Why do we assume that in a dataset the error, presented as the random error term, has mean 0 ?

To me seems impossible that every event we can study in everyday life has an error with 0 mean...


2 Answers 2


In regression settings, we want to approximate the response $y$ by a function $f$ of the input vector $x$ as good as possible by estimating $f$ from the data. This can be written as $$ y = f(x) + \varepsilon $$ where $\varepsilon$ is a random variable with some properties. Assume now that $E(\varepsilon) = \alpha \ne 0$. That is the situation you are fearing. Then we can rewrite above equation as $$ y = \underbrace{f(X) - \alpha}_{g(X)} + \varepsilon' $$ with $E(\varepsilon') = 0$.

Most algorithms that assume mean zero noise automatically estimate $g$, not $f$. So in practice, this is not an issue.

In linear regression, however, the restriction is much stronger. There one assumes (amongst other) that $E(\varepsilon \mid x) = 0$ (strict exogeneity). This assumption is not automatically fulfilled and requires the right regressors to be present in the model.


As you said all the time it is not true. However, as most of the time, the nature of the error is random, we know the random error has a normal distribution. To know more about this, see this post.

Also, you can see this post to see why most of the time, the error is normal:

The error term can be thought of as the composite of a number of minor influences or errors. As the number of these minor influences gets larger, the distribution of the error term tends to approach the normal distribution. This tendency is called the Central Limit Theorem. The t-test and F- test are not applicable unless the error term is normally distributed.

In sum, as the random error has a normal distribution (in most of the time), its mean to be considered to be zero.

  • $\begingroup$ If in the error there is also a feature that influences the classification (for example) but that we haven't considered, than the error is not equally distributed, but it will have some tendency. Am i correct ? $\endgroup$
    – Qwerto
    Commented Jun 25, 2018 at 11:03
  • $\begingroup$ @Qwerto yes. you are. $\endgroup$
    – OmG
    Commented Jun 25, 2018 at 11:07

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