.f. only the terms that depend on $w$ $$p(w|x,t) \propto p(t|x,w)p(w).$$ Introduce back the hyperparameters for the distributions and you have the expression you asked for.
Regarding your second point, yes $p(t,w|x) =p(t|w,x)p(w)$.