Conjugate relation - Marginalizing over parameters

Question

I am trying to show the right term is same as the left term but it's not going well.

Can somebody help me?

Answer 1

let's rewrite the left part:

$ p(y^{\ast} | x^{\ast}, X, Y) = \frac{p(y^{\ast}, x^{\ast}, X, Y)}{p(x^{\ast}, X, Y)} = \frac{\int p(y^{\ast}, x^{\ast}, \theta, X, Y) d\theta}{\int p(y^{\ast}, x^{\ast}, \theta, X, Y) \ d\theta \ dy^{\ast}} = \frac{\int p(y^{\ast} | x^{\ast}, \theta, X, Y) \cdot p(\theta | x^{\ast}, X, Y) \cdot p(x^{\ast}, X, Y) \ d\theta}{\int p(y^{\ast} | x^{\ast}, \theta, X, Y) \cdot p(\theta | x^{\ast}, X, Y) \cdot p(x^{\ast}, X, Y) \ d\theta \ dy^{\ast}} = \{{\text reduce\ some\ conditions\ by\ independence\ assumption}\} = \frac{\int p(y^{\ast} | x^{\ast}, \theta) \cdot p(\theta | X, Y) \cdot p(x^{\ast}, X, Y) \ d\theta}{\int p(y^{\ast} | x^{\ast}, \theta) \cdot p(\theta | X, Y) \cdot p(x^{\ast}, X, Y) \ d\theta \ dy^{\ast}} = \frac{p(x^{\ast}, X, Y) \cdot \int p(y^{\ast} | x^{\ast}, \theta) \cdot p(\theta | X, Y) \ d\theta}{p(x^{\ast}, X, Y) \cdot \underbrace{\int p(y^{\ast} | x^{\ast}, \theta) \ dy^{\ast}}_{1} \cdot \underbrace{\int p(\theta | X, Y) \ d\theta}_{1}} = \int p(y^{\ast} | x^{\ast}, X, Y) \cdot p(\theta | X, Y) \ d\theta $