1+p^{I}_{t}(m_{\emptyset}\mid(m_{\emptyset})^{t})\cdot\mathbb{P}^{I}(\theta\mid(m_{\emptyset})^{t+1})$
$\displaystyle\geq
p^{I}_{t}(m_{\emptyset}\mid(m_{\emptyset})^{t})\cdot\mu_{t+1}(\theta)$
$\displaystyle=\mathbb{P}(\tau>t+1\mid\tau>t)\cdot\mu_{t+1}(\theta),$

which yields the second condition.

($2\Rightarrow 1$) Define a simple information structure $I$ as follows:

$p^{I}_{t}(m_{\theta}\mid(m_{\emptyset})^{t})=\mu_{t}(\theta)-\mathbb{P}(\tau>t+1\mid\tau>t)\cdot\mu_{t+1}(\theta),$

which, from the boundary condition (ii), is weakly greater than $0$. This
means
$p^{I}_{t}(m_{\emptyset}\mid(m_{\emptyset})^{t})=\mathbb{P}(\tau>t+1\mid\tau>t)\in[0,1]$,
so the information structure $I$ is well-defined. Clearly, if the DM always
continues paying attention until seeing $(m_{\theta})$, the (random) stopping
time has the same distribution as that of $\tau$. Therefore, from the first
condition, the IC constraint holds in such a way that the DM always continue
paying attention after seeing $(m_{\emptyset})^{t}$, which implies $I$ is
obedient. Hence, $\tau\overset{d}{=}\tau(I)$. ∎

### B.4 Proof of Proposition 4 (Increasing and extremal belief paths)

Notation. We first introduce some useful notation. Define the set of belief
paths with $\mathcal{W}=(\Delta(\Theta))^{\mathcal{T}}$. Fixing a stopping
time $\tau$, define $\mathcal{W}(\tau)\subset\mathcal{W}$ as the set of belief
paths corresponding to stopping time $\tau$.

######  Definition 10 (Maximal belief path).

Fix stopping time $\tau$. A belief path
$(\mu_{t})_{t\in\mathcal{T}}\in\mathcal{W}(\tau)$ is a maximal belief path
under $\tau$ if there is no
$(\mu^{\prime}_{t})_{t\in\mathcal{T}}\in\mathcal{W}(\tau)$ such that
$\phi(\mu^{\prime}_{t})\geq\phi(\mu_{t})$ for every $t\in\mathcal{T}$ and the
inequality is strict for some $t\in\mathcal{T}$.

The following proposition guarantees the existence of a maximum belief path
for any feasible stopping time $\tau$. The proof is deferred to Online
Appendix IV. This will be a useful object to prove Propositions 4 and 5.

######  Proposition 11.

For every feasible stopping time $\tau$, there exists a maximal belief path
corresponding to $\tau$.

Proof idea. We now outline the ideas behind the proof before presenting it
formally. From Proposition 3, we know that simple information structures are
characterized by two functions over $\mathcal{T}$, and relate to each other
via the IC and Boundary constraints. Notice, however, that _fixing_ the
distribution of $\tau$, the curve $\mathbb{E}[\tau|\tau>t]$ as a function of
$t$ is completely pinned down. Our goal is to show that any maximal belief
path has the property of increasing and extremal belief paths. We prove by a
contradiction that if a maximal belief path $(\mu_{t})_{t\in\mathcal{T}}$ is
not increasing on $\phi$ over $\mathcal{T}$ (or not extremal), we can find
another belief path $(\mu_{t}^{*})_{t}$ fulfilling the boundary and IC
constraints which is increasing in $\phi$ over $\mathcal{T}$ (or extremal),
and $\phi(\mu^{*}_{t})\geq\phi(\mu_{t})$ for every $t\in\mathcal{T}$ with at
least one strict inequality. This contradicts the property of a maximal belief
path, which establishes the proposition.

Figure 13: Illustration of the iterative ironing procedure

(a) Single ironing step

(b) Iterative ironing

To establish the existence of this maximal belief path, we first suppose that
a maximal belief path $(\mu_{t})_{t\in\mathcal{T}}$ is not increasing on
$\phi$ over $\mathcal{T}$. Our proof will proceed inductively. We index the
sequence of beliefs created at each inductive step with superscripts (e.g.,
$(\mu^{i}_{t})_{t}$ is the sequence of beliefs created at inductive step $i$)
and as usual, use subscripts to track time. At each step
$i\in\mathcal{T}\cup\\{-1\\}$, assuming that the sequence of beliefs from the
previous step is $(\mu^{i-1}_{t})_{t}$ and that up to period $i$, the beliefs
are an increasing function of $\phi$ i.e., $\phi(\mu^{i-1}_{t})$ is increasing
over the interval $t\leq i-1$, we construct a new sequence of beliefs,
$(\mu^{i}_{t})_{t}$ by retaining the first $i-1$ beliefs in the previous
sequence (i.e., $(\mu^{i}_{t})_{t=1}^{i-1}=(\mu^{i-1}_{t})_{t=1}^{i-1}$) and
“straightening out” the decreasing portions of the sequence
$(\mu^{i-1}_{t})_{t=i}^{{+\infty}}$.

More explicitly, let $t_{i-1}$ be the first time $\phi(\mu^{i-1}_{t})$ exceeds
$\phi(\mu^{i-1}_{i-1})$ i.e.,
$\phi(\mu^{i-1}_{i-1})\leq\phi(\mu^{i-1}_{t_{i-1}})$ but
$\phi(\mu^{i-1}_{{i-1}})>\phi(\mu^{i-1}_{{s}})$ for all ${i-1}<s<t_{i-1}$. By
choosing weights carefully, we can find an alternate sequence of beliefs
$(\mu^{i}_{t})_{t=i-1}^{t_{i-1}}$ which is are convex combinations of the
beliefs $\mu^{i-1}_{i-1}$ and $\mu^{i-1}_{t_{i-1}}$ while, at the same time,
respecting both boundary and obedience constraints. Call this new sequence of
beliefs $(\mu^{i}_{t})_{t}$ where for $t\leq i-1$, the beliefs are the same as
$(\mu^{i-1}_{t})_{t}$ and for $t>i-1$, the beliefs follow the modification we
performed. However, this new sequence of beliefs is not necessarily increasing
in $\phi$ over $t=i-1,\ldots,t_{i-1}$ so we might have, through this process,
created a new decreasing region. But the sequence is (i) concave; and (ii)
locally strictly increasing i.e., $\phi(\mu^{i}_{i-1})<\phi(\mu^{i}_{i})$. The
outcome of this procedure is illustrated in Figure 13b (a). The solid red line
plots $(\phi(\mu^{i-1}_{t}))_{t}$, the value of full information under the
original path of beliefs; the solid blue line plots the additional expected
costs of waiting until the DM receives full information. Obedience requires
that the red line is, at all times, above the blue line. Following the
modification above, the dotted red line depicts the value of information under
the modified path of beliefs $(\phi(\mu^{i}_{t}))_{t}$. We then iteratively
perform this procedure; the outcome is depicted by the dotted red line in
Figure 13b (b).

We now prove Proposition 4 formally.

###### Proof of Proposition 4 (i) (increasing paths).

Start from an arbitrary simple and obedient dynamic information structure
$I\in\mathcal{I}^{*}$. Denote the DM’s beliefs upon receipt of the null
message up to time $t$ under this information structure with
$\mu^{I}_{t}:=\mu|(m_{\emptyset})^{t}$.

Consider any sequence of
$(\mu^{i}_{t})_{t\in\mathcal{T}}\in(\Delta(\Theta))^{\mathcal{T}}$. We
recursively construct another sequence of beliefs
$(\mu^{i+1}_{t})_{t\in\mathcal{T}}\in(\Delta(\Theta))^{\mathcal{T}}$ which
satisfies the conditions in Lemma 3 for each $i\in\mathcal{T}\cup\\{-1\\}$ as
follows:

  * (i)

If $i=-1$, set $\mu^{i}_{t}=\mu^{I}_{t}$ for every $t\in\mathcal{T}$.

  * (ii)

If $i\geq 0$, given the sequence of
$(\mu^{i}_{t})_{t\in\mathcal{T}}\in(\Delta(\Theta))^{\mathcal{T}}$, define

$t_{i}=\begin{cases}\min\\{t\in\mathcal{T}:t>i,\phi(\mu^{i}_{i})\leq\phi(\mu^{i}_{t})\\}&\,\text{if
the minimum exists,}\\\ +\infty&\,\text{otherwise.}\end{cases}$

    * $\bullet$

If $t_{i}=+\infty$, set $\mu^{i+1}_{t}=\begin{cases}\mu^{i}_{t},&\text{if
}t<i\\\ \mu^{i}_{i},&\text{if }t\geq i\end{cases}$
i.e., the sequence $(\mu^{i+1}_{t})_{t}$ follows the sequence
$(\mu^{i}_{t})_{t}$ up to period $i$, and remains constant at belief
$\mu^{i}_{i}$ thereafter.

    * $\bullet$

If $t_{i}<{+\infty}$, define
$\theta_{i}\in\text{argmax}_{\theta\in\Theta}\frac{\mu_{t_{i}}^{i}(\theta)}{\mu_{i}^{i}(\theta)}$
i.e., the state at which the ratio of the beliefs at time $i$ and time
$t_{i}$—the time at which $\phi$ increases relative to that at time $i$.

It is clear that
$\frac{\mu^{i}_{t_{i}}(\theta_{i})}{\mu^{i}_{i}(\theta_{i})}\geq 1$. Thus,
there exists a sequence $\pi^{i}_{i},\dots,\pi^{i}_{t_{i}-1}\in\mathbb{R}$
such that
$\pi^{i}_{t}\in\big{[}1,\frac{\mathbb{P}(\tau>t)}{\mathbb{P}(\tau>t+1)}\big{]}$
for every $t\in\\{i,\dots,t_{i}-1\\}$ and
$\prod_{t=i}^{t_{i}-1}\pi^{i}_{t}=\frac{\mu^{i}_{t_{i}}(\theta_{i})}{\mu^{i}_{i}(\theta_{i})}$
because the second condition of Lemma 3 for $\mu^{i}$ implies

$\frac{\mu^{i}_{t_{i}}(\theta_{i})}{\mu^{i}_{i}(\theta_{i})}\leq\frac{\mathbb{P}(\tau>i)}{\mathbb{P}(\tau>t_{i})}=\prod_{t=i}^{t_{i}-1}\frac{\mathbb{P}(\tau>t)}{\mathbb{P}(\tau>t+1)}.$

This sequence $(\pi_{s}^{i})_{s=i}^{t_{i}-1}$ simply splits up the ratio of
$\mu_{t_{i}}^{i}(\theta_{i})/\mu^{i}_{i}(\theta_{i})$ into $t_{i}-1-i$ sub-
ratios. We will now construct a new sequence of beliefs to bridge
$\mu^{i}_{i}$ and $\mu^{i}_{t_{i}}$ while ensuring that $\phi$ increases over
the interval. For any $t\in\\{i,\dots,t_{i}\\}$, set

$\mu^{i+1}_{t}=\lambda_{i,t}\mu^{i}_{i}+(1-\lambda_{i,t})\mu^{i}_{t_{i}},$

i.e., a linear combination of the belief at $t_{i}$ and that at $i$ where the
weights are given by

$\displaystyle\lambda_{i,t}=\frac{1}{\mu^{i}_{t_{i}}(\theta_{i})-\mu^{i}_{i}(\theta_{i})}\Big{(}\mu^{i}_{t_{i}}(\theta_{i})-\prod_{s=i}^{t-1}{\pi^{i}_{s}}\cdot\mu^{i}_{i}(\theta_{i})\Big{)}.$

Moreover, if $t\notin\\{i,\dots,t_{i}\\}$, we set $\mu^{i+1}_{t}=\mu^{i}_{t}.$
From the construction of $\mu^{i+1}$, note that $\mu^{i+1}_{i}=\mu^{i}_{i}$
and $\mu^{i+1}_{t_{i}}=\mu^{i}_{t_{i}}$

We will now inductively show that a sequence $(\mu_{t}^{i})_{t\in\mathcal{T}}$
satisfies the conditions in Proposition 3 for each
$i\in\mathcal{T}\cup\\{-1\\}$:

Basic step.
$(\mu^{-1}_{t})_{t\in\mathcal{T}}=(\mu^{I}\mid(m_{\emptyset})^{t})_{t\in\mathcal{T}}$
satisfies the conditions in Proposition 3 because by $I$ was assumed to be
simple and obedient.

Inductive step. Suppose $t\in\mathcal{T}$ such that
$(\mu^{i}_{t})_{t\in\mathcal{T}}$ satisfies the conditions in Proposition 3.
We consider two cases.

Case 1: $t_{i}={+\infty}$. We have that $\phi(\mu^{i}_{i})>\phi(\mu^{i}_{t})$
for every $t>i$. This implies
$\phi(\mu^{i+1}_{t})=\phi(\mu^{i}_{i})>\phi(\mu^{i}_{t})\geq\mathbb{E}[c(\tau)\mid\tau>t]-c(t)$
for every $t>i$. For every $t\leq i$,
$\phi(\mu^{i+1}_{t})=\phi(\mu^{i}_{t})\geq\mathbb{E}[c(\tau)\mid\tau>t]-c(t).$
Therefore, the boundary constraint holds for every time $t\in\mathcal{T}$ for
the sequence of beliefs $(\mu^{i+1}_{t})_{t\in\mathcal{T}}$. Moreover, for
every $t<i$, we know
$\frac{\mu^{i+1}_{t+1}(\theta)}{\mu^{i+1}_{t}(\theta)}=\frac{\mu^{i}_{t+1}(\theta)}{\mu^{i}_{t}(\theta)}\leq\frac{\mathbb{P}(\tau>t)}{\mathbb{P}(\tau>t+1)}.$
For every $t\geq i$, we have
$\frac{\mu^{i+1}_{t+1}(\theta)}{\mu^{i+1}_{t}(\theta)}=\frac{\mu^{i}_{i}(\theta)}{\mu^{i}_{i}(\theta)}=1\leq\frac{\mathbb{P}(\tau>t)}{\mathbb{P}(\tau>t+1)}$.
Thus, the boundary constraint holds for every time $t\in\mathcal{T}$ for the
sequence of beliefs $(\mu^{i+1}_{t})_{t\in\mathcal{T}}$, as required.

Case 2: $t_{i}<+\infty$. From the construction of
$(\pi^{i}_{s})_{s=i}^{t_{i}-1}$, we know

$\displaystyle
1\leq\prod_{s=i}^{t}\pi_{s}^{i}\leq\prod_{s=i}^{t_{i}-1}\pi_{s}^{i}=\frac{\mu^{i}_{t_{i}}(\theta_{i})}{\mu^{i}_{i}(\theta_{i})}\Longrightarrow
0\leq\lambda_{i,t}\leq 1,$

for every $t\in\\{i,\dots,t_{i}-1\\}.$ We now verify each of the constraints
holds.

  1. (i)

(Obedience constraint) The definition of $t_{i}$ implies
$\phi(\mu_{i})>\phi(\mu_{t})$ for every $t\in\\{i+1,\dots,t_{i}-1\\}$ and
$\phi(\mu_{i})\leq\phi(\mu_{t_{i}})$. Because $\phi$ is concave and
$\lambda_{i,t}\in[0,1]$, Jensen’s inequality implies that for every
$t\in\\{i,\dots,t_{i}-1\\},$

$\displaystyle\phi(\mu^{i+1}_{t})$
$\displaystyle=\phi(\lambda_{i,t}\mu^{i}_{i}+(1-\lambda_{i,t})\mu^{i}_{t_{i}})$
(2)
$\displaystyle\geq\lambda_{i,t}\phi(\mu^{i}_{i})+(1-\lambda_{i,t})\phi(\mu^{i}_{t_{i}})$
$\displaystyle\geq\phi(\mu^{i}_{i})\geq\phi(\mu^{i}_{t})\geq\mathbb{E}[c(\tau)\mid\tau>t]-c(t),$
(3)

where the last inequality follows from the first condition in Proposition 3 of
$\mu^{i}$. As such, for every $t\notin\\{i,\dots,t_{i}-1\\}$, we have
$\phi(\mu^{i+1}_{t})=\phi(\mu^{i}_{t})\geq\phi(\mu^{i}_{i})\geq\phi(\mu^{i}_{t})\geq\mathbb{E}[c(\tau)\mid\tau>t]-c(t)$.
Therefore, the obedience constraint holds for every time $t\in\mathcal{T}$ for
the sequence of beliefs $(\mu^{i+1}_{t})_{t\in\mathcal{T}}.$

  2. (ii)

(Boundary constraint) For every $t\in\\{i,\dots,t_{i}-1\\}$ and
$\theta\in\Theta$, we have

$\displaystyle\frac{\mu^{i+1}_{t+1}(\theta)}{\mu^{i+1}_{t}(\theta)}$
$\displaystyle=\frac{\lambda_{i,t+1}\mu^{i}_{i}(\theta)+(1-\lambda_{i,t+1})\mu^{i}_{t_{i}}(\theta)}{\lambda_{i,t}\mu^{i}_{i}(\theta)+(1-\lambda_{i,t})\mu^{i}_{t_{i}}(\theta)}$
$\displaystyle=\frac{\lambda_{i,t+1}+(1-\lambda_{i,t+1})\mu^{i}_{t_{i}}(\theta)/\mu^{i}_{i}(\theta)}{\lambda_{i,t}+(1-\lambda_{i,t})\mu^{i}_{t_{i}}(\theta)/\mu^{i}_{i}(\theta)}$
$\displaystyle\leq\frac{\lambda_{i,t+1}+(1-\lambda_{i,t+1})\mu^{i}_{t_{i}}(\theta_{i})/\mu^{i}_{i}(\theta_{i})}{\lambda_{i,t}+(1-\lambda_{i,t})\mu^{i}_{t_{i}}(\theta_{i})/\mu^{i}_{i}(\theta_{i})}\hfill{\quad\quad\Big{(}\lambda_{i,t}\leq\lambda_{i,t+1}\text{
and
}\frac{\mu^{i}_{t_{i}}(\theta)}{\mu^{i}_{i}(\theta)}\leq\frac{\mu^{i}_{t_{i}}(\theta_{i})}{\mu^{i}_{i}(\theta_{i})}\Big{)}}$
$\displaystyle=\frac{\prod_{s=i}^{t}\pi^{i}_{s}\cdot\mu^{i}_{i}(\theta_{i})}{\prod_{s=i}^{t-1}\pi^{i}_{s}\cdot\mu^{i}_{i}(\theta_{i})}=\pi^{i}_{t}\leq\frac{\mathbb{P}(\tau>t)}{\mathbb{P}(\tau>t+1)}.$

Moreover, for every $t\notin\\{i,\dots,t_{i}-1\\}$ and $\theta\in\Theta$, we
have
$\frac{\mu^{i+1}_{t+1}(\theta)}{\mu^{i+1}_{t}(\theta)}=\frac{\mu^{i}_{t+1}(\theta)}{\mu^{i}_{t}(\theta)}\leq\frac{\mathbb{P}(\tau\geq
t)}{\mathbb{P}(\tau\geq t+1)}$ from the induction hypothesis. Therefore, the
boundary constraint holds for every time $t\in\mathcal{T}$ and state
$\theta\in\Theta$ for the sequence of beliefs
$(\mu^{i+1}_{t})_{t\in\mathcal{T}}.$

We have shown that the sequence $(\mu_{t}^{i})_{t\in\mathcal{T}}$ satisfies
the conditions in Proposition 3 for each $i\in\mathcal{T}\cup\\{-1\\}$. We now
complete the proof of Proposition 4. Define a sequence
$(\mu^{*}_{t})_{t\in\mathcal{T}}\in(\Delta(\Theta))^{\mathcal{T}}$ such that
$\mu^{*}_{t}=\mu^{t}_{t}$ for every $t\in\mathcal{T}$ i.e., taking the
‘diagonal’ by choosing time $t$’s beliefs to be sequence $t$’s beliefs at time
$t$.

  * •

For the obedience constraint, we see that
$\phi(\mu^{*}_{t})=\phi(\mu^{t}_{t})\geq\frac{\mathbb{P}(\tau>t)}{\mathbb{P}(\tau>t+1)}.$

  * •

For the boundary constraint, we have
$\frac{\mu^{*}_{t+1}(\theta)}{\mu^{*}_{t}(\theta)}=\frac{\mu^{t+1}_{t+1}(\theta)}{\mu^{t}_{t}(\theta)}=\frac{\mu^{t+1}_{t+1}(\theta)}{\mu^{t+1}_{t}(\theta)}\leq\frac{\mathbb{P}(\tau>t)}{\mathbb{P}(\tau>t+1)},$
as desired.

  * •

For the increasing property, we see that from Equation 3 that
$\phi(\mu^{*}_{t})=\phi(\mu^{t}_{t})\leq\phi(\mu^{t+1}_{t+1})=\phi(\mu_{t+1}^{*})$.

By Lemma 3, we can therefore find a simple and obedient information structure
$I^{*}\in\mathcal{I}$ such that $\mu^{I}\mid(m_{\emptyset})^{t}=\mu^{*}_{t}$
for which (i) $\phi(\mu^{I}\mid(m_{\emptyset})^{t})=\phi(\mu^{*}_{t})$ is
weakly increasing in $t$; and (ii) preserves the same distribution of stopping
times i.e., $\tau(I)\overset{d}{=}\tau(I^{*})$ (which was held fixed
throughout the proof). From the proof of the obedience constraint, we have
$\phi(\mu^{*}_{t})\geq\phi(\mu_{t})$ for every $t\in\mathcal{T}$, and the
inequality is strict for some $t\in\mathcal{T}$ if
$(\phi(\mu_{t}))_{t\in\mathcal{T}}$ is not weakly increasing. This contradicts
the fact that $(\mu_{t})_{t\in\mathcal{T}}$ is a maximal belief path under
$\tau$. ∎

It remains to show that a maximal belief path $(\mu_{t})_{t\in\mathcal{T}}$
must have a property of extremal belief paths

###### Proof of Proposition 4 (ii) (extremal paths).

Consider feasible stopping time $\tau$. Pick a maximal belief path
$(\mu_{t})_{t\in\mathcal{T}}$ under $\tau$. From the proof before, if
$(\phi(\mu_{t}))_{t\in\mathcal{T}}$ is not weakly increasing, we can find
another belief path $(\mu^{\prime}_{t})_{t\in\mathcal{T}}$ such that
$(\phi(\mu_{t}))_{t\in\mathcal{T}}$ is weakly increasing and
$\phi(\mu_{t})\geq\phi(\mu^{\prime}_{t})$, which contradicts the fact that
$(\mu_{t})_{t\in\mathcal{T}}$ is a maximal belief path under $\tau$.

Next, we will show that $(\mu_{t})_{t\in\mathcal{T}}$ has a property of
extremal paths. Suppose a contradiction that there is a minimum
$t_{0}\in\mathcal{T}$ such that $\mu_{t_{0}+1}\notin\Phi^{*}$ and
$\frac{\mathbb{P}(\tau>t_{0}+1)}{\mathbb{P}(\tau>t_{0})}<\min_{\theta\in\Theta}\frac{\mu_{t_{0}}(\theta)}{\mu_{t_{0}+1}(\theta)}$.
Pick any $\mu^{*}\in\Phi^{*}$. We pick $\lambda\in[0,1)$ as following. If
$\min_{\theta\in\Theta}\frac{\mu_{t_{0}}(\theta)}{\mu^{*}(\theta)}\geq\frac{\mathbb{P}(\tau>t_{0}+1)}{\mathbb{P}(\tau>t_{0})}$,
choose $\lambda=0$. Otherwise, choose $\lambda\in[0,1)$ such that

$\displaystyle\frac{\mathbb{P}(\tau>t_{0}+1)}{\mathbb{P}(\tau>t_{0})}=\min_{\theta\in\Theta}\frac{\mu_{t_{0}}(\theta)}{\lambda\mu_{t_{0}+1}(\theta)+(1-\lambda)\mu^{*}(\theta)}.$

The existence of $\lambda$ follows from the intermediate value theorem. We
define a new belief path $(\mu^{\prime}_{t})_{t\in\mathcal{T}}$ as follows:

$\displaystyle\mu^{\prime}_{t}=\begin{cases}\mu_{t},&\text{if $t\leq
t_{0}$}\\\ \lambda\mu_{t}+(1-\lambda)\mu^{*},&\text{ otherwise.}\end{cases}$

We will show that a pair of a belief path and a distribution of stopping time
$((\mu^{\prime}_{t})_{t\in\mathcal{T}},d(\tau))$ is feasible. The obedience
constraint is still the same for $t\in\\{0,\dots,t_{0}\\}$. For $t\geq t_{0}$,
we have

$\displaystyle\phi(\mu^{\prime}_{t})=\phi(\lambda\mu_{t}+(1-\lambda)\mu^{*})\geq\lambda\phi(\mu_{t})+(1-\lambda)\phi^{*}\geq\phi(\mu_{t})\geq\mathbb{E}[c(\tau)\mid\tau>t]-c(t),$

so the obedience constraint for every $t\in\mathcal{T}$. The boundary
constraint is still the same for $t\in\\{0,\dots,t_{0}-1\\}$. For $t=t_{0}$,
the boundary constraint holds because of the construction of $\lambda$. For
$t>t_{0}$, we have

$\displaystyle\min_{\theta\in\Theta}\frac{\mu^{\prime}_{t}(\theta)}{\mu^{\prime}_{t+1}(\theta)}=\min_{\theta\in\Theta}\frac{\lambda\mu_{t}(\theta)+(1-\lambda)\mu^{*}(\theta)}{\lambda\mu_{t+1}(\theta)+(1-\lambda)\mu^{*}(\theta)}\geq\min_{\theta\in\Theta}\frac{\mu_{t}(\theta)}{\mu_{t+1}(\theta)}\geq\frac{\mathbb{P}(\tau>t+1)}{\mathbb{P}(\tau>t)},$

where the first inequality follows from the fact that $\theta$ that minimizes
the LHS must satisfy $\mu_{t}(\theta)\leq\mu_{t+1}(\theta)$. This concludes
that a pair of a belief path and a distribution of stopping time
$((\mu^{\prime}_{t})_{t\in\mathcal{T}},d(\tau))$ is feasible.

For every $t\leq t_{0}$, $\phi(\mu^{\prime}_{t})=\phi(\mu_{t})$. Moreover, for
every $t>t_{0}$, we have

$\displaystyle\phi(\mu^{\prime}_{t})=\phi(\lambda\mu_{t}+(1-\lambda)\mu^{*})\geq\lambda\phi(\mu_{t})+(1-\lambda)\phi^{*}\geq\phi(\mu_{t}).$

Therefore, $\phi(\mu^{\prime}_{t})\geq\phi(\mu_{t})$ for every
$t\in\mathcal{T}$. Because $(\mu_{t})_{t\in\mathcal{T}}$ is a maximal path
under $\tau$, we must have $\phi(\mu^{\prime}_{t})=\phi(\mu_{t})$ for every
$t\in\mathcal{T}$, which implies that $\phi(\mu_{t})=\phi^{*}$ for every
$t>t_{0}$. This contradicts the fact that $\mu_{t_{0}+1}\notin\Phi^{*}$.
Therefore, the belief path $(\mu_{t})_{t\in\mathcal{T}}$ must have a property
of extremal paths, as desired. ∎

With these two parts of the proof, Proposition 4 follows by considering
maximal belief paths of any feasible stopping time.

### B.5 Proof of Proposition 5

###### Proof of Proposition 5.

Assume that $\Theta=\\{0,1\\}$. We abuse notation of a belief
$\mu\in\Delta(\Theta)$ by $\mu=\mu(0)\in[0,1]$. Pick $\mu^{*}\in\Phi^{*}$. By
the optimality of $\phi(\mu^{*})$ and the concavity of $\phi$, we must have
$\phi(\mu)$ is increasing and decreasing if $\mu<\mu^{*}$ and $\mu>\mu^{*}$,
respectively.

Without loss of generality, assume that $\mu_{0}<\mu^{*}$. Suppose a
contradiction that $\mathbb{P}(\tau>T)>0$ for every $T\in\mathcal{T}$ and
every belief path $\mu_{t}$ corresponding to $\tau$ must satisfy
$\mu_{t}\notin\Phi^{*}$. By Proposition 4, there exists a increasing and
extremal belief path $(\mu_{t})_{t\in\mathcal{T}}$ corresponding to $\tau.$
Therefore, $\phi(\mu_{t})$ is increasing in $t$.

Case 1: $\mu_{t}<\mu^{*}$ for every $t\in\mathcal{T}$. Because $\phi(\mu)$ is
increasing if $\mu<\mu^{*}$ and $\phi(\mu_{t})<\mu^{*}$, we have $\mu_{t}$ is
increasing in $t$. Therefore, $\bar{\mu}\coloneqq\lim_{t\to\infty}\mu_{t}$
exists and less than or equal to $\mu^{*}$. Because
$(\mu_{t})_{t\in\mathcal{T}}$ is extremal, we have

$\displaystyle\frac{\mathbb{P}(\tau>t+1)}{\mathbb{P}(\tau>t)}=\min\bigg{\\{}\frac{\mu_{t}}{\mu_{t+1}},\frac{1-\mu_{t}}{1-\mu_{t+1}}\bigg{\\}}=\frac{\mu_{t}}{\mu_{t+1}}$

for every $t\in\mathcal{T}$, which implies that

$\displaystyle
0=\lim_{T\to\infty}\mathbb{P}(\tau>T)=\prod_{t=0}^{\infty}\frac{\mathbb{P}(\tau>t+1)}{\mathbb{P}(\tau>t)}=\prod_{t=0}^{\infty}\frac{\mu_{t}}{\mu_{t+1}}=\frac{\mu_{0}}{\bar{\mu}}>0,$

which is a contradiction.

Case 2: there exists $t\in\mathcal{T}$ such that $\mu_{t}>\mu^{*}$. Because
$\mu_{0}<\mu^{*}$, we can find $t_{0}\in\mathcal{T}$ such that
$\mu_{t_{0}-1}<\mu^{*}$ but $\mu_{t_{0}}>\mu^{*}$. Therefore,

$\displaystyle\frac{\mathbb{P}(\tau>t_{0})}{\mathbb{P}(\tau>t_{0}-1)}=\min\bigg{\\{}\frac{\mu_{t_{0}-1}}{\mu_{t_{0}}},\frac{1-\mu_{t_{0}-1}}{1-\mu_{t_{0}}}\bigg{\\}}=\frac{\mu_{t_{0}-1}}{\mu_{t_{0}}}<\frac{\mu_{t_{0}-1}}{\mu^{*}}=\min\bigg{\\{}\frac{\mu_{t_{0}-1}}{\mu_{t^{*}}},\frac{1-\mu_{t_{0}-1}}{1-\mu_{t^{*}}}\bigg{\\}}.$

We define a new belief path $(\mu^{\prime}_{t})_{t\in\mathcal{T}}$ such that

$\displaystyle\mu^{\prime}_{t}=\begin{cases}\mu_{t},&\text{if $t<t_{0}$}\\\
\mu^{*},&\text{otherwise.}\end{cases}$

The inequality above implies that the belief path
$(\mu^{\prime}_{t})_{t\in\mathcal{T}}$ is feasible corresponding to $\tau$.
However, $\mu^{\prime}_{t_{0}}=\mu^{*}\in\Phi^{*}$, which is a contradiction.
∎

##  Appendix C Omitted Proofs in Section 4 (designer’s problem)

### C.1 Proof of Lemma 1

###### Proof of Lemma 1.

$(1\Leftrightarrow 2)$ The forward direction is clear by the obedience
condition of Lemma 3. The converse is also clear because we can simply
consider a sequence of beliefs $(\mu_{t})_{t\in\mathcal{T}}$ such that
$\mu_{t}=\mu_{0}$ for every $t\in\mathcal{T}$ and apply Lemma 3 again.

$(2\Leftrightarrow 3)$ For the forward direction, define
$C_{t}:=\mathbb{E}[c(\tau)\mid\tau>t]-c(t)\in[0,\phi^{*}]$, for every
$t\in\mathcal{T}$. Then,

$\displaystyle C_{t}+c(t)$ $\displaystyle=\mathbb{E}[c(\tau)\mid\tau>t]$
$\displaystyle=\mathbb{E}[c(\tau)\mid\tau>t+1]\mathbb{P}(\tau>t+1\mid\tau>t)+c(t+1)\mathbb{P}(\tau=t+1\mid\tau>t)$
$\displaystyle=\mathbb{E}[c(\tau)\mid\tau>t+1]\mathbb{P}(\tau>t+1\mid\tau>t)+c(t+1)(1-\mathbb{P}(\tau>t+1\mid\tau>t))$
$\displaystyle=c(t+1)+C_{t+1}\mathbb{P}(\tau>t+1\mid\tau>t),$

which implies that

$\displaystyle\mathbb{P}(\tau>t+1\mid\tau>t)=\frac{C_{t}+c(t)-c(t+1)}{C_{t+1}},$

as desired. Note here that $\mathbb{P}$ is with respect to the stopping time
we hold fixed. For the converse, we use the same computation to obtain
$\mathbb{E}[c(\tau)\mid\tau>t]-c(t)=C_{t}\leq\phi^{*}$ for every
$t\in\mathcal{T}$. ∎

### C.2 Proof of Proposition 6 (i)

###### Proof of Proposition 6 (i).

$(\Rightarrow)$ First, we show the proposition in the case of simple and
obedient information structures. Suppose a contradiction that there is a
distribution $d\in\mathcal{D}$ on the FOSD frontier such that its
corresponding simple and obedient information structure makes the DM does not
stop paying attention at period 0. This implies that

$\displaystyle\phi(\mu_{0})>\mathbb{E}[c(\tau)].$

Pick a corresponding maximal belief path $(\mu_{t})_{t\in\mathcal{T}}$ and let
$t^{*}\in\mathcal{T}$ be the minimum of $t$ such that
$\mathbb{P}(\tau>t^{*})<1.$ By Proposition 3, we have

$\displaystyle\phi(\mu^{\prime}_{t})\geq\mathbb{E}[c(\tau^{\prime})\mid\tau^{\prime}>t]-c(t)$
$\displaystyle\iff\mathbb{P}(\tau^{\prime}>t)\phi(\mu^{\prime}_{t})\geq\sum_{s=t+1}^{\infty}\big{(}c(s)-c(t)\big{)}\big{(}\mathbb{P}(\tau^{\prime}>s-1)-\mathbb{P}(\tau^{\prime}>s)\big{)}$
$\displaystyle\iff\mathbb{P}(\tau^{\prime}>t)(\phi(\mu^{\prime}_{t})+c(t)-c(t+1))\geq\sum_{s=t+1}^{\infty}(c(s+1)-c(s))\mathbb{P}(\tau^{\prime}>s).$
(4)

We set

$\displaystyle\mathbb{P}(\tau^{\prime}>t)=\begin{cases}\mathbb{P}(\tau>t)+\epsilon,&\text{if
$t=t^{*}$}\\\ \mathbb{P}(\tau>t)&\text{otherwise.}\end{cases}$

for arbitrarily small $\epsilon>0$. We can see that
$\tau^{\prime}\overset{FOSD}{\succ}\tau$. We construct a belief path
$(\mu^{\prime}_{t})_{t\in\mathcal{T}}$ as follows:

$\displaystyle\mu^{\prime}_{t}=\begin{cases}\delta\mu_{0}+(1-\delta)\mu_{t},&\text{if
$t=t^{*}$}\\\ \mu_{t}&\text{otherwise,}\end{cases}$

where $\delta>0$ is a unique solution of

$\displaystyle\delta+\frac{1-\delta}{\mathbb{P}(\tau>t^{*})}=\frac{1}{\mathbb{P}(\tau>t^{*})+\epsilon}\Longrightarrow\delta=\frac{\epsilon}{(\mathbb{P}(\tau>t^{*})+\epsilon)(1-\mathbb{P}(\tau>t^{*}))}$

First, we will show that the boundary constraint holds for a pair of a
stopping time and a belief path
$(\tau^{\prime},(\mu^{\prime}_{t})_{t\in\mathcal{T}})$. Consider that

$\displaystyle\frac{\mu^{\prime}_{t^{*}-1}(\theta)}{\mu^{\prime}_{t^{*}}(\theta)}=\frac{\mu_{t^{*}-1}(\theta)}{\delta\mu_{t^{*}-1}(\theta)+(1-\delta)\mu_{t^{*}}(\theta)}=\frac{\mathbb{P}(\tau>t)}{\delta\mathbb{P}(\tau>t)+(1-\delta)}=\mathbb{P}(\tau^{\prime}>t).$

Moreover,

$\displaystyle\frac{\mu^{\prime}_{t^{*}}(\theta)}{\mu^{\prime}_{t^{*}+1}(\theta)}$
$\displaystyle=\frac{\delta\mu_{t^{*}-1}(\theta)+(1-\delta)\mu_{t^{*}}(\theta)}{\mu_{t^{*}+1}(\theta)}$
$\displaystyle\leq\delta\mathbb{P}(\tau^{\prime}>t^{*}+1)+(1-\delta)\frac{\mathbb{P}(\tau^{\prime}>t^{*}+1)}{\mathbb{P}(\tau>t^{*})}$
$\displaystyle=\mathbb{P}(\tau^{\prime}>t^{*}+1)\Big{(}\delta+\frac{1-\delta}{\mathbb{P}(\tau^{\prime}>t^{*})}\Big{)}=\frac{\mathbb{P}(\tau^{\prime}>t^{*}+1)}{\mathbb{P}(\tau^{\prime}>t)},$

as desired. We will show that the obedient constraint holds for a pair of a
stopping time and a belief path
$(\tau^{\prime},(\mu^{\prime}_{t})_{t\in\mathcal{T}})$ for small $\epsilon>0$.
Because $\phi(\mu_{0})>\mathbb{E}[c(\tau)]$ and
$\big{\lvert}\mathbb{E}[c(\tau^{\prime})]-\mathbb{E}[c(\tau)]\big{\rvert}=\epsilon(c(t^{*}+1)-c(t^{*}))$,
we can find sufficiently small $\epsilon>0$ so that
$\phi(\mu_{0})>\mathbb{E}[c(\tau^{\prime})]$. For $t<t^{*}$, because
$\mathbb{P}(\tau>t)=1$, we have

$\displaystyle\phi(\mu^{\prime}_{t})=\phi(\mu_{0})>\mathbb{E}[c(\tau^{\prime})]=\mathbb{E}[c(\tau^{\prime})\mid\tau^{\prime}>t]>\mathbb{E}[c(\tau^{\prime})\mid\tau^{\prime}>t]-c(t).$

For $t>t^{*}$, the obedience constraints are the same for $\tau$ and
$\tau^{\prime}$. Moreover, the obedience constraint for $t<t^{*}$ is slackened
if the obedience constraint at time $t^{*}$ holds. Thus, it is sufficient to
check the obedience constraint at time $t^{*}$. Consider that

$\displaystyle\mathbb{P}(\tau^{\prime}>t^{*})(\phi(\mu^{\prime}_{t^{*}})+c(t^{*})-c(t^{*}+1))$
$\displaystyle=(\mathbb{P}(\tau>t^{*})+\epsilon)(\phi(\delta\mu_{0}+(1-\delta)\mu_{t^{*}})+c(t^{*})-c(t^{*}+1))$
$\displaystyle\geq(\mathbb{P}(\tau>t^{*})+\epsilon)(\delta\phi(\mu_{0})+(1-\delta)\phi(\mu_{t^{*}})+c(t^{*})-c(t^{*}+1))$
$\displaystyle=(\mathbb{P}(\tau>t^{*})+\epsilon)(\phi(\mu_{t^{*}})+c(t^{*})-c(t^{*}+1))+\delta(\phi(\mu_{0})-\phi(\mu_{t^{*}}))\mathbb{P}(\tau>t^{*})+C\delta\epsilon,$

where $C=\phi(\mu_{0})-\phi(\mu_{t^{*}})$. The first equality comes from the
construction of $\tau^{\prime}$ and $(\mu^{\prime}_{t})_{t}$. Therefore,

$\displaystyle\mathbb{P}(\tau^{\prime}>t^{*})(\phi(\mu^{\prime}_{t^{*}})+c(t^{*})-c(t^{*}+1))-\mathbb{P}(\tau>t^{*})(\phi(\mu_{t^{*}})+c(t^{*})-c(t^{*}+1))$
$\displaystyle\geq\epsilon(\phi(\mu_{t^{*}})+c(t^{*})-c(t^{*}+1))+\delta(\phi(\mu_{0})-\phi(\mu_{t^{*}}))\mathbb{P}(\tau>t^{*})+C\delta\epsilon$
$\displaystyle=\epsilon\bigg{[}\phi(\mu_{t^{*}})-\frac{(\phi(\mu_{t^{*}})-\phi(\mu_{0}))\mathbb{P}(\tau>t^{*})}{(\mathbb{P}(\tau>t^{*})+\epsilon)(1-\mathbb{P}(\tau>t^{*}))}+c(t^{*})-c(t^{*}+1)\bigg{]}+C\delta\epsilon$
$\displaystyle\geq\epsilon\bigg{[}\phi(\mu_{t^{*}})-\frac{\phi(\mu_{t^{*}})-\phi(\mu_{0})}{1-\mathbb{P}(\tau>t^{*})}+c(t^{*})-c(t^{*}+1)\bigg{]}+C\delta\epsilon$
$\displaystyle=\epsilon\bigg{[}\frac{1}{1-\mathbb{P}(\tau>t^{*})}\cdot\Big{(}\phi(\mu_{0})-\mathbb{P}(\tau>t^{*})\phi(\mu_{t^{*}})\Big{)}+c(t^{*})-c(t^{*}+1)\bigg{]}+C\delta\epsilon,$

where the second inequality employs Proposition 4 part (i) which states that
it is without loss to take $\phi(\mu_{t^{*}})\geq\phi(\mu_{0})$. In order to
maintain the sender’s optimality of $\tau$, we require

  1. 1.

$\phi(\mu_{t^{*}})=\mathbb{E}[c(\tau)\mid\tau>t^{*}]-c(t^{*})$ and

  2. 2.

$\frac{1}{1-\mathbb{P}(\tau>t^{*})}\cdot\Big{(}\phi(\mu_{0})-\mathbb{P}(\tau>t^{*})\phi(\mu_{t^{*}})\Big{)}+c(t^{*})-c(t^{*}+1)\leq
0$, which imply that
$\Big{(}\phi(\mu_{0})-\mathbb{P}(\tau>t^{*})\phi(\mu_{t^{*}})\Big{)}+c(t^{*})-c(t^{*}+1)\leq
0$.

To see this, if 1. is false, then we can choose small $\epsilon>0$ in such a
way that the difference of $\phi(\mu_{t^{*}})$ and
$\phi(\mu^{\prime}_{t^{*}})$ is small so that the new obedience constraint is
not violated. Therefore, $\tau^{\prime}$ is a feasible stopping time, which
contradicts the sender’s optimality of $\tau$. On the other hand, if 2. is
false, then choosing arbitrarily small $\epsilon>0$ slackens the LHS of (4).
This implies $\tau^{\prime}$ is a feasible stopping time, which is again a
contradiction.

By these two conditions, we have

$\displaystyle\phi(\mu_{0})>\mathbb{E}[c(\tau)]$
$\displaystyle=\mathbb{P}(\tau>t^{*})\mathbb{E}[c(\tau)\mid\tau>t^{*}]+\mathbb{P}(\tau=t^{*})c(t^{*})$
$\displaystyle=\mathbb{P}(\tau>t^{*})\phi(\mu_{t^{*}})+c(t^{*}).$
$\displaystyle\geq\phi(\mu_{0})+2c(t^{*})-c(t^{*}+1),$

which means that $c(t^{*})<c(t^{*}+1)-c(t^{*})$. From the definition of
smoothing out $f$ and $c$ given in the main text and made formal in Online
Appendix VII, we have
$c_{k}(t^{*}_{k}/k)=c_{k}((t^{*}_{k}+1)/k\big{)}-c_{k}(t^{*}_{k}/k)$, where
$t^{*}_{k}$ is corresponding $t^{*}$ to the model indexed $k$. By smoothing
out $f$ and $c$, for any sufficiently large $k$, we have
$c_{k}(t+1/k)-c(t)<c^{\prime}(1)$ for every $t\in\mathcal{T}$. Thus,
$c_{k}(t^{*}_{k}/k)=c_{k}((t^{*}_{k}+1)/k)-c_{k}(t^{*}_{k}/k)<c^{\prime}(1)=c_{k}(1)$.
Therefore, $t^{*}_{k}<k$. This implies $c_{k}$ is linear at
$[0,(t^{*}_{k}+1)/k]$, which means that $c((t^{*}+1)/k)=c(t^{*}/k)+c(1/k)\leq
2c(t^{*}/k)$, which is a contradiction.

To generalize this result to any arbitrary information structure, we can see
that its induced simple and obedient information structure yields a weakly
better DM’s surplus because its induced stopping time is the same and a belief
at every optimal history is guaranteed with full information. Therefore, the
DM’s surplus under any arbitrary structure is still 0 (or uniformly bounded
above by $\epsilon$ when $k$ is sufficiently large), as desired. ∎

### C.3 Proof of Proposition 6 (ii)

###### Proof of Proposition 6 (ii).

We begin the proof of the second half, which is
$\mathcal{P}=\overline{\mathcal{P}}$ when $\mu_{0}\in\Phi^{*}$. Suppose that
$\mu_{0}\in\Phi^{*}$. Because the set of feasible stopping times is convex, so
is the set of pairs of platform’s expected value and DM’s induced by feasible
stopping times. Therefore, every point on the concave envelope of $(0,f^{*})$
and the curve $F$ is induced by some dynamic information structure. It
suffices to show that there is no information structure $I$ such that $S(I)$
Pareto dominates some point on the convex envelope of $(0,f^{*})$ and the
curve $F$. Suppose a contradiction that there is an information structure
$I\in\mathcal{I}^{*}$ that violates the previous statement. Define

$\text{conv}\big{(}F\cup\\{(0,f^{*})\\}\big{)}:=\big{\\{}s\in\mathbb{R}^{2}_{+}:s\leq
s^{*},\exists s^{*}\in co\big{(}F\cup\\{(0,f^{*})\\}\big{)}\big{\\}}$

as the set of all points on $co(F\cup\\{(0,f^{*})\\})$ and their free-disposal
points. It is clear that $\text{conv}\big{(}F\cup\\{(0,f^{*})\\}\big{)}$ is a
convex set.

The assumption of $I$ implies that
$S(I)\notin\text{conv}\big{(}F\cup\\{(0,f^{*})\\}\big{)}$. From Proposition 8,
it is sufficient to consider the case that $I$ has a block structure generated
by an increasing sequence $(t_{1},\dots,t_{n})$ with terminal stopping times
$(t_{n+1},\bar{t})$ with the initial term $C_{0}<\phi^{*}$ 464646 Note that
Proposition 8 requires that $C_{0}=\phi^{*}.$ However, we can impose another
constraint $C_{0}\leq\bar{C}$ for some constant $\bar{C}\leq\phi^{*}$ to the
designer’s optimization problem. Then the result of Proposition 8 is still the
same but changes from $C_{0}=\phi^{*}$ to $C_{0}=\bar{C}$.. Consider the
following two cases:

Case 1: $n=0$. This means the information structure induces only two possible
stopping times $t_{n+1}$ and $\bar{t}$ Thus, $S(I)$ is a convex combination of
$F(t_{n+1})$ and $F(\bar{t})$, where $F(t)$ is defined as a pair of platform’s
expected value and DM’s if the DM stops at time $t$ with probability $1$ i.e.,

$F(t):=\big{(}\phi^{*}-c(t),f(t)\big{)}.$

The obedient constraint a time $0$ implies that $c(t_{n+1})\leq\phi^{*}$ and
$c(\bar{t})-c(t_{n+1})\leq\phi^{*}$. If $c(\bar{t})\leq\phi^{*}$, then both
$F(t_{n+1})$ and $F(\bar{t})$ are in $F$. This implies that
$S(I)\in\text{co}(F)$, which is a contradiction. Thus, $c(\bar{t})>\phi^{*}$
and $c(t_{n+1})\leq\phi^{*}$. Then there exists $p\in[0,1]$ such that
$\phi^{*}=pc(t_{n+1})+(1-p)c(\bar{t})$. Consider information structure
$I^{\prime}$ that reveals full information at time $t_{n+1}$ with probability
$p$ and at time $\bar{t}$ with probability $p$. $I^{\prime}$ satisfies the
obedient constraint at time $0$ and time $t_{n+1}$ because
$\phi^{*}=pc(\bar{t})+(1-p)c(t_{n+1})$ and
$c(\bar{t})-c(t_{n+1})\leq\phi^{*}$.

Note that

$\phi^{*}\geq\mathbb{E}^{I}\big{[}c\big{(}\tau(I)\big{)}\big{]}=\mathbb{P}(\tau=t_{n+1})c(t_{n+1})+(1-\mathbb{P}(\tau=t_{n+1}))c(\bar{t}),$

which means $\mathbb{P}\big{(}\tau(I)=t_{n+1}\big{)}\geq
p=\mathbb{P}\big{(}\tau(I^{\prime})=t_{n+1}\big{)}$. This implies there is
$\lambda\in[0,1]$ such that $d(I)=\lambda
d(I^{\prime})+(1-\lambda)\delta_{t_{n+1}}$, which means $S(I)=\lambda
S(I^{\prime})+(1-\lambda)F(t_{n+1})$. However, $I^{\prime}$ yields zero DM’s
surplus, which means $(0,f^{*})$ Pareto dominates $S(I^{\prime})$. Therefore,
$S(I^{\prime})$ and $F(t_{n+1})$ are in
$\text{conv}\big{(}F\cup\\{(0,f^{*})\\}\big{)}$. This implies
$S(I)\in\text{conv}\big{(}F\cup\\{(0,f^{*})\\}\big{)}$, which is a
contradiction.

Case 2: $n>0$. Suppose $I^{\prime\prime}$ be an information structure
generated by an increasing sequence $(t_{1},\dots,t_{n})$ with terminal
stopping times $(t_{n+1},\bar{t})$ but with the initial term
$C^{\prime\prime}_{0}=\phi^{*}$. This is well defined because the obedience
constraint at time $0$ of the information structure $I$ implies $\phi^{*}\geq
c(t_{1})$. Therefore, $\tau(I^{\prime\prime})\mid\tau(I^{\prime\prime})>t_{1}$
and $\tau(I)\mid\tau(I)>t_{1}$ have the same distribution because they share
the same block structure and start with
$C_{t_{1}}=C^{\prime\prime}_{t_{1}}=\phi^{*}$. Consider that

$\displaystyle\mathbb{P}(\tau(I)>t_{1})=\frac{C_{0}-c(t_{1})}{\phi^{*}}\leq\frac{\phi^{*}-c(t_{1})}{\phi^{*}}\leq\mathbb{P}(\tau(I^{\prime\prime})>t_{1}).$

This means there exists $\lambda\in[0,1]$ such that $d(I)=\lambda
d(I^{\prime\prime})+(1-\lambda)\delta_{t_{1}}$. However, $I^{\prime\prime}$
yields zero DM’s surplus, which means $(0,f^{*})$ Pareto dominates
$S(I^{\prime\prime})$. Therefore, $S(I)=\lambda
S(I^{\prime\prime})+(1-\lambda)F(t_{1})\in\text{conv}\big{(}F\cup\\{(0,f^{*})\\}\big{)}$,
which is a contradiction.

From these two cases, we can imply that for every $I\in\mathcal{I}$, we must
have $S(I)\in\text{conv}\big{(}F\cup\\{(0,f^{*})\\}\big{)}$, which implies the
second half of the proposition statement.

Next, we will show that $\mathcal{P}\leq\overline{\mathcal{P}}$ for any belief
$\mu_{0}$. Proposition 3 implies that $(c_{0},f_{0})\in\mathcal{P}$ if and
only if

$\displaystyle c_{0}$
$\displaystyle={\max_{\big{(}d(\tau),(\mu_{t})_{t}\big{)}\in\Delta(\mathcal{T})\times(\Delta(\Theta))^{\mathcal{T}}}\mathbb{E}[\phi(\mu_{0})-c(\tau)]}$
$\displaystyle\text{s.t.}\quad\phi(\mu_{t})\geq\mathbb{E}[c(\tau)\mid\tau>t]-c(t)\quad\forall
t\in\mathcal{T}$ (Obedience)
$\displaystyle\quad\quad\mathbb{P}(\tau>t+1)\mu_{t+1}\leq\mathbb{P}(\tau>t)\mu_{t}$
(Boundary) $\displaystyle\quad\quad f_{0}=\mathbb{E}[f(\tau)].$ (Fix the the
designer’s value)

We relax the obedience constraint to
$\phi^{*}\geq\mathbb{E}[c(\tau)\mid\tau>t]-c(t)$ for every $t\in\mathcal{T}$
and eliminate the boundary constraint. Applying $\phi(\mu_{0})\leq\phi^{*}$,
we obtain

$\displaystyle c_{0}$
$\displaystyle\leq{\max_{d(\tau)\in\Delta(\mathcal{T})}\mathbb{E}[\phi^{*}-c(\tau)]}$
$\displaystyle\text{s.t.}\quad\phi^{*}\geq\mathbb{E}[c(\tau)\mid\tau>t]-c(t)\quad\forall
t\in\mathcal{T}$ (Obedience) $\displaystyle\quad\quad
f_{0}=\mathbb{E}[f(\tau)],$ (Fix the the designer’s value)

which is exactly the optimization problem of solving points on the Pareto
frontier when $\mu_{0}\in\Phi^{*}$. The fact that
$\mathcal{P}=\overline{\mathcal{P}}$ when $\mu_{0}\in\Phi^{*}$ implies that
$c_{0}\leq c^{*}$ where $(c^{*},f_{0})\in\overline{\mathcal{P}}$. Therefore,
$(c_{0},f_{0})\leq(c^{*},f_{0})$, which implies
$\mathcal{P}\leq\overline{\mathcal{P}}$, as desired. ∎

### C.4 Proof of Proposition 8

Before we prove Proposition 8, we first state and prove a helpful lemma.

######  Lemma 8.

For every increasing function $f:\mathcal{T}\rightarrow\mathbb{R}$, there is a
feasible stopping time $\tau$ corresponding to an obedient sequence
$(C_{t})_{t\in\mathcal{T}}$ such that

  1. (i)

$\tau\in\text{argmax}_{\tau^{\prime}:d(\tau^{\prime})\in\mathcal{D}}\mathbb{E}[f(\tau^{\prime})]$.

  1. (ii)

For every $r<s\in\mathcal{T}$, the following statement is true:

If all of the following holds:

    1. (ii.a)

$C_{t-1}-C_{t}=c(t)-c(t-1)$ for every $t\in\\{r+1,\dots,s\\}$;

    2. (ii.b)

$C_{r-1}-C_{r}<c(r)-c(r-1)$; and

    3. (ii.c)

$C_{s}-C_{s+1}<c(s+1)-c(s)$,

then $C_{r}=\phi^{*}$ or $C_{t}=c(t+1)-c(t)$ for some $t\in\\{r+1,\dots,s\\}$.

Before we prove Lemma 8, let us briefly interpret it. The lemma states that
fixing our value function $f$, we can find an optimal stopping distribution
corresponding to a obedience sequence $(C_{t}^{*})_{t\in\mathcal{T}}$ which
fulfils property (ii). Condition (ii.a) states that between periods $r$ and
$s$, the DM obtains no information with full probability. Condition (ii.b) and
(ii.c) states that the DM receives full information with some strictly
positive probability at time $r$ and $s$ respectively. If these conditions are
fulfilled, then either the DM is indifferent at time $r$ ($C_{r}=\phi^{*}$) or
the DM must receive full information with probability one at some time between
$r+1$ and $s$.

###### Proof of Lemma 8.

Fixing $r<s$, we denote the if-then condition of Lemma 8 (ii) as a statement
whose truth value depends on $r,s$ (the boundary times) and $\tau$ (the
stopping time in question). Denote this statement with $S(r,s,\tau)$.

For any
$\tau\in\text{argmax}_{\tau^{\prime}:d(\tau^{\prime})\in\mathcal{D}}\mathbb{E}[f(\tau^{\prime})]$,
define $r(\tau)$ as, if it exists, the smallest number $r\in\mathcal{T}$ such
that there exists $s>r$ which violates $S(r,s,\tau)$. If no such number
exists, define $r(\tau)=+\infty$.

Pick any
$\tau\in\text{argmax}_{\tau^{\prime}:d(\tau^{\prime})\in\mathcal{D}}\mathbb{E}[f(\tau^{\prime})]$.
If $r(\tau)=+\infty$, there is nothing to show. Next suppose that
$r(\tau)<+\infty$. Denote $r\coloneqq r(\tau)$, and let $s>r$ be such that the
statement $S(r,s,\tau)$ is false. Thus, $C_{r}<\phi^{*}$ and
$C_{t}>c(t+1)-c(t)$ for every $t\in\\{r+1,\dots,s\\}$. Thus, we can modify
$(C_{r},\dots,C_{s})$ to either $(C_{r}+\epsilon,\dots,C_{s}+\epsilon)$ or
$(C_{r}-\epsilon,\dots,C_{s}-\epsilon)$ without violating the obedience
constraint for sufficiently small $\epsilon>0$.

Now we rewrite $\mathbb{E}[f(\tau)]$ as a function of
$(C_{t})_{t\in\mathcal{T}}$ as follows:

$\displaystyle
F\Big{(}(C_{t})_{t\in\mathcal{T}}\Big{)}\coloneqq\mathbb{E}[f(\tau)]-f(0)$
$\displaystyle=\sum_{t=0}^{+\infty}\mathbb{P}(\tau>t)\Big{(}f(t+1)-f(t)\Big{)}$
$\displaystyle=\sum_{t=0}^{+\infty}\bigg{(}\prod_{u=0}^{t-1}\frac{C_{u}+c(u)-c(u+1)}{C_{u+1}}\bigg{)}\Big{(}f(t+1)-f(t)\Big{)}.$

where the second equality follows by noting that we can rewrite conditional
probabilities of $\tau$ as a fraction of $C_{t}$s, as in part (iii) of Lemma
1.

We define $G(\epsilon)\coloneqq F((C^{\epsilon}_{t})_{t\in\mathcal{T}}),$
where $C^{\epsilon}_{t}=\begin{cases}C_{t}+\epsilon,&\text{for
}t\in\\{r,\dots,s\\}\\\ C_{t},&\text{otherwise}.\end{cases}$.

Denote $\tau^{\epsilon,r}$ as a feasible stopping time corresponding to
$(C^{\epsilon}_{t})_{t\in\mathcal{T}}$ whenever
$(C^{\epsilon}_{t})_{t\in\mathcal{T}}$ is obedient. Then

$\displaystyle(C_{r}+\epsilon)^{2}\cdot\frac{dG}{d\epsilon}$
$\displaystyle=-\bigg{(}\prod_{t=0}^{r-2}\frac{C_{t}+c(t)-c(t+1)}{C_{t+1}}\bigg{)}\Big{(}C_{r-1}+c(r-1)-c(r)\Big{)}\Big{(}f(s)-f(r)\Big{)}$
$\displaystyle\phantom{=\,}+\sum_{t^{\prime}=s}^{+\infty}\bigg{(}\prod_{t\leq
t^{\prime}-1,t\notin\\{r-1,\dots,s\\}}\frac{C_{t}+c(t)-c(t+1)}{C_{t+1}}\bigg{)}\Big{(}c(s+1)-c(r)\Big{)}\Big{(}f(t^{\prime}+1)-f(t^{\prime})\Big{)}$
$\displaystyle\eqqcolon H((C_{t})_{t\in\mathcal{T}})$

which is not a function of $\epsilon$. Therefore, $\frac{dG}{d\epsilon}\geq 0$
if and only if $H((C_{t})_{t\in\mathcal{T}})\geq 0$. Because we can modify
$(C_{t})_{t\in\mathcal{T}}$ to $(C^{\epsilon}_{t})_{t\in\mathcal{T}}$ and
$(C^{-\epsilon}_{t})_{t\in\mathcal{T}}$ without violating the obedience
constraint for small $\epsilon>0$,
$\frac{dG}{d\epsilon}\big{\rvert}_{\epsilon=0}=0$, which, in turn, implies
that $H((C_{t})_{t\in\mathcal{T}})=0$. This implies
$F((C^{\epsilon}_{t})_{t\in\mathcal{T}})=F((C_{t})_{t\in\mathcal{T}})$ for
every $\epsilon\in\mathbb{R}$. Therefore,
$\tau^{\epsilon,r}\in\text{argmax}_{\tau^{\prime}:d(\tau^{\prime})\in\mathcal{D}}\mathbb{E}[f(\tau^{\prime})]$
as long as the sequence $(C^{\epsilon}_{t})_{t\in\mathcal{T}}$ is obedient.

We can then find $\epsilon>0$ so that the statement $S(r,s,\tau^{\epsilon,r})$
is true and the sequence $(C^{\epsilon}_{t})_{t\in\mathcal{T}}$ is obedient by
increasing $\epsilon>0$ until the constraint $C_{r}\leq\phi^{*}$ binds i.e.,
set $\epsilon^{*}=\phi^{*}-C_{r}$. This implies
$r(\tau^{\epsilon^{*},r})>r=r(\tau)$. Note that the sequence
$(C^{\epsilon^{*}}_{t})_{t\in\mathcal{T}}$ is weakly greater than
$(C_{t})_{t\in\mathcal{T}}$ pointwise. We can iteratively apply this procedure
to the stopping time $\tau$ i.e., define a sequence of stopping times
$\\{\tau_{i}\\}_{i=0}^{+\infty}$ such that

  1. (i)

$\tau_{0}=\tau$.

  2. (ii)

Given $\tau_{i}$, we set $\tau_{i+1}=\tau_{i}^{\epsilon^{*}_{i},r(\tau_{i})}$,
where $\epsilon^{*}_{i}=\phi^{*}-C_{r(\tau(i))}.$

Therefore, the sequence $\big{(}r(\tau_{i})\big{)}_{i=0}^{+\infty}$ is
strictly increasing, which means that
$\lim_{i\to{+\infty}}r(\tau_{i})={+\infty}$. Moreover,
$\tau_{i}\in\text{argmax}_{\tau^{\prime}:d(\tau^{\prime})\in\mathcal{D}}\mathbb{E}[f(\tau^{\prime})]$
for every $i\in\mathbb{N}$. Suppose an obedient sequence
$(C^{i}_{t})_{t\in\mathcal{T}}$ correspond to the feasible stopping time
$\tau_{i}$. For every $t\in\mathcal{T}$, the sequence
$(C^{i}_{t})_{i=0}^{+\infty}$ is increasing and bounded above by $\phi^{*}$.
Thus, we can define $C^{*}_{t}\coloneqq\lim_{t\to{+\infty}}C^{i}_{t}$. Note
that $(C^{*}_{t})_{t\in\mathcal{T}}$ is still obedient by putting limit to the
obedient constraint. Suppose $(C^{*}_{t})_{t\in\mathcal{T}}$ induce a stopping
time $\tau^{*}$. Because $\lim_{i\to{+\infty}}r(\tau_{i})=+\infty$, we obtain
$r(\tau^{*})=+\infty$ by putting limit to conditions (ii.a)-(ii.c). Moreover,

$\displaystyle\mathbb{E}[f(\tau^{*})]-f(0)$
$\displaystyle=\sum_{t=0}^{+\infty}\mathbb{P}(\tau^{*}>t)\Big{(}f(t+1)-f(t)\Big{)}$
$\displaystyle=\sum_{t=0}^{+\infty}\lim_{i\to{+\infty}}\mathbb{P}(\tau_{i}>t)\Big{(}f(t+1)-f(t)\Big{)}$
$\displaystyle=\lim_{i\rightarrow{+\infty}}\sum_{t=0}^{+\infty}\mathbb{P}(\tau_{i}>t)\Big{(}f(t+1)-f(t)\Big{)}$
$\displaystyle=\max_{\tau^{\prime}:d(\tau^{\prime})\in\mathcal{D}}\mathbb{E}[f(\tau^{\prime})]-f(0),$

where the limit interchange follows from the dominated convergence theorem and

$\displaystyle\lim_{i\rightarrow{+\infty}}\sum_{t=0}^{+\infty}\mathbb{P}(\tau_{i}>t)\Big{\lvert}f(t+1)-f(t)\Big{\rvert}$
$\displaystyle=\lim_{i\rightarrow{+\infty}}\sum_{t=0}^{+\infty}\mathbb{P}(\tau_{i}>t)\Big{(}f(t+1)-f(t)\Big{)}$
$\displaystyle=\max_{\tau^{\prime}:d(\tau^{\prime})\in\mathcal{D}}\mathbb{E}[f(\tau^{\prime})]-f(0)<{+\infty}.$

Therefore, $\tau=\tau^{*}$ makes Lemma 8 hold. ∎

###### Proof of Proposition 8.

Fix an increasing function $f:\mathcal{T}\rightarrow\mathbb{R}$, we can find a
feasible stopping time $\tau$ corresponding to an obedient sequence
$(C_{t})_{t\in\mathcal{T}}$ as shown in Lemma 8. Define

$\displaystyle\mathcal{T}_{\text{no}}\coloneqq\big{\\{}t\in\mathcal{T}:C_{t-1}-C_{t}<c(t)-c(t-1)\big{\\}}.$

Then we arrange all members of the set $\mathcal{T}_{no}$ as an increasing
sequence $(t_{i})_{i=1}^{n+1}$. Note that $n$ can be $+\infty$. Lemma 8
implies that $C_{t_{i}}=\phi^{*}$ for every $i\in\\{1,\dots,n\\}$, and
Proposition 6 implies $C_{0}=\phi^{*}$, which corresponds to indifferent
blocks. Suppose that $n<{+\infty}$. If $t>t_{n+1}$, we have

$\displaystyle
C_{t}-C_{t_{n+1}}=\sum_{t^{\prime}=t_{n+1}+1}^{t}(C_{t^{\prime}}-C_{t^{\prime}-1})=\sum_{t^{\prime}=t_{n+1}}^{t-1}(c(t^{\prime})-c(t^{\prime}+1))=c(t_{n+1})-c(t),$
(5)

which converges to $-\infty$ as $t\to{+\infty}$. To make the constraint
$C_{t}\geq c(t+1)-c(t)>0$, the DM must receive full information at some
deterministic time, says $\overline{t}$, with probability one. Therefore,
$C_{\overline{t}-1}=c(\overline{t})-c(\overline{t}-1)$. By (5), we obtain
$C_{t_{n+1}}=c(\overline{t})-c(t_{n+1})$, which concludes a terminal block. ∎

### C.5 Proof of Proposition 9

###### Proof of Proposition 9.

Consider a stopping time $\tau$ generated by an increasing sequence
$(t_{i})_{i=1}^{n+1}$ and a terminal time $\overline{t}>t_{n+1}$. If
$t_{1}>1$, take a new increasing sequence

$(t^{\prime}_{i})_{i}=(1,2,\dots,t_{1}-1,t_{1},t_{2},\dots,t_{n+1})$

and the same terminal time $\overline{t}$ to construct a stopping time
$\tau^{\prime}$ with our algorithm. We see that

$\displaystyle\mathbb{E}[c(\tau)]$
$\displaystyle=\mathbb{P}(\tau=t_{1})c(t_{1})+\mathbb{P}(\tau>t_{1})\mathbb{E}[c(\tau)\mid\tau>t_{1}]$

and

$\displaystyle\mathbb{E}[c(\tau^{\prime})]$
$\displaystyle=\sum_{t=1}^{t_{1}}\mathbb{P}(\tau^{\prime}=t)c(t)+\mathbb{P}(\tau^{\prime}>t_{1})\mathbb{E}[c(\tau^{\prime})\mid\tau^{\prime}>t_{1}]$
$\displaystyle=\sum_{t=1}^{t_{1}}\mathbb{P}(\tau^{\prime}=t)c(t)+\mathbb{P}(\tau^{\prime}>t_{1})\mathbb{E}[c(\tau)\mid\tau>t_{1}]$

because $\tau\rvert\tau>t_{1}$ and $\tau^{\prime}\rvert\tau>t_{1}$ have the
same distribution. Consider that

$\displaystyle\mathbb{P}(\tau^{\prime}>t_{1})=\prod_{t=1}^{t_{1}}\bigg{(}1-\frac{c(t)-c(t-1)}{\phi^{*}}\bigg{)}$
$\displaystyle\geq
1-\frac{\sum_{t=1}^{t_{1}}\big{(}c(t)-c(t-1)\big{)}}{\phi^{*}}$
$\displaystyle=1-\frac{c(t_{1})}{\phi^{*}}=\mathbb{P}(\tau>t_{1}),$

where the inequality follows from the fact that $(1-x)(1-y)\geq 1-x-y$ for
every $x,y\geq 0$. We know that
$\mathbb{E}[c(\tau^{\prime})]=\mathbb{E}[c(\tau)]=\phi^{*}$. Thus,

$\displaystyle 0$
$\displaystyle=\mathbb{E}[c(\tau^{\prime})]-\mathbb{E}[c(\tau)]$
$\displaystyle=\sum_{t=1}^{t_{1}}\mathbb{P}(\tau^{\prime}=t)c(t)+\Big{(}\mathbb{P}(\tau^{\prime}>t_{1})-\mathbb{P}(\tau>t_{1})\Big{)}\mathbb{E}[c(\tau)\mid\tau>t_{1}]-\mathbb{P}(\tau=t_{1})c(t_{1}),$

and rearranging, we have

$\displaystyle\bigg{(}\sum_{t=1}^{t_{1}}\frac{\mathbb{P}(\tau^{\prime}=t)}{\mathbb{P}(\tau=t_{1})}c(t)\bigg{)}+\bigg{(}\frac{\mathbb{P}(\tau^{\prime}>t_{1})-\mathbb{P}(\tau>t_{1})}{\mathbb{P}(\tau=t_{1})}\cdot\mathbb{E}[c(\tau)\mid\tau>t_{1}]\bigg{)}=c(t_{1}).$

Because
$\frac{\mathbb{P}(\tau^{\prime}>t_{1})-\mathbb{P}(\tau>t_{1})}{\mathbb{P}(\tau=t_{1})}\geq
0$ and $f$ is more convex than $c$, $f\circ c^{-1}$ is convex and applying
Jensen’s inequality, we have

$\displaystyle\bigg{(}\sum_{t=1}^{t_{1}}\frac{\mathbb{P}(\tau^{\prime}=t)}{\mathbb{P}(\tau=t_{1})}f(t)\bigg{)}+\bigg{(}\frac{\mathbb{P}(\tau^{\prime}>t_{1})-\mathbb{P}(\tau>t_{1})}{\mathbb{P}(\tau=t_{1})}\cdot\mathbb{E}[f(\tau)\mid\tau>t_{1}]\bigg{)}\geq
f(t_{1}).$

This implies $\mathbb{E}[f(\tau^{\prime})]\geq\mathbb{E}[f(\tau)]$. Iterating
on the above argument, we see that to maximize $\mathbb{E}[f(\tau)]$, it
weakly better to choose the increasing sequence $(t_{i})_{i=1}^{n+1}$ such
that $t_{i}=i$ for every $i\in\\{1,\dots,n\\}$.

We will now show that having choosing a terminal time (i.e., $n<+\infty$) is
dominated. It is without loss to consider the case with $n=0$ i.e., $t_{n}=0$.
Suppose that the sequence $(t_{1},\overline{t})$ generates a feasible stopping
time $\tau$. Now consider an alternate feasible stopping time
$\tau^{\prime\prime}$ generated the sequence
$(t^{\prime\prime}_{i})_{i\in\mathcal{T}}=(t_{1},t_{1}+1,\dots)$. Observe that

$\displaystyle\mathbb{P}(\tau^{\prime\prime}=t_{1})=\frac{c(t_{1})}{\phi^{*}}\geq\frac{(c(\overline{t})-\phi^{*})}{c(\overline{t})-c(t_{1})}=\mathbb{P}(\tau=t_{1})$

Since $\mathbb{E}[c(\tau^{\prime\prime})]=\mathbb{E}[c(\tau)]=\phi^{*}$,

$\displaystyle 0$
$\displaystyle=\mathbb{E}[c(\tau^{\prime\prime})]-\mathbb{E}[c(\tau)]$
$\displaystyle=(\mathbb{P}(\tau^{\prime\prime}=t_{1})-\mathbb{P}(\tau=t_{1}))c(t_{1})+\mathbb{E}[c(\tau^{\prime\prime})\mid\tau^{\prime\prime}>t_{1}]\mathbb{P}(\tau^{\prime\prime}>t_{1})-\mathbb{P}(\tau=\overline{t})c(\overline{t}),$

which then implies

$\displaystyle\bigg{(}\frac{\mathbb{P}(\tau^{\prime\prime}=t_{1})-\mathbb{P}(\tau=t_{1})}{\mathbb{P}(\tau=\overline{t})}\cdot
c(t_{1})\bigg{)}+\bigg{(}\frac{\mathbb{P}(\tau^{\prime\prime}>t_{1})}{\mathbb{P}(\tau=\overline{t})}\cdot\mathbb{E}[c(\tau^{\prime\prime})\mid\tau^{\prime\prime}>t_{1}]\bigg{)}=c(\overline{t}).$

Because
$\frac{\mathbb{P}(\tau^{\prime}=t_{1})-\mathbb{P}(\tau=t_{1})}{\mathbb{P}(\tau=\overline{t})}\geq
0$ and $f$ is more convex than $c$, $f\circ c^{-1}$ is convex and applying
Jensen’s inequality,

$\displaystyle\bigg{(}\frac{\mathbb{P}(\tau^{\prime\prime}=t_{1})-\mathbb{P}(\tau=t_{1})}{\mathbb{P}(\tau=\overline{t})}\cdot
f(t_{1})\bigg{)}+\bigg{(}\frac{\mathbb{P}(\tau^{\prime\prime}>t_{1})}{\mathbb{P}(\tau=\overline{t})}\cdot\mathbb{E}[f(\tau^{\prime\prime})\mid\tau^{\prime\prime}>t_{1}]\bigg{)}\geq
f(\overline{t}).$

Therefore, $\mathbb{E}[f(\tau^{\prime\prime})]\geq\mathbb{E}[f(\tau)]$. We
have shown that choosing any terminal time is always weakly dominated. But we
have also shown that it is always weakly better to choose sequences which are
consecutive in $\mathcal{T}$. This implies that every feasible stopping time
$\tau^{\prime\prime}$ must satisfy

$\mathbb{E}[f(\tau^{\prime\prime})]\leq\mathbb{E}^{\mathcal{A}(\mathcal{T})}[f(\tau)]$

which concludes the proof. ∎

Online Appendix to ‘Attention Capture’
Andrew Koh and Sivakorn Sanguanmoo333MIT Department of Economics; email:
<EMAIL_ADDRESS>and<EMAIL_ADDRESS>
FOR ONLINE PUBLICATION ONLY

##  Appendix IV Maximal belief paths and attainment of sup

In this online appendix we introduce a topology on the space of belief paths
and feasible stopping times. This will establish the existence of a maximal
belief path used in the Proof of Proposition 4 in Appendix B. It will also
ensure that the supremum of the sender’s problem can be attained, as claimed
in the main text.

### IV.1 Topological space of belief paths and the existence of a maximal
belief path

Our first step is, fixing a distribution of stopping time $d(\tau)$, to study
the set of belief paths that satisfy Proposition 3 in order to verify that
$\tau$ is a feasible stopping time. Recall that the set of belief paths
$\mathcal{W}=(\Delta(\Theta))^{\mathcal{T}}$. We endow $\mathcal{W}$ with the
product topology of the weak topology on $\Delta(\Theta)$. Because
$\Delta(\Theta)$ endowed with the weak topology is compact and metrizable, so
is $\mathcal{W}$ by Tychonoff theorem. Recall that we defined
$\mathcal{W}(\tau)\subset\mathcal{W}$ as the set of belief paths corresponding
to stopping time $\tau$.

######  Lemma 9.

$\mathcal{W}(\tau)$ is compact for every feasible stopping time $\tau$.

###### Proof.

We show that $\mathcal{W}(\tau)$ is sequentially compact instead. Consider
$((\mu^{i}_{t})_{t\in\mathcal{T}})_{i=0}^{\infty}\subset\mathcal{W}(\tau)$.
Because $\mathcal{W}$ is compact, there exists a convergent subsequence
$((\mu^{i_{n}}_{t})_{t\in\mathcal{T}})_{n=0}^{\infty}$ as $n\to\infty$.
Suppose that
$(\mu^{i_{n}}_{t})_{t\in\mathcal{T}}\to(\mu^{*}_{t})_{t\in\mathcal{T}}\in\mathcal{W}$
as $n\to\infty$. This implies $\mu_{t}^{i_{n}}\to\mu^{*}_{t}$ as $n\to\infty$
under the weak topology on $\Delta(\Theta)$ for every $t\in\mathcal{T}$. It
suffices to show that $(\mu^{*}_{t})_{t\in\mathcal{T}}\in\mathcal{W}(\tau).$
Because $\phi$ is continuous under the weak topology on $\Delta(\Theta)$, for
every $t\in\mathcal{T}$, we have

$\displaystyle\phi(\mu^{*}_{t})=\lim_{n\to\infty}\phi(\mu^{i_{n}}_{t})\geq\mathbb{E}[c(\tau)\mid\tau>t]-c(t),$

where the inequality follows by
$(\mu^{i_{n}}_{t})_{t\in\mathcal{T}}\in\mathcal{W}(\tau)$. This implies the
obedience constraint holds for the belief path
$(\mu^{*}_{t})_{t\in\mathcal{T}}$. Moreover, we have
$\mathbb{P}(\tau>t+1)\mu^{i_{n}}_{t+1}(\theta)\leq\mathbb{P}(\tau>t)\mu^{i_{n}}_{t}(\theta)$
for every $t\in\mathcal{T}$. Therefore,

$\displaystyle\mathbb{P}(\tau>t+1)\mu^{*}_{t+1}(\theta)=\lim_{n\to\infty}\mathbb{P}(\tau>t+1)\mu^{i_{n}}_{t+1}(\theta)\leq\mathbb{P}(\tau>t)\mu^{i_{n}}_{t}(\theta)=\mathbb{P}(\tau>t)\mu^{*}_{t}(\theta),$

which implies the boundary constraint. This implies
$(\mu^{*}_{t})_{t\in\mathcal{T}}\in\mathcal{W}(\tau)$, as desired. ∎

We now prove Proposition 11 that maximal belief paths exist.

###### Proof of Proposition 11.

Define $F:\mathcal{W}\to\mathbb{R}$ such that
$F((\mu_{t})_{t\in\mathcal{T}})=\sum_{t=0}^{\infty}\frac{1}{2^{t}}\phi(\mu_{t})$.
$F$ is well-defined because
$|F\big{(}(\mu_{t})_{t\in\mathcal{T}}\big{)}|\leq\sum_{t=0}^{\infty}\frac{1}{2^{t}}\phi^{*}\leq
2\phi^{*}$. Moreover, for every $\epsilon>0$ and every
$T>1+\log(\phi^{*}/\epsilon)/\log 2$, we have
$\sum_{t=T}^{\infty}\frac{1}{2^{t}}\phi(\mu_{t})\leq\frac{1}{2^{T-1}}\phi^{*}<\epsilon$.
This implies $\sum_{t=0}^{\infty}\frac{1}{2^{t}}\phi(\mu_{t})$ is uniformly
convergent. Therefore, $F$ is continuous under $\mathcal{W}$. By Lemma 9 and
$\mathcal{W}(\tau)$ is nonempty, we can find

$\displaystyle(\mu^{*}_{t})_{t\in\mathcal{T}}\in\text{argmax}_{(\mu_{t})_{t\in\mathcal{T}}\in\mathcal{W}(\tau)}F((\mu_{t})_{t\in\mathcal{T}}).$

Suppose that there exists $(\mu_{t})_{t\in\mathcal{T}}\in\mathcal{W}(\tau)$
such that $\phi(\mu_{t})\geq\phi(\mu^{*}_{t})$ for every $t\in\mathcal{T}$ and
the inequality is strict for some $t\in\mathcal{T}$. This directly implies
$F((\mu_{t})_{t\in\mathcal{T}})>F((\mu^{*}_{t})_{t\in\mathcal{T}})$, which
contradicts the optimality of $(\mu^{*}_{t})_{t\in\mathcal{T}}$. Therefore,
$(\mu^{*}_{t})_{t\in\mathcal{T}}$ is a maximal belief path under $\tau$. ∎

### IV.2 Topological space of feasible stopping times and the existence of a
solution to the sender’s optimization problem

We will create a topology of feasible stopping times by using properties of
the set $\mathcal{W}$. However, we need the regularity assumption of $c$ being
smooth. We start with the following lemma.

######  Lemma 10.

Suppose $c$ be a smooth function. For any $\epsilon>0$, there exists
$T\in\mathcal{T}$ such that for every $t>T$, we have
$\mathbb{E}[c(\tau)1\\{\tau>t\\}]<\epsilon$ for every feasible stopping time
$\tau$.

###### Proof of Lemma 10.

We know from Lemma 1 that given feasible stopping time $\tau$ we can find a
sequence $(C_{t})_{t\in\mathcal{T}}$ such that

$\displaystyle\mathbb{P}(\tau>t+1\mid\tau>t)=\frac{C_{t}+c(t)-c(t+1)}{C_{t+1}}$

for every $t\in\mathcal{T}$, which means $C_{t}\geq
c(t+1)-c(t)\geq\underline{$c$}$. Thus,

$\displaystyle\mathbb{P}(\tau=t)\leq\mathbb{P}(\tau>t-1)$
$\displaystyle=\prod_{s=0}^{t-2}\mathbb{P}(\tau>s+1\mid\tau>s)$
$\displaystyle=\prod_{s=0}^{t-2}\frac{C_{s}+c(s)-c(t+1)}{C_{s+1}}$
$\displaystyle=\frac{C_{0}-c(1)}{C_{t-1}}\prod_{s=1}^{t-2}\frac{C_{s}+c(s)-c(s+1)}{\phi^{*}}$
$\displaystyle\leq\frac{\phi^{*}-c(1)}{\underline{$c$}}\prod_{s=1}^{t-2}\bigg{(}1-\frac{c(s+1)-c(s)}{\phi^{*}}\bigg{)}$
$\displaystyle\leq\frac{\phi^{*}-c(1)}{\underline{$c$}}\bigg{(}1-\frac{\underline{$c$}}{\phi^{*}}\bigg{)}^{t-2}.$

Note that $c(s)\leq\bar{c}\cdot s$. Therefore,

$\displaystyle\mathbb{E}[c(\tau)1\\{\tau>t\\}]=\sum_{s>t}c(s)\mathbb{P}(\tau=s)\leq\frac{\bar{c}(\phi^{*}-c(1))}{\underline{$c$}}\sum_{s>t}s\bigg{(}1-\frac{\underline{$c$}}{\phi^{*}}\bigg{)}^{s-2}.$

We know that
$\sum_{s=0}^{\infty}s\Big{(}1-\frac{\underline{$c$}}{\phi^{*}}\Big{)}^{s-2}$
converges. For any $\epsilon>0$, there exists $T\in\mathcal{T}$ such that, for
every $t>T$, we have

$\mathbb{E}[c(\tau)1\\{\tau>t\\}]\leq\frac{\bar{c}(\phi^{*}-c(1))}{\underline{c}}\sum_{s>t}s\bigg{(}1-\frac{\underline{c}}{\phi^{*}}\bigg{)}^{s-2}\leq\epsilon,$

for every feasible stopping time $\tau$, as desired. ∎

We introduce a metric $\Delta$ of $\mathcal{D}$ as follows: for any
probability measures $d_{1},d_{2}\in\mathcal{D}$,

$\Delta(d_{1},d_{2})\coloneqq\sum_{t=0}^{\infty}\big{\lvert}d_{1}(t)-d_{2}(t)\big{\rvert}c(t).$

This metric is well-defined because the obedience constraint at $t=0$ implies

$\sum_{t=0}^{\infty}\big{\lvert}d_{1}(t)-d_{2}(t)\big{\rvert}c(t)\leq\sum_{t=0}^{\infty}d_{1}(t)c(t)+\sum_{t=0}^{\infty}d_{2}(t)c(t)\leq
2\phi^{*}<\infty.$

It is easy to verify that this is indeed a metric. We use the metric $\Delta$
to construct a topological space of $\mathcal{D}$ and obtain the following
proposition

######  Proposition 12.

$\mathcal{D}$ is compact under the metric $\Delta$.

###### Proof of Proposition 12.

Define $\Psi:\mathcal{D}\to l^{1}$ such that
$\big{(}\Psi(d)\big{)}_{t}=c(t)d(t)$ for every $d\in\mathcal{D}$ and
$t\in\mathcal{T}$. It is easy to see that $\Psi(d)\in l^{1}$ and
$\|\Psi(d)\|_{1}\leq\phi^{*}$ by the obedience constraint at $t=0$. Thus,
$\Psi(\mathcal{D})$ is bounded.

We show that $\Psi(\mathcal{D})$ is closed. Consider any convergent sequence
$(\Psi(d_{i}))_{i\in\mathbb{N}}$ under $l^{1}$ norm where
$d_{i}\in\mathcal{D}$ for every $i\in\mathbb{N}$. Suppose that
$\Psi(d_{i})\to(s^{*}_{t})_{t\in\mathcal{T}}$ under $l^{1}$ norm. The function
$(s_{t})_{t\in\mathbb{N}}\in l^{1}\mapsto\sum_{t=0}^{\infty}s_{t}/c(t)$ is
continuous because, for any $\epsilon>0$ and
$(s_{t})_{t\in\mathbb{N}},(s^{\prime}_{t})_{t\in\mathbb{N}}\in l^{1}$, if
$\|s^{\prime}-s\|_{1}\leq\epsilon c(1)$, then

$\bigg{\lvert}\sum_{t=1}^{\infty}\frac{s^{\prime}_{t}-s_{t}}{c(t)}\bigg{\rvert}\leq\frac{1}{c(1)}\|s^{\prime}-s\|_{1}\leq\epsilon.$

Consider that
$\sum_{t=0}^{\infty}(\Psi(d_{i}))_{t}/c(t)=\sum_{t=0}^{\infty}d_{i}(t)=1$ for
every $i\in\mathbb{N}$. This implies $\sum_{t=0}^{\infty}s^{*}_{t}/c(t)=1$.
Therefore, we can construct a distribution $d^{*}\in\Delta(\mathcal{T})$ in
such a way that $d^{*}(t)=s^{*}_{t}/c(t)$ for every $t\in\mathcal{T}$, which
means $\Psi(d^{*})=(s^{*}_{t})_{t\in\mathcal{T}}$. Suppose the corresponding
belief path to the distribution of stopping time $d_{i}$ is
$(\mu^{i}_{t})_{t\in\mathcal{T}}\in\mathcal{W}$. Because $\mathcal{W}$ is
sequentially compact, there exists a subsequence
$\big{(}(\mu^{i_{n}}_{t})_{t\in\mathcal{T}}\big{)}_{n\in\mathbb{N}}$
converging to some belief path $(\mu_{t}^{*})_{t\in\mathcal{T}}$. We will show
that $(d^{*},(\mu_{t}^{*})_{t\in\mathcal{T}})\in\mathcal{V}$. Let $\tau_{n}$
be the stopping time corresponding to
$(d_{n},(\mu^{n}_{t})_{t\in\mathcal{T}})$.

  1. 1.

Obedience constraint: for any $n\in\mathbb{N}$ and $t\in\mathcal{T}$,

$\displaystyle\quad\quad\mathbb{P}(\tau_{i_{n}}>t)\phi(\mu^{i_{n}}_{t})\geq\mathbb{E}[c(\tau_{i_{n}})1\\{\tau_{i_{n}}>t\\}]-\mathbb{P}(\tau_{i_{n}}>t)c(t)$
$\displaystyle\Longrightarrow\sum_{s>t}d_{i_{n}}(s)\phi(\mu^{i_{n}}_{t})\geq\sum_{s>t}d_{i_{n}}(s)(c(s)-c(t))$
$\displaystyle\Longrightarrow\lim_{n\to\infty}\sum_{s>t}d_{i_{n}}(s)\phi(\mu^{i_{n}}_{t})\geq\lim_{n\to\infty}\sum_{s>t}d_{i_{n}}(s)(c(s)-c(t))$
$\displaystyle\Longrightarrow\sum_{s>t}\lim_{n\to\infty}d_{i_{n}}(s)\phi(\mu^{i_{n}}_{t})\geq\sum_{s>t}\lim_{n\to\infty}d_{i_{n}}(s)(c(s)-c(t))$
$\displaystyle\Longrightarrow\sum_{s>t}d^{*}(s)\phi(\mu^{*}_{t})\geq\sum_{s>t}d^{*}(s)(c(s)-c(t)),$

which implies the obedient constraint for
$(d^{*},(\mu^{*}_{t})_{t\in\mathcal{T}})$. The summation interchange follows
by the fact that $\phi(\mu)\leq\mu^{*}$ and Lemma 10, which implies both
summations are uniformly convergent.

  2. 2.

Boundary constraint: for any $n\in\mathbb{N}$, $t\in\mathcal{T},$ and
$\theta\in\Theta$,

$\displaystyle\quad\quad\mathbb{P}(\tau_{i_{n}}>t)\mu^{i_{n}}_{t}(\theta)\geq\mathbb{P}(\tau_{i_{n}}>t+1)\mu^{i_{n}}_{t+1}(\theta)$
$\displaystyle\Longrightarrow\sum_{s>t}d_{i_{n}}(s)\mu^{i_{n}}_{t}(\theta)\geq\sum_{s>t+1}d_{i_{n}}(s)\mu^{i_{n}}_{t+1}(\theta).$

With the same argument, applying limit $n$ to $\infty$ implies the boundary
constraint for $(d^{*},(\mu^{*}_{t})_{t\in\mathcal{T}})$.

Therefore, $(d^{*},(\mu_{t}^{*})_{t\in\mathcal{T}})\in\mathcal{V},$ which
means $d^{*}\in\mathcal{D}$. Thus,
$(s^{*}_{t})_{t\in\mathcal{T}}=\Psi(d^{*})\in\Psi(\mathcal{D})$, so
$\Psi(\mathcal{D})$ is closed.

Next, We show that $\Psi(\mathcal{D})$ is equismall at infinity. From Lemma
10, for any $\epsilon>0$, there exists $T\in\mathcal{T}$ such that for every
$t>T$ and every $I\in\mathcal{I}^{*}$, we have

$\displaystyle\epsilon>\mathbb{E}[c(\tau)1\\{\tau(I)>t\\}]=\sum_{s>t}c(s)d(I)(s)=\sum_{s>t}\big{(}\Psi(d(I))\big{)}_{s},$

which implies that $\Psi(\mathcal{D})$ is equismall at infinity. Therefore, by
Theorem 44.2 from Treves (1990), we have $\Psi(\mathcal{D})$ is compact under
$l^{1}$ norm. Note that, for any $d_{1},d_{2}\in\mathcal{D}$,
$\Delta(d_{1},d_{2})=\|\Psi(d_{1})-\Psi(d_{2})\|_{1}$, which means that $\Psi$
is continuous. Therefore, because $\Psi(\mathcal{D})$ is compact,
$\mathcal{D}$ is compact under the metric $\Delta.$ ∎

This proposition directly implies the existence of the sender’s optimization
problem under the regularity condition of $f$. To see this, the function
$d(\tau)\mapsto\sum_{t\in\mathcal{T}}d(t)f(t)$ is continuous under the
topology of $\mathcal{D}$ because, for any $\epsilon>0$ and
$d_{1},d_{2}\in\mathcal{D}$, if $\Delta(d_{1},d_{2})<\epsilon/L$, then

$\bigg{\lvert}\sum_{t\in\mathcal{T}}d_{1}(t)f(t)-\sum_{t\in\mathcal{T}}d_{2}(t)f(t)\bigg{\rvert}\leq\sum_{t\in\mathcal{T}}|d_{1}(t)-d_{2}(t)|f(t)\leq
L\Delta(d_{1},d_{2})=\epsilon.$

This implies there is a solution to the optimization problem of the sender.

##  Appendix V Proof of Proposition 10

Before we prove Proposition 10, we first develop several lemmas establishing
properties of $cconv_{t}(f)$ and $s^{*}$.

######  Lemma 11.

For every $t\in\mathcal{T}$, $s^{*}(t)>t^{*}$, $cconv_{t}(f)$ is concave, and
$cconv_{t}(f)\geq f$ over the domain of $cconv_{t}(f)$.

###### Proof of Lemma 11.

We will show that $cconv_{t}(f)(s)$ has decreasing differences. Within this
proof, we abuse notation $s^{*}(t)$ by $s^{*}$. If $f(s^{*}+1)-f(s^{*})\geq
f(s^{*})-f(s^{*}-1)$, then

$\displaystyle(s^{*}-1-t)(f(s^{*})-f(s^{*}-1))-f(s^{*}-1)$
$\displaystyle=(s^{*}-t)(f(s^{*})-f(s^{*}-1))-f(s^{*})$
$\displaystyle\leq(s^{*}-t)(f(s^{*}+1)-f(s^{*}))-f(s^{*})$
$\displaystyle<-f(t),$

which implies that $f(s^{*})-f(s^{*}-1)<\frac{f(s^{*}-1)-f(t)}{s^{*}-1-t}$,
which contradicts the definition of $s^{*}$. Thus,
$f(s^{*}+1)-f(s^{*})<f(s^{*})-f(s^{*}-1)$, which means $s^{*}>t^{*}$. This
implies for every $s>s^{*}$, we have $f(s+1)-f(s)<f(s)-f(s-1)$. The definition
of $cconv_{t}(f)$ implies
$cconv_{t}(f)(s+1)-cconv_{t}(f)(s)<cconv_{t}(f)(s)-cconv_{t}(f)(s-1)$ for
every $s>s^{*}$. The last inequality is also true for every $s<s^{*}$ because
$cconv_{t}(f)$ is linear over the domain $\\{t,\dots,s^{*}\\}$. It suffices to
show that
$cconv_{t}(f)(s^{*}+1)-cconv_{t}(f)(s^{*})<cconv_{t}(f)(s^{*})-cconv_{t}(f)(s^{*}-1)$,
which is followed by

$\displaystyle cconv_{t}(f)(s^{*})-cconv_{t}(f)(s^{*}-1)$
$\displaystyle=\frac{f(s^{*})-f(t)}{s^{*}-t}$
$\displaystyle>f(s^{*}+1)-f(s^{*})$
$\displaystyle=cconv_{t}(f)(s^{*}+1)-cconv_{t}(f)(s^{*}),$

as desired. To see that $cconv_{t}(f)\geq f$ over the domain of
$cconv_{t}(f)$, it suffices to consider the domain of $\\{t,\dots,s^{*}\\}$.
We will show by induction on $\Delta\in\\{0,\dots,s^{*}-t\\}$ that
$cconv_{t}(f)(s^{*}-\Delta)\geq f(s^{*}-\Delta)$. It is clear that the
inequality is true when $\Delta=0$. Assume that the inequality is true for
some $\Delta$. By the definition of $s^{*}$, we must have

$\displaystyle
f(s^{*}-\Delta)-f(s^{*}-\Delta-1)\geq\frac{f(s^{*}-\Delta)-f(t)}{s^{*}-\Delta-t},$

which implies

$\displaystyle f(s^{*}-\Delta-1)$
$\displaystyle\leq\Big{(}1-\frac{1}{s^{*}-\Delta-t}\Big{)}f(s^{*}-\Delta)+\frac{1}{s^{*}-\Delta-t}f(t)$
$\displaystyle\leq\Big{(}1-\frac{1}{s^{*}-\Delta-t}\Big{)}cconv_{t}(f)(s^{*}-\Delta)+\frac{1}{s^{*}-\Delta-t}cconv_{t}(f)(t)$
$\displaystyle=cconv_{t}(f)\bigg{(}\Big{(}1-\frac{1}{s^{*}-\Delta-t}\Big{)}(s^{*}-\delta)+\frac{t}{s^{*}-\Delta-t}\bigg{)}$
$\displaystyle=cconv_{t}(f)(s^{*}-\Delta-1),$

as desired. ∎

We introduce properties of $s^{*}$ that will be useful in the proof of
Proposition 10.

######  Lemma 12 (Properties of $s^{*}$).

  1. (i)

For every $t,t^{\prime}\in\mathcal{T}$ such that $t\leq t^{\prime}<s^{*}(t)$,
we have $s^{*}(t^{\prime})\leq s^{*}(t)$.

  2. (ii)

For every $t<t^{\prime}\in\mathcal{T}$, we have
$s^{*}(t^{\prime})-t^{\prime}\leq s^{*}(t)-t$. This becomes a strict
inequality if $t<t^{*}.$

  3. (iii)

For every $t,t_{1},t_{2}\in\mathcal{T}$ such that $t\leq t_{1}<t_{2}\leq
s^{*}(t)$, we have

$\displaystyle
f(t_{1})<\frac{t_{2}-t_{1}}{t_{2}-t}f(t)+\frac{t_{1}-t}{t_{2}-t}f(t_{2}).$

###### Proof.

(i) From Lemma 11, we know $cconv_{t}(f)(t^{\prime})\geq f(t^{\prime})$. Thus,

$\displaystyle f(s^{*}(t)+1)-f(s^{*}(t))$
$\displaystyle<\frac{f(s^{*}(t))-f(t)}{s^{*}(t)-t}$
$\displaystyle=\frac{cconv_{t}(f)(s^{*}(t))-cconv_{t}(f)(t^{\prime})}{s^{*}(t)-t^{\prime}}$
$\displaystyle\leq\frac{f(s^{*}(t))-f(t^{\prime})}{s^{*}(t)-t^{\prime}},$

which implies that $s^{*}(t^{\prime})\leq s^{*}(t)$, as desired.

(ii) It is sufficient to show the statement in the case of $t^{\prime}=t+1$.
If $s^{*}(t)-t>1$, then $t<t+1<s^{*}(t)$. From (i), we have $s^{*}(t+1)\leq
s^{*}(t)$, which implies that $s^{*}(t+1)-(t+1)<s^{*}(t)-t$. On the other
hand, if $s^{*}(t)-t=1$, then $f(t+2)-f(t+1)<f(t+1)-f(t)$, which means $t\geq
t^{*}$. This implies that $f(t+3)-f(t+2)<f(t+2)-f(t+1)$, so $s^{*}(t+1)=t+2$.
Therefore, $s^{*}(t+1)-(t+1)=1=s^{*}(t)-t$, as desired. From the proof, we get
the strict inequality when $t<t^{*}$.

(iii) Consider any $t^{\prime}\in\\{t+1,\dots,s^{*}(t)\\}$. The definition of
$s^{*}(\cdot)$ implies that
$f(t^{\prime})-f(t^{\prime}-1)>\frac{f(t^{\prime}-1)-f(t+1)}{t^{\prime}-t-2}$.
Therefore,

$\displaystyle(t^{\prime}-t-1)f(t^{\prime}-1)-(t^{\prime}-t-2)f(t^{\prime})<f(t+1).$

If $t^{\prime}<s^{*}(t)$, we also have
$(t^{\prime}-t)f(t^{\prime})-(t^{\prime}-t-1)f(t^{\prime}+1)<f(t+1)$, which
implies

$\displaystyle(t^{\prime}-t)f(t^{\prime}-1)-(t^{\prime}-t-2)f(t^{\prime}+1)<2f(t+1).$

With a simple induction, we obtain the equation in (iii). ∎

We are now ready to prove Proposition 10.

###### Proof of Proposition 10.

Choose a feasible stopping time $\tau$ that maximizes $\mathbb{E}[f(\tau)]$.
Suppose that $\tau$ is generated by an increasing sequence $t_{1},\dots,t_{n}$
with a pair of terminal times $(t_{n+1},\bar{t})$. Without loss of generality,
assume $t_{n+1}>t_{n}$. If $\bar{t}-t_{n+1}=\phi^{*}/c$, then the DM is
indifferent between continuing paying attention and stopping at time
$t_{n+1}$. Thus, an increasing sequence $t_{1},\dots,t_{n+1}$ with a pair of
terminal times $(\bar{t},\bar{t})$ also generates $\tau$. It is sufficient to
consider the case that $\bar{t}-t_{n+1}<\phi^{*}/c$. We will show that
$t_{i}=i$ for every $i\in\\{1,\dots,n\\}$ and $t_{n+1}=\bar{t}=n+\phi^{*}/c$
The proof consists of two following steps:

Step 1: Terminal Stage. We will show that $t_{n+1}=\bar{t}$. Suppose a
contradiction that $t_{n+1}<\bar{t}$. Define stopping times
$\tau_{1},\tau_{2}$ generated by the same increasing sequence
$t_{1},\dots,t_{n}$ but with different terminal stopping times
$(t_{n+1}-1,\bar{t})$ and $(t_{n+1}+1,\bar{t})$. the stopping time $\tau_{1}$
is well-defined because $t_{n+1}-1\geq t_{n}$ and
$\bar{t}-(t_{n+1}-1)\leq\phi^{*}/c$. The stopping time $\tau_{2}$ is also
well-defined because $\bar{t}>t_{n+1}$. We obtain the following equations:

$\displaystyle\mathbb{E}[f(\tau)\mid\tau>t_{n}]$
$\displaystyle=\frac{\bar{t}-\phi^{*}/c-t_{n}}{\bar{t}-t_{n+1}}f(t_{n+1})+\frac{\phi^{*}/c+t_{n}-t_{n+1}}{\bar{t}-t_{n+1}}f(\bar{t})$
$\displaystyle\mathbb{E}[f(\tau_{1})\mid\tau_{1}>t_{n}]$
$\displaystyle=\frac{\bar{t}-\phi^{*}/c-t_{n}}{\bar{t}-t_{n+1}+1}f(t_{n+1}-1)+\frac{\phi^{*}/c+t_{n}-t_{n+1}+1}{\bar{t}-t_{n+1}+1}f(\bar{t}),$
$\displaystyle\mathbb{E}[f(\tau_{2})\mid\tau_{2}>t_{n}]$
$\displaystyle=\frac{\bar{t}-\phi^{*}/c-t_{n}}{\bar{t}-t_{n+1}-1}f(t_{n+1}+1)+\frac{\phi^{*}/c+t_{n}-t_{n+1}-1}{\bar{t}-t_{n+1}-1}f(\bar{t}),.$

Note that, if $t_{n+1}=t_{n}+1$, the second equation still holds. To see this,
we have $\phi^{*}/c+t_{n+1}>\bar{t}\geq\phi^{*}/c+t_{n}$, so
$\bar{t}=\phi^{*}/c+t_{n}$, which implies that the coefficient of
$f(t_{n+1}-1)$ is equal to $0$. Because $\tau,\tau_{1},$ and $\tau_{2}$ are
identical until time $t_{n}$, the optimality of $\tau$ implies

$\displaystyle\mathbb{E}[f(\tau_{1})]\leq\mathbb{E}[f(\tau)]$
$\displaystyle\Longrightarrow\mathbb{E}[f(\tau_{1})\mid\tau>t_{n}]\leq\mathbb{E}[f(\tau)\mid\tau>t_{n}]$
$\displaystyle\Longrightarrow(\bar{t}-t_{n+1})f(t_{n+1}-1)+f(\bar{t})\leq(\bar{t}-t_{n+1}+1)f(t_{n+1})$

and

$\displaystyle\mathbb{E}[f(\tau_{2})]\leq\mathbb{E}[f(\tau)]$
$\displaystyle\Longrightarrow\mathbb{E}[f(\tau_{2})\mid\tau>t_{n}]\leq\mathbb{E}[f(\tau)\mid\tau>t_{n}]$
$\displaystyle\Longrightarrow(\bar{t}-t_{n+1})f(t_{n+1}+1)\leq(\bar{t}-t_{n+1}-1)f(t_{n})+f(\bar{t}).$

Because $\bar{t}-t_{n+1}>0$, these two inequalities altogether imply
$f(t_{n+1}+1)-f(t_{n+1})<f(t_{n+1})-f(t_{n+1}-1)$. By the definition of
S-shaped, we obtain $t^{*}<t_{n+1}<\bar{t}$, so $f$ is concave in the interval
$[t_{n+1},\bar{t}]$. Consider a stopping time $\tau^{\prime}$ generated by the
same increasing sequence $t_{1},\dots,t_{n}$ but with a pair of terminal
stopping times $(t_{n}+\phi^{*}/c,t_{n}+\phi^{*}/c)$. We obtain
$\mathbb{E}[f(\tau^{\prime})\mid\tau^{\prime}>t_{n}]=f(t_{n}+\phi^{*}/c),$
which implies that

$\displaystyle\mathbb{E}[f(\tau^{\prime})\mid\tau^{\prime}>t_{n}]-\mathbb{E}[f(\tau)\mid\tau^{\prime}>t_{n}]$
$\displaystyle=f(t_{n}+\phi^{*}/c)-\bigg{(}\frac{\bar{t}-\phi^{*}/c-t_{n}}{\bar{t}-t_{n+1}}f(t_{n+1})+\frac{\phi^{*}/c+t_{n}-t_{n+1}}{\bar{t}-t_{n+1}}f(\bar{t})\bigg{)}>0$

by the concavity of $f$ in the interval $[t_{n+1},\bar{t}]$. Because $\tau$
and $\tau^{\prime}$ are identical up until time $t_{n}$, we have
$\mathbb{E}[f(\tau^{\prime})]>\mathbb{E}[f(\tau)]$, which contradicts the
optimality of $\tau$. Therefore, we must have $t_{n+1}=\bar{t}$, and this
implies $t_{n+1}=\bar{t}=t_{n}+\phi/c$.

Step 2: Indifferent Stage. Consider any $i\in\\{0,\dots,n-1\\}$. We will show
that $t_{i+1}-t_{i}=1$, where $t_{0}$ is treated as $0$. Suppose a
contradiction that $t_{i+1}-t_{i}>1$. We consider the following three cases:

Case 1: $t_{i}\geq\tilde{t}$. We will show that it is sufficient to ignore
this case. Consider a stopping time $\tau^{\prime\prime}_{1}$ generated by an
increasing sequence $t_{1},\dots,t_{i}$ with a pair of terminal times
$(t_{i}+\phi^{*}/c,t_{i}+\phi^{*}/c).$ We can see that the DM is indifferent
at time $t_{i}$ for both information structures corresponding to stopping
times $\tau$ and $\tau^{\prime\prime}_{1}$. Thus,

$\displaystyle\mathbb{E}[\tau\mid\tau>t_{i}]=\mathbb{E}[\tau^{\prime\prime}\mid\tau>t_{i}]=t_{i}+\frac{\phi^{*}}{c}.$

Therefore, by Jensen’s inequality, we have

$\displaystyle\mathbb{E}[f(\tau)\mid\tau>t_{i}]\leq\mathbb{E}[cconv_{t_{i}}(f)(\tau)\mid\tau>t_{i}]\leq
cconv_{t_{i}}(f)\Big{(}t_{i}+\frac{\phi^{*}}{c}\Big{)}=f\Big{(}t_{i}+\frac{\phi^{*}}{c}\Big{)}=\mathbb{E}[f(\tau^{\prime\prime}_{1})\mid\tau>t_{i}],$

where the equality follows from the fact that $t_{i}\geq\tilde{t}$, which
implies $s^{*}(t_{i})-t_{i}\leq\frac{\phi^{*}}{c}$. Because $\tau$ and
$\tau^{\prime\prime}_{1}$ are identical up until time $t_{i}$, we can conclude
that $\mathbb{E}[f(\tau)]\leq\mathbb{E}[f(\tau^{\prime\prime}_{1})]$. Consider
that all indifferent blocks generated by $\tau^{\prime\prime}$ start before
time $\tilde{t}.$ This implies we only need to consider that $t_{i}<\tilde{t}$
for every $i\in\\{0,\dots,n-1\\}$

Case 2: $t_{i}<\tilde{t}<t_{i+1}$. By Step 1 and the previous case, we can
modify the stopping time so that all consequent indifference stages start
before time $\tilde{t}$, and the terminal stage has a single terminal time.
This implies the stage $[t_{i},t_{i+1}]$ is the last indifferent stage
followed by the terminal stage $[t_{i+1},t_{i+1}+\phi^{*}/c]$. Note that
$t_{i+1}<t_{i}+\phi^{*}/c$. Therefore,

$\displaystyle\mathbb{E}[f(\tau)\mid\tau>t_{i}]$
$\displaystyle=\frac{(t_{i+1}-t_{i})c}{\phi^{*}}f(t_{i+1})+\frac{\phi^{*}-(t_{i+1}-t_{i})c}{\phi^{*}}f\Big{(}t_{i+1}+\frac{\phi^{*}}{c}\Big{)}$
$\displaystyle\leq\frac{(t_{i+1}-t_{i})c}{\phi^{*}}cconv_{t_{i+1}}(f)(t_{i+1})+\frac{\phi^{*}-(t_{i+1}-t_{i})c}{\phi^{*}}cconv_{t_{i+1}}(f)\Big{(}t_{i+1}+\frac{\phi^{*}}{c}\Big{)}$

Consider a stopping time $\tau^{\prime\prime}_{2}$ generated by an increasing
sequence $t_{1},\dots,t_{i}$ with a pair of terminal times
$(t_{i+1},t_{i+1}+\phi^{*}/c-1)$. This satisfies the conditions of a terminal
stage because $t_{i+1}+\phi^{*}/c-t_{i}>\phi^{*}/c$. This implies

$\displaystyle\mathbb{E}[f(\tau^{\prime\prime}_{2})\mid\tau>t_{i}]=\frac{c(t_{i+1}-t_{i}-1)}{\phi^{*}-c}f(t_{i+1})+\frac{\phi^{*}-(t_{i+1}-t_{i})c}{\phi^{*}-c}f\Big{(}t_{i+1}+\frac{\phi^{*}}{c}-1\Big{)}.$

From Lemma 12 (ii), we have
$s^{*}(t_{i+1})-t_{i+1}<s^{*}(\tilde{t})-\tilde{t}\leq\phi^{*}/c$, where the
strict inequality follows by the fact that $\tilde{t}<t^{*}$. Thus,
$s^{*}(t_{i+1})\leq t_{i+1}+\phi^{*}/c-1$. Therefore,

$\displaystyle\mathbb{E}[f(\tau^{\prime\prime}_{2})\mid\tau>t_{i}]=\frac{c(t_{i+1}-t_{i}-1)}{\phi^{*}-c}cconv_{t_{i+1}}(f)(t_{i+1})+\frac{\phi^{*}-(t_{i+1}-t_{i})c}{\phi^{*}-c}cconv_{t_{i+1}}(f)\Big{(}t_{i+1}+\frac{\phi^{*}}{c}-1\Big{)}.$

By Jensen’s inequality, we have

$\displaystyle
cconv_{t_{i+1}}(f)\Big{(}t_{i+1}+\frac{\phi^{*}}{c}-1\Big{)}\geq\frac{c}{\phi^{*}}cconv_{t_{i+1}}(f)(t_{i+1})+\frac{\phi^{*}-c}{\phi^{*}}cconv_{t_{i+1}}(f)\Big{(}t_{i+1}+\frac{\phi^{*}}{c}\Big{)}.$

The above equalities and inequality imply that
$\mathbb{E}[f(\tau^{\prime\prime}_{2})\mid\tau>t_{i}]\geq\mathbb{E}[f(\tau)\mid\tau>t_{i}]$.
However, this is contradiction because $\tau^{\prime\prime}_{2}$ has two
different terminal times and $t_{i+1}-t_{i}<\phi^{*}/c$ and Step 1 says it is
impossible to have sender’s optimal stopping time induced by two different
terminal times.

Case 3: $t_{i}<t_{i+1}\leq\tilde{t}$. The first two cases imply that it is
sufficient to consider the case that all indifference blocks end weakly before
time $\tilde{t}$.

Suppose a contradiction that $t_{i+1}-t_{i}>1$. Consider a stopping time
$\tau^{\prime\prime}_{3}$ generated by an increasing sequence $t_{1},\dots
t_{i},t_{i}+1,t_{i+1},\dots,t_{n}$ with a pair of terminal stopping times
$(t_{n}+\phi^{*}/c,t_{n}+\phi^{*}/c)$. We have

$\displaystyle
t_{i}+\phi^{*}/c=\mathbb{E}[\tau\mid\tau>t_{i}]=\mathbb{E}[\tau^{\prime\prime}_{3}\mid\tau^{\prime\prime}_{3}>t_{i}].$

Consider that

$\displaystyle 0$
$\displaystyle=\mathbb{P}(\tau^{\prime\prime}_{3}>t_{i})\big{(}\mathbb{E}[\tau^{\prime\prime}_{3}\mid\tau^{\prime\prime}_{3}>t_{i}]-\mathbb{E}[\tau\mid\tau>t_{i}]\big{)}$
$\displaystyle=\mathbb{P}(\tau^{\prime\prime}_{3}=t_{i}+1)\cdot(t_{i}+1)+\big{(}\mathbb{P}(\tau^{\prime\prime}_{3}=t_{i+1})-\mathbb{P}(\tau=t_{i+1})\big{)}\cdot
t_{i+1}$ $\displaystyle\hskip
14.22636pt+\big{(}\mathbb{P}(\tau^{\prime\prime}_{3}>t_{i+1})-\mathbb{P}(\tau>t_{i+1})\big{)}\mathbb{E}[\tau\mid\tau>t_{i+1}].$

Using the same argument shown in Proposition 9, we have
$\mathbb{P}(\tau^{\prime\prime}_{3}>t_{i+1})>\mathbb{P}(\tau>t_{i+1})$. We can
rewrite the above equation as

$\displaystyle\mathbb{P}(\tau=t_{i+1})\cdot t_{i+1}$
$\displaystyle=\mathbb{P}(\tau^{\prime\prime}_{3}=t_{i}+1)\cdot(t_{i}+1)+\mathbb{P}(\tau^{\prime\prime}_{3}=t_{i+1})\cdot
t_{i+1}$ $\displaystyle\hskip
28.45274pt+\big{(}\mathbb{P}(\tau^{\prime\prime}_{3}>t_{i+1})-\mathbb{P}(\tau>t_{i+1})\big{)}\mathbb{E}[\tau\mid\tau>t_{i+1}].$

Consider any $t\in\\{t_{i+1}+1,\dots,s^{*}(\tilde{t}-1)\\}$. Because
$t_{i}+1\leq t_{i+1}-1\leq\tilde{t}-1$, we have

$\displaystyle t\leq s^{*}(\tilde{t}-1)\leq s^{*}(t_{i}+1).$

Therefore, by Lemma 12 (iii), we

$\displaystyle(t-t_{i}-1)f(t_{i+1})<(t-t_{i+1})f(t_{i}+1)+(t_{i+1}-t_{i}-1)f(t),$

for every $t\in\\{t_{i+1}+1,\dots,s^{*}(\tilde{t}-1)\\}$. Because the support
of $\tau\mid\tau>t_{i+1}$ is a subset of
$\\{t_{i+1}+1,\dots,s^{*}(\tilde{t}-1)\\}$, a linear combination of the above
inequality for $t\in\\{t_{i+1}+1,\dots,s^{*}(\tilde{t}-1)\\}$ implies

$\displaystyle\mathbb{P}(\tau=t_{i+1})\cdot f(t_{i+1})$
$\displaystyle<\mathbb{P}(\tau^{\prime\prime}_{3}=t_{i}+1)\cdot
f(t_{i}+1)+\mathbb{P}(\tau^{\prime\prime}_{3}=t_{i+1})\cdot f(t_{i+1})$
$\displaystyle\hskip
28.45274pt+\big{(}\mathbb{P}(\tau^{\prime\prime}_{3}>t_{i+1})-\mathbb{P}(\tau>t_{i+1})\big{)}\mathbb{E}[f(\tau)\mid\tau>t_{i+1}].$

This implies
$\mathbb{E}[f(\tau^{\prime\prime}_{3})\mid\tau^{\prime\prime}_{3}>t_{i}]>\mathbb{E}[f(\tau)\mid\tau>t_{i}]$,
which contradicts the optimality of $\tau$.

These two steps imply that we can we choose an sender’s optimal stopping time
such that it is generated by an increasing sequence $1,\dots,t_{*}$ with a
terminal time $t_{*}+\phi/c$, where $t_{*}\leq\tilde{t}$. We will show that
$t_{*}\geq\tilde{t}-1$. Suppose a contradiction that $t_{*}<\tilde{t}-1$
Consider a stopping time $\tau^{\prime\prime\prime}$ generated by an
increasing sequence $1,2,\dots,t_{*}+1$ with a terminal time
$t_{*}+1+\phi^{*}/c$. Therefore,

$\displaystyle 0$
$\displaystyle=\mathbb{E}[\tau^{\prime\prime\prime}\mid\tau^{\prime}>t_{*}]-\mathbb{E}[\tau\mid\tau>t_{*}]$
$\displaystyle=\frac{c}{\phi^{*}}(t_{*}+1)+\frac{\phi_{*}-c}{\phi^{*}}\big{(}t_{*}+1+\frac{\phi^{*}}{c}\big{)}-\big{(}t^{*}+\frac{\phi^{*}}{c}\big{)}$

Because $t_{*}+1<\tilde{t}$, we have
$t_{*}+1,t_{*}+1+\phi^{*}/c\in[t_{*}+1,s^{*}(t_{*}+1)].$ Therefore,

$\displaystyle\mathbb{E}[f(\tau)\mid\tau>t_{*}]$
$\displaystyle=f\Big{(}t_{*}+\frac{\phi^{*}}{c}\Big{)}$
$\displaystyle<cconv_{t_{*}+1}(f)\Big{(}t_{*}+\frac{\phi^{*}}{c}\Big{)}$
$\displaystyle=\frac{c}{\phi^{*}}cconv_{t_{*}+1}(f)(t_{*}+1)+\frac{\phi^{*}-c}{\phi^{*}}cconv_{t_{*}+1}(f)\Big{(}t_{*}+1+\frac{\phi^{*}}{c}\Big{)}$
$\displaystyle=\frac{c}{\phi^{*}}f(t_{*}+1)+\frac{\phi^{*}-c}{\phi^{*}}f\Big{(}t_{*}+1+\frac{\phi^{*}}{c}\Big{)}$
$\displaystyle=\mathbb{E}[f(\tau^{\prime\prime\prime})\mid\tau^{\prime}>t_{*}].$

This implies $\mathbb{E}[f(\tau)]<\mathbb{E}[f(\tau^{\prime\prime\prime})]$,
which contradicts the optimality of $\tau$. This concludes that the sender’s
feasible stopping time is induced by an increasing sequence $1,2,\dots,t_{*}$
with a terminal time $t_{*}+\phi/c$ such that
$t_{*}\in\\{\tilde{t}-1,\tilde{t}\\}$, as desired. ∎

##  Appendix VI Convergence to $\Phi^{*}$ for $|\Theta|>2$

Here we develop an analog of Proposition 5 for multiple states.

######  Proposition 13.

Let $|\Theta|=n>2$. Suppose that

$\displaystyle\sum_{t=0}^{\infty}(\log\mathbb{P}(\tau>t)-\log\mathbb{P}(\tau>t+1))^{n-1}=+\infty.$

Moreover, assume that
$\\{\mu\in\Delta(\Theta):\phi(\mu)\geq\phi(\mu_{0})\\}\subset\text{int
}\Delta(\Theta)$. Then, there exists $T\in\mathcal{T}$ such that
$\mu_{t}\in\Phi^{*}$ for every $t>T$.

Additional assumptions that we need for this proposition is (i) tail of the
probability measure of stopping time does not decay too fast (but still allows
decay with a constant exponential rate) and (ii) the prior belief does not
stay too close to the boundary of $\Delta(\Theta)$ so that the value of full
information at $\mu_{0}$ is greater than that at any belief on the boundary of
$\Delta(\Theta)$.

Before proving Proposition 13, we introduce a definition of a feasible set of
$\mu_{t+1}$ given $\mu_{t}$ and stopping time $\tau$ derived by the boundary
constraint.

######  Definition 11.

For every nonnegative number $r\leq 1$ and belief $\mu\in\Delta(\Theta)$,
define $F(\mu,r)\subset\Delta(\Theta)$ such that

$\displaystyle F(\mu,r)=\\{\mu^{\prime}\in\Delta(\Theta)\mid
r\mu^{\prime}(\theta)\leq\mu(\theta)\quad\forall\theta\in\Theta\\}.$

With this definition, the boundary constraint is equivalent to $\mu_{t+1}\in
F\Big{(}\mu_{t},\frac{\mathbb{P}(\tau>t_{0}+1)}{\mathbb{P}(\tau>t)}\Big{)}$
for every $t\in\mathcal{T}$. We now begin the proof of Proposition 13

###### Proof of Proposition 13.

Consider any feasible stopping time $\tau$.

Choose a maximal belief path and a distribution of stopping time
$(\mu_{t})_{t\in\mathcal{T}}$. From the proof of Proposition 4, we must have
$\phi(\mu_{t})$ is increasing in $t\in\mathcal{T}$ and the boundary constraint
binds. Consider any $t_{0}\in\mathcal{T}$. We will show that, for every
$t\in\mathcal{T}$ and $\lambda\in[0,1)$ such that $t>t_{0}+1$, we have
$\lambda\mu_{t_{0}+1}+(1-\lambda)\mu_{t}\notin\text{ int
}F\Big{(}\mu_{t_{0}},\frac{\mathbb{P}(\tau>t_{0}+1)}{\mathbb{P}(\tau>t_{0})}\Big{)}.$

Assumes a contradiction that there are $t_{1}\in\mathcal{T}$ and
$\lambda\in[0,1)$ such that $t_{1}>t_{0}+1$ and
$\lambda\mu_{t_{0}+1}+(1-\lambda)\mu_{t_{1}}\in\text{ int
}F\Big{(}\mu_{t_{0}},\frac{\mathbb{P}(\tau>t_{0}+1)}{\mathbb{P}(\tau>t_{0})}\Big{)}$.
This means $\frac{\mathbb{P}(\tau>t_{0}+1)}{\mathbb{P}(\tau>t_{0})}<1$, so the
property of an extremal path implies $\mu_{t_{0}+1}\in\text{Bd
}F\Big{(}\mu_{t_{0}},\frac{\mathbb{P}(\tau>t_{0}+1)}{\mathbb{P}(\tau>t_{0})}\Big{)}$,
so $\mu_{t_{0}+1}\neq\mu_{t_{0}}$.

We define a new belief path $(\mu^{\prime}_{t})_{t\in\mathcal{T}}$ as follows:

$\displaystyle\mu^{\prime}_{t}=\begin{cases}\lambda\mu_{t}+(1-\lambda)\mu_{t_{1}},&\text{if
$t_{0}<t\leq t_{1}$}\\\ \mu_{t},&\text{otherwise.}\end{cases}$

We will show that a pair of a belief path and a distribution of stopping time
$((\mu^{\prime}_{t})_{t\in\mathcal{T}},d(\tau))$ is feasible. The obedience
constraint is still the same for $t\notin\\{t_{0}+1,\dots,t_{1}\\}$. If
$t\in\\{t_{0}+1,\dots,t_{1}\\}$, we have

$\displaystyle\phi(\mu^{\prime}_{t})\geq\lambda\phi(\mu_{t})+(1-\lambda)\phi(\mu_{t_{1}})\geq\phi(\mu_{t}),$

where the inequality follows by the fact that $\phi(\mu_{t})$ is increasing in
$t\in\mathcal{T}$. This directly implies the obedience constraint for
$t\in\\{t_{0}+1,\dots,t_{1}\\}$. The boundary constraint is still the same for
$t\in\\{0,\dots,t_{0}-1\\}\cup\\{t_{1},\dots\\}$. The boundary constraint
holds when $t=t_{0}$ by the construction of $\lambda$. For any
$t\in\\{t_{0}+1,\dots,t_{1}-1\\}$, we have

$\displaystyle\min_{\theta\in\Theta}\frac{\mu^{\prime}_{t}(\theta)}{\mu^{\prime}_{t+1}(\theta)}=\min_{\theta\in\Theta}\frac{\lambda\mu_{t}(\theta)+(1-\lambda)\mu_{t_{1}}(\theta)}{\lambda\mu_{t+1}(\theta)+(1-\lambda)\mu_{t_{1}}(\theta)}\geq\min_{\theta\in\Theta}\frac{\mu_{t}(\theta)}{\mu_{t+1}(\theta)}\geq\frac{\mathbb{P}(\tau>t+1)}{\mathbb{P}(\tau>t)},$

This concludes that the belief path $(\mu^{\prime}_{t})_{t\in\mathcal{T}}$ is
also a maximal belief path corresponding to the stopping time $\tau$ because
$\phi(\mu^{\prime}_{t})\geq\phi(\mu_{t})$ for every $t\in\mathcal{T}$.
However, the above inequality us strict for $t=t_{0}$ because
$\mu_{t_{0}+1}\neq\mu_{t_{0}}$, which implies that
$(\mu^{\prime}_{t})_{t\in\mathcal{T}}$ does not satisfy the property of an
extremal path. This contradicts with the fact that
$(\mu^{\prime}_{t})_{t\in\mathcal{T}}$ is a maximal belief path. Therefore,
for every $t_{0},t_{1}\in\mathcal{T}$ such that $t_{1}>t_{0}+1$, we must have
$\lambda\mu_{t_{0}+1}+(1-\lambda)\mu_{t_{1}}\notin\text{int
}F\Big{(}\mu_{t_{0}},\frac{\mathbb{P}(\tau>t+1)}{\mathbb{P}(\tau>t)}\Big{)}$.

For any $\Theta^{\prime}\subset\Theta$, define

$\displaystyle\mathcal{T}_{\Theta^{\prime}}=\bigg{\\{}t\in\mathcal{T}\bigg{\rvert}\Theta^{\prime}=\Big{\\{}\theta\in\Theta\Big{\rvert}\mu_{t+1}(\theta)=\frac{\mathbb{P}(\tau>t)}{\mathbb{P}(\tau>t+1)}\mu_{t}(\theta)\Big{\\}}\bigg{\\}}$

Note that $\mathcal{T}_{\emptyset}=\emptyset$ because the boundary constraint
must bind. Moreover, for every $t\in\mathcal{T}_{\Theta}$, we must have
$\mathbb{P}(\tau>t)=\mathbb{P}(\tau>t+1)$ and $\mu_{t}=\mu_{t+1}$. Therefore,

$\displaystyle\sum_{\Theta^{\prime}\subsetneq\Theta}\sum_{t\in\mathcal{T}_{\Theta^{\prime}}}(\log\mathbb{P}(\tau>t)-\log\mathbb{P}(\tau>t+1))^{n-1}$
$\displaystyle=\sum_{t=0}^{\infty}(\log\mathbb{P}(\tau>t)-\log\mathbb{P}(\tau>t+1))^{n-1}=+\infty.$

Because $2^{|\Theta|}$ is finite, there is a nonempty set
$\Theta^{\prime}\subset\Theta$ such that
$\sum_{t\in\mathcal{T}_{\Theta^{\prime}}}(\log\mathbb{P}(\tau>t)-\log\mathbb{P}(\tau>t+1))^{n-1}=+\infty$.
Consider any $t_{0}<t_{1}\in\mathcal{T}_{\Theta^{\prime}}$, we will show that
there exists $\theta\in\Theta^{\prime}$ such that
$\mu_{t_{1}}(\theta)\geq\frac{\mathbb{P}(\tau>t_{0})}{\mathbb{P}(\tau>t_{0}+1)}\mu_{t_{0}}(\theta)$.
Suppose a contradiction that
$\mu_{t_{1}}(\theta)<\frac{\mathbb{P}(\tau>t_{0})}{\mathbb{P}(\tau>t_{0}+1)}\mu_{t_{0}}(\theta)$
for every $\theta\in\Theta^{\prime}$. Because
$\mu_{t_{0}+1}(\theta)<\frac{\mathbb{P}(\tau>t_{0})}{\mathbb{P}(\tau>t_{0}+1)}\mu_{t_{0}}(\theta)$
for every $\theta\notin\Theta^{\prime}$ we can find a sufficiently small
$1-\lambda>0$ such that
$\lambda\mu_{t_{0}+1}(\theta)+(1-\lambda)\mu_{t_{1}}(\theta)<\frac{\mathbb{P}(\tau>t_{0})}{\mathbb{P}(\tau>t_{0}+1)}\mu_{t_{0}}(\theta)$
for every $\theta\notin\Theta^{\prime}$. For $\theta\in\Theta$, we have

$\displaystyle\frac{\mathbb{P}(\tau>t_{0})}{\mathbb{P}(\tau>t_{0}+1)}\mu_{t_{0}}(\theta)>\lambda\mu_{t_{0}+1}(\theta)+(1-\lambda)\mu_{t_{1}}(\theta).$

This implies $\lambda\mu_{t_{0}+1}+(1-\lambda)\mu_{t_{1}}\in\text{int
}F\Big{(}\mu_{t_{0}},\frac{\mathbb{P}(\tau>t+1)}{\mathbb{P}(\tau>t)}\Big{)}$,
which is a contradiction. Thus, there exists $\theta\in\Theta^{\prime}$ such
that
$1\geq\mu_{t_{1}}(\theta)\geq\frac{\mathbb{P}(\tau>t_{0})}{\mathbb{P}(\tau>t_{0}+1)}\mu_{t_{0}}(\theta)$.
This implies

$\displaystyle\varprod_{\theta\in\Theta^{\prime}}\bigg{(}\log\mu_{t_{0}}(\theta),\log\bigg{(}\frac{\mathbb{P}(\tau>t_{0})}{\mathbb{P}(\tau>t_{0}+1)}\mu_{t_{0}}(\theta)\bigg{)}\bigg{)}\cap\varprod_{\theta\in\Theta^{\prime}}\bigg{(}\log\mu_{t_{1}}(\theta),\log\bigg{(}\frac{\mathbb{P}(\tau>t_{1})}{\mathbb{P}(\tau>t_{1}+1)}\mu_{t_{1}}(\theta)\bigg{)}\bigg{)}=\emptyset,$

for every $t_{0}<t_{1}\in\mathcal{T}_{\Theta^{\prime}}$.

Because
$\big{\\{}\mu\in\Delta(\Theta)\mid\phi(\mu)\geq\phi(\mu_{0})\big{\\}}\subset\text{int
}\Delta(\Theta)$ and $\phi(\mu_{t})$ is increasing in $t$, we have
$\mu_{t}\in\big{\\{}\mu\in\Delta(\Theta)\mid\phi(\mu)\geq\phi(\mu_{0})\big{\\}}\subset\text{int
}\Delta(\Theta).$ Therefore, for each $\theta\in\Theta$, there is
$\underline{$\mu$}_{\theta}>0$ such that
$\mu_{t}(\theta)\geq\underline{$\mu$}_{\theta}$ for every $t\in\mathcal{T}$.
This implies that for every $t\in\mathcal{T}_{\Theta^{\prime}}$ we have

$\displaystyle\varprod_{\theta\in\Theta^{\prime}}\bigg{(}\log\mu_{t}(\theta),\log\bigg{(}\frac{\mathbb{P}(\tau>t)}{\mathbb{P}(\tau>t+1)}\mu_{t}(\theta)\bigg{)}\bigg{)}\subset\varprod_{\theta\in\Theta^{\prime}}[\log\underline{$\mu$}_{\theta},0],$

This implies

$\displaystyle\bigcup_{t\in\mathcal{T}_{\Theta^{\prime}}}\varprod_{\theta\in\Theta^{\prime}}\bigg{(}\log\mu_{t}(\theta),\log\bigg{(}\frac{\mathbb{P}(\tau>t)}{\mathbb{P}(\tau>t+1)}\mu_{t}(\theta)\bigg{)}\bigg{)}\subset\varprod_{\theta\in\Theta^{\prime}}[\log\underline{$\mu$}_{\theta},0].$

We showed that two different boxes inside the union have disjoint interiors.
This implies

$\displaystyle
Vol\bigg{(}\varprod_{\theta\in\Theta^{\prime}}[\log\underline{$\mu$}_{\theta},0]\bigg{)}$
$\displaystyle\geq\sum_{t\in\mathcal{T}_{\Theta^{\prime}}}Vol\bigg{(}\varprod_{\theta\in\Theta^{\prime}}\bigg{(}\log\mu_{t}(\theta),\log\bigg{(}\frac{\mathbb{P}(\tau>t)}{\mathbb{P}(\tau>t+1)}\mu_{t}(\theta)\bigg{)}\bigg{)}\bigg{)}$
$\displaystyle=\sum_{t\in\mathcal{T}_{\Theta^{\prime}}}(\log\mathbb{P}(\tau>t)-\log\mathbb{P}(\tau>t+1))^{|\Theta|^{\prime}}$
$\displaystyle=\infty,$

where the last equality follows from the fact that $|\Theta^{\prime}|\leq
n-1$. This is a contradiction because the LHS is finite. Therefore, a maximal
belief path must stay at $\Phi^{*}$ for large enough $t\in\mathcal{T}$, as
desired. ∎

##  Appendix VII Details of smoothing out $f$ and $c$

In this appendix we explicitly define a sequence of models as follows.

######  Definition 12.

The model indexed by $k\in\mathbb{N}_{>0}$ has times
$\mathcal{T}_{k}=\big{\\{}\frac{1}{k},\frac{2}{k},\ldots\big{\\}}$.
Furthermore, define

  1. (i)

$c_{k}:\mathcal{T}_{k}\to\mathbb{R}$ such that for all $t\in\mathcal{T}$, (i)
$c_{k}(t)=c^{\prime}(t)$; and (ii) $c_{k}$ is linear between $t$ and $t+1$.

  2. (ii)

$f_{k}:\mathcal{T}_{k}\to\mathbb{R}$ such that for all $t\in\mathcal{T}$, (i)
$f_{k}(t)=f^{\prime}(t)$; and (ii) $f_{k}$ is linear between $t$ and $t+1$.

This simply ‘fills in’ the original functions through linear interpolation. In
particular, $c_{k}$ ($f_{k}$) and $c^{\prime}$ ($f^{\prime}$) agree on
$\mathcal{T}$. This is illustrated in Figure 14 below.

Figure 14: Illustration of $f_{k=3}$

Now define the deterministic time change $s_{k}:\mathcal{T}_{k}\to\mathcal{T}$
such that $s(t)=t\cdot k$ and define $f$, $c$ such that for all
$t\in\mathcal{T}_{k}$,

$f(s_{k}(t))=f_{k}(t)\quad\text{and}\quad c(s_{k}(t))=c_{k}(t),$

noting that the domain of $f$ and $c$ is $\mathcal{T}$. Finally, observe that
by taking $k$ large, $|f(t+1)-f(t)|$ and $|c(t+1)-c(t)|$ can be made
arbitrarily small.

##  Appendix VIII Robustness to multiple senders

In this appendix, we outline a simple setting with multiple (possibly
competing) senders and show that equilibrium distributions of stopping times
must remain within $\mathcal{D}$ which we characterize in the main text.

Index the set of information providers with $\mathcal{P}:=\\{1,2,\ldots n\\}$
and suppose that each provider $i\in\mathcal{P}$ has some value function
$v_{i}:A\times\mathcal{T}\to\mathbb{R}$ i.e., they care jointly about what
action the DM ends up taking (“persuasion”) and how long the DM pays attention
for (“attention”). Information provider $i$ has finite message space $M_{i}$
and denote $\mathcal{H}^{i}_{t}=\prod^{t}M_{i}$ as the space of possible time
$t$ histories for provider $i$. Suppose that all information providers have
(costless) access to the full set of dynamic information structures defined on
their respective message spaces i.e., each provider $i\in\mathcal{P}$ chooses
$I_{i}\in\mathcal{I}(M_{i})$ where $\mathcal{I}(M_{i})$ is the set of all
dynamic information structures defined on $M_{i}$.

At each period $t\in\mathcal{T}$ the DM observes the vector
$(m^{i})_{i=1}^{n}\in\prod^{n}_{i=1}M_{i}$ generated by
$(I_{i})_{i\in\mathcal{P}}\in\prod_{i=1}^{n}\mathcal{I}(M_{i})$, the profile
of dynamic information structures each provider chooses. The DM then decides
whether to stop and take action, or continue paying attention for another
period.

An equilibrium of this game is a profile of dynamic information structures
$\\{I_{i}\\}_{i\in\mathcal{P}}\in\prod_{i=1}^{n}\mathcal{I}(M_{i})$ such that
for each provider $i\in\mathcal{P}$ and all
$I^{\prime}_{i}\in\mathcal{I}(M_{i})$,

$\displaystyle\mathbb{E}^{(I_{-i},I_{i})}[v_{i}(a,\tau)]\geq\mathbb{E}^{(I_{-i},I_{i}^{\prime})}[v_{i}(a,\tau)]$

where we denote $I_{-i}:=(I_{j})_{j\neq i}$ and the expectation operator is
taken with respect to the all providers’ information structures under the
assumption that for every time $t\in\mathcal{T}$ the DM observes all
providers’ time $t$ messages.

Now notice that this is simply another dynamic information structure for which
messages are vector valued, and the history dependence of the $i$-th message
at time $t$ only enters through the path of $i$-th messages at times $s<t$.
More formally, define $M:=\prod_{i\in\mathcal{P}}M_{i}$ and
$\mathcal{H}_{t}=\prod_{s=1}^{t}\Big{(}\prod_{i=1}^{n}M_{i}\Big{)}$. Then
construct a single dynamic information structure $I\in\mathcal{I}(M)$ by
choosing

$\displaystyle
p^{I}\Big{(}(m_{t}^{i})_{i=1}^{n}\Big{|}\theta,H_{t}\in\mathcal{H}_{t}\Big{)}=\prod_{i=1}^{n}p^{I_{i}}\Big{(}m^{i}_{t}\Big{|}\theta,H_{t}^{i}\in\mathcal{H}_{t}^{i}\Big{)}.$

for all $\theta\in\Theta$. This replicates the the distribution of each path
of messages the DM received under $(I_{1},\ldots I_{n})$. As such, if an
equilibrium exists in this game, the corresponding equilibrium distribution of
stopping times must remain within $\mathcal{D}$. In fact, observe that because
each provider can only condition their information on the realization of their
own messages realizations (i.e., the distribution of $m^{i}_{t}$ can only
depend on $H^{i}_{t-1}\in\mathcal{H}^{i}_{t-1}$), the set of distributions of
stopping times might be strictly smaller than $\mathcal{D}$.

There are, of course, richer or more realistic settings to which our
characterization result applies. Our goal is not to offer an exhaustive
account, but to highlight that set or feasible stopping times we study holds
more broadly than the case with a single designer with full commitment. For
instance, we might imagine that at time $t$, the DM receives message
$m_{t}^{X_{t}}$ where $X_{t}\in\mathcal{P}$ is the provider whose message the
DM observes at time $t$. $X_{t}$ might be chosen by a separate player, and can
depend on the history of all past message realizations from each provider.
This might correspond to the case in which a news feed aggregator (e.g.,
Facebook) can choose the information provider (news outlet) whose message the
DM observes, but not the message itself. It should be evident that the
distributions of equilibrium stopping times remain within $\mathcal{D}$.

##  Appendix IX Supporting calculations for Example 1

In this Appendix we supply the supporting calculations omitted in Example 1.

### IX.1 Calculation verifying the stopping nodes on the original structure
$I$

Observe that for any $c>0$, the DM stops after the histories
$(m_{1}=1,m_{2}=1)$ and $(m_{1}=0,m_{2}=0)$ since on those histories, she
becomes so convinced that the state is either $1$ or $0$ that no matter what
signal she receives in $t=3$, she will take the same action and no information
is provided after $t=4$. On the other hand, DM continues after the histories
$(m_{1}=1,m_{2}=0)$ and $(m_{1}=0,m_{2}=1)$ since her belief after those
histories is $1/2$ and information provided at $t=3$ remains valuable. It is
then easy to see that there must exist $\underline{c}>0$ such that for all
$c<\underline{c}$, the DM optimally stops at the nodes depicted in Figure 1c
(a) in the main text. (Indeed, the reader can verify that $c=2/39$.)

### IX.2 Calculation that full information can be chosen to replicate non-
stopping beliefs

We illustrate this for the history $(m_{1}=1)$ by setting

$\displaystyle\mathbb{P}(m_{2}=1|\theta=1)=p_{1}\quad\mathbb{P}(m_{2}=\emptyset|\theta=1)=1-p_{1}$
$\displaystyle\mathbb{P}(m_{2}=0|\theta=0)=p_{0}\quad\mathbb{P}(m_{2}=\emptyset|\theta=0)=1-p_{0}.$

We want to set the posterior belief conditional on not learning about the
state perfectly to $1/2$, the same belief as the corresponding node in $I$:

$\displaystyle\mathbb{P}(\theta=1|m_{2}=\emptyset,m_{1}=1)=\cfrac{(1-p_{1})\cdot(2/3)}{(2/3)(1-p_{1})+(1/3)(1-p_{0})}=1/2$

on the other hand, we want to ensure that on history $(m_{1}=1)$, DM learns
the state fully with probability $5/9$ to preserve incentives to continue
paying attention:

$\displaystyle 2p_{1}/3+p_{0}/3=5/9$

and we have two equations and two unknowns to pin down the information
structure after history $(m_{1}=1)$.

### IX.3 Calculation that the DM prefers to continue paying attention upon
receipt of the null message

Let $V$ denote the DM’s continuation value from paying attention after
histories $\\{m_{1}=1\\}$ or $\\{m_{1}=0\\}$ on information structure
$I^{\prime}$:

$V=\underbrace{(5/9)\cdot(1-c)}_{\begin{subarray}{c}\text{learn state
perfectly}\\\ \begin{subarray}{c}\text{at
$t=2$}\end{subarray}\end{subarray}}+\underbrace{(4/9)\cdot(1-2c)}_{\begin{subarray}{c}\text{learn
state perfectly}\\\ \begin{subarray}{c}\text{at
$t=3$}\end{subarray}\end{subarray}},$

noting here that because the modification from $I^{\prime}$ to
$I^{\prime\prime}$ preserves incentives from time $t=2$ onward, this also
corresponds to the continuation value after history $\\{m_{1}=\emptyset\\}$ on
the information structure $I^{\prime\prime}$. Now let
$u^{*}(\mu):=\max_{a\in\\{0,1\\}}\mathbb{E}_{\mu}[-(a-\theta)^{2}]$ be the
expected payoff from stopping and taking the optimal action under belief
$\mu$. We have

$\displaystyle u^{*}(1/2)$
$\displaystyle=u^{*}\big{(}(2/3)\cdot(1/2)+(1/3)\cdot(1/2)\big{)}$
$\displaystyle\leq(1/2)\big{[}u^{*}(2/3)+u^{*}(1/3)\big{]}$ $\displaystyle\leq
V/2+V/2=V$

where the first inequality is from the convexity of $u^{*}$ and the second
inequality is because the DM did indeed find it optimal to continue paying
attention after histories $(m_{1}=1)$ and $(m_{1}=0)$ on information structure
$I^{\prime}$. We have thus verified that under the modification
$I^{\prime\prime}$, the DM’s incentive to continue paying attention at non-
terminal nodes must once again weakly increase.

### IX.4 Calculation verifying properties of $I_{1}$ and $I_{2}$

Obedience. We argued in the main text that the DM facing $I_{1}$ is
indifferent at $t=0$. This implies she must strictly prefer to pay attention
at $t=1$.

For $I_{2}$, observe that the DM’s value of information is $1/2$. We
conjecture that it is optimal for her to continue paying attention whenever
she does not learn the state and verify that this is in fact optimal. Under
the conjectured stopping rule, the DM is only uncertain as to when she learns
the state. At $t=0$, we have
$1/2=c\cdot\mathbb{P}(\tau=1)+3c\mathbb{P}(\tau>1)$ since $c=1/4$. At $t=1$
and $t=2$, the argument is identical to that of $I_{1}$, but conditioning on
the event that $\\{\tau>1\\}$.

Optimality. Apply Proposition 10.