# Conditional probability

1. Sep 23, 2010

### mariank

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=$$\sqrt{q}Y+\sqrt{1-q}X-\Phi^{-1}(p).$$
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)=$$\int_{0}^\infty \int_{0}^\infty \phi_{zy}(z,y) dz dy/0.5$$ where $$\phi_{zy}(z,y)$$ is the joint normal birvariate distribution of z and y, where the covariance between z and y is$$\sqrt{q}.$$
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)=$$\int_0^\infty P(Z>0|Y=y) \phi(y) dy$$$$(P(Z>0|Y=y)=P(X>\phi^{-1}(p)-y\sqrt{q}/{\sqrt{1-q}}))$$....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. Sep 25, 2010

### SW VandeCarr

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
3. Sep 30, 2010

### SW VandeCarr

For the continuous case, I think you want:

$$f_{Z,Y}(z,y)=f_{Z|Y}(z|y)f_{Y}(y); Y>0, Z>0$$

Last edited: Sep 30, 2010