maximum likelihood and mse(mean square error)

Question

thisis the equation of maxilum likelihood

Then this is maximum likelihgood when the distribution is Gaussian distribution.
I want to know to to derive the bottom equation from the left ?

Answer 1

The PDF of the (univariate) normal distribution is

$$\frac{1}{\sigma\sqrt{2\pi}} \exp\left[-\frac{1}{2} \left(\frac{y - \hat{y}}{\sigma}\right)^2\right]$$

This is your $P_{model}(y_i\ | \ x_i ; \theta)$ for a single observation. Assuming each observation is independent, the likelihood (probability of observing $y_1...y_m$ together) is $$\prod_{i = 1}^m P_{model}(y_i\ | \ x_i ; \theta).$$ Taking the log of this whole expression gives your result.