Conditional probability

Discussion in 'Probability' started by Vedran, Jun 15, 2010.

  1. Vedran

    Vedran Guest

    How can p(A|B|C) be expanded?

    Are p(A|(B|C)), p((A|B)|C) and p(A|B|C) all different or are they the
    same thing.

    Thanks in advance!
     
    Vedran, Jun 15, 2010
    #1
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  2. Do you know of any of them what it means?

    Cheers,
    Bastian
     
    Bastian Erdnuess, Jun 15, 2010
    #2
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  3. I know what p(X|Y) means but what does X|Y mean?
     
    Frederick Williams, Jun 15, 2010
    #3
  4. Vedran

    Vedran Guest


    Here is the explanation of what x,y and z mean.

    ****

    Mutual information I(x;y) in information theory is defined as:

    I(x;y) = p(x|y)/p(x) (1)

    Conditional mutual information is written as: I(x;y|z)

    By that rationale I would guess that conditional mutual information would
    be:
    I(x;y|z) = p(x|(y|z))/p(x) (2)

    Actual definition is:
    I(x;y|z) = p(x|y,z)/p(x|z) (3)

    And I really do not have I clue how (3) is obtained. I(x;y|z) is defined
    as "the information provided about the event x by the occurrence of the
    event y given z".
     
    Vedran, Jun 16, 2010
    #4
  5. Vedran

    Henry Guest

    If p(X|Y) is short for p(X|Y=y), e.g. the conditional density
    of the random variable X given that another random variable Y
    takes the value y, e.g. the derivative of the conditional cdf
    Prob(X<=x|Y=y), then I would assume X|Y is short for X|Y=y
    and so is a (conditional/constrained?) random variable which
    has the conditional density.

    Going back to p(A|B|C), I would take that to be short for
    p(A|B=b|C=c) which looks as if it should be p(A|B=b,C=c).
    Isn't this the same as p(A|(B=b|C=c)), p((A|B=b)|C=c)?

    The only thing which makes me hesitate is the
    Borel–Kolmogorov paradox
    http://en.wikipedia.org/wiki/Borel–Kolmogorov_paradox
     
    Henry, Jun 16, 2010
    #5
  6. = p(x,y)/[p(x)*p(y)]
    Read this as I((x;y)|z) not as I(x;(y|z)).
    That doesn't really make sense to me.
    = p(x,y|z)/[p(x|z)*p(y|z)]

    That makes sense to me.
    Read this as "the information provided about the event x by the
    occourrence of the event y, assuming that all the time z was already
    given".

    Cheers,
    Bastian
     
    Bastian Erdnuess, Jun 16, 2010
    #6
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