Theorem: Conflicting-task and ambiguous-cue bounds

For the task distribution above:

1. $$q_{\mathrm{a}}+q_{\mathrm{w}}\le 1,\Delta \ge q_{\mathrm{v}}+q_{\mathrm{a}}-1.$$

2. If the classifier is cue invariant on the test distribution, then

   $$q_{\mathrm{v}}+q_{\mathrm{a}}\le 1,\Delta =0.$$

   If it is also invariant under cue removal, then ${\Delta }_0=0$.

3. At the midpoint cue, the task-balanced accuracy satisfies

   $$\frac{1}{2}\mathbb{P}\{g_{\theta }(X,c_{1/2};0)=Y\}+\frac{1}{2}\mathbb{P}\{g_{\theta }(X,c_{1/2};0)=\pi (Y)\}\le \frac{1}{2}.$$

The same bounds hold for a randomized classifier if its randomization does not reveal the intended task.

Proof. For every realization of $(X,Y)$, the two events

are disjoint because $\pi (Y)\ne Y$. Their indicator sum is at most one. Taking expectations proves $q_{\mathrm{a}}+q_{\mathrm{w}}\le 1$, and rearrangement gives

If the predictions under $c_{\mathrm{v}}$ and $c_{\mathrm{a}}$ agree almost surely, then $q_{\mathrm{w}}=q_{\mathrm{v}}$, so $\Delta =0$, and the same disjointness argument gives $q_{\mathrm{v}}+q_{\mathrm{a}}\le 1$. Agreement with the zero-cue prediction gives ${\Delta }_0=0$.

At the midpoint cue a single prediction can equal at most one of $Y$ and $\pi (Y)$. Multiplying the corresponding indicator sum by $1/2$ proves (eq:balancedambiguous). Conditioning on the classifier's randomization proves the last assertion.