In bayes version for continuous case, what does it mean to integrate with respect to $d\theta$ when $\theta$ is a vector not a a scalar value?

$$p(\theta|D) = \frac{p(D|\theta)p(\theta)}{p(D)}$$

Where $D$ is a set of observed data points, and $\theta$ is a vector of parameters to be estimated.

$$ p(D) = \int p(D|\theta)p(\theta)d\theta $$

Is this integral a surface integral, component-wise integral or what ?

  • $\begingroup$ You are computing the marginal distribution of $p(D)$ over the range of your parameter space $\theta \in \Theta$. The choice of integral depends on how your parameter space is defined. $\endgroup$ – Maxtron Sep 22 '18 at 16:48
  • $\begingroup$ Can you give an example with a specific parameter space? I'm not sure I fully understand your answer $\endgroup$ – Loai Ghoraba Sep 22 '18 at 17:27
  • 2
    $\begingroup$ Say your parameter $\theta$ is a 2-D vector and bounded, i.e., $\theta_1 \in [a_1, b_1]$ and $\theta_2 \in [a_2, b_2]$, then $\Theta \in \{ [a_1, b_1] \times [a_2, b_2] \}$, which is a surface. $\endgroup$ – Maxtron Sep 22 '18 at 18:01

That will be the n-dimensional integral, like the 2D case below: $$p(D) = \int_\Theta\int p(D | \theta_1,\theta_2)p(\theta_1,\theta_2)d\theta_1d\theta_2$$ Because it does not matter if you put them in a vector or not; if you formulate your likelihood and joint PDF correctly. You just have $n$ random variables.

  • $\begingroup$ Just to make sure, so it is not a surface integral, right? That is , $d\theta$ is just a compact representation of multiple deltas multiplied together ? $\endgroup$ – Loai Ghoraba Sep 22 '18 at 21:03
  • $\begingroup$ Well, what would you call integrals with $dxdy$? $\endgroup$ – gunes Sep 22 '18 at 21:12
  • $\begingroup$ You are right, my bad :) $\endgroup$ – Loai Ghoraba Sep 23 '18 at 17:45

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.