The Gaussian covariance likelihood ratio has its fixed-dimensional Wilks limit

By [F2] and [F3], one observation has density ${\varphi }_{\mu ,\mathrm{\Sigma }}(x)=(2\pi {)}^{-p\mathrm{/}2}(\det \mathrm{\Sigma }{)}^{-1\mathrm{/}2}\exp [-(x-\mu {)}^T{\mathrm{\Sigma }}^{-1}(x-\mu )\mathrm{/}2]$: the independent-coordinate product density follows from [F4], its continuous Gaussian formula is Borel by [F15], and the invertible affine map $z\mapsto \mu +{\mathrm{\Sigma }}^{1\mathrm{/}2}z$ has determinant $\mathrm{\mid }\det {\mathrm{\Sigma }}^{1\mathrm{/}2}\mathrm{\mid }=(\det \mathrm{\Sigma }{)}^{1\mathrm{/}2}$, so [F5] gives the displayed density. Independence in [F1] then makes the sample density its product over $i=1,\dots ,n$. The finite-dimensional Lebesgue measure is sigma-finite by [F13], so the product law is dominated by it and the displayed function is a likelihood density by [F14].

Under $H_0$, the positive root ${\mathrm{\Sigma }}_0^{1\mathrm{/}2}$ is invertible by [F12], so define $Y_i={\mathrm{\Sigma }}_0^{-1\mathrm{/}2}(X_i-\mu )$. By [F1]–[F2] these are iid $N_p(0,I_p)$. Write $C_n=n^{-1}{\sum }_i{Y}_i{Y}_i^T$ and $A_n={\mathrm{\Sigma }}_0^{-1\mathrm{/}2}M{\mathrm{\Sigma }}_0^{-1\mathrm{/}2}=C_n-\bar{Y}{\bar{Y}}^T$. For a symmetric matrix, let $\mathrm{vech}$ list its upper-triangular entries in a fixed order; put $V_i=\mathrm{vech}(Y_i{Y}_i^T-I_p)$. These vectors have mean zero and diagonal covariance: diagonal coordinates have variance $2$, off-diagonal coordinates variance $1$, and distinct coordinates have covariance zero. Indeed, independence and symmetry give zero for any product with an odd coordinate power; integration by parts against [F3]'s density gives $E{Z}^2=1$ and $E{Z}^4=3$ (using ${\varphi }^{\mathrm{\prime }}(z)=-z\varphi (z)$ and vanishing polynomial-times-Gaussian boundary terms), while independence factors mixed even products. These finite moments make $V_i$ square-integrable, so [F7] yields $\sqrt{n}\mathrm{vech}(C_n-I_p)\Rightarrow N(0,D)$ with those diagonal variances. Also $\sqrt{n}\bar{Y}=O_P(1)$ by [F7]–[F8], whence $\sqrt{n}\bar{Y}{\bar{Y}}^T=n^{-1\mathrm{/}2}(\sqrt{n}\bar{Y})(\sqrt{n}\bar{Y}{)}^T=o_P(1)$ by [F8]. Thus $H_n=\sqrt{n}(A_n-I_p)$ converges in distribution to a symmetric Gaussian matrix $H$ whose entries on and above the diagonal are independent, with variances $2$ on the diagonal and $1$ off it.

Taking logarithms of that positive product density gives, up to the parameter-free constant $-np\log (2\pi )\mathrm{/}2$, ${\mathrm{\ell }}_n(\mu ,\mathrm{\Sigma })=-\frac{n}{2}\log \det \mathrm{\Sigma }-\frac{1}{2}{\sum }_{i=1}^n(X_i-\mu {)}^T{\mathrm{\Sigma }}^{-1}(X_i-\mu )\mathrm{.}$ Expanding $X_i-\mu =(X_i-\bar{X})+(\bar{X}-\mu )$ and using ${\sum }_i(X_i-\bar{X})=0$ shows the quadratic sum equals $n(\bar{X}-\mu {)}^T{\mathrm{\Sigma }}^{-1}(\bar{X}-\mu )+\mathrm{tr}({\mathrm{\Sigma }}^{-1}R)$. Since ${\mathrm{\Sigma }}^{-1}\succ 0$, this is minimized uniquely at $\mu =\bar{X}$ for every fixed $\mathrm{\Sigma }$.

By [F6], $M=R\mathrm{/}n\succ 0$ almost surely. The square root $M^{1\mathrm{/}2}$ exists by [F12] and is invertible because $M\succ 0$. At $\mu =\bar{X}$, compare the covariance log likelihood at any $\mathrm{\Sigma }\succ 0$ with its value at $M$. Let $B=M^{1\mathrm{/}2}{\mathrm{\Sigma }}^{-1}{M}^{1\mathrm{/}2}\succ 0$. The trace and determinant identities give ${\mathrm{\ell }}_n(\bar{X},\mathrm{\Sigma })-{\mathrm{\ell }}_n(\bar{X},M)=-\frac{n}{2}\{\mathrm{tr}B-\log \det B-p\}\mathrm{.}$ By [F11], the eigenvalues ${\lambda }_1,\dots ,{\lambda }_p$ of $B$ are positive and the brace is ${\sum }_j({\lambda }_j-\log {\lambda }_j-1)\ge 0$, since $x-\log x-1\ge 0$ for $x>0$ by differentiation and its unique minimum at $1$. Equality holds precisely when all ${\lambda }_j=1$, hence $B=I$ and $\mathrm{\Sigma }=M$. Thus the unrestricted MLE is $(\bar{X},M)$; under $H_0$ the maximizing mean remains $\bar{X}$ by step 2.1.

Substituting the unrestricted and null maxima into [F10], and using $R=nM$, yields $-2\log {\mathrm{\Lambda }}_n=n\{\mathrm{tr}({\mathrm{\Sigma }}_0^{-1}M)+\log \det {\mathrm{\Sigma }}_0-\log \det M-p\}\mathrm{.}$ Since $\det ({\mathrm{\Sigma }}_0^{-1\mathrm{/}2}M{\mathrm{\Sigma }}_0^{-1\mathrm{/}2})=\det M\mathrm{/}\det {\mathrm{\Sigma }}_0$, this is the asserted finite-sample identity.

Since $H_n=O_P(1)$ by [F8], $E_n=A_n-I_p=H_n\mathrm{/}\sqrt{n}$ has Frobenius norm $\mathrm{\parallel }{E}_n{\mathrm{\parallel }}_F=O_P(n^{-1\mathrm{/}2})$, where $\mathrm{\parallel }B{\mathrm{\parallel }}_F=({\sum }_{i,j}{B}_{ij}^2{)}^{1\mathrm{/}2}$. On an event whose probability tends to one, every eigenvalue $e_j$ of the symmetric $E_n$ has $\mathrm{\mid }{e}_j\mathrm{\mid }\le 1\mathrm{/}2$. Taylor's theorem on $[-1\mathrm{/}2,1\mathrm{/}2]$ gives $e-\log (1+e)=e^2\mathrm{/}2+O(\mathrm{\mid }e{\mathrm{\mid }}^3)$ with a uniform finite remainder constant there. Hence the statistic in step 4.1 equals $n{\sum }_{j=1}^p(e_j-\log (1+e_j))=\frac{1}{2}\mathrm{tr}(H_n^2)+O_P(n\mathrm{\parallel }{E}_n{\mathrm{\parallel }}_F^3)=\frac{1}{2}\mathrm{tr}(H_n^2)+o_P(1),$ because $p$ is fixed and $n\mathrm{\parallel }{E}_n{\mathrm{\parallel }}_F^3=O_P(n^{-1\mathrm{/}2})$.

The map $H\mapsto \frac{1}{2}\mathrm{tr}(H^2)$ is continuous, so [F8] and step 1.2 give convergence to $\frac{1}{2}\mathrm{tr}(H^2)$. For the stated independent entries this is ${\sum }_{i=1}^p(H_{ii}\mathrm{/}\sqrt{2}{)}^2+{\sum }_{1\le i<j\le p}{H}_{ij}^2,$ a sum of $p+p(p-1)\mathrm{/}2=p(p+1)\mathrm{/}2$ independent squared standard normals, which has law ${\chi }_{p(p+1)\mathrm{/}2}^2$ by [F9]. Step 5.1 and Slutsky [F8] prove the asserted limit. When $p=1$, the off-diagonal sum is empty and the same count gives one degree of freedom; $p$ is fixed throughout, and no finite-sample calibration follows from this limit.