Mathematical Foundations

Under either convention, $V_1^c\ge {\underline{v}}_1$, the guarantee computed against all feasible follower behaviour with the same admissible leader strategies. If every ${\Gamma }_1$ announcement (every mixture) has an outcome-equivalent constant rule in ${\Gamma }_2$ (the follower responds to the whole rule; observation, announcement costs and feasible actions unchanged), then $V_2^c\ge V_1^c$ for each fixed convention $c$.

For every experiment $\pi$ with signal $s$, $\mathrm{VoI}(\pi )\le V(\mathrm{I}(\theta ;s))$, and $V(I)$ is the supremum of $\mathrm{VoI}(\pi )$ over experiments with $\mathrm{I}(\theta ;s)\le I$. If $\pi {⪰}_{\mathrm{B}}{\pi }^{\prime }$ then $\mathrm{I}(\theta ;s)\ge \mathrm{I}(\theta ;s^{\prime })$, so the bound is monotone along Blackwell's order.

Consider a finite zero-sum game in which nature draws $\omega$, player $A$ observes ${\eta }_A(\omega )$, player $B$ observes ${\eta }_B(\omega )$, and both then act once, with payoff $g(\omega ,a,b)$. If $A$'s observation is refined — ${\eta }_A=f\circ {\eta }_A^{\prime }$ for some $f$, with ${\eta }_B$ and everything else unchanged — then the value for $A$ weakly increases.

In a finite zero-sum game the two sides of the bracket coincide: ${\max }_{p}L(p)={\min }_{q}U(q)=v$.

In a finite zero-sum game, for a pair $(p^{\star },q^{\star })$ the following are equivalent:

• saddle point: $p^{\top }G{q}^{\star }\le p^{\star \top }G{q}^{\star }\le p^{\star \top }Gq$ for all $p,q$;

• both guarantees attained and equal: $L(p^{\star })=U(q^{\star })$;

• Nash equilibrium of the two-player game $(G,-G)$.

In that case $p^{\star \top }G{q}^{\star }=v$.

Let the leader's and the follower's action sets $X$, $Y$ be finite, with payoffs $F(x,y)$ for the leader and $G(x,y)$ for the follower, and let the leader announce a rule $\phi :Y\to X$. Put $\ell ={\max }_y{\min }_{x}G(x,y)$, the follower's guaranteed payoff;

• Under $(\mathrm{P})$ the leader's ${\Gamma }_2$ value is $\max \{K_0,M_0\}$, and it is attained.

• Under $(\mathrm{L})$ it is $\max \{F(x,y):G(x,y)\ge \ell \}$.

Let $\Theta$ and the signal sets be finite. $\pi {⪰}_{\mathrm{B}}{\pi }^{\prime }$ if and only if, for every finite action set, every utility $u(\theta ,a)$ and every prior on $\Theta$ (the same prior for both experiments), the optimal expected utility of a Bayesian decision-maker observing $\pi$ is at least that of one observing ${\pi }^{\prime }$, actions, payoffs and observation costs being the same.

Consider a two-player zero-sum stochastic game with finite state set $\mathcal{X}$, finite stage action sets $X_A$, $X_B$ and horizon $T$. Before every simultaneous move both players observe the date $t$ and the state $x$; payoffs are the expected sum of stage rewards $(r_t,-r_t)$; the stage model and the transition kernel $P_t(x^{\prime }\mid x,a,b)$ are common knowledge; independently randomised actions are admissible. Define $v_T\equiv 0$ and, for $t<T$,

each $\mathrm{val}$ a finite matrix-game value (Theorem 0.6 (p. 7)). Then the game starting at $x$ at date $0$ has value $v_0(x)$, and the policies that play an optimal mixture of the stage matrix at each $(t,x)$ are optimal; they are Markov, and need not be stationary.

For all $p,q$: $L(p)\le p^{\top }Gq\le U(q)$, hence ${\max }_{p}L(p)\le {\min }_{q}U(q)$.

In a finite zero-sum game, for any pair $(p,q)$: ${\mathrm{expl}}_A+{\mathrm{expl}}_B=U(q)-L(p)=\mathrm{gap}\ge 0$. A certificate is precisely a proof that the total exploitability of the pair is zero.