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 \((1-s)\). ↩︎