Ordinary least square solution of linear regression can be derived from the assumption of normally distributed residuals: $$ e_i=y_i-\hat{y_i}\\ e_i\sim N(0, \sigma^2) $$

What I don't quite understand from reading a lot of derivations is how we go from $MLE=\arg \max_{\theta}\sum{log(p_{\theta}^i)}$ to the sum of logarithms of PDFs of normal distribution. PDFs are called "probabilities" everywhere I read (e.g. https://machinelearningmastery.com/linear-regression-with-maximum-likelihood-estimation), which is incorrect. PDFs can be > 1. Only integrals of PDFs can be interpreted as probabilities of X falling into the integration interval.

Do I understand correctly, that the proper derivation should introduce a small interval $\epsilon$, so that we could first rewrite probabilities as:

$$ p_{\theta,\epsilon}^i=\int_{y_i-\epsilon}^{y_i+\epsilon}PDF(N(\hat{y^i}, \sigma^2))dy $$

Then we could assume that with small enough $\epsilon$ $$ p_{\theta,\epsilon}^i\approx 2*\epsilon*[\frac 1 {\sqrt {2*\pi}*\sigma} e^{-(\hat{y_\theta^i} - y_i)^2/\sigma^2}] $$

Now MLE can finally be approximated with $$ \require{cancel} MLE=\arg\max_{\theta}\sum log(2*\epsilon*[\frac 1 {\sqrt {2*\pi}*\sigma} e^{-(\hat{y_\theta^i} - y_i)^2/\sigma^2}])\\ =\arg\max_{\theta}\sum \cancel{log(2*\epsilon)} + \cancel{log(\frac 1 {\sqrt {2*\pi}*\sigma})} -(\hat{y_\theta^i} - y_i)^2/\sigma^2\\ =\arg\min_{\theta}\sum (\hat{y_\theta^i} - y_i)^2 $$

And the last expression is the OLS.

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    $\begingroup$ you would be quicker to open a textbook than write all that. you are incorrect you just need the pdf directly. see eg en.wikipedia.org/wiki/Maximum_likelihood_estimation $\endgroup$
    – seanv507
    Aug 5, 2022 at 13:10
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    $\begingroup$ Welcome to CV! You're correct about the sloppy language that conflates a probability density with a probability. Concerning the introduction of a "small interval," please see stats.stackexchange.com/a/397166/919. Then read the other answers in that thread (especially Ben's post) to see why that small interval does not need to be introduced. $\endgroup$
    – whuber
    Aug 5, 2022 at 13:51

1 Answer 1


You seem to mix two issues: first, how to interpret maximum likelihood estimation and second the derivation of OLS as an ML estimator.

Regarding the first point: Leaving mathematical rigour aside and thinking about the likelyhood about the joint probabilities for all possible small intervals can be useful. Let $\mathbf{y}=\{y_1, y_2..., y_n\}$ the data we observe. For each datapoint, we want to choose an estimator which maximises the probability to observe data in a small neighbourhood of $y_1$ and in a small neighbourhood of $y_2$ and in a small neighbourhood of $y_3$ and so on... We impose that these probabilites are compatible with the assumption of all $e_i=y_i-\hat{y}_i$ being drawn form a normally distributed random variable.

Regardingthe second point, i.e. estimation I agree with seanv507: Following wikipedia notation you are maximising the likelihood function which is the product of individual densities:

\begin{equation} f_n(\mathbf{y,\theta})=\Pi_{k=1}^n f_k^{univar}(y_k;\theta) \end{equation} plugging in univariate normal densities, taking logs and simplifying along the lines you do gives you the desired result.


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