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

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

  • 1
    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.2012-02-27

1 Answers 1

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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.