If you minimize RSS, does that minimize MSE?

Question

Does minimizing the $RSS$ in your model always equate to minimizing the $MSE$ as $MSE=\frac{1}{n}\cdot RSS$?

Answer 1

Yes, yes it does: This is a trivial application of optimisation rules. So long as $n$ is constant (i.e., does not depend on $\theta$) then for any objective function $F$ and any positive function $h$ you have:

$$\underset{\theta \in \Theta}{\text{arg min }} F(\theta) = \underset{\theta \in \Theta}{\text{arg min }} h(n) F(\theta).$$

For your particular case you have:

$$\underset{\theta \in \Theta}{\text{arg min }} \text{RSS}(\theta) = \underset{\theta \in \Theta}{\text{arg min }} \frac{1}{n} \cdot \text{RSS}(\theta) = \underset{\theta \in \Theta}{\text{arg min }} \text{MSE}(\theta).$$

(Note that in both cases it is the argument that minimises the function that remains unchanged. The actual minimised function value is changed by the corresponding multiplicative constant.)

Answer 2

MSE is just RSS divided by the sample size, so, for a given sample, anything that optimizes one of them will also optimize the other