How to show that OLS and MLE are the same for $\beta$

Question

I have several questions regarding this proof:

1. Shouldn't the $\propto$ be used instead of the $=$ when we leave out the $\frac{1}{\sqrt{2\pi\sigma²}}$ ?
2. Is a maximization problem simply inverted to a minimization problem by $\cdot (-1)$?
3. Is the second last line correct? Why is $(-(y_i - (\beta_0+x_{i1} \beta_1 + ...))²$ equal to $(y_i - (\beta_0+x_{i1} \beta_1 + ...)²)$? It seems the closing paranthese is misplaced.
4. In the end they are basically the same because we take the derivative and set it to zero: $\frac{d}{d\beta}\sum(y_i - x_i \beta) \stackrel{!}{=} 0$ and it doesn't matter if we maximize or minimize, or not? Also $-\frac{d}{d\beta}\sum(y_i - x_i \beta) \stackrel{!}{=} 0$ would yield the same since we can simply multiply by $\cdot (-1)$

Thanks in advance! :)

Answer 1

1. No, because the maximization happens over $\beta$, not $\sigma$, thus changing $\sigma$ would not influence the optimal choice of $\beta$.

2. Yes. To give another simple example of this, think of parabolas: finding the minimum of $f(x) = x^2$ is equivalent to finding the maximum of $f(x) = - x^2$.

3. No, this is what you mean in point 2) actually: The parentheses are indeed wrong in the last 2 steps.

It should be:

arg max$_{\beta} \frac{1}{2\sigma^2} \sum_{i=1}^{n} (-(y_i - \mu_i)^2)$

= arg max$_{\beta} \frac{1}{2\sigma^2} \sum_{i=1}^{n} (-[y_i - (\beta_0 + \beta_1x_{i1} + \beta_2x_{i2} + ... + \beta_px_{ip})]^2)$

And now you use that minus sign in front of the $y_i$ to flip from arg max to arg min:

= arg min$_{\beta} \frac{1}{2\sigma^2} \sum_{i=1}^{n} (y_i - (\beta_0 + \beta_1x_{i1} + \beta_2x_{i2} + ... + \beta_px_{ip}))^2$

4. Choosing to maximize or minimize depends on the estimator you use, or you minimize the least squares, or you maximize the likelihood, but your proof just says that both will indeed give the same solution.