Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

share|cite|improve this question
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

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.