We consider the residuals $\varepsilon_i$ as random variables drawn independently from some distribution with mean zero. In other words, for each value of $x$, the corresponding value of $y$ is generated as a mean response $\alpha + \beta x$ plus an additional random variable $\varepsilon$ called the error term, equal to zero on average. Under such interpretation, the least-squares estimators $\hat \alpha$ and $\hat \beta$ will themselves be random variables whose means will equal the "true values" $\alpha$ and $\beta$.


Is there a conceptual way to understand how is it that the normal distribution of the error (which is somewhat acceptable) turns $\alpha$ and $\beta$ into normally distributed?

  • 1
    $\begingroup$ The quote does not say they are normally distributed. Nonetheless, they are normally distributed and the proof can be found here en.wikipedia.org/wiki/Proofs_involving_ordinary_least_squares $\endgroup$
    – Tim
    Jun 25, 2023 at 11:21
  • $\begingroup$ I have read that page, but I do not see any proof for that specific statement @Tim $\endgroup$
    – Minsky
    Jun 25, 2023 at 14:58
  • $\begingroup$ Check the “Finite-sample distribution” section $\endgroup$
    – Tim
    Jun 25, 2023 at 17:16
  • $\begingroup$ As Tim and Dave pointed out, assuming that $\epsilon$ is normally distributed means that the parameters $\hat{\alpha},\hat{\beta}$ are also normal in finite-samples by a property of normal distributions (any affine transformation of a normal is normal). If $\epsilon$ is not normal, if the data is i.i.d. then the parameters are asymptotically normal as $n\to\infty$. This can be shown by a simple proof with the Central Limit Theorem. $\endgroup$
    – Adam
    Jun 25, 2023 at 22:51

1 Answer 1


For a general parameter vector $\beta$ and model matrix $X$, the OLS solution for a linear model is given by $\hat\beta = (X^TX)^{-1}X^Ty$. Assume that $\varepsilon_i\overset{iid}{\sim}N(0, \sigma^2)$.

By the assumptions of the model, $y = X\beta + \varepsilon$.

$$ (X^TX)^{-1}X^Ty\\ =(X^TX)^{-1}X^T(X\beta + \varepsilon)\\ =(X^TX)^{-1}(X^TX)\beta + (X^TX)^{-1}X^T\varepsilon\\ =\beta + ((X^TX)^{-1}X^T)\varepsilon $$

A linear transformation of a multivariate normal vector, such as $((X^TX)^{-1}X^T)\varepsilon$, is multivariate normal. Thus, the entire $\hat\beta$ has a normal distribution with mean $\beta$ $($shift the normal $((X^TX)^{-1}X^T)\varepsilon$ by $\beta$, which will, again, be normal$)$. That the mean is $\beta$ gives the unbiasedness.

Simple linear regression is a special case of this, where the parameter vector $\beta$ just has a slope and an intercept parameter, often called $\beta$ and $\alpha$, respectively, and the model matrix is a column of $1$s next to a column of the observations of the explanators variable, $x$.

(This is all taken from the "Finite-sample distribution" section of the Wikipedia article that Tim mentioned, and it should be found as a derivation or an exercise in a linear regression textbook.)


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