A while ago I was trying (not entirely successfully) to figure out the definition of a regression model. Now I am narrowing it down to a simple linear regression and trying to identify (loosely speaking) where the basic model ends and the optional assumptions begin. Wooldridge "Introductory Econometrics: A Modern Approach" (6th edition, 2016) states the following at the beginning of Chapter 2:

In writing down a model that will “explain $y$ in terms of $x$,” we must confront three issues. First, since there is never an exact relationship between two variables, how do we allow for other factors to affect $y$? Second, what is the functional relationship between $y$ and $x$? And third, how can we be sure we are capturing a ceteris paribus relationship between $y$ and $x$ (if that is a desired goal)? We can resolve these ambiguities by writing down an equation relating $y$ to $x$. A simple equation is $$ y=\beta_0+\beta_1 x_1+u. \quad [2.1] $$ Equation $(2.1)$, which is assumed to hold in the population of interest, defines the simple linear regression model. (Emphasis is mine.)

This does not look complete enough to indicate anything about $y$ and $x$ probabilistically. As far as I understand, one could write such an equation for any variables $y$ and $x$, with any chosen constants $\beta_0$ and $\beta_1$, and it would hold as long as we choose the right values of $u$. So this does not look like a good candidate for a definition of a regression model to me.

Now on the other end of the spectrum we could have a linear model $$ y|x_1\sim i.i.D(\beta_0+\beta_1 x_1,\sigma^2) $$ for some density $D$ characterized by a location and a scale parameter. This is probably too restrictive as a definition of a simple linear regression [model], because I think we could still call it one if the scale above was a function of $x_1$ or if the distribution was something else than the specific $D$.

In between too loose and too restrictive, we can have models like $$ \mathbb{E}(y|x_1)=\beta_0+\beta_1 x_1, $$ or $$ \mathbb{E}(y|x_1)=\beta_0+\beta_1 x_1, \quad \text{Var}(y|x_1)=\sigma^2 $$ and other. Somewhere there I expect to find a definition that makes the most sense in the context of the term "a simple linear regression [model]".

So what is the definition of a simple linear regression [model]?

  • $\begingroup$ I began my answer at stats.stackexchange.com/a/148713/919 with a general formulation of a probabilistic regression model. The discussion that follows covers all your examples and more, so I suspect it may be a complete answer to your questions here. $\endgroup$
    – whuber
    Feb 12, 2020 at 15:13
  • $\begingroup$ (2.1) looks pretty much like a simple linear regression model, though it would be usual to have $u$ have mean $0$ and be uncorrelated with the $x_i$. This allows finding in some sense optimal estimates of $\beta_0$ and $\beta_1$. A stronger assumption would be that $u$ is independent of the $x_i$ and i.i.d. (or even be normally distributed). $\endgroup$
    – Henry
    Feb 12, 2020 at 15:26
  • $\begingroup$ @whuber, in your answer, A linear model of a linear relationship with additive errors as well as the other items in that list suffer from the same problem as $(2.1)$ above. Or is the problem I have indicated not really a problem? You do write A model has *additive errors when $f$ is linear in $\varepsilon$. In such cases it is always assumed that $\mathbb{E}(\varepsilon)=0$*. After this addition, is the linear regression model complete? If so, what does it tell us probabilistically about $y$ and $x$? $\endgroup$ Feb 12, 2020 at 15:28
  • $\begingroup$ I urge you to look at the initial formulation of the model as $Y = f(X,\theta,\epsilon).$ Everything else is just special cases of that presented in order to address the specific question of what "linear" might mean. This general formulation has none of the issues you attach to (2.1). $\endgroup$
    – whuber
    Feb 12, 2020 at 15:30
  • 1
    $\begingroup$ I'm sorry? Given $x_1,$ the distribution of $y$ is the distribution of $u,$ shifted by $3+2x_1:$ you know everything about that conditional distribution that you know about $u.$ In some circumstances you might specify that $u$ has a Normal$(0,\sigma^2)$ distribution; it other circumstances you might only require that $u$ have zero mean and finite variance; but regardless, it is obvious you do have information about the conditional distribution of $y.$ $\endgroup$
    – whuber
    Feb 12, 2020 at 16:27

1 Answer 1


Davidson and MacKinnon in ("Econometric Theory and Methods") address head on the flaw you point out in Wooldridge's presentation: "At this stage as long as we say nothing about the unobserved quantity $u_{t}$, equation (1.01) [$y_t=\beta_0+\beta_1x_t+u_t$] does not tell us anything. In fact, we can allow the parameters $\beta_0$ and $\beta_1$ to be quite arbitrary, since for any given $\beta_0$ and $\beta_1$, the model can always be made to be true by defining $u_t$ suitably. If we wish to make sense of the regression model (1.01), then we must make some assumptions about the properties of the error term $u_t$... Most commonly it is assumed that, whatever the value of $x_t$, the expectation of the random variable $u_t$ is zero. This assumption usually serves to identify the unknown parameters $\beta_0$ and $\beta_1$ in the sense that, under this assumption, equation (1.01) can be true only for specific values of those parameters".

Only because it addresses the same issue, I'll mention that in his book "Extending the Linear Model with R" (page 7), Faraday claims "The construction of the least squares estimates does not require any assumptions about $\epsilon$". This statement is false. The above paragraph explains why. Both authors are world-class experts, so I would file these as typos and nothing more.

Why that particular assumption? What is definitional and what do we need to assume? The CEF error $\epsilon_i$ in $Y_i=E[Y_i|X_i]+\epsilon_i$ is always mean independent of $X_i$. Agrist and Pischke "Mostly Harmless Econometrics": $E[\epsilon_i|X_i]=E[Y_i-E[Y_i|X_i]|X_i]=E[Y_i|X_i]-E[Y_i|X_i]=0.$ That is definitional. But once we assume linearity for the CEF, the linear projection error $u_t$ in $y_t=\beta_0+\beta_1x_t+u_t$ by definition is only uncorrelated with $x_t$, which is a weaker condition than mean-independence. (See Hansen "Econometrics" for the proof). And since regression is at its heart about the conditional mean function $E[y_t|x_t]=\beta_0+\beta_1x_t+E[u_t|x_t]$ we need to assume mean-independence $E[u_t|x_t]=E[u_t]=0$ in order to identify the conditional mean function. In the case when $x_t$ is a non-random/fixed regressor, the assumption $E[u_t|x_t]=E[u_t]$ is superfluous; it is definitional, hence we only need to assume that $E[u_t]=0$.

Most of the times we are looking to do something with our model and thus we'd want the estimators to have some minimal desirable properties. No other assumptions are necessary to derive the OLS estimators or to show that they are unbiased. To derive the variance of the OLS estimators we need to assume that the errors have a constant variance and are uncorrelated. To construct hypothesis tests or confidence intervals we need to either assume the errors are normally distributed or have a large enough sample so that the OLS estimators are approximately normally distributed.

  • $\begingroup$ Thank you. So it is $[2.1]$ combined with $\mathbb{E}(u|X)=0$? $\endgroup$ Feb 8, 2021 at 19:22
  • $\begingroup$ I think for the purpose of identifying $\beta_0$ and $\beta_1$ it is sufficient to assume $E(u_i|x_i)=0$ (as in Weisberg), or only contemporaneous\same observation mean independence +[2.1]. Hayashi makes the stronger assumption $E(u_i|x_1,x_2,...,x_n)=0$ but seems unnecessary for the simplest model that just wants to solve the problem of "combination of observations" (as in Stigler). In practice we may want to add the rather vacuous assumption that there is sample variation in the regressor (as in Wooldridge SLR.3) otherwise we end up with zero in the denominator of $\hat{\beta_{1}}$. $\endgroup$
    – sergiu
    Feb 8, 2021 at 20:42

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