Log likelihood from multi-variate Normal distribution?

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If we consider a linear mixed model that consider fixed effect and random effect: $$y = X\beta + W_uu + e,$$

where

  • $n$ is the sample size
  • $y \in \mathbb{R}^{n \times 1}$ is the response vector
  • $X \in \mathbb{R}^{n \times d}$ is the fixed effect design matrix,
  • $\beta \in \mathbb{R}^{d \times 1}$ is the vector of coefficients corr. to fixed effects
  • $W_u \in \mathbb{R}^{n \times c}$ is the random effect design matrix
  • $u \in \mathbb{R}^{c \times 1}$ is the vector of coefficients corr. to random effects
  • $e \in \mathbb{R}^{n \times 1}$ is the error term,

and $u \sim N(0, {\sigma_u}^2 I), e \sim N(0, {\sigma_e}^2 I)$ while $\beta, {\sigma_u}^2$ and ${\sigma_e}^2$ are all not known, then the likelihood of $y$ without random effects should be $$Pr(y|\beta, {\sigma_u}^2, {\sigma_e}^2) = N(X\beta, W_u{W_u}^T {\sigma_u}^2 + {\sigma_e}^2 I) = \dfrac{\exp(-0.5(y - X\beta)^T \Sigma^{-1} (y - X\beta))}{\sqrt{(2\pi)^n \det \Sigma}},$$

where $\Sigma = W_u{W_u}^T {\sigma_u}^2 + {\sigma_e}^2 I = {\sigma_u}^2 (W_u{W_u}^T + (\sigma_e/\sigma_u)^2 I) := {\sigma_u}^2 (K + \delta_e I)$.

Now if the above formula I came up with is correct, I came up with the log-likelihood to be $$-\dfrac{1}{2} (\color{red}{n \ln (2\pi) + 2 \ln (\sigma_u)} + \ln \det (K + \delta_e I) + \dfrac{1}{(\sigma_u)^2}(y - X\beta)^T\det (K + \delta_e I) (y - X\beta))$$

but from the lecture slides I have, I find to log-likelihood to be $$-\dfrac{1}{2} (\color{red}{n \ln (2\pi{\sigma_u}^2)} + \ln \det (K + \delta_e I) + \dfrac{1}{(\sigma_u)^2}(y - X\beta)^T\det (K + \delta_e I) (y - X\beta)).$$

My concerns are

  • How to come up with the latter expression?
  • Are the terms colored in red equivalent?
  • Is either one, especially my own deduction, wrong?
  • If yes then why/how?

Thanks a lot in advance.