The Gaussian profile likelihood reduces to the trace score

Each independent observation with the common law in [F3] has the density in [F2]. On the one-point source $\{\ast \}$, the constant law $K_r(\ast ,\cdot )=N_p(\mu ,\mathrm{\Sigma })$ is a probability kernel: its sections are probability measures and each evaluation map is constant, hence measurable [F4]. The density $f_r(\ast ,x)=f_{\mu ,\mathrm{\Theta }}(x)$ is continuous in $x$ and constant in $\ast$, hence jointly measurable; [F5] gives the product law density ${\prod }_{r=1}^N{f}_{\mu ,\mathrm{\Theta }}(x_r)$. By [F6], independence identifies this product law with the joint observation law. Its log-likelihood for fixed $\mu$ and $\mathrm{\Theta }$ is $L(\mu ,\mathrm{\Theta })=-\frac{Np}{2}\log (2\pi )+\frac{N}{2}\log \det \mathrm{\Theta }-\frac{1}{2}{\sum }_{r=1}^N(x_r-\mu {)}^T\mathrm{\Theta }(x_r-\mu )\mathrm{.}$

Write $y_r=x_r-\bar{x}$, so ${\sum }_r{y}_r=0$ and $x_r-\mu =y_r+(\bar{x}-\mu )$. Expanding and summing removes the cross term: ${\sum }_{r=1}^N(x_r-\mu {)}^T\mathrm{\Theta }(x_r-\mu )={\sum }_{r=1}^N{y}_r^T\mathrm{\Theta }{y}_r+N(\mu -\bar{x}{)}^T\mathrm{\Theta }(\mu -\bar{x})\mathrm{.}$

Positive definiteness of $\mathrm{\Theta }$ makes the last quadratic term nonnegative and equal to zero exactly at $\mu =\bar{x}$. Therefore $\bar{x}$ uniquely maximizes $L(\mu ,\mathrm{\Theta })$ over $\mu$.

By [F1], ${\sum }_r{y}_r{y}_r^T=NS$. Cyclicity of the finite matrix trace gives ${\sum }_r{y}_r^T\mathrm{\Theta }{y}_r={\sum }_r\mathrm{tr}(\mathrm{\Theta }{y}_r{y}_r^T)=N\mathrm{tr}(S\mathrm{\Theta })$. Substituting this identity and $\mu =\bar{x}$ into step 1.1 yields the displayed profile log-likelihood and its normalization.