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Topic: Conditional probability problem in the context of a simulation
Replies: 10   Last Post: Jan 27, 2011 5:01 AM

 Messages: [ Previous | Next ]
 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 p-value 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(A|B=0), or similarly Prob(A|C=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(A|B=0) and Prob(A|C=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(A|B=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(A|B=0)*P(B=0) + P(A|C=0)*P(C=0) -
P(A|B=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, B|C=0, C|B=0 are continuous random
variables then you can ignore the possibility of B=0 and C=0
simultaneously and get

P(A|B=0 or C=0)
= [P(A|B=0)*p_B(0) + P(A|C=0)*p_C(0)] / [p_B(0) + p_C(0)]

Date Subject Author
1/20/11 Mike Lacy
1/20/11 Dan Heyman
1/21/11 Mike Lacy
1/21/11 Dan Heyman
1/21/11 Henry
1/25/11 Mike Lacy
1/25/11 Paul
1/25/11 Ray Koopman
1/27/11 Richard Ulrich
1/26/11 Luis A. Afonso
1/27/11 Luis A. Afonso