# What does “the denominator does not contain any theta dependence” mean in Bayes' Rule? [duplicate]

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.

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