Paying for precision

Suppose my payoff $u(a,\mu )\equiv -(a-\mu {)}^2$ from taking an action $a\in R$ depends on an unknown state $\mu \in R$.¹ I can learn about $\mu$ by collecting data $X=\{x_1,x_2,\dots ,x_n\}$, where the observations $x_i$ are iid normally distributed with mean $\mu$ and variance ${\sigma }^2$:² $$x_i\mid \mu \sim N(\mu ,{\sigma }^2).$$ I use these data, my prior belief $$\mu \sim N({\mu }_0,{\sigma }_0^2),$$ and Bayes’ rule to form a posterior belief $$\mu \mid X\sim N(\frac{{\tau }_0}{{\tau }_0+n\tau }{\mu }_0+\frac{n\tau }{{\tau }_0+n\tau }\overline{x},\frac{1}{{\tau }_0+n\tau }),$$ where ${\tau }_0\equiv 1/{\sigma }_0^2$ is the precision of my prior, $\tau \equiv 1/{\sigma }^2$ is the precision of the $x_i$, and $$\overline{x}\equiv \frac{1}{n}\sum _{i=1}^n{x}_i$$ is their arithmetic mean. Then my expected payoff from taking action $a$ equals $$E\left[u\right(a,\mu )\mid X]=-(a-E[\mu \mid X]{)}^2-\mathrm{Var}(\mu \mid X).$$ I maximize this payoff by choosing $a^{\ast }\equiv E[\mu \mid X]$. This yields expected payoff $$E\left[u\right(a^{\ast },\mu )\mid X_n]=-\frac{1}{{\tau }_0+n\tau },$$ which is increasing in $n$. Intuitively, collecting more data makes me more informed and makes my optimal action more likely to be “correct.” But data are costly: I have to pay $\kappa n\tau$ to collect $n$ observations, where $\kappa >0$ captures the marginal cost of information.³ I choose $n$ to maximize my total payoff $$\begin{array}{cc}U\left(n\right)& \equiv E\left[u\right(a^{\ast },\mu )\mid X]-\kappa n\tau ,\end{array}$$ which has maximizer $$n^{\ast }=max\{0,\frac{1}{\tau }(\frac{1}{\sqrt{\kappa }}-{\tau }_0)\}.$$ If $1\le \sqrt{\kappa }{\tau }_0$ then $n^{\ast }=0$ because the cost of collecting any data isn’t worth the variance reduction they deliver. Whereas if $1>\sqrt{\kappa }{\tau }_0$ then $n^{\ast }$ is strictly positive and gives me total payoff $$U\left(n^{\ast }\right)=-2\sqrt{\kappa }+\kappa {\tau }_0.$$ Both $n^{\ast }$ and $U\left(n^{\ast }\right)$ are decreasing in $\kappa$. Intuitively, making the data more expensive makes me want to collect less, leaving me less informed and worse off. In contrast, making my prior more precise (i.e., increasing ${\tau }_0$) makes me want to collect less data but leaves me better off. This is because being well-informed means I can pay for less data and still be well-informed.

Curiously, making the $x_i$ more precise (i.e., increasing $\tau$) makes me want to collect more data but does not change my welfare. This is because the cost $\kappa \tau$ of each observation $x_i$ scales with its precision. This cost exactly offsets the value of the information gained, leaving my total payoff $U\left(n^{\ast }\right)$ unchanged.