# Why does $P(\theta_1\mid D, \theta_2) \propto P(D \mid \theta_1, \theta_2)P(\theta_1)$ hold?

Suppose that in a Bayesian framework we have observed data $D$, using independent prior distributions on the parameters of the model, denoted by $\theta_1, \theta_2$. Then, the joint posterior distribution of $\theta_1, \theta_2$ can be written as: $$P(\theta_1, \theta_2 \mid D) \propto P(D \mid \theta_1, \theta_2)p(\theta_1)p(\theta_2)$$ One way to obtain estimates from the posterior is use Gibbs Sampling. This sampler requires simulating from the full conditional distributions for both $\theta_1$, $\theta_2$. One relation for the full conditional that caught my eye in a book is the following: $$P(\theta_1\mid D, \theta_2) \propto P(D \mid \theta_1, \theta_2)p(\theta_1)$$ I am wondering how exactly this proportional relation is derived. Would anyone have any ideas? It seems that there are extra assumptions here that are implicit.

The starting point is the proportionality symbol$$\propto$$as it takes different meanings in the few formulas in the question. In $$P(\theta_1, \theta_2 \mid D) \propto P(D \mid \theta_1,\theta_2)p(\theta_1)p(\theta_2)$$ the function of $(\theta_1, \theta_2)$ on the l.h.s. is proportional to the function of $(\theta_1, \theta_2)$ on the r.h.s., meaning that they differ by a multiplicative constant $\kappa$. This is a constant w.r.t. the functions of $(\theta_1, \theta_2)$, meaning that $\kappa$ may depend on (and hence be a function of) other factors, like $D$: $$\kappa=\kappa(D)$$ In fact, since $P(\theta_1, \theta_2 \mid D)$ is a probability density, this constant $\kappa$ is uniquely defined as the normalisation of the r.h.s. that turns it into a probability density, with total mass equal to one: $$\kappa=\kappa(D)=1\Big/\int_\Theta P(D \mid \theta_1,\theta_2)p(\theta_1)p(\theta_2)\,\text{d}\lambda(\theta_1,\theta_2)$$ for the appropriate measure $\text{d}\lambda$.
In $$P(\theta_1\mid D, \theta_2) \propto P(D \mid \theta_1, \theta_2)p(\theta_1)\qquad\qquad(1)$$ the function of $\theta_1$ on the l.h.s. is proportional to the function of $\theta_1$ on the r.h.s., meaning that $\kappa$ may depend on (and hence be a function of) other factors, like $D$ and now $\theta_2$: $$\kappa=\kappa(D,\theta_2)$$ Note that $$P(\theta_1 \mid \theta_2, D) \propto P(D \mid \theta_1,\theta_2)p(\theta_1)p(\theta_2)$$ is equally correct.