Estimating linear regression with OLS vs. ML Assume that I'm going to estimate a linear regression where I assume $u\sim N(0,\sigma^2)$. What is the benefit of OLS against ML estimation? I know that we need to know a distribution of $u$ when we use ML Methods, but since I assume $u\sim N(0,\sigma^2)$ whether I use ML or OLS this point seems to be irrelevant. Thus the only advantage of OLS should be in the asymptotic features of the $\beta$ estimators. Or do we have other advantages of the OLS method?
 A: You are focusing on the wrong part of the concept in your question.  The beauty of least squares is that it gives a nice easy answer regaurdless of the distribution, and if the true distribution happens to be normal, then it is the maiximum likelihood answer as well (I think this is the Gauss-Markov thereom).  When you have a distribution other than the normal then ML and OLS will give different answers (but if the true distribution is close to normal then the answers will be similar).
A: Using the usual notations, the log-likelihood of the ML method is
$l(\beta_0, \beta_1 ; y_1, \ldots, y_n) = \sum_{i=1}^n \left\{ -\frac{1}{2} \log (2\pi\sigma^2) - \frac{(y_{i} - (\beta_0 + \beta_1 x_{i}))^{2}}{2 \sigma^2} \right\}$.
It has to be maximised with respect to $\beta_0$ and $\beta_1$. 
But, it is easy to see that this is equivalent to minimising
$\sum_{i=1}^{n}  (y_{i} - (\beta_0 + \beta_1 x_{i}))^{2} $.
Hence, both ML and OLS lead to the same solution.
More details are provided in these nice lecture notes.
A: the only difference for finite samples is, that the ML-estimator for the residual variance is biased. It does not account for the number of regressors used in the model.
