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I am being asked to produce an equation for some characteristic $y$ based on measurements provided for explanatory variables $x_1,x_2,x_3,x_4$. The first thing I did was (naively) fit the linear regression, $y=\beta_0+\beta_1x_1 +\beta_2x_2 +\beta_3x_3 +\beta_4x_4 + \varepsilon$. I got the following results:

$$ \begin{array}{|l|c|} \hline & y \\ \hline x_1 & 0.227^{**} \\ & (0.100) \\ x_2 & 0.554 \\ & (0.370) \\ x_3 & -0.150^{***} \\ & (0.029) \\ x_4 & 0.155^{***} \\ & (0.006) \\ \text{const.} & -6.821 \\ & (10.123) \\ \hline \text{Observations} & 32 \\ \text{R}^{2} & 0.962 \\ \text{Adjusted R}^{2} & 0.957 \\ \text{Residual Std. Error} & 2.234 (\mathrm{df} = 27) \\ \text{F Statistic} & 171.713^{***} (\mathrm{df} = 4; 27) \\ \hline \hline \end{array} $$

Overall decent results. But when I check the fitted values against residuals I get the quadratic-ish relation.

enter image description here

Since the residuals are not randomly distributed and the trend reminds some quadratic function, it means that the way I assumed linearity at the very beginning was not actually correct.

In the previous example, changing $y$ to $log(y)$ solved the problem. However, here when I tried a similar approach I ended up with a switched direction of the curvature.

enter image description here

My question is what else can be done in order to find the "correct" non-linear model? Should I try squaring and logging each term to see how it affects the model? Or is there a more concise way of doing so? What should I do if I cannot find a combination that would solve this problem?


EDIT.

I have added the matrix of scatterplots so to show the relationship between each of the variables. It is hard for me to observe anything specific, to be honest. The only thing that is somehow transparent to me is the positive relation between $y$ and $x_4$. What I also find quite interesting is the relation between $y$ and $x_1$ but not entirely sure what that tells me.

enter image description here

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    $\begingroup$ What is the purpose of the model ? Prediction or inference ? The first plot in the question does not strike me as an obvious case of nonlinearity. What do you know about $y$ and the $x$ variables ? How are they all related ? $\endgroup$ Feb 14, 2021 at 20:44
  • $\begingroup$ The exercise states that I should provide the equation for predicting $y$ from measurements of explanatory variables $x$. But I don't think about it as in building a prediction model in machine learning class. I think it's more like: "based on data on some physics phenomenon find the equation that would describe it". The predicting bit then would simply mean: "what would happen when you input for $x$ some values, what would then the $y$ be?". Perhaps the question is a bit ambiguous. $\endgroup$
    – bajun65537
    Feb 15, 2021 at 10:03
  • $\begingroup$ I have edited my question and added the matrix of scatterplots so that we can see how the $y$ and $x$ variables are related. $\endgroup$
    – bajun65537
    Feb 15, 2021 at 10:08

1 Answer 1

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It is important not to over-interpret these plots. The first plot of residuals vs fitted values is a little misleading in my opinion if you only focus on the red line, partly due to the fairly small sample size.

enter image description here

Yes, the red line has a curved shape, but looking at the data points, it is not clear at all that there is nonlinearity. This type of pattern can easily occur simply through random variation. We would like the line to be perfectly horizontal, but in practice this will never happen.

Here is a plot from a very simple simulation of a linear model:

set.seed(2)
X <- rnorm(30)
Y <- 10 + X + rnorm(30)

plot(lm(Y~X))

enter image description here

As you can see, even when sampling from a normal distribution with a linear relationship, the plot of residuals vs fitted values can still appear nonlinear if we only focus on the red line.

If you have reason to believe that all the $x$ variables are related to $y$ and there is no interdependence among the $x$s (as it appears from the corelation plots), then I would stick with the fist model.

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