Full conditionals - Gibbs Sampler i want to draw samples from a 5-dimensional posterior distribution $f(k,\theta,\lambda,b_1,b_2|Y=y)$. From Bayes-Theorem there is the following relationship between posterior and likelihood:
posterior $\propto$ likelihhod $\times$ prior
so the full conditional of $\theta$ is
$f(\theta|k, \lambda, b_1, b_2, Y=y) \propto f(Y|\theta,\lambda,k,b_1,b_2) \cdot f(\theta)$ ?
i can now transform this to a Gamma distribution by by ignoring all terms that are constant with respect to the parameter right ?
 A: Let me set $\theta_1=\theta$ and $\theta_2=(k,\lambda,b_1,b_2)$ for the sake of lightness. 
Bayes formula: 
$$\pi(\theta_1,\theta_2\mid y) \overset{\theta_1,\theta_2}{\propto} f(y \mid \theta_1,\theta_2) \pi(\theta_1,\theta_2)$$.
The notation $\overset{\theta_1,\theta_2}{\propto}$ is not standard. I invented myself after I faced a situation for which the naked symbol $\propto$ was ambiguous. It means that both members are proportional as functions of $(\theta_1,\theta_2)$.
By standard measure theory, the full conditional distribution of $\theta_1$, that is to say $\pi(\theta_1,\mid y, \theta_2)$ is given by the proportionality relation
$$\pi(\theta_1,\mid y, \theta_2) \overset{\theta_1}{\propto} \pi(\theta_1,\theta_2\mid y),$$
then by Bayes formula previously recalled:
$$\boxed{\pi(\theta_1,\mid y, \theta_2) \overset{\theta_1}{\propto} f(y \mid \theta_1,\theta_2) \pi(\theta_1,\theta_2)}.$$
Thus your formula is wrong in general. But if you use independent priors on $\theta_1$ and $\theta_2$ then the prior distribution has form $\pi(\theta_1,\theta_2)=\pi(\theta_1)\pi(\theta_2)$ (this is not the same "$\pi$" occuring three times here : "$\pi$" means "pdf of") and it is proportional to the prior $\pi(\theta_1)$ as a function of $\theta_1$: $$\pi(\theta_1,\theta_2)=\pi(\theta_1)\pi(\theta_2) \overset{\theta_1}{\propto}  \pi(\theta_1)$$ 
and the general boxed formula reduces to your formula in this case:
$$\pi(\theta_1,\mid y, \theta_2) \overset{\theta_1}{\propto} f(y \mid \theta_1,\theta_2) \pi(\theta_1).$$
