The D-equivalence violation bounds the finite-budget design's log-determinant gap

By [F1] and [F5], ${\sum }_x{\xi }_x{s}_x={\sum }_x{\xi }_x\mathrm{tr}(M^{-1}{f}_x{f}_x^T)=\mathrm{tr}\text{\hspace{0.17em}⁣}\left(M^{-1}{\sum }_x{\xi }_x{f}_x{f}_x^T\right)=\mathrm{tr}(M^{-1}M)=\mathrm{tr}(I_p)=p\mathrm{.}$ The first rank-one trace identity follows by expanding the diagonal sum: $\mathrm{tr}(M^{-1}{f}_x{f}_x^T)={\sum }_{i,j}(M^{-1}{)}_{ij}(f_x{)}_j(f_x{)}_i=f_x^T{M}^{-1}{f}_x$. Since the ${\xi }_x$ are nonnegative and sum to one, their weighted average of the $s_x$ is at most their maximum. The displayed equality therefore gives ${\max }_x{s}_x\ge p$ and $v\ge 0$.

Let $\eta$ be any nonsingular approximate competitor and put $N=M(\eta )$ and $M_t=(1-t)M+tN$. The segment is positive definite because for each nonzero $u$, $u^T{M}_{t}u=(1-t)u^{T}Mu+t{u}^{T}Nu>0$. Define $g(t)=\log \det M_t$. By [F3], $g$ is concave, and [F4] gives $g^{\mathrm{\prime }}(0)=\mathrm{tr}(M^{-1}(N-M))$. For $0<s<1$, concavity gives $g(s)\ge (1-s)g(0)+sg(1)$, hence $g(1)-g(0)\le (g(s)-g(0))\mathrm{/}s$. Letting $s\downarrow 0$ and using the right derivative $g^{\mathrm{\prime }}(0)$ proves $\log \det N-\log \det M\le \mathrm{tr}(M^{-1}(N-M))$.

Expand $N={\sum }_x{\eta }_x{f}_x{f}_x^T$ and $M={\sum }_x{\xi }_x{f}_x{f}_x^T$. By the rank-one trace identity proved in step 1.1, $\mathrm{tr}(M^{-1}N)={\sum }_x{\eta }_x{s}_x$ and $\mathrm{tr}(M^{-1}M)=p$. Since $\eta$ is an approximate design, ${\sum }_x{\eta }_x{s}_x\le {\max }_x{s}_x=p+v$. Thus step 1.2 yields $\log \det M(\eta )-\log \det M(\xi )\le v$.

If $v=0$, then $s_x\le p$ for every $x$, so [F6] makes $\xi$ D-optimal. Conversely, if $\xi$ is D-optimal, [F6] gives $s_x\le p$ for every $x$; step 1.1 says their $\xi$-weighted average equals $p$, so their maximum is exactly $p$ and $v=0$. This proves both directions.