OLS estimator derivation: second-order condition to prove global minimum? In deriving our ordinary least squares estimates, we can partially differentiate the sum of squared errors $\sum_{i=1}^{n} {e_i^2} = \sum_{i=1}^{n} {(Y_i- \hat{\alpha}-\hat{\beta}X_i )^2}$ with respect to our estimators $\hat{\alpha}$ and $\hat{\beta}$ to get the following first order conditions:

$$\frac{\delta SSR}{\delta \hat{\alpha}}= -2\sum_{i=1}^{n} {e_i}=-2\sum_{i=1}^{n} {(Y_i- \hat{\alpha}-\hat{\beta}X_i)}=0$$
$$\frac{\delta SSR}{\delta \hat{\beta}}= -2\sum_{i=1}^{n} {X_i e_i}=-2\sum_{i=1}^{n} X_i {(Y_i- \hat{\alpha}-\hat{\beta}X_i)}=0$$

How can we then prove that the sum of squared residuals is a global minimum?

Firstly, can I confirm that the Hessian looks like this?

\begin{array}{cc}
   2n &amp; 2\sum_{i=1}^{n}X_i  \\
   2\sum_{i=1}^{n}X_i  &amp; 2\sum_{i=1}^{n}(X_i)^2  \\
  \end{array}

Secondly, how can I show that the determinant of this Hessian is positive definite, i.e. $ 4[n \sum_{i=1}^{n}(X_i)^2 - (\sum_{i=1}^{n}X_i)^2]&gt;0?$ The presentation bears some resemblance to the population variance formula, and I know that variance must be non-negative, but I'm wondering how to move on. Is there some way to factorise this? Thank you.
It seems that you working on the right lines. By multiplying out the brackets, it can easily be verified that $n\sum{X}_i^2 - (\sum{X}_i)^2 = n\sum{(X_i - \bar{X}})^2$ where $\bar{X}$ is the mean. As you yourself spotted, this is basically a standard expression for the variance. From this, it follows that $n\sum{X}_i^2 - (\sum{X}_i)^2 &gt; 0$ provided that $X_i \neq \bar{X}$ for some $i$.

Incidentally, you also need to verify that one of the second derivatives is positive (say $SSR{_{\alpha\alpha}}&gt;0$) which follows directly from your Hessian.