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.

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- Nov 29th 2012, 06:53 AMPeterPan2009Equivalence of distributions (chain rule)
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. - Nov 29th 2012, 07:17 AMHallsofIvyRe: Equivalence of distributions (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". - Nov 29th 2012, 08:10 AMPeterPan2009Re: Equivalence of distributions (chain rule)
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? - Nov 29th 2012, 08:22 AMScopurRe: Equivalence of distributions (chain rule)
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)} $ - Nov 29th 2012, 12:45 PMSorobanRe: Equivalence of distributions (chain rule)
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".

Quote:

$\displaystyle \text{Prove: }\:P(x\,|\,y,z) \:=\: \frac{P(x, y\,|\,z)}{P(y\,|\,z)}$

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)} $

- Nov 29th 2012, 11:04 PMPeterPan2009Re: Equivalence of distributions (chain rule)
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!