Re: Bayes problem - error?
From: Rob Johnson (rob_at_trash.whim.org)
Date: 09/16/04
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Date: Thu, 16 Sep 2004 14:22:48 +0000 (UTC)
In article <5133a979.0409142324.55509c42@posting.google.com>,
ikaterina@yoak.com (Kate Yoak) wrote:
>I am working to solve a problem surrounding an application of the
>bayes theorem. After many hours, I still cannot come up with the
>correct solution. I wonder if there is something I am not seeing. I
>have checked all the calculations very carefully with mathematica at
>every step. The error has to lie somewhere with probability
>manipulations, not plugging in the algebra.
>
>This is a little long... Thank you in advance if you choose to take
>the time to go through it.
>
>Brief Problem Summary:
>Regular blue eye/brown eye setup [blue: xx; heterozygote Xx or xX):
>
>blue population= p^2.
>heterozygote population=2p(1-p)
>
>0<p<1
>
>Assuming random mating, show that among brown-eyed children of
>brown-eyed parents, the expected proportion of heterozygotes is
>2p/(1+2p).
>
>Solution:
>
>The quantity we are trying to find is P(HC|BC,BP)
>I use HC to mean Hetero Children, BC Brown Children, BP brown parents,
>HP hetero parents, NHP non-hetero(but brown) parents, XX
>
>Using the following corollary to bayes theorem:
>
>P(A,B,C)=P(A|B,C)*P(B|C)*P(C) =>
>P(A|B,C)=P(A|C)/P(B|C) *P(B|A,C)
>
>We get P(HC|BC,BP)=P(BC|HC,BP)*P(HC|BP)/P(BC|BP)
>P(BC|HC,BP)=1 (all hetero children are brown)
>
>So P(HC|BC,BP)=P(HC|BP)/P(BC|BP)
>
>To get the numerator:
>
>P(HC|BP) is more completely expressed as P(HC|2BP) (two brown parents)
>P(HC|2BP)=P(HC|2HP)*P(2HP|2BP)+P(HC|1HP,1NHP); (2NHP will produce no
>HC)
>
>P(HC|2HP)=P(HC|1HP,1NHP)=1/2
>If you have Xx + Xx the probability of Xx is 1/2.
>XX+Xx => same 1/2
>
>P(2HP|2BP)=P(2HP,2BP)/P(2BP) (directly from bayes formula)
> = P(2HP)/P(2BP)=hetero^2/brown^2 (population-wide proportion)
>
>similarly
>
>P(1HP,1NHP|2BP)=P(1HP,1NHP)/P(2BP)=(brown-hetero)*hetero/brown^2.
>
>Going back to the numerator:
>P(HC|2BP)=1/2*(hetero^2/brown^2+(brown-hetero)*hetero/brown^2)=
> 1/(2brown^2)*brown*hetero=hetero/(2brown)
>
>The denominator can be simplified as follows:
>
>P(BC|2BP)=1-P(bc|2BP) (bc=blue child)
> = 1-P(bc|2HP)*P(2HP|2BP) (only 2 hetero parents can produce a blue
>child)
> = 1-1/4* hetero^2/brown^2 (from above)
> = (4 brown^2-hetero^2) /4brown^2.
>Putting it all together:
>P(HC|BC,BP)=[hetero/(2brown)]/[(4 brown^2-hetero^2) /4brown^2]=
> = [hetero * 2brown/(4 brown^2-hetero^2)
>
>Plugging in: brown=1-p^2, hetero=2p(1-p)
>
> = p(1+p)/(1+2p) (this I've checked with mathematica, and am certain.
>
>Close.. but not quite!
The probability that one is brown-eyed and heterozygotic is 2p(1-p)
The probability that one is brown-eyed and monozygotic is (1-p)^2
Parents\Child -> Heterozygote Monozygote
-------------
Heterozygote 4(p(1-p))^2 x 1/2 or 1/4
Mixed 4p(1-p)^3 x 1/2 or 1/2
Monozygote (1-p)^4 x 0 or 1
The probability that the parents are brown-eyed and the child is
brown-eyed is
3/4 4(p(1-p))^2 + 1 4p(1-p)^3 + 1 (1-p)^4 = (1-p)^2 (1+2p)
The probability that the parents are brown-eyed and the child is
heterozygotic is
1/2 4(p(1-p))^2 + 1/2 4p(1-p)^3 + 0 (1-p)^4 = (1-p)^2 2p
Thus, given that the parents are brown-eyed and the child is brown-eyed,
the probability that the child is heterozygotic is 2p/(1+2p).
Rob Johnson <rob@trash.whim.org>
take out the trash before replying
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