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Archiver > GENEALOGY-DNA > 2008-04 > 1208095475


From: "Ken Nordtvedt" <>
Subject: Re: [DNA] underestimating variance
Date: Sun, 13 Apr 2008 08:04:35 -0600
References: <40037.65124.qm@web28303.mail.ukl.yahoo.com><007d01c8997b$2a314bc0$6400a8c0@Ken1><000001c89d57$56159630$0201a8c0@owner8151f88a9>


Yes, I have a program (if I saved it) to do this, and have done it often in
the past. I just run the difference equations

p(n,G+1) = -m p(n,G) + .5 m [ p(n+1,G) + p(n-1,G) ] with initial conditions
p(0,1)) = 1 and all other p(n,1) = 0

By the gut, your numbers look about right, but I'll see if I can find my
program.

Ken

----- Original Message -----
From: "Sandy Paterson" <>
To: <>
Sent: Sunday, April 13, 2008 5:12 AM
Subject: Re: [DNA] underestimating variance


> Hi Ken
>
>>
> But if many markers are used in the haplotype, the sum of the variances of
> each marker approach a nice symmetric Gaussian distribution about the sum
> of
> the expected values --- by the central limit theorem.
>>
>
> Using some Heath Robinson-esque maths I think I've found a way of
> determining the expected distribution of repeats about the start value.
>
> Here's an example, using a mutation rate of .002 and 1000 generations.
>
> +7 .0000
> +6 .0002
> +5 .0013
> +4 .0069
> +3 .0288
> +2 .0933
> +1 .2153
> +0 .3084
> -1 .2153
> -2 .0933
> -3 .0288
> -4 .0069
> -5 .0013
> -6 .0002
> -7 .0000
>
> 1.0000
>
> So .3084 should be modal, .2153 should be modal +-1, and so on.
>
> Are you able to simulate and compare? (I'm not that confident that my
> logic
> is correct).
>
>
> Sandy Paterson
>
>
>
>
>
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