Identification and welfare evaluation in sequential sampling models

Abstract

Consider an agent who faces choice problems and learns information about an objective state of the world through a technology of sequential experiments. We consider two cases of learning costs. In the first, the agent discounts future payoffs geometrically. In the second, she incurs a constant flow cost of time. If the observable data consist only of the joint distributions over chosen actions and decision times, an analyst can uniquely identify the discount factor in the first case and the flow cost of time in the second case. Moreover, we show how an analyst can recover the agent’s ex ante welfare in both cases, besides identifying her prior belief. Our approach does not rely on any knowledge about the underlying sequential experiment.

Appendices

Addendum to the proof of theorem 1 for SeSa-GD

1.1 Proof of fact 3.

Given a history $e^1$, a menu A and an act $f\in A$, let $P_A^{e^1}(f,t)$ be the probability that the agent chooses f from A at period $t\ge 1$ after history $e^1$ occurs. Note that $W_1(A\cup f^u_r)=\mathbb {E}_{e^1}[\tilde{W}_1(A\cup f^u_r,e^1)]$. By the induction hypothesis,

$$\begin{aligned} \tilde{W}_1(A\cup f^u_r,e^1)=\int _0^\infty \left( 1-\sum _{t=1}^T\delta ^{t-1}P^{e^1}_{A\cup \{f^u_r,f^u_{r'}\}}(f^u_{r'},t)\right) dr'=r+\int _r^\infty \left( 1-\sum _{t=1}^T\delta ^{t-1}P^{e^1}_{A\cup \{f^u_{r'}\}}(f^u_{r'},t)\right) dr'. \end{aligned}$$

So we have

$$\begin{aligned} \frac{d}{dr}\tilde{W}_1(A\cup f^u_r,e^1)=\sum _{t=1}^T\delta ^{t-1}P^{e^1}_{A\cup \{f^u_{r}\}}(f^u_{r},t)\le \sum _{t=1}^TP^{e^1}_{A\cup \{f^u_{r}\}}(f^u_{r},t)=P^{e^1}_{A\cup \{f^u_{r}\}}(f^u_{r})\le 1. \end{aligned}$$

Hence $\frac{d}{dr}\delta W_1(A\cup \{f^u_r\})=\delta \mathbb {E}_{e^1}[\frac{d}{dr}\tilde{W}_1(A\cup f^u_r,e^1)]<1$.

1.2 Proof that $P_{A\cup \{f_r^u\}}(f_r^u,0) = 1$ for $r>\bar{r}$, and $P_{A\cup \{f^u_r\}}(f^u_r,0) = 0$ for $r<\bar{r}$.

When $r>\bar{r}\ge \underline{r}$ it holds $P_{A\cup \{f_r^u\}}(f_r^u,0) = 1$, because of Fact 2 and

$$\begin{aligned} V(A,u,\pi _0)\le \delta W_1(A\cup \{f_r^u\})< r. \end{aligned}$$

When $\underline{r}<r<\bar{r}$, we have

$$\delta W_1(A\cup \{f^u_r\})>\max \{r,V(A,u,\pi _0)\}=V(A\cup \{f^u_r\},u,\pi _0),$$

and thus $P_{A\cup \{f^u_r\}}(A\cup \{f^u_r\},0) = 0$ (this ensures $\mathbb {E}_{e^1\sim \mu (\pi _0)}[P^{e^1}_{A\cup \{f^u_r\}}(f^u_r,t)]=P_{A\cup \{f^u_r\}}(f^u_r,t)$). When $r<\underline{r}\le \bar{r}$, we have $r<\delta W_1(A\cup \{f_r^u\})< V(A,u,\pi _0)$, implying

$$\begin{aligned} P_{A\cup \{f^u_r\}}(A\cup \{f^u_r\},0) = 0. \end{aligned}$$

Proof of Theorem 1 for SeSa-LC

Just as in the case of SeSa-GD, the result is trivially true if the menu A contains only a single constant act which yields only the worst prize. Hence, we exclude this case in the following.

The proof proceeds by induction on the number of periods T.

For $T=0$ the agent’s choice follows Subjective Expected Utility, up to tie-breaking. In particular, Theorem 2 from Lu (2016) applies, and immediately gives the statement.

Suppose the statement is true for $T=n$, and consider a model with $T=n+1$. Fix a menu $A\in \mathcal {A}$ which is not the constant menu containing only the worst prize. Define $\underline{r} = \inf \{r\in \mathbb {R}_+:W_1(A\cup \{f_r^u\})-c\ge V(A,u,\pi _0)\}$ and $\bar{r} = \sup \{r\in \mathbb {R}_+: W_1(A\cup \{f_r^u\})-c\ge r\}$.

The following easily proven Facts are true for $\underline{r},\bar{r}$.

• Fact 1: $\bar{r}<\infty$, and $\underline{r} = 0$ if and only if $W_1(A)-c\ge V(A,u,\pi _0)$.

• Fact 2: For every $r>\bar{r}$, it holds $\tau (A\cup \{f_r^u\}, e^0)=1$. The same is true for any $r\in (0,\underline{r})$ by definition of $\underline{r}$.

• Fact 3: $r\mapsto \delta W_1(A\cup \{f_r^u\})$ is continuous, non-decreasing with derivative almost everywhere strictly smaller than 1. For $r<\bar{r}$ it holds $\delta W_1(A\cup \{f_r^u\})> r$ and for $r>\bar{r}$ it holds $\delta W_1(A\cup \{f_r^u\})< r$.

The first statement of Fact 3 can be proven as in the case of SeSa-GD.

There are two cases to consider for the proof.

Case 1: Menu A has $\underline{r}>\bar{r}$. Since $\underline{r}\ge \bar{r}>0$ we have $V(A,u,\pi _0)>W_1(A)-c$, so that $W_c(A) = V(A,u,\pi _0)$. From Fact 2 it follows overall $\tau (A\cup \{f_r^u\}, e^0)=1$ for every $r\in \mathbb {R}_+$. Conditioning on the value of r, it follows that $P_{A\cup \{f_r^u\}}(f_r^u,0) = 0$ for $r<V(A,u,\pi _0)$ and $P_{A\cup \{f_r^u\}}(f_r^u,0) = 1$ for $r>V(A,u,\pi _0)$. This implies for such a menu A

$$\begin{aligned} \int _0^\infty P_{A\cup \{f_r^u\}}(A)dr&= \int _0^{V(A,u,\pi _0)}dr \\&= V(A,u,\pi _0)\\&= W_c(A). \end{aligned}$$

Case 2: Menu A has $\underline{r}\le \bar{r}$. Note that in this case $\bar{r} = 0$ cannot happen, due to Fact 1. Continuity implies that $\bar{r} = W_1(A\cup \{f_{\bar{r}}^u\})-c$. Given a history $e^1$, a menu A and an act $f\in A$, let $P_A^{e^1}(f,t)$ be the probability that the agent chooses f from A at period $t\ge 1$ after history $e^1$ occurs. By the induction hypothesis, as well as similar arguments to the discounting case, we have

$$\begin{aligned} \bar{r}&= W_1(A\cup \{f_{\bar{r}}^u\}) - c\\&= \mathbb {E}_{e^1\sim \mu (\pi _0)}\left[ \tilde{W}_1(A\cup \{f_{\bar{r}}^u\}, e^1)\right] -c\\&= \mathbb {E}_{e^1\sim \mu (\pi _0)}\left[ \int _0^\infty P^{e^1}_{A\cup \{f^u_r,f^u_{\bar{r}}\}}(A\cup \{f_r^u\})dr\right] -c\\&=\mathbb {E}_{e^1\sim \mu (\pi _0)}\left[ \bar{r} + \int _{\bar{r}}^{\infty }P^{e^1}_{A\cup \{f^u_r,f^u_{\bar{r}}\}}(A\cup \{f_r^u\})dr\right] -c \\&= \bar{r}-c + \mathbb {E}_{e^1\sim \mu (\pi _0)}\left[ \int _{\bar{r}}^{\infty }P^{e^1}_{A\cup \{f^u_r\}}(A)dr\right] . \end{aligned}$$

Overall,

$$\begin{aligned} c = \mathbb {E}_{e^1\sim \mu (\pi _0)}\left[ \int _{\bar{r}}^{\infty }P^{e^1}_{A\cup \{f^u_r\}}(A)dr\right] . \end{aligned}$$

(10)

Analogously to the discounting case, it holds $W_c(A) = W_1(A\cup \{f_{\underline{r}}^u\}) -c$. The induction hypothesis and calculations analogous to the derivation of (10) deliver

$$\begin{aligned} W_c(A)&= W_1(A\cup \{f_{\underline{r}}^u\})-c \\&= \mathbb {E}_{e^1\sim \mu (\pi _0)}\left[ \tilde{W}_1(A\cup \{f_{\underline{r}}^u\}, e^1)\right] -c \\&= \mathbb {E}_{e^1\sim \mu (\pi _0)}\left[ \underline{r} + \int _{\underline{r}}^{\infty }P^{e^1}_{A\cup \{f^u_r,f^u_{\underline{r}}\}}(A\cup \{f_r^u\})dr\right] -c \\&= \mathbb {E}_{e^1\sim \mu (\pi _0)}\left[ \underline{r} + \int _{\underline{r}}^{\infty }P^{e^1}_{A\cup \{f^u_r\}}(A)dr\right] -c \\&= \underline{r}-\bar{r} + \mathbb {E}_{e^1\sim \mu (\pi _0)}\left[ \bar{r}+ \int _{\underline{r}}^{\bar{r}}P^{e^1}_{A\cup \{f^u_r\}}(A)dr + \int _{\underline{r}}^{\infty }P^{e^1}_{A\cup \{f^u_r\}}(A)dr\right] - c \\&=\bar{r} - \mathbb {E}_{e^1\sim \mu (\pi _0)}\left[ \int _{\bar{r}}^{\infty }P^{e^1}_{A\cup \{f^u_r\}}(f_r^u)dr\right] \\&= \underline{r} + \int _{\underline{r}}^{\bar{r}}P_{A\cup \{f^u_r\}}(A)dr \\&= \int _{0}^{\infty }P_{A\cup \{f^u_r\}}(A)dr. \end{aligned}$$

Here, in the sixth equality, we have used (10), and in the seventh, we have interchanged the expectation with the integral. In the last equality we have used that for $r>\bar{r}$ it holds $V(A,u,\pi _0)<W_1(A\cup \{f_r^u\})-c <r$, and for $r<\underline{r}$ it holds $r\le W_1(A\cup \{f_r^u\})-c<V(A,u,\pi _0)$. These last statements can be proven by the same type of arguments as for the case of SeSa-GD.