I am looking for some help on the decomposition of effects. I am forming a ratio (R) of a to b at t=1 and t=0. I want to express the change in this ratio (Delta R) as the sum of a "contribution of a" and a "contribution of b" component. These two components should be additive in nature.
Here is a numeric example:
t=1 t=0
a 200 150
b 175 100
R 1.14 1.5
Delta R -0.36 = "contribution of a" + "contribution of b"
Is it possible to solve this without an approximation? If not, which would be the easiest approximation to solve these calculations irrespective of magnitude?
Many thanks!
For the exact change in the ratio we have, $$\Delta\left(\frac ab\right) = {a+\Delta a \over b+\Delta b} - \frac ab = {b \Delta a - a \Delta b \over b(b-\Delta b)}.$$ Although you can get a term that only contains $\Delta b$ from this, there’s no way that I can see to get one that involves only $\Delta a$.
For small values of $\Delta a$ and $\Delta b$, the change in the ratio can be approximated by the differential: $$D\left(\frac ab\right)[\Delta a,\Delta b] = \left(\frac1b,-\frac a{b^2}\right)\cdot\left(\Delta a,\Delta b\right) = {b\Delta a-a\Delta b\over b^2}.$$ For your example, this approximation predicts a change in the ratio of $0.625$, almost double the actual change, so is unfortunately not likely to be of use to you.
A possible way to separate the effects of $\Delta a$ and $\Delta b$ is to look at the change in the logarithm of the ratio instead: $$\ln\left({a+\Delta a \over b+\Delta b}\right)-\ln\left(\frac ab\right) = \ln\left(1+\frac{\Delta a}a\right)-\ln\left(1+\frac{\Delta b}b\right).$$