Hi,

I have this equivalence:

, 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, 07:53 AMPeterPan2009Equivalence of distributions (chain rule)
Hi,

I have this equivalence:

, 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, 08:17 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, 09:10 AMPeterPan2009Re: Equivalence of distributions (chain rule)
Yes, they are probability distributions.

Well, in the case of conditional probabilities that would be defined as:

, right? But I don't see how that can be used to underline the validity of the equivalance? - Nov 29th 2012, 09:22 AMScopurRe: Equivalence of distributions (chain rule)
Write out individually, what

and you will see the equivalence quite easily. Here ill give you the first

- Nov 29th 2012, 01:45 PMSorobanRe: Equivalence of distributions (chain rule)
Hello, PeterPan2009!

I assume that these are conditional probabilities,

. . so we can apply Bayes' Theorem: .

I further assume that the comma represrnts "and".

Quote:

The left side is: .

The right side is: .

- Nov 30th 2012, 12:04 AMPeterPan2009Re: 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!