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Conditional Probability

  1. Oct 27, 2014 #1
    Basically I am wondering how you deal with a conditional cdf and turning that into a conditional pdf when the random variables are independent. I know that f(X|Y) =f(X)f(Y)/f(Y)=f(X)

    I tried to derive this in a nice attached laTex document but it does not seem right to me.

    Note(this is for a homework problem but this is only a derivation I am trying to use to solve it so I decided to post it here because it is not a textbook problem)
     

    Attached Files:

  2. jcsd
  3. Oct 28, 2014 #2

    RUber

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    Homework Helper

    I have only seen this explained as
    ##f(X|Y) = \frac{f(X \cap Y)}{f(Y)} ## where ##f(X \cap Y) = f(X)f(Y)## by the definition of independence.
    In your work, it seems like in part (6) you were taking the integral with respect to y, where you should be considering a fixed y and taking the integral with respect to x.
    I have not put pen to paper, but it looks like that could get you something in a more recognizable form.
     
  4. Oct 29, 2014 #3

    Stephen Tashi

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    The inequality [itex] g(x,y) < z [/itex] can't necessarily be rewritten in the form [itex] x < h(y,z) [/itex].

    For example, the solution [itex] x^2 + y < z [/itex] might require that [itex] x [/itex] be in an interval of the form [itex] -a < x < a [/itex] rather than in an interval of the form [itex] x < a [/itex].
     
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