Bayesian Inversion review.

Question

I am looking over a paper currently and am stuck on a particular step seeing how the intuition works. I know this is quite elementary but it is known:

$$P(ab\mid c)=P(a\mid c)P(b\mid c)$$

This being said I am confused how

$${P(a\mid c,d)\over{P(b\mid c,d)}}={P(c\mid a,d)\over{P(c\mid b,d)}}{P(a\mid d) \over {P(b\mid d)}}$$

If possible I would like to see the intuition in with how the above Bayesian inversion is done.

Thanks.

Answer 1

I know this is quite elementary but it is known:

$$P(ab|c)=P(a|c)P(b|c)$$

It's not elementary.   That is true only when $a,b$ are conditionally independent given $c$.

It is also irrelevant to the following:

This being said I am confused how

$${P(a|c,d)\over{P(b|c,d)}}={P(c|a,d)\over{P(c|b,d)}}{P(a|d)\over{P(b|d)}}$$

Bayes' Rule: $P(a\mid c,d) = P(a\mid d)\,P(c\mid a,d)\div P(c\mid d)$

Because: $~~~~~P(a\mid c,d) ~{= P(a,c,d)\,\div P(c,d) \\ = P(a,d)\,P(c\mid a,d)\div P(c, d) \\ = P(d)\,P(a\mid d)\,P(c\mid a,d)\div (P(d)\,P(c\mid d)) \\= P(a\mid d)\,P(c\mid a,d)\div P(c\mid d)}$

Likewise : $~~~~P(b\mid c,d) = P(b\mid d)P(c\mid b,d)\div P(c\mid d)$

Then: $~~~~~~~~\require{cancel}\dfrac{P(a\mid c,d)}{P(b\mid c,d)}=\dfrac{P(a\mid d)\,P(c\mid a,d)\cancel{\div P(c\mid d)}}{P(b\mid d)\,P(c\mid b,d)\cancel{\div P(c\mid d)}}$

Answer 2

$$P(a|c,d) = \frac{P(a,c,d)}{P(c,d)} = \frac{P(a,c,d)P(a,d)}{P(c,d)P(a,d)} = \frac{P(c|a,d)P(a,d)}{P(c,d)} = \frac{P(c|a,d)P(a,d)P(d)}{P(c,d)P(d)} = \frac{P(c|a,d)P(a|d)P(d)}{P(c,d)}$$

Similarly,

$$P(b|c,d) = \frac{P(c|b,d)P(b|d)P(d)}{P(c,d)}$$