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
$$\begin{align} q_s &\equiv \Pr(\omega=1\mid s) \\ &= \frac{\Pr(s\mid\omega=1)\Pr(\omega=1)}{\Pr(s)} \\ &= \begin{cases} \frac{\beta p}{(1\alpha)(1p)+\beta p} & \text{if}\ s=0 \\ \frac{(1\beta)p}{\alpha(1p)+(1\beta)p} & \text{if}\ s=1 \end{cases} \end{align}$$
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
$$\DeclareMathOperator{\E}{E} \begin{align} \E[c(a,\omega)\mid s] &= (1q_s)c(a,0)+q_sc(a,1) \\ &= (1q_s)a^2+q_s(a1)^2 \end{align}$$
given \(s\)
, which leads me to choose \(a=q_s\)
.
Then my minimized expected cost
$$\begin{align} \E[c(q_s,\omega)\mid s] &= q_s(1q_s) \\ &= p(1p)\times\begin{cases} \frac{(1\alpha)\beta}{\left((1\alpha)(1p)+\beta p\right)^2} & \text{if}\ s=0 \\ \frac{\alpha(1\beta)}{\left(\alpha(1p)+(1\beta)p\right)^2} & \text{if}\ s=1 \end{cases} \end{align}$$
equals the posterior variance in my belief about \(\omega\)
after receiving \(s\)
.
The expected value of this variance before receiving \(s\)
equals
$$\begin{align} V(p,\alpha,\beta) &\equiv q_0(1q_0)\Pr(s=0)+q_1(1q_1)\Pr(s=1) \\ &= p(1p)\times\frac{\alpha(1\alpha)(1p)+\beta(1\beta)p}{\left((1\alpha)(1p)+\beta p\right)\left(\alpha(1p)+(1\beta)p\right)}, \end{align}$$
which depends on my prior \(p\)
as well as the error rates \(\alpha\)
and \(\beta\)
.
For example, the chart below plots
$$V(p,\epsilon,\epsilon)=p(1p)\times\frac{\epsilon(1\epsilon)}{p(1p)+\epsilon(1\epsilon)(12p)^2}$$
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 \(\omega=0\)
.

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