Randomization lowers worst-case risk under zero-one loss

The one-point data law assigns mass one to its only point under either parameter, so it is a probability measure by [F1]. For each $q\in [0,1]$, define ${\delta }_q(\ast ,B)=q{1}_{\{1\in B\}}+(1-q)1_{\{0\in B\}}$ for $B\subseteq \{0,1\}$; its two singleton masses are nonnegative and sum to one, hence it is a probability measure. Evaluations are measurable on the one-point data space, so ${\delta }_q$ is a probability kernel by [F2]. Deterministic rules are the cases $q=0$ and $q=1$.

By the finite expectation formula [F4] and the loss-risk definition [F3], the rule with probability $q$ of action $1$ has risks $R(0)=q$ and $R(1)=1-q$. Its worst-case risk is therefore $\max (q,1-q)\ge (q+1-q)\mathrm{/}2=1\mathrm{/}2$, with equality at $q=1\mathrm{/}2$.

The two deterministic rules have $q=0$ or $q=1$, so each has worst-case risk $1$. The fair randomized rule has worst-case risk $1\mathrm{/}2$, and step 1.2 shows no rule can do better. Thus the minimax value is $1\mathrm{/}2$ and randomization strictly improves worst-case risk over every deterministic rule.