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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
    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.2012-02-27
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    just one thing remember for ever,that integral of product does not equal to product of integrals2012-02-27
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    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

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