I have been working on linear regression for forecasting purposes.

I have a model where I have used a BoxCox transformation on the $Y$ variable (Sales) with $\lambda = 0.3$. The model is as follows:

$$Y = 13 - 0.2 X_1 + 3 X_2$$

How can I interpret the effect of sales from each variable, given the transformation?

For instance, assume $X_1=5$ and $X_2=0.3$ then we get:

$$Y = 13 - 0.2 \times 5 + 3 \times 0.3 = 12.9$$

Reverse BoxCox of 12.9 with $\lambda = 0.3$ is 196. So it forecasts 196 sales. But if we increase $X_2$ from 0.3 to 1 and increase $X_1$ from 5 to 10 then we get:

$$Y = 13 - 0.2 \times 10 + 3 \times 1 = 14$$

Reverse BoxCox then gives 244 sales forecasted sales. An increase of 48 sales. But how much of that is down to $X_1$ and $X_2$?

If we calculate them separately the answer doesn't add up!

EDIT: to make the above point clear:

If $X_1$ went from 5 to 10 but $X_2$ doesn't change:

$Y = 13 - 0.2 \times 10 + 3 \times 0.3 = 11.9$, transform to 159 sales. A -37 decrease from the starting position (196-159)

However, if $X_2$ went from 0.3 to 1 but $X_1$ doesn't change: $Y = 13 - 0.2 \times 5 + 3 \times 1 = 15$, transforms to 294 sales. A +98 increase in sales (294-196).

But if $X_2$ increases sales by 98 and $X_1$ decreases sales by 37 you would expect the net increase to be +61 when you include both changes. But from earlier we found that by applying both we actually get a +48 increase in sales!

How can we work out the actual sales change resulting in changes in $X_1$ and $X_2$?


1 Answer 1


By using BoxCox transformation on the Y variable, you are turning your model from linear regression into non-linear regression. Note that although it seems you are training a linear regression, it’s only linear with respect to the transformed version of Y but non-linear regarding the original Y. Therefore you can not expect a linear behaviour from a non-linear model which means you can not work out the actual sales change resulting in changes in X1 and X2. That’s because unlike linear models, in non-linear models f(x+b) is not equal to f(x) + f(b)


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