Results 1 to 6 of 6

Math Help - Equivalence of distributions (chain rule)

  1. #1
    Newbie
    Joined
    Sep 2012
    From
    Netherlands
    Posts
    8

    Equivalence of distributions (chain rule)

    Hi,

    I have this equivalence:

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,372
    Thanks
    1314

    Re: 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".
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2012
    From
    Netherlands
    Posts
    8

    Re: Equivalence of distributions (chain rule)

    Yes, they are probability distributions.

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

    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?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Oct 2008
    Posts
    77
    Thanks
    1

    Re: Equivalence of distributions (chain rule)

    Write out individually, what

    P(x|y,z)
    P(y, x|z)
    P(y|z)

    and you will see the equivalence quite easily. Here ill give you the first
    P(x|y,z) =\frac{P(X=x,Y=y,Z=z)}{P(Y=y,Z=z)}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,678
    Thanks
    611

    Re: Equivalence of distributions (chain rule)

    Hello, PeterPan2009!

    I assume that these are conditional probabilities,
    . . so we can apply Bayes' Theorem: . P(a\,|\,b) \:=\:\frac{P(a\wedge b)}{P(b)}
    I further assume that the comma represrnts "and".


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

    The left side is: . P(x\,|\,y\wedge z) \;=\;\frac{P(x\wedge y\wedge z)}{P(y\wedge z)}

    The right side is: . \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)}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Sep 2012
    From
    Netherlands
    Posts
    8

    Re: 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!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 9
    Last Post: November 9th 2010, 01:40 AM
  2. Chain Rule Inside of Chain Rule
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 22nd 2009, 08:50 PM
  3. Replies: 5
    Last Post: October 19th 2009, 01:04 PM
  4. Replies: 3
    Last Post: May 25th 2009, 06:15 AM
  5. Replies: 2
    Last Post: December 13th 2007, 05:14 AM

Search Tags


/mathhelpforum @mathhelpforum