In a regression setting, consider the issue of model bias (rather than sampling bias or other statistical biases). If the true data generating process looks like $Y = f(X) + \mathrm{noise}$ and our goal is to estimate $f$ well, we might fit a particular statistical model or machine learning algorithm to the available dataset and get an estimate $\hat f$. For some procedures, perhaps at many points $X=x_0$ it turns out that $\mathbb{E}(\hat f(x_0)) \approx f(x_0)$, and we say the model bias is low. In other words: on average across many repetitions of taking a sample and fitting the model, the fitted models tend to match the true regression function across most of the range of $X$.

For other procedures, perhaps it turns out that $\mathbb{E}(\hat f(x_0)) \neq f(x_0)$ at many points $X=x_0$, and we say the model bias is high(er). This concept is often illustrated with the example of a nonlinear $f$ and a single feature $X$. For instance, if the true function is $f(X) = X^2$ on the range $X \in (-1, 1)$, but we try to fit $\hat f$ with a simple linear regression, there's no way a straight line can match the true $f$ for most of the relevant range of $X$, not even in expectation. So a simple linear model would have relatively high model bias for the true regression function here, compared to (for instance) polynomial regression or spline models which could have low or no model bias.


It is clear to me how adding polynomials (or other nonlinear transformations) of the features could reduce model bias. So far so good. But there is also the claim that adding more features (entirely new measurements, not transformations of features already in the model) can also reduce model bias. I don't see why. In my humble opinion it will not fix it since linear regression will stay linear if a new feature is added. With one feature we will have a straight line in a plane. By adding a totally new feature, we will still have a "straight" fitted function (a plane with no nonlinearities) in a 3 dimensional space.

So in what sense is it possible for model bias to be reduced when we add new features, not just nonlinear transformations on existing features?

  • $\begingroup$ Consider changing the topic of your question to "Why adding polynomial features..." instead of "additional". $\endgroup$
    – user209249
    Apr 14, 2020 at 21:33

2 Answers 2


You wrote:

With one feature we will have a straight line in a plain, and by adding one feature we will have a straight plane in a 3 dimensional space.

Imagine that your response variable, $Y$, truly depends on two features, $X_1$ and $X_2$. Let's even say that a linear model is true:

$Y = f(X_1, X_2) + \mathrm{noise} = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \mathrm{noise}$

If these betas are nonzero, then in the 3D space, $f(X_1, X_2)$ is a tilted plane. It's not curved (no polynomials, interaction terms, etc.). But it sits at an angle to each of the $X$ axes.

Plane in 3D space that is tilted with respect to both explanatory variables.

Now, let's say that you start by ignoring $X_2$ and just trying to fit the model

$Y = \beta_0 + \beta_1 X_1 + \mathrm{noise}$

Then you will also get a plane as you pointed out. This plane is tilted with respect to the $X_1$ axis, but not with respect to the $X_2$ axis.

Plane in 3D space that is tilted with respect to only one explanatory variable.

This is a biased model. Since you don't include $X_2$ in your model, there's no way for the fitted plane $Y = \hat\beta_0 + \hat\beta_1 X_1$ to match the true plane $Y = \beta_0 + \beta_1 X_1 + \beta_2 X_2$, not even on average across different samples from the population.


If you think about it, adding polynomial features is somewhat equivalent to polynomial regression. Instead of adding a polynomial term to the model, you add it to the features. You will end up with the same equations. A polynomial model has less bias because it can fit more complex functions.

Linear regression with two components: $$ LinReg(x_1, x_2) = a_0 + a_1 x_1 + a_2 x_2$$

Polynomial regressions with one component: $$ PolyReg(x_1) = a_0 + a_1 x_1 + a_2 x_1^2$$

If you set $x_2 = x_1^2$ you end up with the same expression.

  • $\begingroup$ Sören: You misinterpreted my question. I understand why adding polynomial features fix high bias. I don't undertstand why non-polynomial features do so. $\endgroup$
    – EL Dendo
    Apr 14, 2020 at 21:57
  • $\begingroup$ I find it impossible even to interpret your question because it has no context and could mean so many different things. It's also predicated on a questionable assertion that attempts to connect "polynomial features" (presumably, polynomial functions of data) with "high bias" (whatever that might mean: what is "high" and with respect to what?). $\endgroup$
    – whuber
    Nov 28, 2023 at 18:57

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