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Given that linear models can be solved exactly via calculus, how is it possible to define a variance for the parameters ($\mathbf{a}$) which minimize some error function? say $Err=(o_i-f(x_i; \mathbf{a}))^2$ There no distribution of $\mathbf{a}$ to find the variance of, no?

Craig
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    If you think as a frequentist, there is the distribution of the data $x$, as the actual sample is just one possible sample from the population. If you think as a Bayesian, in addition to the above you formalize your knowledge about parameters by a probability distribution as if the parameters were random variables. – Richard Hardy Sep 02 '21 at 09:31
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    "Solved exactly" is misleading, because unless the fit is perfect, the data do not "exactly" fit the model. Random variables are introduced to analyze that lack of fit. This is done explicitly in many formulations, such as https://stats.stackexchange.com/a/148713/919, but in other formulations it is only implicit. – whuber Sep 02 '21 at 12:27

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