Henry
Posts:
1,089
Registered:
12/6/04


Re: Conditional probability problem in the context of a simulation
Posted:
Jan 21, 2011 11:18 AM


On Jan 21, 3:44 pm, "danhey...@yahoo.com" <danhey...@yahoo.com> wrote: > On Jan 21, 8:45 am, Mike Lacy <mgl...@gmail.com> wrote: > > > > > > > On Jan 20, 3:42 pm, "danhey...@yahoo.com" <danhey...@yahoo.com> wrote: > > > > On Jan 20, 9:34 am, Mike Lacy <mgl...@gmail.com> wrote: > > > > > I'm trying to implement a simulation for a hypothesis testing problem > > > > in which there is more than one way that the null hypothesis might be > > > > true. The null hypothesis would be true if, say, either B = 0 or C = > > > > 0. Using "A" to represent the particular empirical evidence against > > > > the null that is of interest, the pvalue of interest would be > > > > something like: > > > > > Prob[A  (B=0 or C=0)] > > > > > I can easily enough set up a simulation that constrains B = 0 and > > > > simulate Prob(AB=0), or similarly Prob(AC=0), but I can't see anyway > > > > to represent the constraint (B=0 or C = 0). > > > > > So, I'm wondering, as purely a matter of conditional probability, if > > > > there is anyway to say anything about the value of > > > > Prob[A  (B=0 or C=0)] in terms of Prob(AB=0) and Prob(AC=0)? > > > > > I tried proceeding from Prob[A and (B or C)] / Prob(B or C), but > > > > didn't get anywhere. > > > > > I certainly can't see anything possible here, and I'd appreciate a > > > > confirmation of this or (better yet, of course :}) some kind of > > > > solution. > > > > > Regards, > > > > Mike Lacy > > > > If B and C are independent you don't have a problem, so I'll assume > > > they're dependent. Can you arrange for both B=0 and C=0? If so, use > > > P{B or C}=P{B}+P{C}P{BC}. > > > Thanks for your response. Yes, I can impose conditions that make B = 0 > > and C = 0 for the entire set of simulation repetitions. But B and C > > are actually population values imposed as conditions of the > > simulation, rather than random variables. So, one can structure the > > simulation (i.e., choose which variable is shuffled w.r.t. which other > > variable), so as to corresponds to a population value of B = 0 or C = > > 0 or B = C = 0. > > > Regards, > > Mike Lacy > > Ft. Collins CO > > I am now confused about what your problem is. I'm afraid I can't help. Hide quoted text  > >  Show quoted text 
Are you looking for something like this?
If A, B=0, C=0 are events with positive probabilities then
P(AB=0 or C=0) = P(A and (B=0 or C=0)) / P(B=0 or C=0) = [P(A and B=0) + P(A and C=0)  P(A and B=0 and C=0)] / [P(B=0) + P(C=0)  P(B=0 and C=0)] = [P(AB=0)*P(B=0) + P(AC=0)*P(C=0)  P(AB=0 and C=0)*P(B=0 and C=0) ] / [P(B=0) + P(C=0)  P(B=0 and C=0)]
If on the other hand B, C, BC=0, CB=0 are continuous random variables then you can ignore the possibility of B=0 and C=0 simultaneously and get
P(AB=0 or C=0) = [P(AB=0)*p_B(0) + P(AC=0)*p_C(0)] / [p_B(0) + p_C(0)]

