Every lecture and book says that the denominator in Bayes' Rule does not depend on the parameter $\theta$. However, the denominator also includes $\theta$ in the formula of Bayes' Rule. I just cannot understand what it really says.


In the Bayesian formula:

$$\text{posterior} = \,\frac{\text{likelihood} \cdot \text{prior}}{\text{normalizing constant}}$$

If we call the observations $y$ and the parameters $\theta$, then this equates:

$$p(\theta | y) = \, \frac{p(y | \theta) \cdot p(\theta)}{p(y)}$$

Here, the normalizing constant $p(y)$ is calculated as:

$$p(y) = \int p(y | \theta) \cdot p(\theta) \,\mathrm{d}\theta$$

Since you integrate out $\theta$ (the parameters), the denominator no longer depends on it.

  • 1
    $\begingroup$ Ohhh, I now understand why the denominator does not contain $ \theta $ dependency. $\endgroup$ – StoryMay Dec 1 '19 at 12:35

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