0
$\begingroup$

What's the between $P(x|y)P(y|z)$ and $P(x|y,z)P(y|z)$, it seems to me they both equal to $P(x,y|z)$.

Does any condition should be satisfied if they both equal to $P(x,y|z)$.

$\endgroup$

2 Answers 2

1
$\begingroup$

By definition $\mathsf P(x,y\mid z)~=~\mathsf P(x\mid y,z)~\mathsf P(y\mid z)$ always holds for any random variables $x,y,z$ where $\mathsf P(y\mid z)\neq 0$.

Does any condition should be satisfied if they both equal to P(x,y|z).

Yes.   $\mathsf P(x\mid y,z)=\mathsf P(x\mid y)$ exactly when $x\perp z\mid y$ (that is, $x$ and $z$ are conditionally independent when given $y$).

Only when one this is so can we make the substitution to claim $\mathsf P(x,y\mid z)~=~\mathsf P(x\mid y)~\mathsf P(y\mid z)$.

$\endgroup$
0
$\begingroup$

Your second term rightly equals p(x,y|z):

$ P(x|y,z) P(y|z) = \frac{P(x,y,z)}{P(y,z)} \frac{P(y,z)}{P(z)} = P(x,y|z)$

I don't quite see, how your first term derives from P(y|x,z), though.

$\endgroup$
1
  • $\begingroup$ Thank you. But, can you tell me if $P(x|y)P(y|z)=P(x,y|z)$, I am confused😂. If it's wrong, then the first term should be wrong. $\endgroup$
    – Macer
    Mar 11, 2019 at 23:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .