5.3.3 - Marginal Independence

5.3.3 - Marginal Independence

Marginal independence has already been discussed before. Variables \(X\) and \(Y\) are marginally independent if:

\(\pi_{ij+}=\pi_{i++}\pi_{+j+}\)

They are marginally independent if they are independent within the marginal table. In other words, we consider the relationship between \(X\) and \(Y\) only and completely ignore \(Z\). Controlling for, or adjusting for different levels of \(Z\) would involve looking at the partial tables.

We saw this already with the Berkeley admission example. Recall from the marginal table between sex and admission status, where the estimated odds-ratio was 1.84. We also had significant evidence that the corresponding odds-ratio in the population was different from 1, which indicates a marginal relationship between sex and admission status.

 Question: How would you test the model of marginal independence between scouting and SES status in the boy-scout example? See the files boys.sas (boys.lst) or boys.R (boys.out) to answer this question. Hint: look for the chi-square statistic \(X^2=172.2\).

Recall, that joint independence implies marginal, but not the other way around. For example, if the model (\(XY\), \(Z\)) holds, it will imply \(X\) independent of \(Z\), and \(Y\) independent of \(Z\). But if \(X\) is independent of \(Z\) , and \(Y\) is independent of \(Z\), this will NOT necessarily imply that \(X\) and \(Y\) are jointly independent of \(Z\).

 Stop and Think!
Can you see why this is the case? Can you construct a simple example?

If \(X\) is independent of \(Z\) and \(Y\) is independent of \(Z\) we can’t tell if there is an association between \(X\) and \(Y\) or not; it could go either way, that is in our graphical representation there could be a link between \(X\) and \(Y\) but there may not be one either.

For example, let’s consider a \(2\times2\times2\) table.

The following \(XZ\) table has \(X\) independent of \(Z\) because each cell count is equal to the product of the corresponding margins divided by the total (e.g., \(n_{11}=2=8(5/20)\), or equivalently, OR = 1)

  C=1 C=2 Total
A=1 2 6 8
A=2 3 9 12
Total 5 15 20

The following \(YZ\) table has \(Y\) independent of \(Z\) because each cell count is equal to the product of the corresponding margins divided by the total (e.g., \(n_{11}=2=8(5/20)\)).

 

C=1

C=2

Total

B=1

2

6

8

B=2

3

9

12

Total

5

15

20

The following \(XY\) table is consistent with the above two tables, but here \(X\) and \(Y\) are NOT independent because all cell counts are not equal to the product of the corresponding margins divided by the total (e.g., \(n_{11}=2\)), whereas \(8(8/20)=16/5\)). Or you can see this via the odds ratio, which is not equal to 1.

  B=1 B=2 Total
A=1 2 6 8
A=2 6 6 12
Total 8 12 20

Furthermore, the following \(XY\times Z\) table is consistent with the above tables but here \(X\) and \(Y\) are NOT jointly independent of \(Z\) because all cell counts are not equal to the product of the corresponding margins divided by the total (e.g., \(n_{11}=1\), whereas \(2(5/20)=0.5\)).

  C=1 C=2 Total
A=1, B=1 1 1 2
A=1, B=2 1 5 6
A=2, B=1 1 5 6
A=2, B=2 2 4 6
Total 5 15 20
So this example shows, that we could have (1) \(X\) independent of \(Z\), and (2) \(Y\) independent of \(Z\), but (3) \(X\) and \(Y\) are not jointly independent of \(Z\).

Other marginal independence models in a three-way table would be, \((X,Y)\) while ignoring \(Z\), and \((Y,Z)\), ignoring any information on \(X\).


Legend
[1]Link
Has Tooltip/Popover
 Toggleable Visibility