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Conditional probability

  1. Sep 23, 2010 #1
    Dear all,

    I am in need of an advice regarding a conditional probability... Although it is very simple, I haven't found anywhere anything similar so I am not sure I get it right...I tried numerically calculating this conditional probability using 2 approaches (in Mathematica) and I get 2 completely different results, which are both very different from my intuition... I would be very happy if you could help me figure it out.

    Let X, Y be 2 standard normal r.v. and Z=[tex]\sqrt{q}Y+\sqrt{1-q}X-\Phi^{-1}(p). [/tex]
    s.t he unconditional probability of P(Z>0)=1-p

    What I would like to compute is P(Z>0|Y>0).

    here is what I have tried:

    1. P(Z>0|Y>0)=P(Z>0, Y>0)/P(Y>0)=[tex]\int_{0}^\infty \int_{0}^\infty \phi_{zy}(z,y) dz dy/0.5[/tex] where [tex]\phi_{zy}(z,y)[/tex] is the joint normal birvariate distribution of z and y, where the covariance between z and y is[tex] \sqrt{q}. [/tex]
    Weird enough, computing this numerically turns out that P(Z>0|Y>0) is at most 0.5(even when I put p=0.01, that means P(Z>0)=0.99) (which is very weird, right?)

    2. The second thing I was advised to do is to write P(Z>0|Y>0)=[tex]\int_0^\infty P(Z>0|Y=y) \phi(y) dy [/tex][tex](P(Z>0|Y=y)=P(X>\phi^{-1}(p)-y\sqrt{q}/{\sqrt{1-q}}))[/tex]....these results are even smaller than the previous one....

    So what is that I am doing wrong?

    I would very much appreciate your help. Thanks a lot,folks!!
     
    Last edited: Sep 23, 2010
  2. jcsd
  3. Sep 25, 2010 #2
    I don't understand this expression. The correct expression for Bayes Theorem is:

    P(Z>0|Y>0)=P(Y>0|P(Z>0)P(Z>0)/P(Y>0)
     
    Last edited: Sep 25, 2010
  4. Sep 30, 2010 #3
    For the continuous case, I think you want:

    [tex]f_{Z,Y}(z,y)=f_{Z|Y}(z|y)f_{Y}(y); Y>0, Z>0[/tex]
     
    Last edited: Sep 30, 2010
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