\le \Vert {\theta{'}}{{\cal V}}_n \Vert \cdot
\Vert  {{\cal V}}_n^{-1} {{\cal S}}_n \Vert$$ and $$\Vert {\theta{'}}{{\cal V}}_n \Vert^2 ={\theta{'}}{{\cal V}}_n^2 \theta \le 
\Vert \theta\Vert^2 \lambda_{\max,n}^2.$$ Hence $$|{\theta{'}}{{\cal S}}_n| \le  \Vert \theta\Vert \lambda_{\max,n}\Vert  {{\cal V}}_n^{-1} {{\cal S}}_n \Vert .$$ Furthermore $${\theta{'}}{{\cal V}}_n \theta e^{-L_{c,n}\Vert \theta\Vert} \ge 
\lambda_{\min,n} \Vert\theta\Vert^2  e^{-L_{c,n}\Vert \theta\Vert} .$$ Therefore $${\theta{'}}\nabla u(\theta) \le
\lambda_{\min,n}\Vert \theta\Vert \big( L_{\lambda,n} \Vert  {{\cal V}}_n^{-1} {{\cal S}}_n \Vert - \Vert\theta\Vert e^{-L_{c,n}\Vert \theta\Vert} \big).$$ For $\Vert\theta\Vert=  3 L_{\lambda,n} \Vert  {{\cal V}}_n^{-1} {{\cal S}}_n \Vert$ $$L_{\lambda,n} \Vert  {{\cal V}}_n^{-1} {{\cal S}}_n \Vert - \Vert\theta\Vert e^{-L_{c,n}\Vert \theta\Vert} 
= L_{\lambda,n} \Vert  {{\cal V}}_n^{-1} {{\cal S}}_n \Vert ( 1-  3e^{-3  L_{c,n} L_{\lambda,n} \Vert{{\cal V}}_n^{-1} {{\cal S}}_n\Vert})$$ Then for $\Vert {{\cal V}}_n^{-1} {{\cal S}}_n \Vert \le 1/(3L_{c,n} L_{\lambda,n})$ $$3e^{-3 L_{\lambda,n} L_{c,n}\Vert{{\cal V}}_n^{-1} {{\cal S}}_n\Vert}\ge 3 e^{-1}>1.$$ Hence, if $\Vert {{\cal V}}_n^{-1} {{\cal S}}_n \Vert \le 1/(3L_{c,n} L_{\lambda,n})$, we have ${\theta{'}}\nabla{u}(\theta) < 0$ for all $\theta$ with $\Vert\theta\Vert=  3 L_{\lambda,n} \Vert  {{\cal V}}_n^{-1} {{\cal S}}_n \Vert$. By the remark just after Proposition \[prop:mle0\], this completes the proof of Lemma \[lem:small-mle\].

### Behavior of the hindsight strategy {#sec:hindsight}

We summarize properties of $\log{{\cal K}}_n^{{\hat\theta_n^*}}$ in view of the standard theory of exponential families ([@barndorff-nielsen]) in statistical inference. Define $$\psi_i(\theta)=\log(1+p_i (e^{\theta c_i}-1)),
\quad
\psi(\theta)= 