# Can absolute or relative contributions from X be calculated for a multiplicative model? $\log{ y}$ ~ $\log {x_1} + \log{x_2}$

(How) can absolute or relative contributions be calculated for a multiplicative (log-log) model?

### Relative contributions from a linear (additive) model

E.g., there are 3 contributors to $$y$$ (given by the three additive terms):

$$y = \beta_1 x_{1} + \beta_2 x_{2} + \alpha$$

In this case, I would interpret the absolute contribution of $$x_1$$ to $$y$$ to be $$\beta_1 x_{1}$$, and the relative contribution of $$x_1$$ to $$y$$ to be:

$$\frac{\beta_1 x_{1}}{y}$$

(assuming everything is positive)

### Relative contributions from a log-log (multiplicative) model

In log-space, a model could take the following form:

$$\log{y} = \beta_1 \log{x_{1}} + \beta_2 \log{x_{2}} + \alpha$$

which in the 'real-world', assumes the following form:

$$y = e^\alpha x_{1}^{\beta_1}x_{2}^{\beta_2}$$

But (how) can absolute or relative contributions be calculated from such a multiplicative (log-log) model? i.e., (how) can we calculate how much $$x_1$$ contributes to $$y$$? For example if $$e^\alpha=10$$, and $$x_1^{\beta_1} = 100$$ and $$x_2^{\beta_2} = 1000$$, then $$y = 10^6$$, but what portion of that 1,000,000 was contributed to by $$x_1$$?

One way I can think of is the following (beware that relative importance can mean different things and be counted in different ways):

The relative contribution of log-log model, seen as a linear model, is found by setting everything else to zero and taking the ratio:

Sp:

$$1/r_{x_1} = \frac{\beta_1 \log x_{1}}{\log y}$$

or:

$$\log y \approx r_{x_1} \beta_1 \log x_1$$

so the relative contribution of the multiplicative model can be derived from:

$$y \approx x_1^{r_{x_1} \beta_1}$$

$$m_{x_1} = x_1^{r_{x_1} \beta_1}/y$$

One can even use the original variable and its exponent as:

$$m_{x_1} = x_1^{\beta_1}/y$$

Still another way to quantify importance for multiplicative models (inspired by geometric means) is:

Given $$y = ax_1^{\beta_1}x_2^{\beta_2} \cdots x_n^{\beta_n}$$, then:

$$m_{x_k} = \frac{x_k}{\sqrt[\beta_1 + \beta_2 \cdots + \beta_n]{y}}$$

So if $$y = x^2$$ then $$m_x = 1$$ since $$x$$ has $$100$$% importance on $$y$$.

Is there a standard way to make the relative contributions add to unity? Or alternatively, to make the absolute contributions to add to y?

When one wants to determine the percent a certain variable affects another variable, one uses a ratio between the two variables. Only for linear additive models do the percents add up to the original value. For multiplicative models these two properties of relative importance diverge and require different approaches (what I explicitly mentioned in the beginning).

So a percent of $$x_k$$ over $$y$$ is always some ratio like $$\frac{x_k}{y}$$ on the other hand for multiplicative models these ratios no longer can be added and don't add up. So one will have to modify their requirements, for example instead of adding up to original value, maybe the product equals the original value.

In this sense one can do a variation of the above proposals (eg the geometric mean inspired formula) and use:

$$y = x_1^{\beta_1} \cdots x_n^{\beta_n}$$

$$m_{x_k} = \frac{x_k^{\beta_k}}{y^{\frac{\beta_k}{\beta_1+\cdots+\beta_n}}}$$

With this definition of relative importance one still uses a ratio (so percent of importance of some value on some other value) while at the same time the product of all $$m_{x_k}$$ equals $$1$$, or:

$$m_{x_1} \cdots m_{x_n} = 1$$

Some references which follow another route:

Measures of relative importance:

• Hey @Nikos, thanks for taking the time to answer. I'm still unsure whether the $r$ or $m$ values should be used. In my example of $y = 10 .10^2 .10^3$, what contributions from $\alpha$, $x_1$, and $x_2$ would you come up with? Is there a standard way to make the relative contributions add to unity? Or alternatively, to make the absolute contributions to add to $y$? – Ben Mar 23 at 9:25
• see updated answer, in summary the 2 requirements you ask, coincide only for linear additive models. For multiplicative models these 2 requirements are not represented by the same formula and one has to choose which route to follow – Nikos M. Mar 24 at 9:04