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

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

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

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Could you perhaps be persuaded to use English prose to connect your formulas rather than those little constellation of dots that nobody remembers what are supposed to mean? You're not taking dictation here; you have time to write things out. – Henning Makholm Feb 27 '12 at 14:31
just one thing remember for ever,that integral of product does not equal to product of integrals – dato datuashvili Feb 27 '12 at 14:35
What do you think is wrong? If $x_1$ and $x_2$ are independent random variables, does that imply that they are also conditionally independent given $\theta$? If not, the last part of your first sentence is not necessarily true. – Dilip Sarwate Feb 27 '12 at 14:41

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