Bayes problem - error?

From: Kate Yoak (ikaterina_at_yoak.com)
Date: 09/15/04 

Date: 15 Sep 2004 00:24:22 -0700

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!

Relevant Pages

  • Re: Bayes problem - error?
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  • Re: Bayes problem - error?
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