What does "the denominator does not contain any theta dependence" mean in Bayes' Rule?

Question

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.

Answer 1

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.