# Theory question: How to use Mean Absolute Error properly in a log scaled linear regression

First of all, I had a look here and in a couple of other questions: I couldn't find what I am looking for.

So my question is purely theoretical (although I have an example by my hands).

Suppose I have some data $(x_i,y_i)$ for $i=1,..,n$. Suppose I fit the following models with IID $\epsilon_i \sim N(0, \sigma^2)$ for $i=1,..,n$

• $M_1: \log(y_i)= \beta_0+\beta_1x_i+\epsilon_i$
• $M_2: \log(y_i)= \beta_0+\beta_1x_i+\beta_2x_i^2+\epsilon_i$
• $M_3: \log(y_i)= \beta_0+\beta_1x_i+\beta_2x_i^2+\beta_3x_i^3+\epsilon_i$

Now I want to see which of these models is better, so I use the following (maybe weird, but stay with me) method, to evaluate their "predictive powers":

1. Use $(x_i, \log(y_i))$ for $i=1,..,\frac{n}{2}$, to fit $M_1, M_2, M_3$ respectively.
2. Now use the fitted model (so $M_1, M_2,M_3$ respectively), to predict $y_i$'s using the $x_i$'s from the remaining $\frac{n}{2}$ data , so from $i = \frac{n}{2}+1, .., n$ (careful, predict $y_i$ not $\log(y_i)$)
3. Use MAE or Mean Absolute Error (here) $MAE = \frac{1}{\frac{n}{2}}\sum_{i=\frac{n}{2}+1}^{n}|y_i-\hat{y}_i|$, being careful that $\hat{y}_i$ is in the original scale of values!

So now my question:

If I do point $1.$ and I fit the three models (hence obtaining estimates for the parameters, their standard errors etc..) and then use these parameters (respectively of course!) to predict the responses of the other $x_i$'s:

1. Will I be predicting $\log(y_i)$'s right? And this is true... Is it also true that in order to get $\hat{y}_i$'s , instead of $\widehat{\log{(y)}}_i$, I should just take the exponential of those terms? So in general, is it true $\hat{y}_i = e^{\widehat{\log{(y)}}_i}$?
2. Once I find the three MAE's, how do I judge the models? Should I be looking for the one with smaller MAE?

For 1, the answer here really depends on what you want to predict. The value $\hat{y}_i$ is the thing you want to predict. I'll elucidate with an example: If you want to predict the mean value of the $y_i$s then what you've proposed with $\hat{y}_i = e^{\widehat{log(y_i)}}$ will be an under estimate of what you want (because of the convexity of the map $x \mapsto e^{x}$). However, because the exponential map is monotonic, the map preserves quantiles and using your normality distribution error assumption the prediction $e^{\widehat{log(y_i)}}$ will be a prediction of the median of $y_i | x_i$. To estimate the mean of $y_i \vert x_i$ you'll need an estimate of $\sigma$ because the mean of $y_i$ in your model will be $e^{\widehat{log(y_i)} +\hat{\sigma}^2/2}$. This follows from the Moment Generating Function (MGF) of a normal distribution and setting $t=1$ (the variable of the MGF).
One thing to keep in mind with the three models propose is that there is likely a substantial degree of colinearity because the covariates are polynomial functions of one another, e.g. $x_i$, $x_i^2$, and $x_i^3$.