s {{\cal S}}_n) - \psi({\hat\theta_n^*}(\bar {{\cal T}}_0 + s{{\cal S}}_n)).$$ Then $\log {{\cal K}}_n^{{\hat\theta_n^*}({{\cal T}}_n)} = g(1)$. It is easily seen that $g(0)=0$. By $$g'(s)= {\hat\theta_n^*}(\bar {{\cal T}}_0 + s {{\cal S}}_n){'}{{\cal S}}_n.$$ Again it is easily seen that $g'(0)=0$, since ${\hat\theta_n^*}(\bar {{\cal T}}_0)=0$. Then $$g(1)=\int_0^1 \int_0^s g''(u)du  ds.$$ Now $$g''(u)={{\cal S}}_n{'}H_\psi({\hat\theta_n^*}(\bar {{\cal T}}_0 + u{{\cal S}}_n))^{-1} {{\cal S}}_n.$$ By $$\begin{aligned}
e^{-\Vert {\hat\theta_n^*}(\bar {{\cal T}}_0 + u{{\cal S}}_n) \Vert L_{c,n}} {{\cal S}}_n{'}{{\cal V}}_n^{-1} {{\cal S}}_n 
&\le {{\cal S}}_n{'}H_\psi({\hat\theta_n^*}(\bar{{\cal T}}_0 + u{{\cal S}}_n))^{-1} {{\cal S}}_n \\
& \le e^{\Vert {\hat\theta_n^*}(\bar{{\cal T}}_0 + u{{\cal S}}_n) \Vert L_{c,n}} {{\cal S}}_n{'}{{\cal V}}_n^{-1}  {{\cal S}}_n. \end{aligned}$$ Also $\int_0^1 \int_0^s 1 du  ds = 1/2$. Furthermore by Lemma \[lem:small-mle\], if $\Vert {{\cal V}}_n^{-1} {{\cal S}}_n \Vert \le 1/(3L_{c,n} L_{\lambda,n})$ then $\Vert{\hat\theta_n^*}(\bar{{\cal T}}_0 + u{{\cal S}}_n) \Vert \le 3 L_{\lambda,n} \Vert {{\cal V}}_n^{-1} {{\cal S}}_n \Vert$ for all $0\le u\le 1$. Combining these results we have the proposition.

As in Proposition \[prop:small-mle\] we have the following corollary.

\[cor:hindsight\] In the same setting as in Theorem \[thm:usual-SLLN\] Skeptic can weakly force $$E_1 \ \Rightarrow \ 
\lim_n \frac{\log {{\cal K}}_n^{{\hat\theta_n^*}}}{{{\cal S}}_n{'}{{\cal V}}_n^{-1} {{\cal S}}_n/2}  = 1.$$

### Laplace method for evaluating the difference of the hindsight strategy and the 