I am trying to find a predictive model for my dataset with the following properties: I have 52 predictors, 1 response and 53 records. The predictors differ in scaling and type: 7 of them are categorical variables and the rest is numeric (with discrete and continuous values with different range).

My first Idea was to use a Random Forest because it can handle both with categorical and continuous data and is useful for datasets with more variables than records.

The result of the fit of the learning data is a 0.85 R^2. But consider the plot of the response variable against the modelled data:

enter image description here

I think the model overrates the little values of the response and overrates the bigger ones.

My questions:

1) How should I deal with this under-/overrating? What could be the reason for this and how could I improve the result?

2) Which other models can I use (I don't know much models which can deal with the described properties of my data)?

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    $\begingroup$ This is a common problem of random forests. It often happens that random forests have problems predicting the tails of the outcome's distribution. I've read that one way to improve upon this is to perform a linear regression on each node instead of averaging, though I wasn't able to find any implementation. I don't recall where I read this. If I find the source I will post it here. $\endgroup$
    – George
    Oct 7, 2015 at 12:30
  • $\begingroup$ Actually that RF predictions "flattens out" / "gets a smaller slope", when the model regression is not perfect, is a very sensible 'Baysian-kinda' property. Think of your mean response as your prior. The RF model will effectively incorporate the uncertainty in its predictions, such that these will move closer to the mean response. The same is true for classification. Only when the regressor or classifier is performing 100%, will the predictions not be effected by the prior. $\endgroup$ Oct 7, 2015 at 14:01
  • $\begingroup$ I agree with George and DJohnson also. You may be able to tune your model with variable filtering, I would personally favor highest absolute spearman correlation or variable importance. You need to wrap the entire process in a e.g. 10-fold cross validation to assess over fitting. $\endgroup$ Oct 7, 2015 at 14:07
  • $\begingroup$ Can I also use the OOB estimates for variable importance to reduce dimension? $\endgroup$
    – Fabi_92
    Oct 8, 2015 at 8:11

1 Answer 1


There are several components to your question. These include (but are not limited to): 1) constrained variable selection when the number of observations (n) is small relative to the number of predictors (p), 2) heuristic selection vs optimization, 3) dealing with mixtures of distributions among the predictors, 4) comparison of the fit between predicted and actual, 5) finding an appropriate model for y, and 6) separating statistical understanding from pure, machine learning prediction.

I'm not an advocate of optimizing approaches to variable selection as wasteful of CPU. Moreover, given the smallish n and p which is not so big anyway, I think leveraging an RF would be methodological overkill. A more useful model-building step (assuming some exploratory work has been done to assess whether or not applying transformations improves the fit) would be heuristic evaluation of pair-wise relationships between y and the candidate predictors. The idea here is that if a potential predictor does not have at least a modestly significant relationship with y, then it probably can be eliminated. This could be done in an ANOVA-type context using a relaxed significance of p<=.15 or so for inclusion. Of course, causal purists would argue that tertiary (masking) and/or interaction effects can be lost this way but, in practice, these tertiary effects are usually small if they are significant at all. Besides, a better guide to including tertiary effects is prior theoretical insight. The advantage of using ANOVA is that it is invariant to the scale (mixture) of distributions, providing a measure (the F-statistic) of relative effect sizes leading to an preliminary importance ranking in selecting predictors. Of course, ANOVAs use linear assumptions. If you think the relationships are nonlinear then there are many tools now for evaluating nonlinear dependence but these require a level of sophistication that your question belies.

The under/over weighting you point out in the scatterplot is a bit of a red herring since it's benchmarked against an orthogonal, 45-degree line. The better comparison would be to an average line of best fit, which would be demonstrably balanced.

In terms of appropriate models for your data, I don't see any reason why classic OLS estimation wouldn't provide reasonable insight. Of course, there are other methods such as partial least squares which are designed specifically for situations where p>>n but your mixture of distributions precludes their use.

You haven't indicated what your "high-level" goal is. Are you simply trying to find a good, predictive fit or are you trying to uncover some underlying process in order to gain insight into causality? Either way, by choosing predictors that maximize the predictive fit, you have put a stake in the ground in terms of understanding causality.

Final model variable selection would be based on those variables that passed the threshold of relaxed significance and could be identified using the lasso, a widely available variable selection method.

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    $\begingroup$ Why is comparing $y$ and $\hat y$ against diagonal line is a "red herring" is not clear to me. $\endgroup$
    – amoeba
    Oct 7, 2015 at 12:22
  • $\begingroup$ I merely suggested that it was a "bit" of a red herring, particularly when compared with the actual line of best fit. $\endgroup$
    – user78229
    Oct 7, 2015 at 12:29
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    $\begingroup$ I just don't understand the logic here. If $\hat y$ is always $y/2-1000$, then a linear fit would be perfect with R^2=1, but obviously prediction in this situation is strongly biased. I don't see how "best fit" is relevant at all. $\endgroup$
    – amoeba
    Oct 7, 2015 at 12:30
  • $\begingroup$ Are you questioning whether or not the line of best fit in this case would not be "balanced" wrt under-/over-fitting? $\endgroup$
    – user78229
    Oct 7, 2015 at 12:39
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    $\begingroup$ Of course not, "best fit" is always "balanced". I am saying that this is irrelevant for this discussion. $\endgroup$
    – amoeba
    Oct 7, 2015 at 12:40

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