Numerical identifiability and optimal experiment design — Examples

Differentiation gives $F^{\mathrm{\prime }}(\theta )=3{\theta }^2$, so $F^{\mathrm{\prime }}(0)=0$ and the Jacobian has rank $0<1$.

If $u^3=v^3$, then $(u-v)(u^2+uv+v^2)=0$. The second factor is $(u+v\mathrm{/}2{)}^2+3v^2\mathrm{/}4$, which is nonnegative and vanishes only when $u=v=0$. Hence either $u=v$ directly or the second factor vanishes and again $u=v=0$. Thus $F$ is injective on $\mathbb{R}$, in particular on a neighborhood of $0$, so local identifiability holds despite the rank-deficient Jacobian.

Put $\lambda =k_1+k_2$. The function $A(t)=A_0{e}^{-\lambda t}$ satisfies the rate equation and initial value. If $\tilde{A}$ is any other differentiable solution, then $(e^{\lambda t}\tilde{A}(t){)}^{\mathrm{\prime }}=0$; applying [F3] on every compact time interval shows this product is constant, and its value at $0$ is $A_0$. Thus the observation map is $F(k_1,k_2)=(A_0{e}^{-(k_1+k_2)t}{)}_{t\in \mathcal{T}}$. Now fix any neighborhood $U$ of $(k_1,k_2)$ inside $(0,\mathrm{\infty }{)}^2$. Since $U$ is open, it contains a ball of some radius $r>0$. Choose $0<\epsilon <\min \{k_2,r\mathrm{/}2\}$. Then $(k_1+\epsilon ,k_2-\epsilon )$ is positive, distinct from $(k_1,k_2)$, and lies in that ball. The sums of the rates agree, so these two parameters have identical observations. Every neighborhood therefore contains a distinct parameter with identical observations; the model is not locally identifiable by [F1], and hence is not globally identifiable.

For each $t\in \mathcal{T}$, differentiating the observation gives the row $-A_{0}t{e}^{-(k_1+k_2)t}(1,1)$. Every row is a scalar multiple of $(1,1)$, so the stacked Jacobian has rank at most one and has the nonzero kernel vector $(1,-1)$. By [F2], every supported design's information matrix is singular. This rank failure agrees with, but is weaker than, the exact aliases proved in step 1.1.

The map is C1 and $F^{\mathrm{\prime }}(1)=2\mathrm{\ne }0$, so its one-dimensional Jacobian has full column rank. By [F2], $F$ is injective on some open neighborhood of $1$, proving local identifiability.

The distinct parameters $1$ and $-1$ both lie in $\mathrm{\Theta }$ and satisfy $F(1)=1=F(-1)$. Hence $F$ is not injective on all of $\mathrm{\Theta }$, so it is not globally identifiable by [F1].

The exact information values are $M(d_1)=(101\mathrm{/}100{)}^2=10201\mathrm{/}10000$ and $M(d_2)=1$, so the true log-determinant criterion ranks $d_1$ above $d_2$. The computed values are $M_h(d_1)=(99\mathrm{/}100{)}^2=9801\mathrm{/}10000$ and $M_h(d_2)=(51\mathrm{/}50{)}^2=10404\mathrm{/}10000$, so the computed criterion ranks $d_2$ above $d_1$ by strict monotonicity of $\log$ from [F6].

The two Jacobian errors have absolute value $1\mathrm{/}50$, and for one-dimensional matrices this is the singular-value error, consistent with [F2]. Here $K=101\mathrm{/}100$, $\eta =1\mathrm{/}50$, and $w=1$, so [F3] gives the uniform bound $\delta =2(101\mathrm{/}100)(1\mathrm{/}50)+(1\mathrm{/}50{)}^2=51\mathrm{/}1250$. For any approximate design $\xi =(\lambda ,1-\lambda )$ with $0\le \lambda \le 1$, $M(\xi )$ is a convex combination of $10201\mathrm{/}10000$ and $1$, hence is at least $1$; $M_h(\xi )$ is a convex combination of $9801\mathrm{/}10000$ and $10404\mathrm{/}10000$, hence is at least $\alpha =9801\mathrm{/}10000$. Their segment also has this lower bound.

By [F4] each criterion error is at most $\delta \mathrm{/}\alpha =(51\mathrm{/}1250)\mathrm{/}(9801\mathrm{/}10000)=408\mathrm{/}9801$. The computed maximizer is $d_2$, so [F5] bounds its true log-determinant gap from the true maximizer $d_1$ by $816\mathrm{/}9801$. Directly, that gap is $\log (10201\mathrm{/}10000)\le 201\mathrm{/}10000$ by the integral in [F6], and $201\mathrm{/}10000<816\mathrm{/}9801$ by cross-multiplication. Thus the ranking flip is consistent with the stated uniform error guarantee.

For an arbitrary approximate design $\xi$, write $\mu ={\sum }_t{\xi }_{t}t$ and $\nu ={\sum }_t{\xi }_t{t}^2$. Direct multiplication gives $M(\xi )=\left[\begin{array}{cc}1& \mu \\ \mu & \nu \end{array}\right]$ and $\det M(\xi )=\nu -{\mu }^2$. Since $0\le t\le 2$, $(t-1{)}^2\le 1$, so $\nu -{\mu }^2={\sum }_t{\xi }_t(t-1{)}^2-(\mu -1{)}^2\le 1$. The equal endpoint design ${\xi }^{\ast }=(1\mathrm{/}2,0,1\mathrm{/}2)$ has $M({\xi }^{\ast })=\left[\begin{array}{cc}1& 1\\ 1& 2\end{array}\right]$ and determinant $1$, so it maximizes the determinant and hence $\log \det M$ among nonsingular designs because $\log$ is strictly increasing by [F5].

For ${\xi }^{(0)}$, $M_0=\left[\begin{array}{cc}1& 1\\ 1& 5\mathrm{/}3\end{array}\right]$ and $M_0^{-1}=\left[\begin{array}{cc}5\mathrm{/}2& -3\mathrm{/}2\\ -3\mathrm{/}2& 3\mathrm{/}2\end{array}\right]$. Thus $(s_0,s_1,s_2)=(5\mathrm{/}2,1,5\mathrm{/}2)$ and $v_0=1\mathrm{/}2$. One update gives ${\xi }^{(1)}=(5\mathrm{/}12,1\mathrm{/}6,5\mathrm{/}12)$; the update is defined by [F2] and remains nonsingular by [F3].

The updated matrix is $M_1=\left[\begin{array}{cc}1& 1\\ 1& 11\mathrm{/}6\end{array}\right]$, with inverse $M_1^{-1}=\left[\begin{array}{cc}11\mathrm{/}5& -6\mathrm{/}5\\ -6\mathrm{/}5& 6\mathrm{/}5\end{array}\right]$. Therefore $(s_0,s_1,s_2)=(11\mathrm{/}5,1,11\mathrm{/}5)$ and $v_1=11\mathrm{/}5-2=1\mathrm{/}5$. With budget $N=1$, the procedure returns the updated design with status budget-exhausted and violation $1\mathrm{/}5$.

For the endpoint design ${\xi }^{\ast }$, $M({\xi }^{\ast }{)}^{-1}=\left[\begin{array}{cc}2& -1\\ -1& 1\end{array}\right]$, so its sensitivities at $0,1,2$ are $2,1,2$ and its violation is zero. By [F4] this is an exact D-optimality certificate; [F4] also bounds the log-determinant gap of the one-update output by $1\mathrm{/}5$. The weights are approximate-design weights; integer run counts are a separate problem by [F1].