Theorem: Exact uncued accuracy and a memoryless optimizer

For the task in Section sec:task,

at every assessment step. The memoryless predictor

attains this optimum.

Conditional quadrant reference accuracies are

for the positive-positive, negative-negative, positive-negative, and negative-positive quadrants, respectively. Oracle rule access gives accuracy one.

Proof. Outside the null boundaries $x_1=0,x_2=0,x_2=x_1,x_2=-x_1$, the mixture is invariant under coordinate sign changes and exchange. Each quadrant has probability $1/4$, and its two diagonal subregions have equal probability.

In a same-sign quadrant, $A=B$. Three rule IDs therefore vote for $B$. On one half of that quadrant, $D=B$, giving posterior correctness one; on the other half it gives $3/4$. The conditional average is $7/8$.

In an opposite-sign quadrant, $A\ne B$, and two IDs vote for $B$. On one half, $D=B$, giving correctness $3/4$. On the other half, $D$ differs from both $A$ and $B$, so the largest label probability is $1/2$, again attained by $B$. The conditional average is $5/8$.

Hence $B$ is a maximizing rule at every nonboundary point, and

The target is deterministic when $R_t,X_t$ are both known, so the oracle accuracy is one.