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 ?

  • 1
    $\begingroup$ Yes you can ignore terms not involving the parameter. About the Gamma distribution we cannot answer without seeing $f$. $\endgroup$ – Stéphane Laurent Mar 26 '15 at 11:08
  • $\begingroup$ hey stephane thanks, yeah u are right, u cant answer it without seeing f :). but the formula of the full conditional is correct ? $\endgroup$ – user2016445 Mar 26 '15 at 11:10
  • 1
    $\begingroup$ This formula is correct assuming prior independence between $\theta$ and the other parameters. $\endgroup$ – Stéphane Laurent Mar 26 '15 at 11:42
  • $\begingroup$ can u please explain why do i need there the independence of the priors to get this formula ? $\endgroup$ – user2016445 Mar 26 '15 at 11:51
  • 2
    $\begingroup$ You have a prior on $(\theta,k,\lambda,b_1,b_2)$. If there is independence then it has form $\pi(\theta)\pi(k,\lambda,b_1,b_2)$ and you can drop the second factor when ignoring things not involving $\theta$. $\endgroup$ – Stéphane Laurent Mar 26 '15 at 11:57

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: $$\pi(\theta_1,\mid y, \theta_2) \overset{\theta_1}{\propto} f(y \mid \theta_1,\theta_2) \pi(\theta_1).$$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.