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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 ?

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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