# 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 $$X$$.

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