Please explain to me why $ \int_b p(a|b) db \neq p(a) $

Question

I have one question that bugs me. How is it that:

$ \int_a p(a|b) da = \int_a \frac{p(a,b)}{p(b)} da = 1 $

but

$ \int_b p(a|b) db = \int_b \frac{p(a,b)}{p(b)} db \neq p(a) $

I don't understand this because $ p(a,b) = p(b,a) $ and therefore

$ p(a|b) = \frac{p(a,b)}{p(b)} \\ \Rightarrow p(a,b) = p(a|b) p(b) = p(b|a) p(a) = p(b,a) $

and if I'm not mistaken:

$ p(a|b)p(b) = p(b|a)p(a) \\ \frac{p(a,b)}{p(b)} p(b) = \frac{p(a,b)}{p(a)} p(a) \\ p(a,b) = p(a,b) $

But how is it, that this is not correct:

$ \int_b p(a|b) db = \int_b \frac{p(a,b)}{p(b)} db = \int_b p(a,b) \frac{1}{p(b)} db = \int_b p(a,b) db \int_b \frac{1}{p(b)} db \\ \int_b p(a,b) db \int_b \frac{1}{p(b)} db = \int_b p(a,b) db *1 = \int_b p(a,b) =p(a) $

which should hold because of the marginalization rule:

$ p(a) =\int_b p(a,b) $

Answer 1

What is true is that $\int_bp(a,b)db = p(a)$. Roughly speaking the probability that $a$ and $b$ occur, summed over all $b$, is just the probability that $a$ occurs.

I don't really understand what you are asserting to be true and what you are asserting to be false and what you are asking. But there are several mistakes in your post. Firstly, the equality $p(a|b) p(b) = p(b|a) p(b)$ is wrong, in general, unless $p(a)=p(b)$ or one of $p(a)$, $p(b)$ is 0. So is $\int_b p(a,b) \frac{1}{p(b)} db = \int_b p(a,b) db \int_b \frac{1}{p(b)} db$. And so is $\int1/p(b)db=1$, as you can easily convince yourself by considering a finite probability space. I am not sure: does that resolve your difficulties?