# Is there value in assigning a dimension-specific weighting when calculating the main effect?

I've been looking into factorial designs and noticed an interesting property. Is this simply a mathematical trick or can we attribute some 'meaning' to it?

Say we have a $$2^k$$ design where k = 3 here. This gives the following possible combinations (the inclusion of a variable represents the high value, and the omission the low value):

$$f(a,b,c), f(a, b), f(a,c), f(b,c), f(a), f(b), f(c), f()$$

To get the main effect of $$a$$, you would find the difference between the average of values with $$a$$ and without:

$$f(a,b,c)-f(b,c)$$ [dimensions = 3]

$$f(a,b)-f(b)$$ [dimensions = 2]

$$f(a,c)-f(c)$$ [dimensions = 2]

$$f(a)-f()$$ [dimensions = 1]

When you express the dependent variable $$z = f(...)$$ as a function of the independent variables $$a,b,c$$ we could think of a plot in 4 dimensions where we are calculating the average length of the projection in the z direction.

The number of projections we have in each dimension is well known (pascal's triangle). If we assigned the inverse of this as the weighting for each dimension we get the 'weighted Main effect' $$WM$$ (for want of a better name).

$$WM_a = \frac{1}{3}[(f(a,b,c) - f(b, c)) + \frac{1}{2} ((f(a,b) - f(b)) + (f(a,c) - f(c))) + (f(a) - f())]$$

It is the average of the average length in each dimension. To me this intuitively makes some sense if you want each dimension to have an equal weighting and when you look at the result of A+B (see comment). (And some similarities with a discrete finite taylor expansion).

If we sum up the 'weighted Main effect' of $$a,b,c$$ we find we are left with f(a,b,c). This is true for k of any value.

It is pleasing to me that this sums up - is this just a mathematical trick or could we attribute some meaning to this value?

• What I also find nice about this is if we look at $\frac{WM_a}{2}$ + $\frac{WM_b}{2}$: $\frac{1}{6}[f(a,b,c) + (f(a,b,c)-f(b, c)) + \frac{1}{2}((f(a) -f(b,c)) + (f(b) - f(a,c))) + (f(a,b) - f()) -f()]$We see that all terms containing c are subtracted. – fiz_buz May 7 '20 at 21:39
• Which intuitively makes sense if you were estimating the effect AB. – fiz_buz May 7 '20 at 21:45