Paying for precision

Suppose my payoff u(a,μ)≡−(a−μ)2$u(a,\mu )\equiv -(a-\mu {)}^2$ from taking an action a∈R depends on an unknown state μ∈R.¹ I can learn about μ by collecting data X={x1,x2,…,xn}, where the observations xi are iid normally distributed with mean μ and variance σ2:² xi∣μ∼N(μ,σ2). I use these data, my prior belief μ∼N(μ0,σ20), and Bayes’ rule to form a posterior belief μ∣X∼N(τ0τ0+nτμ0+nττ0+nτˉx,1τ0+nτ), where τ0≡1/σ20 is the precision of my prior, τ≡1/σ2 is the precision of the xi, and ˉx≡1nn∑i=1xi is their arithmetic mean. Then my expected payoff from taking action a equals E[u(a,μ)∣X]=−(a−E[μ∣X])2−Var(μ∣X). I maximize this payoff by choosing a∗≡E[μ∣X]. This yields expected payoff E[u(a∗,μ)∣Xn]=−1τ0+nτ, 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 κnτ to collect n observations, where κ>0 captures the marginal cost of information.³ I choose n to maximize my total payoff U(n)≡E[u(a∗,μ)∣X]−κnτ, which has maximizer n∗=max{0,1τ(1√κ−τ0)}. If 1≤√κτ0 then n∗=0 because the cost of collecting any data isn’t worth the variance reduction they deliver. Whereas if 1>√κτ0 then n∗ is strictly positive and gives me total payoff U(n∗)=−2√κ+κτ0. Both n∗ and U(n∗) are decreasing in κ. 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 τ0) 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 xi more precise (i.e., increasing τ) makes me want to collect more data but does not change my welfare. This is because the cost κτ of each observation xi scales with its precision. This cost exactly offsets the value of the information gained, leaving my total payoff U(n∗) unchanged.