Log Likelihood for a Gaussian process regression model

According to Bishop, the author from "Statistical Pattern Recognition", we can optimize the hyperparameters of a Gaussian process by maximizing the likelihood function

$$p(\textbf{t}|\theta),$$ where $$\textbf{t}$$ denotes the target vectors $$(t_1, .. ,t_N)$$ of the corresponding input values $$x_1, ..., x_N$$ and $$\theta$$ the hyperparameters.

He then claims, that the log likelihood function is given by the standard form for a multivariate Gaussian distribution:

$$\ln{p(\textbf{t}|\theta}) = -\frac{1}{2}\ln{|C_N|}-\frac{1}{2} \textbf{t}^T C_N^{-1}\textbf{t}-\frac{N}{2}\ln{2\pi}$$

Considering the multivariate Gaussian distribution for $$X = [X_1, ...., X_N]^T$$ is given by: (https://cs229.stanford.edu/section/gaussians.pdf)

$$p(x|\mu,\Sigma) = \frac{1}{(2\pi)^{\frac{n}{2}}|\Sigma|^{\frac{1}{2}} }\exp\left(-\frac{1}{2}(x-\mu)\Sigma^{-1}(x-\mu)\right)$$

To me, the first equation is derived by simply the $$\log$$ of the second equation (keep in mind that the mean is zero in the Gaussian Process described by Bishop) without taking the product into account, which would be $$\prod_N{\ln{p(\textbf{t}|\theta})}$$. I am not sure, what I am missing here. As far as I know, taking the log of a gaussian distribution is not enough to maximize a wanted parameter.

To be more precise, would the MLE of equation 2 not be:

$$\prod_K{p(x|\mu,\Sigma) = \frac{1}{(2\pi)^{\frac{kn}{2}}|\Sigma|^{\frac{k}{2}} }\exp\left(-\frac{1}{2}\sum_K(x-\mu)\Sigma^{-1}(x-\mu)\right)}$$

which is not equal to the first equation, after taking the log. Notice the $$k$$ in the denumerator and the sum of the term in the exp term. Thus, I would expect something like:

$$\ln{p(\textbf{t}|\theta}) = -\frac{K}{2}\ln{|C_N|}-\frac{1}{2} \sum \textbf{t}^T C_N^{-1}\textbf{t}-\frac{KN}{2}\ln{2\pi}$$

This assumption is further verified by the accepted answer in this post: Maximum Likelihood Estimators - Multivariate Gaussian

For me, the first equation is just the log of a gaussian multivariate normal distribution with zero mean, not the MLE.

$$\log_k\left(\prod x_i\right)=\sum\log_k\left(x_i\right)$$

$$\text{AND}$$

$$\log_k\left(k^x\right)=x$$

$$\implies$$

$$\log_k\left(\prod k^{x_i}\right) =\sum x_i$$

You’re taking the base-$$e$$ logarithm of a product of $$e^x$$ expressions, turning the base-$$e$$ logarithm of the product into the sum of the exponents.

• Please see my edit, I think I phrased my question poorly Aug 14, 2022 at 15:52