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AMOVA

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Created
3/5/2021, 1:17:00 PM
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Analysis

Analysis of Molecular Variance (AMOVA)

Law of total variance

Suppose that X and Y are random variables on the same probability space, and the variance of Y is finite. Then,
Var(Y)=E[Var(YX)]+Var(E[YX])Var(Y) = E[Var(Y|X)] + Var(E[Y|X])
Intuitively, we can think of a bivariate Gaussian distribution P(X, Y):
In this case, we can get a distribution P(YX=Xi)P(Y|X=X_i), as in the figure below:
It intuitively makes sense that Var(Y)Var(Y) should be the average of all their individual variance, i.e. Var(Y)=E[Var(YX)]Var(Y) = E[Var(Y|X)].
Now, let's see what happens if we rotate the bivariate Gaussian distribution:
We can see that Var(Y) doesn't only depend on the individual variances of the P(YX=Xi)P(Y|X=X_i) distribution, but that it also depends on how spread out the distributions themselves are along the Y axis. That is to say, it depends on how spread out P(YX=Xi)P(Y|X=X_i) are from each mean of that E[YX=Xi]E[Y|X=X_i], which is equivalent to Var(E[YX=Xi])Var(E[Y|X=X_i]).
As an aside, there is a general variance decomposition formula for one or more components:
Var[Y]=E[Var(YX1,X2)]+E[Var(E[YX1,X2]X1)]+Var(E[YX1])Var[Y] = E[Var(Y|X_1, X_2)] + E[Var(E[Y|X_1, X_2]|X_1)] + Var(E[Y|X_1])
which follows from the law of total conditional variance:
Var(YX1)=E[Var(YX1,X2)X1]+Var(E[YX1,X2]X1).Var(Y|X_1) = E[Var(Y|X_1, X_2)|X_1] + Var(E[Y|X_1, X_2] | X_1).
Table of AMOVA result
Search
.
Df
Sum Sq
Mean Sq
Between samples Within pop
Open
226
3469268.633
15350.74616
Within samples
Open
229
10709
46.76419214
Total
Open
457
5375347.707
11762.24881
Between pop: 각 population 평균의 분산, Var(E[YX])Var(E[Y|X])
Between samples within pop: 각 population 내 sample의 분산, Var(E[YX])Var(E[Y|X])
Within samples: Diploid인 경우 생성됨. Allele에 따른 sample 내 분산(?)
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