Theorem: Conditional non-monotonic trade-off and maximizing curve

Under Assumption ass:comparison:

1. For $e\le e_0$, $A_{\mathrm{lin}}(e,f)$ is maximized at $f=0$.

2. For $e>e_0$, its unique maximizer is

   $$f^{\ast }(e)=\frac{k(e-e_0)}{2\lambda }\in (0,F).$$

   This curve is strictly increasing.

3. For $e>e_0$,

   $$A_{\mathrm{lin}}(e,f)>A_{\mathrm{lin}}(e,0)$$

   whenever

   $$0<f<\min \left\{F,\frac{k(e-e_0)}{\lambda }\right\}.$$

   The ambiguity surface is non-monotonic in $f$, with its peak at (eq:explicitpeak).

4. For every fixed $\sigma >0$, $S_{\mathrm{lin}}(e,f,\sigma )$ is strictly decreasing in $f$. It is also strictly decreasing as $e$ increases along the maximizing curve $f=f^{\ast }(e)$, $e>e_0$.

5. Endpoint cue use remains nonzero. More precisely, if

   $$\gamma =\underset{e,f}{\min }\{h(f)-\text{abs}b(e,f)\}>0,$$

   then the zero-stress cue drop satisfies

   $${\Delta }_{\mathrm{lin}}(e,f)\ge 2\Phi (\gamma /\tau )-1>0.$$

   The bound can be made arbitrarily close to one by increasing the separation $h_0/\tau$.

Proof. Because $\Phi$ is strictly increasing, maximizing $A_{\mathrm{lin}}$ is equivalent to maximizing

Its derivative is $k(e-e_0)-2\lambda f$, and its second derivative is $-2\lambda <0$. For $e\le e_0$, the maximum is at zero. For $e>e_0$, the stationary point is (eq:explicitpeak), and the first parameter condition puts it inside $(0,F)$. Also,

which proves the strict-gain interval.

Put

Condition (eq:cuepreservation) gives $\mu >0$. Furthermore,

Therefore ${\partial }_f(\mu /D)<0$, proving the fixed-$e$ stress ordering.

Along the maximizing curve,

The denominator $D(f^{\ast }(e))$ is strictly increasing. The ratio, and consequently $S_{\mathrm{lin}}$, is strictly decreasing along that curve.

Finally, the two endpoint probabilities against the visual reference are

The first is at least $\Phi (\gamma /\tau )$, and the second is at most $\Phi (-\gamma /\tau )$. Their difference proves (eq:comparisoncue).