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Archiver > GENEALOGY-DNA > 2011-02 > 1297351712


From: "Sandy Paterson" <>
Subject: Re: [DNA] variance quiz game
Date: Thu, 10 Feb 2011 15:28:32 -0000
References: <002701cbc7a9$8c29ef30$c2482dae@Ken1> <000001cbc842$3e876b10$bb964130$@com><000c01cbc86c$2641c9e0$c2482dae@Ken1> <000301cbc875$0ed2dca0$2c7895e0$@com><00c901cbc891$ae5ec290$c2482dae@Ken1> <000001cbc895$b479baf0$1d6d30d0$@com><010c01cbc8a7$460557c0$c2482dae@Ken1> <000b01cbc8ff$76c21e40$64465ac0$@com><006a01cbc932$3fb47550$c2482dae@Ken1>
In-Reply-To: <006a01cbc932$3fb47550$c2482dae@Ken1>


> Or am I on a different planet? [[[ Maybe; I'm here on Klingon I'm talking

> about doing the clade/haplogroup TMRCA estimate using the N variances from

> each haplotype to the assumed founding haplotype. ]]


I think that explains it. I haven't posted anything on haplogroup TMRCA
estimation for probably a year now.

Everything I've posted in the past couple of weeks has been about pair-wise
TMRCA estimation. So yes, we're on different planets. I'm still interested
in haplogroup TMRCA estimation though, so I'll re-read your recent postings
in that light.


Sandy








-----Original Message-----
From:
[mailto:] On Behalf Of Ken Nordtvedt
Sent: 10 February 2011 14:53
To:
Subject: Re: [DNA] variance quiz game


----- Original Message -----
From: "Sandy Paterson" <>
To: <>
Sent: Thursday, February 10, 2011 1:49 AM
Subject: Re: [DNA] variance quiz game


> I'm with you up to the point where you consider the entire N(N-1)/2
> possible
> pair-wise comparisons.
>
> And yes, I can see that some of them will have a low (young) TMRCA, some
> will be middle of the road, some would be close to the group TMRCA and
> some
> would equal that of the group.
>
> But surely the weightings are obvious? The TMRCA of the pair judged to
> have
> the largest TMRCA must have a weighting of 1 with a weighting of 0 for all
> others.

[[[[ Up to half of the haplotype pairs could have the full clade TMRCA as
their TMRCA ]]]


: [[[ We are not understanding each other. The weightings would be for
individual (N of them) haplotypes. The symmetric correlation matrix will
have N^2 entries and need the N(N-1)/2 pairs of haplotypes .for its
determination. ]]]>

> Or am I on a different planet? [[[ Maybe; I'm here on Klingon I'm talking

> about doing the clade/haplogroup TMRCA estimate using the N variances from

> each haplotype to the assumed founding haplotype. ]]



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