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Given the Bayes formula:

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

If there is a distribution (let's say $g$) over the parameter $\theta$, how should one rewrite the Bayes formula?

$D$ is the evidence and $\theta$ is the prior (model parameters).

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  • $\begingroup$ Isn't $p(\theta)$ the distribution over the parameter $\theta$? $\endgroup$
    – mhdadk
    Commented Jun 25, 2021 at 19:36
  • $\begingroup$ if $ \theta $ is fixed, then no. If $D$ is fixed then yes. Do you mean that $ \theta$ is a random variable @mhdadk? $\endgroup$
    – Ash
    Commented Jun 25, 2021 at 19:40
  • $\begingroup$ Could you please edit your question and explain what $\theta$ and $D$ are? $\endgroup$
    – mhdadk
    Commented Jun 25, 2021 at 19:49
  • $\begingroup$ So here, by saying that $g$ is a distribution over $ \theta $ I'm assuming that $\theta$ is a random variable. @mhdadk $\endgroup$
    – Ash
    Commented Jun 25, 2021 at 19:54
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    $\begingroup$ You might want Bayes' Theorem for continuous random variables which is stated here: en.wikipedia.org/wiki/… $\endgroup$
    – fblundun
    Commented Jun 25, 2021 at 20:51

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Bayes formula assumes that there is a distribution over every quantity, so there is no need to rewrite it.

Bayes formula applies to both discrete and continuous random quantities. If the quantities are discrete then the $p()$ are probability mass functions. If the quantities are continuous, then the $p()$ are probability density functions.

You refer to the possibility of "fixed" quantities in your comments, but there are no fixed quantites in Bayes formula. If $\theta$ was fixed (i.e., known) then the formula would be pointless. If $D$ was fixed, then it could not provide evidence about $\theta$. In Bayesian applications, $D$ is observed and $\theta$ is an unknown parameter that determines the distribution of $D$. In this context, $p(D|\theta)$ is the likelihood function, the probability (i.e., the likelihood) of observing that particular $D$ value as a function of $\theta$.

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