# Setup

I'm trying to find condition(s) that characterize the solution to a statistical decision problem. The environment is as follows.

• $\Omega$ is a finite set of states of the world.
• A decision maker has prior $f_0$ about the distribution of states.
• $g(\cdot|\omega)$ is a conditional distribution of signals when the realized state is $\omega$

The decision problem has two stages:

• Stage 1: the decision maker decides how many observations to draw from $g(\cdot|\omega)$; each observation incurs a constant cost of $c$.
• Stage 2: the decision maker makes a decision to maximize an objective function $\phi(a,\omega)$, where $\phi$ is strictly concave in $a$ for each $\omega$.

Let $n\in\mathbb Z_+$ be the number of observations, and denote the posterior (obtained using Bayes' rule) by $f_n$. Given the concanvity assumption, there is a unique decision $a^*(n)$ that is optimal in stage 2. Then the decision problem can be formulated as $$\max_{\;n\in\mathbb Z_+} \sum_{\omega\in\Omega}f_n(\omega)\phi(a^*(n),\omega)-cn.\tag{1}$$

# Question

As stated, the problem is pretty general. I was wondering if there are the conditions (analogous to first and second order conditions for optimization problems in calculus courses) that characterize the solution to $(1)$? If not, what minimum extra structures does the problem need in order to have one?

This is a mixed integer and convex problem. The general way to solve problems of your type is to solve for the optimum value $val(n)$ for each $n$, and then take the best $n$. However, I think your problem is not clearly specified and is actually more complicated than you give (see below).

The definition of $f_n(\omega)$ is unclear to me. I think this would actually depend on the observations, call them $\{S_1, \ldots, S_n\}$ and assume they are discrete for simplicity. In that case, the problem is more complicated since the choice of $a$ does not just depend on $n$, it also depends on the observations $\{S_1, \ldots, S_n\}$.

So you would have:

\begin{align} &f_n(x, s_1, \ldots, s_n) \\ &=Pr[\omega=x|S_1=s_1, \ldots, S_n=s_n] \\ &= \frac{Pr[S_1=s_1, \ldots, S_n=s_n|\omega=x]Pr[\omega = x]}{\sum_{y\in\Omega} Pr[S_1=s_1, \ldots, S_n=s_n|\omega=y]Pr[\omega=y]} \end{align}

So if you choose a given $n \in \{0, 1, 2, \ldots\}$, you observe $S_1, \ldots S_n$ and then choose $a\in\mathcal{A}$ to maximize $\sum_{x \in \Omega} f_n(x, S_1, \ldots, S_N)\phi(a, x)$. In that case, the expected utility given you choose $n$ is:

$$val(n) = E\left[\max_{a \in \mathcal{A}}\left[\sum_{x\in\Omega}f_n(x, S_1, \ldots, S_n)\phi(a,x)\right]\right] - nc$$

where the expectation is with respect to the random $S_1, \ldots, S_n$ observations given you choose to look at $n$ of them (and so the choice of the optimal $a$ depends on the realizations of $S_1, \ldots, S_n$).

If the $\phi(a,x)$ function satisfies $0 \leq \phi(a,x) \leq \phi_{max}$ for all $(a,x) \in \mathcal{A} \times \Omega$, then you can restrict attention to values $n$ that satisfy $\phi_{max} - nc \geq 0$. So you only need to consider non-negative integers $n$ that are less than or equal to $\phi_{max}/c$. So compute $val(n)$ for $n \in \{0, 1, \ldots, \lfloor \phi_{max}/c\rfloor\}$, then pick the maximizing $n^*$.

• Thanks for the answer :) I have a follow-up question: In the equation for $val(n)$ where the expectation is over $S_1,\dots,S_n$, given how $f_n$ is defined, do we not get the following result: \begin{aligned}val(n)&=E\left[\max_{a\in\mathcal A}\left[\sum_{x\in\Omega} f_n(x,S_1,\dots,S_n)\phi(a,x)\right]\right]-nc\\&=\max_{a\in\mathcal A}\left[\sum_{x\in\Omega}E[f_n(x,S_1,\dots,S_n)]\phi(a,x)\right]-nc\\&=\max_{a\in\mathcal A}\left[\sum_{x\in\Omega}f_0(x)\phi(a,x)\right]-nc,\end{aligned} which would suggest optimal $n^*=0$? Did I err somewhere in my reasoning? – Herr K. Jan 2 '15 at 0:56
• You cannot exchange the expectation and the $\max$. For example, let $S$ be uniformly distributed in $[0,1]$. Then $E[\max_{a\in\mathbb{R}} (5 - (a-S)^2)] = E = 5$. But $\max_{a\in\mathbb{R}} E[5 - (a-S)^2] = \max_{a\in\mathbb{R}} [5 - a^2 + a - 1/3] = 59/12 \approx 4.91667 < 5$. In general, since $f(a,S) \leq \max_{x\in A} f(a,S)$, we have $E[f(a,S)] \leq E[\max_{x \in A} f(a,S)]$ for all $a$, and so $\max_{a\in A}E[f(a,S)] \leq E[\max_{x\in A} f(a,S)]$. – Michael Jan 2 '15 at 5:50
• Typo in comment above, I meant "since $f(a,S) \leq \max_{x\in A} f(x,S)$, we have $E[f(a,S)] \leq E[\max_{x\in A} f(x,S)]$ for all $a$ and so $\max_{a\in A} E[f(a,S)] \leq E[\max_{x \in A} f(x,S)]$. – Michael Jan 3 '15 at 10:59
• Oh yes. Thank you very much! That typo did not cause any confusion though. – Herr K. Jan 3 '15 at 18:48

This problem is studied in Stigler (1961), "The Economics of Information," Journal of Political Economy, Vol. 69.

The literature quickly moved away from this model because in most settings the agent does not choose a fixed number of draws at the outset but rather sequentially chooses whether to incur cost c to get an additional draw.

• Stigler (1961) studies how search and advertising may lead to dispersion in prices. I can see the relevance to my problem in general, but I'm not sure how his paper informs the current problem in particular. Would you please kindly help me see the connection? – Herr K. Dec 30 '14 at 22:02