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I A conditional probability

  1. Apr 25, 2016 #1
    Is it possible to solve something like this generally or does it depend on the pdf's of the variables?

    P(x < f(y) | x > -f(y))
     
  2. jcsd
  3. Apr 25, 2016 #2

    mathman

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    Your expression can be given as [itex]\frac{P(-f(y)<x<f(y))}{P(-f(y)<x<\infty)}[/itex].
     
  4. Apr 26, 2016 #3
    As a step in using the CDF method for a random variable as a function of X where i have X_PDF, I came from:

    P(x > -f(y) AND x < f(y)) =
    P(x > -f(y)) * P(x < f(y) | x > -f(y))

    The aim is to convert the P()'s to X_CDF()'s.

    Your answer led me back a step, which made me think that maybe P(x > -f(y) AND x < f(y)) might be expressed like (1-X_CDF(-f(y))) - X_CDF(f(y))

    It seems to be correct for my case, so thank you :)
     
  5. Apr 26, 2016 #4
    By the way..

    In the CDF method, I understand that I need to reformulate expressions to get something like P(X < y) which equals X_CDF(y) or P(X > y) which equals (1-X_CDF(y)), since I know the expression of X_PDF(x) = X_CDF'(x).

    What if I have P(A + B < y), knowing A_PDF(a) and B_PDF(b)?
    Would that require that I know AplusB_PDF(a,b) and some transformation from y to a and y to b?
     
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