Is the conclusion(last line) correct? If not, what's wrong with this proof ? (Probability Density Estimation, Bayesian inference)

Question

(Probability Density Estimation, Bayesian inference)

$x_1$ and $x_2$ are independent random variables, so $p\left(x_1,x_2\right)=p\left(x_1\right)\cdot p\left(x_2\right)$ and $p\left(x_1,\left.x_2\right|\theta \right)=p\left(\left.x_1\right|\theta \right)\cdot p\left(\left.x_2\right|\theta \right)$,

from $p\left(x_1,x_2\right)=p\left(x_1\right)\cdot p\left(x_2\right)$,

I get $ \int p\left(x_1,\left.x_2\right|\theta \right)p(\theta )d\theta =\int p\left(\left.x_1\right|\theta \right)p(\theta )d\theta \cdot \int p\left(\left.x_2\right|\theta \right)p(\theta )d\theta$

And because $p\left(x_1,\left.x_2\right|\theta \right)=p\left(\left.x_1\right|\theta \right)p\left(\left.x_2\right|\theta \right)$

$\therefore \int p\left(\left.x_1\right|\theta \right)p\left(\left.x_2\right|\theta \right)p(\theta )d\theta =\int p\left(\left.x_1\right|\theta \right)p(\theta )d\theta \cdot \int p\left(\left.x_2\right|\theta \right)p(\theta )d\theta$

Answer 1

As indicated by others, you are considering that two different hypotheses are equivalent although they are not. These two hypotheses are:

1. Independence per se, which translates as $p(x_1,x_2)=p(x_1)p(x_2)$.
2. Conditional independence, which translates as $p(x_1,x_2\mid\theta)=p(x_1\mid\theta)p(x_2\mid\theta)$.

None of these implies the other, hence assuming in effect that both hold at the same time can only yield odd assertions.