Hi,
I have this equivalence:
$\displaystyle P(x|y,z) = \frac{P(y, x|z)}{P(y|z)}$
, but I don't understand how these are the same thing. Is there anybody who knows? It may be an application of the chain rule.
I assume these are probability distributions. In that case, no, this has nothing to do with the chain rule (unless you are given a specific form for P that you did not tell us). It has to do with the definition of "conditional probability".
Yes, they are probability distributions.
Well, in the case of conditional probabilities that would be defined as:
$\displaystyle P(b|a) = \frac{P(a,b)}{P(a)}$
, right? But I don't see how that can be used to underline the validity of the equivalance?
Write out individually, what
$\displaystyle P(x|y,z)$
$\displaystyle P(y, x|z)$
$\displaystyle P(y|z)$
and you will see the equivalence quite easily. Here ill give you the first
$\displaystyle P(x|y,z) =\frac{P(X=x,Y=y,Z=z)}{P(Y=y,Z=z)} $
Hello, PeterPan2009!
I assume that these are conditional probabilities,
. . so we can apply Bayes' Theorem: .$\displaystyle P(a\,|\,b) \:=\:\frac{P(a\wedge b)}{P(b)}$
I further assume that the comma represrnts "and".
$\displaystyle \text{Prove: }\:P(x\,|\,y,z) \:=\: \frac{P(x, y\,|\,z)}{P(y\,|\,z)}$
The left side is: .$\displaystyle P(x\,|\,y\wedge z) \;=\;\frac{P(x\wedge y\wedge z)}{P(y\wedge z)}$
The right side is: .$\displaystyle \frac{P(x\wedge y\,|\,z)}{P(y\,|\,z)} \;=\; \dfrac{\dfrac{P(x\wedge y\wedge z)}{P(z)}} {\dfrac{P(y\wedge z)}{P(z)}} \;=\;\frac{P(x\wedge y\wedge z)}{P(y\wedge z)} $
Hi guys!
Thanks a lot! This was most helpful!
Scopur: I did the "derivation" myself before verifying it with Soroban's.
Previously, I've got a little confused by the '|' character, but it's not a mathematical symbol so I shouldn't be. This little exercise helped my understanding such wise.
Have a great day!