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According to the Rob Hyndman's book on forecasting, "section 5.3 - Selecting predictors", when selecting predictors in a regression model:

Adjusted $R^2$

Computer output for regression will always give the $R^2$ value, discussed in Section 5/1. However, it is not a good measure of the predictive ability of a model. Imagine a model which produces forecasts that are exactly 20% of the actual values. In that case, the $R^2$ value would be 1 (indicating perfect correlation), but the forecasts are not very close to the actual values.

In addition, $R^2$ does not allow for "degrees of freedom''. Adding any variable tends to increase the value of $R^2$ even if that variable is irrelevant. For these reasons, forecasters should not use $R^2$ to determine whether a model will give good predictions.

An equivalent idea is to select the model which gives the minimum sum of squared errors (SSE), given by

$SSE = \sum e_i^2$ , (with $e_i = y_i - \hat{y_i}$ )

Minimizing the SSE is equivalent to maximizing $R^2$ and will always choose the model with the most variables, and so is not a valid way of selecting predictors.

Now based on the wikipedia page on $R^2$:

$R^2 = 1 - \frac{SSE}{SST}$ , with $SST= \sum (y_i - \bar{y})^2$

My questions:

  1. I can't understand, based on the the definitions of $R^2$ and $SSE$, why $R^2$ would always increase when the number of variables in the model increases? and why does the $SSE$ would also decrease? Especially if this happens as he says even when the variable is irrelevant?

  2. Does this relation between SSE and number of variables hold only for linear models?

  3. We use the $SSE$ for training ML models in general all the time, but from the above quoted reasons, that would be a bad idea. Is this conclusion (maximizing $R^2$/minimizing $SSE$ is not a good measure of the predictive ability of a model) specific to forecasting problems (i.e. when using regression models specifically for forecasting purposes) or does it apply to any type of regression model? And if so, why is the use of $SSE$ so relevant if it is such a bad idea?

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    $\begingroup$ @klumbard the question you referenced answers my first question, but not my second. $\endgroup$ – Skander H. - Reinstate Monica Jan 8 '18 at 21:10
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    $\begingroup$ The problem in (2) doesn't lie with using SSE: it would persist using any loss function for training the models. That should make it obvious that using SSE isn't inherently a bad idea: it's misusing it for variable selection that may be problematic. $\endgroup$ – whuber Jan 8 '18 at 23:33
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As noted by @klumbard in the comment I think your first question is addressed by this question, so I'll try to address your second question.

I would interpret the quoted text not as saying that SSE is bad as a loss function for training models, but rather it says that it's bad for model selection and it doesn't tell you if your model fit is good if you don't measure model complexity. I think that agrees with my ML experience because you don't believe generally that your model will have good predictive performance just because the training loss is small. A primary reason for this is overfitting, for example suppose your model was an interpolating polynomial in which case $SSE = 0$. There are several ways to try and fix things, one common way to is to add a cost associate with large regression coefficients e.g.

$ \min_{ \beta } \left\{ \frac{1}{N} \left\| y - X \beta \right\|_2^2 + \lambda \| \beta \|_2 \right\}. $

where $\left\|\right\|_2 $ is the $l_2$ norm. I think this strategy is called regularisation or weight decay in ML and ridge regression in statistics.

The idea to minimise MSE arises naturally when using conditional expectation such as when we train a discriminative model for prediction. Suppose $f(X)$ is any function of $X$, then the conditional mean minimises the MSE:

$E(Y_i | X =x) = argmin_{f(X)} E((y - f(X))^2)$.

That is, we can use the conditional mean to get the best prediction of $Y$ when $X=x$, if we measure best by MSE. There are quite a few strategies for approximating $E(Y_i | X =x)$, so this is very convenient.

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