Suppose I receive a noisy signal $$s\in\{0,1\}$$ about an unknown state $$\omega\in\{0,1\}$$. The signal has false positive rate $$\renewcommand{\epsilon}{\varepsilon} \Pr(s=1\mid\omega=0)=\alpha$$ and false negative rate $$\Pr(s=0\mid\omega=1)=\beta$$ with $$\alpha,\beta\in[0,0.5]$$.1 I use these rates, my prior belief $$p=\Pr(\omega=1)$$, and Bayes’ rule to form a posterior belief \begin{align} q_s &\equiv \Pr(\omega=1\mid s) \\ &= \frac{\Pr(s\mid\omega=1)\Pr(\omega=1)}{\Pr(s)} \\ &= \begin{cases} \frac{\beta p}{(1-\alpha)(1-p)+\beta p} & \text{if}\ s=0 \\ \frac{(1-\beta)p}{\alpha(1-p)+(1-\beta)p} & \text{if}\ s=1 \end{cases} \end{align} that depends on the signal I receive.

Now suppose I take an action $$a\in[0,1]$$ with cost $$c(a,\omega)\equiv(a-\omega)^2$$. I want to minimize my expected cost \DeclareMathOperator{\E}{E} \begin{align} \E[c(a,\omega)\mid s] &= (1-q_s)c(a,0)+q_sc(a,1) \\ &= (1-q_s)a^2+q_s(a-1)^2 \end{align} given $$s$$, which leads me to choose $$a=q_s$$. Then my minimized expected cost \begin{align} \E[c(q_s,\omega)\mid s] &= q_s(1-q_s) \\ &= p(1-p)\times\begin{cases} \frac{(1-\alpha)\beta}{\left((1-\alpha)(1-p)+\beta p\right)^2} & \text{if}\ s=0 \\ \frac{\alpha(1-\beta)}{\left(\alpha(1-p)+(1-\beta)p\right)^2} & \text{if}\ s=1 \end{cases} \end{align} equals the posterior variance in my belief about $$\omega$$ after receiving $$s$$. The expected value of this variance before receiving $$s$$ equals \begin{align} V(p,\alpha,\beta) &\equiv q_0(1-q_0)\Pr(s=0)+q_1(1-q_1)\Pr(s=1) \\ &= p(1-p)\times\frac{\alpha(1-\alpha)(1-p)+\beta(1-\beta)p}{\left((1-\alpha)(1-p)+\beta p\right)\left(\alpha(1-p)+(1-\beta)p\right)}, \end{align} which depends on my prior $$p$$ as well as the error rates $$\alpha$$ and $$\beta$$. For example, the chart below plots $$V(p,\epsilon,\epsilon)=p(1-p)\times\frac{\epsilon(1-\epsilon)}{p(1-p)+\epsilon(1-\epsilon)(1-2p)^2}$$ against $$\epsilon$$ when $$p\in\{0.5,0.7,0.9\}$$. If $$\epsilon=0$$ then the signal is fully informative because it always matches the state $$\omega$$. Larger values of $$\epsilon\le0.5$$ lead to less precise posterior beliefs. Indeed if $$\epsilon=0.5$$ then the signal is uninformative because $$\Pr(s=1)=0.5$$ (and, hence, $$q_0=q_1=p$$) independently of $$\omega$$. The slope $$\partial V(p,\epsilon,\epsilon)/\partial\epsilon$$ falls as my prior $$p$$ moves away from $$0.5$$ because having a more precise prior makes my beliefs less sensitive to the signal. The next chart shows the contours of $$V(p,\alpha,\beta)$$ in the $$\alpha\beta$$-plane. These contours are symmetric across the diagonal line $$\alpha=\beta$$ when my prior $$p$$ equals $$0.5$$ but asymmetric when $$p\not=0.5$$. Intuitively, if I have a strong prior that $$\omega=1$$ then positive signals $$s=1$$ are less surprising, and shift my belief less, than negative signals $$s=0$$. So if $$p>0.5$$ then I need to increase the false positive rate $$\alpha$$ by more than I decrease the false negative rate $$\beta$$ to keep $$V(p,\alpha,\beta)$$ constant. One consequence of this asymmetry is that the constrained minimization problem $$\min_{\alpha,\beta}V(p,\alpha,\beta)\ \text{subject to}\ 0\le\alpha,\beta\le0.5\ \text{and}\ \alpha+\beta\ge B$$ has a corner solution $$(\alpha^*,\beta^*)=\begin{cases} (0,B) & \text{if}\ p\le1/2 \\ (B,0) & \text{if}\ p>1/2 \end{cases}$$ for all lower bounds $$B\in[0,0.5]$$ on the sum of the error rates. Intuitively, if I can limit my exposure to false positives and negatives then I should prevent whichever occur in the state that’s most likely under my prior. For example, if $$p>0.5$$ then I’m best off allowing some false positives but preventing any false negatives. This makes negative signals fully informative because they only occur when $$\omega=0$$.

1. There is no loss in generality from assuming $$\alpha,\beta\le0.5$$ because observing $$s$$ is the same as observing $$(1-s)$$. ↩︎