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From: "Ken Nordtvedt" <>
Subject: Re: [DNA] variance quiz game
Date: Wed, 9 Feb 2011 08:15:09 -0700
References: <002701cbc7a9$8c29ef30$c2482dae@Ken1><000001cbc842$3e876b10$bb964130$@com>
----- Original Message -----
From: "Sandy Paterson" <>
> I've forgotten the general form. Is it (for 67 markers)
>
> G = [ Var(1) + Var(2) + ....Var(67) ] / [ 67 x 2 x m ]
>
> where m is the mean of the mutation rates?
>
> Sandy
That's the text book formula, although "67 times mean m" should probably be
stated as "M = Sum of marker rates". Although the two quantities are
obviously identical I have always thought it a "crime" what the former did
to most all newcomers in sending them down the wrong conceptual path.
When the TMRCAs get to be sufficiently old so that some of the m(i)G values
are not much less than 1, the estimator for N markers should read, however:
G = [ Var(1) w(1) + ....... Var(N) w(N) ]
divided by [ m(1) w(1) + ....... m(N) w(N) ]
with weight factors w(i) = 1 / [ 1 + 4 m(i) G ]
This gives tighter sigmas to the G distribution.
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