The full-affine-span Gaussian-mixture M-step maximizes Q

By [F1] and [F2], every numerator and denominator in ${\gamma }_{ik}$ is finite and strictly positive. Hence ${\gamma }_{ik}>0$. Summing the responsibility formula over $k$ gives ${\sum }_k{\gamma }_{ik}=1$, so $N_k>0$ and ${\sum }_k{N}_k=n$. Thus all displayed weighted means and scatters are defined.

For each nonzero $v\in {\mathbb{R}}^p$, $v^{\mathsf{T}}{S}_{k}v=N_k^{-1}{\sum }_i{\gamma }_{ik}(v^{\mathsf{T}}(x_i-{\bar{x}}_k){)}^2$. If it were zero, positivity of each ${\gamma }_{ik}$ would force $v^{\mathsf{T}}{x}_i=v^{\mathsf{T}}{\bar{x}}_k$ for every $i$. That would put all observations in the proper affine hyperplane $\{x:v^{\mathsf{T}}x=v^{\mathsf{T}}{\bar{x}}_k\}$, contradicting their full affine span. Since $S_k$ is visibly symmetric, it is positive definite.

Expanding the component density in [F1,F2] in the definition of $Q$ gives $Q(\vartheta \mid \theta )={\sum }_k{N}_k\log {\rho }_k-\frac{np}{2}\log (2\pi )-\frac{1}{2}{\sum }_k(N_k\log \det V_k+{\sum }_i{\gamma }_{ik}(x_i-m_k{)}^{\mathsf{T}}{V}_k^{-1}(x_i-m_k))\mathrm{.}$ Here [F13] supplies $\log (\exp u)=u$ and the product and reciprocal laws. For each positive $a$, $\log (a^{-1\mathrm{/}2})=-\frac{1}{2}\log a$ follows by applying those laws to $(a^{-1\mathrm{/}2}{)}^2=1\mathrm{/}a$; applying the product law $p$ times gives $\log ((2\pi {)}^{-p\mathrm{/}2})=-\frac{p}{2}\log (2\pi )$. All logarithms are finite: weights are positive, and [F2] gives finite positive component densities for each positive-definite covariance, so their determinants are positive.

Define ${\alpha }_k=N_k\mathrm{/}n$, and for each subset $B\subseteq \{1,\dots ,K\}$ set $P(B):={\sum }_{k\in B}{\alpha }_k$ and $R(B):={\sum }_{k\in B}{\rho }_k$. Finite sums make these set functions additive on disjoint subsets; a disjoint sequence has at most $K$ nonempty members, so they are countably additive. Step 1.1 and the candidate weight constraint give total mass one, hence [F3] makes $P,R$ probability measures. Each ${\rho }_k>0$, so $P\ll R$. For every subset $B$, ${\sum }_{k\in B}({\alpha }_k\mathrm{/}{\rho }_k){\rho }_k=P(B)$, hence $dP\mathrm{/}dR(k)={\alpha }_k\mathrm{/}{\rho }_k$. The KL definition [F4] therefore gives $D(P\mathrm{\parallel }R)={\sum }_k{\alpha }_k\log ({\alpha }_k\mathrm{/}{\rho }_k)\ge 0$ by [A1,F5]. The weight-dependent part of $Q$ at $\alpha$ minus that at $\rho$ is ${\sum }_k{N}_k\log ({\alpha }_k\mathrm{/}{\rho }_k)=nD(P\mathrm{\parallel }R)\ge 0$, with equality only when $P=R$, hence ${\rho }_k={\alpha }_k$ for every $k$ by [F5].

For fixed candidate covariance $V_k$, write $d_i=x_i-{\bar{x}}_k$ and $e={\bar{x}}_k-m_k$. Since ${\sum }_i{\gamma }_{ik}{d}_i=0$, expanding the quadratic form gives ${\sum }_i{\gamma }_{ik}(x_i-m_k{)}^{\mathsf{T}}{V}_k^{-1}(x_i-m_k)={\sum }_i{\gamma }_{ik}{d}_i^{\mathsf{T}}{V}_k^{-1}{d}_i+N_k{e}^{\mathsf{T}}{V}_k^{-1}e\mathrm{.}$ The last term is nonnegative and is zero exactly when $m_k={\bar{x}}_k$, because $V_k^{-1}$ is positive definite by [F12]. It occurs with a negative coefficient in $Q$ by step 2.2, so the unique maximizing mean is ${\bar{x}}_k$.

At this mean, coordinate expansion of the finite trace gives ${\sum }_i{\gamma }_{ik}{d}_i^{\mathsf{T}}{V}_k^{-1}{d}_i=N_k\mathrm{tr}(V_k^{-1}{S}_k)$. Therefore the covariance-dependent part of $Q$ is $-\frac{N_k}{2}(\log \det V_k+\mathrm{tr}(V_k^{-1}{S}_k))$ by step 2.2. Set $A=V_k^{-1}$; determinant identity [F11], the logarithm reciprocal law [F13], and the finite trace identity [F10] rewrite this as $\frac{N_k}{2}(\log \det A-\mathrm{tr}(A{S}_k))$. The candidate $A_{\ast }=S_k^{-1}$ is symmetric positive definite by step 2.1 and [F12]. Conversely, every symmetric positive-definite $A$ has positive-definite inverse, so this reparameterization ranges over the full covariance domain.

Fix any other positive-definite $A$ and put $H=A-A_{\ast }$, $A_t=A_{\ast }+tH$, and $h(t)=\log \det A_t-\mathrm{tr}(A_t{S}_k)$. Each $A_t$ is positive definite by convexity of the cone. By [F6], $h$ is strictly concave when $H\mathrm{\ne }0$. Jacobi's formula [F7], the scalar logarithm derivative [F8], and the chain rule [F9] give $h^{\mathrm{\prime }}(0)=\mathrm{tr}(A_{\ast }^{-1}H)-\mathrm{tr}(H{S}_k)=0$, since $A_{\ast }^{-1}=S_k$ and [F10] permits cyclic rearrangement of the finite trace. If $h(1)\ge h(0)$, strict concavity gives $h(s)>h(0)$ for each $0<s<1$. For $0<u<s$, concavity gives $(h(u)-h(0))\mathrm{/}u\ge (h(s)-h(0))\mathrm{/}s>0$. Letting $u\downarrow 0$ contradicts $h^{\mathrm{\prime }}(0)=0$. Hence $h(1)<h(0)$ for every $A\mathrm{\ne }{A}_{\ast }$, so $V_k=S_k$ is the unique maximizing covariance.

The expression for $Q$ is a sum over components of a weight term, a mean-dependent quadratic term, and a covariance-dependent term, with no cross-component parameter terms. Steps 3.1, 3.2, and 3.4 show each is uniquely maximized by the displayed parameters; step 2.1 ensures each maximizing covariance remains in the permitted positive-definite domain. The product choice is therefore an admissible attained global maximizer, and equality forces every component parameter to equal the displayed update formulas. Thus the displayed E-step responsibilities and M-step formulas are well-defined and attain the claimed maximum, without using a procedure-level property.