When to sell an indivisible object:

Optimal timing with Markovian buyers*  *  This research was supported by Basic
Science Research Program through the National Research Foundation of Korea
(NRF) funded by the Ministry of Education (2021R1$|$1A4A01059254).

Kiho Yoon

Department of Economics, Korea University

145 Anam-ro, Seongbuk-gu, Seoul, Korea 02841

<EMAIL_ADDRESS>
http://econ.korea.ac.kr/~kiho

Abstract

We study the problem of when to sell an indivisible object. There is a
monopolistic seller who owns an indivisible object and plans to sell it over a
given span of time to the set of potential buyers whose valuations for the
object evolve over time. We formulate the seller’s problem as a dynamic
mechanism design problem. We provide the procedure for finding the optimal
solution and show how to check incentive compatibility. We also examine
sufficient conditions for the optimality of myopic stopping rule and ex-post
individual rationality. In addition, we present some comparative static
results regarding the seller’s revenue and the selling time.

Keywords: mechanism design, dynamic pricing, revenue management, stopping
time, Markov process

JEL Classification: C73; D42; D82

1 Introduction

We study the problem of when to sell an indivisible object. There is a
monopolistic seller who owns an indivisible object and plans to sell it over a
given span of time to the set of potential buyers whose valuations for the
object evolve over time. For instance, the seller is a wholesale company and
the buyers are retailers whose final demands fluctuate due to changing
retailer-specific circumstances and general economic conditions. To find the
optimal selling procedure, we adopt the dynamic mechanism design approach.1 1
The literature on dynamic mechanism design is growing rapidly, as evidenced by
the literature overviews of Bergemann and Said (2011), Vohra (2012), Bergemann
and Pavan (2015), Pavan (2017), and Bergemann and Välimäki (2019).

In the next section, we formulate a general dynamic mechanism design problem
and present a necessary and sufficient condition as well as tractable
sufficient conditions for incentive compatibility. This material is closely
related to Pavan et al. (2014). We then set up the monopolistic seller’s
problem as a dynamic mechanism design problem in section 3. The seller chooses
the stochastic selling time to maximize his revenue subject to the constraints
that (i) he can sell at most one indivisible object over the entire span of
time, (ii) the buyers have non-negative payoffs from participating in the
mechanism, and (iii) the buyers have an incentive to tell the true valuations
in each period. The seller commits to the mechanism.

As is standard in the mechanism design literature, we solve for the relaxed
problem as ignoring the last constraint, and then check whether the solution
satisfies the ignored constraint. The seller observes Markovian random
variables, which are buyers’ valuations, in each period and decides when to
stop and sell the object. Hence, the seller’s problem is essentially an
optimal stopping problem. However, we also have to check the participation
constraint and the incentive constraint of buyers’ telling the truth.
Therefore, our problem is more complicated than the standard optimal stopping
problem.

We provide the procedure for finding the optimal solution and show how to
check incentive compatibility by utilizing the results obtained in section 2.
We first analyze several two-period examples, and then move on to the general
case with an arbitrary number of periods. It may sometimes be a difficult task
to find an optimal allocation as well as to check incentive compatibility.
Hence, it would be nice if the myopic stopping rule, which compares the
seller’s current revenue only with the next period’s expected revenue, is
optimal. Proposition 1 presents a sufficient condition for the optimality of
myopic stopping rule. We also examine ex-post participation constraint, that
is, whether buyers have an incentive to participate in the mechanism for all
periods, not only for the first period. Proposition 2 shows that ex-post
individual rationality holds if the transition kernel regarding buyers’
valuations satisfies the first-order stochastic dominance. Next, Propositions
3 and 4 present comparative static results regarding the seller’s revenue and
the selling time. The final section contains a discussion on straightforward
extensions and future research agenda.

As far as we know, this is the first paper that investigates the optimal
selling time against a fixed set of buyers whose valuations evolve over time.
The revenue management literature also studies the optimal sales of
homogeneous goods over time, but the potential buyers in this framework do not
live in all periods but arrive randomly in each period and, more importantly,
each buyer’s valuation for the good stays the same over his/her life span.
See, in particular, Gallien (2006), Board and Skrzypacz (2016), and Gershkov
et al. (2018) that employ the mechanism design approach and study environments
in which arrival times are private information and buyers act strategically.
These paper also contain good literature reviews on revenue management. Other
notable papers that study optimal dynamic mechanisms for settings with
randomly arriving buyers are Pai and Vohra (2013) and Mierendorff (2016).

As mentioned above, this paper is close in spirit to the theory of optimal
stopping time. See the excellent books of Chow et al. (1971), Shiryaev (1978),
and Furguson (2006) for general theory. See also Stigler (1961), McCall
(1965), and the survey paper of Lippman and McCall (1976) for this line of
research in the economics literature. Compared to these works, the present
paper is a work of mechanism design with its focus on incentive compatibility
and individual rationality.

2 The dynamic mechanism design formulation

We consider a single-player setting without loss of generality.2 2 It is
straightforward to extend the results to the multi-player setting. We focus on
the single-player setting for notational convenience. Let
$t\in\\{1,2,\ldots,T\\}$ denote a period, where $T$ may be infinite. The
player’s type in period $t$, which is private information, is
$\theta_{t}\in\Theta=[\underline{\theta},\overline{\theta}]$ with
$0\leq\underline{\theta}<\overline{\theta}\leq\infty$. After $\theta_{t}$ is
realized in period $t$, a public action $a_{t}\in A$ is determined. In
addition, let $z_{t}\in I\\!\\!R$ be a monetary transfer from the player in
period $t$. Given sequences $(\theta_{1},\ldots,\theta_{T})$ of types and
$(a_{1},\ldots,a_{T})$ of actions, together with ($z_{1},\ldots,z_{T})$ of
monetary transfers, the player’s total payoff is

$\sum_{t=1}^{T}\delta^{t-1}\bigl{(}v(\theta_{t},a_{t})-z_{t}\bigr{)},$

where $\delta\in[0,1]$ is the discount factor and $v(\cdot)$ is a (one-period)
valuation function.3 3 We exclude $\delta=1$ when $T=\infty$. Let
$F_{1}(\theta_{1})$ denote the distribution of $\theta_{1}$, with
$f_{1}(\theta_{1})$ being the corresponding density function. Define
$\theta^{t}=(\theta_{1},\ldots,\theta_{t})$ and $a^{t}=(a_{1},\ldots,a_{t})$,
and let $F_{t}(\theta_{t}|\theta^{t-1},a^{t-1})$ denote the conditional
distribution of $\theta_{t}$, with $f_{t}(\theta_{t}|\theta^{t-1},a^{t-1})$
being the corresponding density function. We impose the following Markov
property throughout the paper:

$F_{t}(\theta_{t}|\theta^{t-1},a^{t-1})=F_{t}(\theta_{t}|\theta_{t-1},a_{t-1}),$

that is, $F_{t}$ does not depend on $\theta_{s}$ or $a_{s}$ for
$s=1,\ldots,t-2$.4 4 We may alternatively impose the Markov assumption as
$F_{t}(\theta_{t}|\theta^{t-1},a^{t-1})=F_{t}(\theta_{t}|\theta_{t-1},a^{t-1})$,
i.e., $F_{t}$ does not depend on $\theta_{s}$ but depends on $a_{s}$ for
$s=1,\ldots,t-2$. This alternative assumption does not affect the following
results.

A dynamic (direct) mechanism is given by the decision rule
$\alpha_{t}:\Theta^{t-1}\times A^{t-1}\times\Theta\rightarrow A$ and the
transfer scheme $\tau_{t}:\Theta^{t-1}\times A^{t-1}\times\Theta\rightarrow
I\\!\\!R$ for $t=1,\ldots,T$. Thus,
$\alpha_{t}(\hat{\theta}^{t-1},a^{t-1},\hat{\theta}_{t})$ and
$\tau_{t}(\hat{\theta}^{t-1},a^{t-1},\hat{\theta}_{t})$ are the action chosen
and the transfer, respectively, in period $t$ when the player’s reports and
the actions chosen in previous periods are $\hat{\theta}^{t-1}$ and $a^{t-1}$,
respectively, and the report in this period is $\hat{\theta}_{t}$.5 5 To be
precise, we have to define $\alpha_{1}:\Theta\rightarrow A$ and
$\tau_{1}:\Theta\rightarrow I\\!\\!R$ differently from those for $t\geq 2$,
and to have $\alpha_{1}(\hat{\theta}_{1})$ and $\tau_{1}(\hat{\theta}_{1})$.
We chose to dispense with this extra distinction. Similar convention applies
below. The player’s strategy is
$\sigma_{t}:\Theta^{t-1}\times\Theta^{t-1}\times
A^{t-1}\times\Theta\rightarrow\Theta$ for $t=1,\ldots,T$. Thus,
$\hat{\theta}_{t}=\sigma_{t}(\theta^{t-1},\hat{\theta}^{t-1},a^{t-1},\theta_{t})$
is the report in period $t$ when the type, report, and action in period $s$
for $s=1,\ldots,t-1$ are $\theta_{s},\hat{\theta}_{s}$, and $a_{s}$,
respectively, and the type in this period is $\theta_{t}$.

Define a period-$t$ history

$\hat{h}^{t}=(\hat{\theta}_{1},\alpha_{1}(\hat{\theta}_{1}),\hat{\theta}_{2},\alpha_{2}(\hat{\theta}_{1},\alpha_{1}(\hat{\theta}_{1}),\hat{\theta}_{2}),\cdots,\hat{\theta}_{t-1},\alpha_{t-1}(\hat{\theta}_{1},\alpha_{1}(\hat{\theta}_{1}),\cdots,\hat{\theta}_{t-1})).$

In recursive form,
$\hat{h}^{t}=(\hat{h}^{t-1},\hat{\theta}_{t-1},\alpha_{t-1}(\hat{h}^{t-1},\hat{\theta}_{t-1}))$
with $\hat{h}^{1}=\emptyset$. Thus, $\hat{h}^{t}$ is the sequence of reports
$\hat{\theta}^{t-1}$ and corresponding actions given by the decision rule.
Define recursively

$\eqalign{&U_{t}(\hat{\theta}_{t},\theta_{t};\hat{\theta}^{t-1})=\
v(\theta_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))-\tau_{t}(\hat{h}^{t},\hat{\theta}_{t})\cr+&\delta\int_{\theta_{t+1}=\underline{\theta}}^{\overline{\theta}}U_{t+1}(\tilde{\theta}_{t+1};\hat{\theta}^{t-1},\hat{\theta}_{t})dF_{t+1}(\tilde{\theta}_{t+1}|\theta_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t})),}$

and define with a slight abuse of notation that

$U_{t}(\theta_{t};\hat{\theta}^{t-1})=U_{t}(\theta_{t},\theta_{t};\hat{\theta}^{t-1}).$

$U_{t}(\hat{\theta}_{t},\theta_{t};\hat{\theta}^{t-1})$ is the player’s
continuation payoff in period $t$ when the report and the action chosen in
period $s=1,\ldots,t-1$ are $\hat{\theta}_{s}$ and
$\alpha_{s}(\hat{h}^{s},\hat{\theta}_{s})$ respectively and the true type and
the report in this period are $\theta_{t}$ and $\hat{\theta}_{t}$
respectively. Note that
$U_{t}(\hat{\theta}_{t},\theta_{t};\hat{\theta}^{t-1})$ does not depend on
$\theta^{t-1}$, the true types in periods $1,\ldots,t-1$.6 6 Observe that,
under the general transition kernel, $U_{t}$ depends on
$\theta_{1},\ldots,\theta_{t-1}$ as well as
$\hat{\theta}_{1},\ldots,\hat{\theta}_{t-1}$ and we have
$U_{t}(\hat{\theta}_{t},\theta_{t};\theta^{t-1},\hat{\theta}^{t-1})=v(\theta_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))-\tau_{t}(\hat{h}^{t},\hat{\theta}_{t})+\delta\int_{\theta_{t+1}=\underline{\theta}}^{\overline{\theta}}U_{t+1}(\tilde{\theta}_{t+1};\theta^{t},\hat{\theta}^{t})dF_{t+1}(\tilde{\theta}_{t+1}|$
$\theta^{t},\alpha^{t}(\hat{h}^{t},\hat{\theta}^{t}))$. In this case,
$F_{t+1}(\tilde{\theta}_{t+1}|\theta^{t},\alpha^{t}(\hat{h}^{t},\hat{\theta}^{t}))$
depends on the true types $\theta^{t}$ and the chosen actions
$\alpha^{t}(\hat{h}^{t},\hat{\theta}^{t})$ (that depend on the reports
$\hat{\theta}^{t}$). On the other hand, $F_{t+1}$ does not depend on the true
types $\theta^{t-1}$ under the Markov assumption.

Incentive compatibility is

$U_{t}(\theta_{t};\hat{\theta}^{t-1})\geq
U_{t}(\hat{\theta}_{t},\theta_{t};\hat{\theta}^{t-1}),$ $None$

for $t=1,\ldots,T$.

Observe that $(IC_{t})$ is equivalent to

$\eqalign{&U_{t}(\theta_{t};\hat{\theta}^{t-1})-U_{t}(\hat{\theta}_{t};\hat{\theta}^{t-1})\geq
v(\theta_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))-v(\hat{\theta}_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))\cr+&\
\delta\int_{\theta_{t+1}=\underline{\theta}}^{\overline{\theta}}U_{t+1}(\tilde{\theta}_{t+1};\hat{\theta}^{t-1},\hat{\theta}_{t})d\bigl{(}F_{t+1}(\tilde{\theta}_{t+1}|\theta_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))-F_{t+1}(\tilde{\theta}_{t+1}|\hat{\theta}_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))\bigr{)}.}$
$None$

Interchanging the roles of $\theta_{t}$ and $\hat{\theta}_{t}$, we have

$\eqalign{&U_{t}(\hat{\theta}_{t};\hat{\theta}^{t-1})-U_{t}(\theta_{t};\hat{\theta}^{t-1})\geq
v(\hat{\theta}_{t},\alpha_{t}(\hat{h}^{t},\theta_{t}))-v(\theta_{t},\alpha_{t}(\hat{h}^{t},\theta_{t}))\cr+&\
\delta\int_{\theta_{t+1}=\underline{\theta}}^{\overline{\theta}}U_{t+1}(\tilde{\theta}_{t+1};\hat{\theta}^{t-1},\theta_{t})d\bigl{(}F_{t+1}(\tilde{\theta}_{t+1}|\hat{\theta}_{t},\alpha_{t}(\hat{h}^{t},\theta_{t}))-F_{t+1}(\tilde{\theta}_{t+1}|\theta_{t},\alpha_{t}(\hat{h}^{t},\theta_{t}))\bigr{)}.}$

Combining these inequalities, we get

$\eqalign{&v(\hat{\theta}_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))-v(\theta_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))\cr+&\
\delta\int_{\theta_{t+1}=\underline{\theta}}^{\overline{\theta}}U_{t+1}(\tilde{\theta}_{t+1};\hat{\theta}^{t-1},\hat{\theta}_{t})d\bigl{(}F_{t+1}(\tilde{\theta}_{t+1}|\hat{\theta}_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))-F_{t+1}(\tilde{\theta}_{t+1}|\theta_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))\bigr{)}\cr\geq&U_{t}(\hat{\theta}_{t};\hat{\theta}^{t-1})-U_{t}(\theta_{t};\hat{\theta}^{t-1})\cr\geq&v(\hat{\theta}_{t},\alpha_{t}(\hat{h}^{t},\theta_{t}))-v(\theta_{t},\alpha_{t}(\hat{h}^{t},\theta_{t}))\cr+&\
\delta\int_{\theta_{t+1}=\underline{\theta}}^{\overline{\theta}}U_{t+1}(\tilde{\theta}_{t+1};\hat{\theta}^{t-1},\theta_{t})d\bigl{(}F_{t+1}(\tilde{\theta}_{t+1}|\hat{\theta}_{t},\alpha_{t}(\hat{h}^{t},\theta_{t}))-F_{t+1}(\tilde{\theta}_{t+1}|\theta_{t},\alpha_{t}(\hat{h}^{t},\theta_{t}))\bigr{)}.}$

We will assume that both $v(\cdot)$ and $f_{t}(\cdot)$ for $t=1,\ldots,T$ are
continuously differentiable and that $\alpha_{t}(\cdot)$ for $t=1,\ldots,T$ is
differentiable almost everywhere. Then, dividing the previous inequalities by
$\hat{\theta}_{t}-\theta_{t}$ and taking limits, we get

$\eqalign{&{dU_{t}(\theta_{t};\hat{\theta}^{t-1})\over
d\theta_{t}}=v_{\theta}(\theta_{t},\alpha_{t}(\hat{h}^{t},\theta_{t}))\cr+&\delta\int_{\underline{\theta}}^{\overline{\theta}}U_{t+1}(\tilde{\theta}_{t+1};\hat{\theta}^{t-1},\theta_{t})\
{\partial
f_{t+1}(\tilde{\theta}_{t+1}|\theta_{t},\alpha_{t}(\hat{h}^{t},\theta_{t}))\over\partial\theta_{t}}d\tilde{\theta}_{t+1}}$

almost everywhere. Note that the notation $v_{\theta}(\theta,a)$ is the
partial derivative of $v(\theta,a)$ with respect to $\theta$ and that
$\partial f_{t+1}/\partial\theta_{t}$ is only with respect to $\theta_{t}$ in
$f_{t+1}(\theta_{t+1}|\theta_{t},a_{t})$. We have

$\eqalign{{dU_{t}(\theta_{t};\hat{\theta}^{t-1})\over d\theta_{t}}=&\
v_{\theta}(\theta_{t},\alpha_{t}(\hat{h}^{t},\theta_{t}))+\delta\Bigl{[}U_{t+1}(\tilde{\theta}_{t+1};\hat{\theta}^{t-1},\theta_{t}){\partial
F_{t+1}(\tilde{\theta}_{t+1}|\theta_{t},\alpha_{t}(\hat{h}^{t},\theta_{t}))\over\partial\theta_{t}}\Bigr{]}_{\underline{\theta}}^{\overline{\theta}}\cr-&\delta\int_{\underline{\theta}}^{\overline{\theta}}{dU_{t+1}(\tilde{\theta}_{t+1};\hat{\theta}^{t-1},\theta_{t})\over
d\theta_{t+1}}\ {\partial
F_{t+1}(\tilde{\theta}_{t+1}|\theta_{t},\alpha_{t}(\hat{h}^{t},\theta_{t}))\over\partial\theta_{t}}d\tilde{\theta}_{t+1}\cr=\
v_{\theta}(\theta_{t},\alpha_{t}(\hat{h}^{t},&\theta_{t}))-\delta\int_{\underline{\theta}}^{\overline{\theta}}{dU_{t+1}(\tilde{\theta}_{t+1};\hat{\theta}^{t-1},\theta_{t})\over
d\theta_{t+1}}\ {\partial
F_{t+1}(\tilde{\theta}_{t+1}|\theta_{t},\alpha_{t}(\hat{h}^{t},\theta_{t}))\over\partial\theta_{t}}d\tilde{\theta}_{t+1},}$

where the first equality follows from integration by parts and the second
equality follows from the fact that
$F_{t+1}(\underline{\theta}|\theta_{t},\alpha_{t}(\hat{h}^{t},\theta_{t}))=0$
and
$F_{t+1}(\overline{\theta}|\theta_{t},\alpha_{t}(\hat{h}^{t},\theta_{t}))=1$
for all $\theta_{t}$ and so $\partial F_{t+1}/\partial\theta_{t}=0$ when
$\tilde{\theta}_{t+1}=\underline{\theta}$ or $\overline{\theta}$.

By recursion, we have

$\eqalign{&{dU_{t}(\theta_{t};\hat{\theta}^{t-1})\over
d\theta_{t}}=v_{\theta}(\theta_{t},\alpha_{t}(\hat{h}^{t},\theta_{t}))\cr+&\sum_{s=t+1}^{T}(-\delta)^{s-t}\underbrace{\int_{\underline{\theta}}^{\overline{\theta}}\cdots\int_{\underline{\theta}}^{\overline{\theta}}}_{(s-t)-{\eightrm
times}}v_{\theta}(\tilde{\theta}_{s},\alpha_{s}(\hat{h}^{t},\theta_{t},\alpha_{t}(\hat{h}^{t},\theta_{t}),\tilde{\theta}_{t+1},\alpha_{t+1}(\hat{h}^{t},\theta_{t},\alpha_{t}(\hat{h}^{t},\theta_{t}),\tilde{\theta}_{t+1}),\cdots,\tilde{\theta}_{s}))\cr\times&{\partial
F_{t+1}(\tilde{\theta}_{t+1}|\theta_{t},\alpha_{t}(\hat{h}^{t},\theta_{t}))\over\partial\theta_{t}}\cdots{\partial
F_{s}(\tilde{\theta}_{s}|\tilde{\theta}_{s-1},\alpha_{s-1}(\hat{h}^{t},\theta_{t},\alpha_{t}(\hat{h}^{t},\theta_{t}),\tilde{\theta}_{t+1},\cdots,\tilde{\theta}_{s-1}))\over\partial\theta_{s-1}}d\tilde{\theta}_{s}\cdots
d\tilde{\theta}_{t+1}}$ $None$

almost everywhere.

Define

$h_{+}^{s}(\hat{h}^{t},\theta_{t})=(\hat{h}^{t},\theta_{t},\alpha_{t}(\hat{h}^{t},\theta_{t}),\tilde{\theta}_{t+1},\alpha_{t+1}(\hat{h}^{t},\theta_{t},\alpha_{t}(\hat{h}^{t},\theta_{t}),\tilde{\theta}_{t+1})\cdots,\tilde{\theta}_{s}),$

that is, $h_{+}^{s}(\hat{h}^{t},\theta_{t})$ is the period-$s$ history plus
period-$s$ report $\tilde{\theta}_{s}$, conditional on period-$t$ history
$\hat{h}^{t}$ and period-$t$ report $\theta_{t}$. Then,

$h_{+}^{s}(\hat{h}^{t},\hat{\theta}_{t})=(\hat{h}^{t},\hat{\theta}_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}),\tilde{\theta}_{t+1},\alpha_{t+1}(\hat{h}^{t},\hat{\theta}_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}),\tilde{\theta}_{t+1})\cdots,\tilde{\theta}_{s})$.

Define

$\eqalign{&D_{t}(\hat{\theta}_{t},\theta_{t};\hat{\theta}^{t-1})=v_{\theta}(\theta_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))\cr+&\sum_{s=t+1}^{T}(-\delta)^{s-t}\underbrace{\int_{\underline{\theta}}^{\overline{\theta}}\cdots\int_{\underline{\theta}}^{\overline{\theta}}}_{(s-t)-{\rm
times}}v_{\theta}(\tilde{\theta}_{s},\alpha_{s}(h_{+}^{s}(\hat{h}^{t},\hat{\theta}_{t})))\
{\partial
F_{t+1}(\tilde{\theta}_{t+1}|\theta_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))\over\partial\theta_{t}}\cr\times&{\partial
F_{t+2}(\tilde{\theta}_{t+2}|\tilde{\theta}_{t+1},\alpha_{t+1}(h_{+}^{t+1}(\hat{h}^{t},\hat{\theta}_{t})))\over\partial\theta_{t+1}}\cdots{\partial
F_{s}(\tilde{\theta}_{s}|\tilde{\theta}_{s-1},\alpha_{s-1}(h_{+}^{s-1}(\hat{h}^{t},\hat{\theta}_{t})))\over\partial\theta_{s-1}}d\tilde{\theta}_{s}\cdots
d\tilde{\theta}_{t+1}.}$ $None$

Note that this essentially captures the future impact of the change in
$\theta_{t}$ when the current true type is $\theta_{t}$, the current report is
$\hat{\theta}_{t}$, and the past reports are $\hat{\theta}^{t-1}$. Note also
that (2) can be written in recursive form as

$\eqalign{D_{t}(\hat{\theta}_{t},\theta_{t};\hat{\theta}^{t-1})=&\
v_{\theta}(\theta_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))\cr-&\delta\int_{\underline{\theta}}^{\overline{\theta}}D_{t+1}(\tilde{\theta}_{t+1};\hat{\theta}^{t})\
{\partial
F_{t+1}(\tilde{\theta}_{t+1}|\theta_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))\over\partial\theta_{t}}d\tilde{\theta}_{t+1}.}$
$None$

Define with a slight abuse of notation that
$D_{t}(\theta_{t};\hat{\theta}^{t-1})=D_{t}(\theta_{t},\theta_{t};\hat{\theta}^{t-1})$.
Then, (1) can be rewritten as

${dU_{t}(\theta_{t};\hat{\theta}^{t-1})\over
d\theta_{t}}=D_{t}(\theta_{t};\hat{\theta}^{t-1})$

almost everywhere. Since $D_{t}(\theta_{t};\hat{\theta}^{t-1})$ is continuous
almost everywhere on the interval $[\underline{\theta},\overline{\theta}]$, it
is Riemann integrable and we have

$U_{t}(\theta_{t};\hat{\theta}^{t-1})=U_{t}(\underline{\theta};\hat{\theta}^{t-1})+\int_{\underline{\theta}}^{\theta_{t}}D_{t}(\tilde{\theta}_{t};\hat{\theta}^{t-1})d\tilde{\theta}_{t}.$

In particular, we have

$U_{1}(\theta_{1})=U_{1}(\underline{\theta})+\int_{\underline{\theta}}^{\theta_{1}}D_{1}(\tilde{\theta}_{1})d\tilde{\theta}_{1},$

where

$\eqalign{&D_{1}(\tilde{\theta}_{1})=v_{\theta}(\tilde{\theta}_{1},\alpha_{1}(\tilde{\theta}_{1}))+\sum_{t=2}^{T}(-\delta)^{t-1}\underbrace{\int_{\underline{\theta}}^{\overline{\theta}}\cdots\int_{\underline{\theta}}^{\overline{\theta}}}_{(t-1)-{\rm
times}}v_{\theta}(\tilde{\theta}_{t},\alpha_{t}(\tilde{h}^{t},\tilde{\theta}_{t}))\cr&\times{\partial
F_{2}(\tilde{\theta}_{2}|\tilde{\theta}_{1},\alpha_{1}(\tilde{\theta}_{1}))\over\partial\theta_{1}}\cdots{\partial
F_{t}(\tilde{\theta}_{t}|\tilde{\theta}_{t-1},\alpha_{t-1}(\tilde{h}^{t-1},\tilde{\theta}_{t-1}))\over\partial\theta_{t-1}}d\tilde{\theta}_{t}\cdots
d\tilde{\theta}_{2}.}$

We now present a necessary and sufficient condition for incentive
compatibility.

Theorem 1. Incentive compatibility holds if and only if

$U_{t}(\theta_{t};\hat{\theta}^{t-1})=U_{t}(\underline{\theta};\hat{\theta}^{t-1})+\int_{\underline{\theta}}^{\theta_{t}}D_{t}(\tilde{\theta}_{t};\hat{\theta}^{t-1})d\tilde{\theta}_{t}\
\ \ \forall t,\forall\theta_{t}{\rm\ and\ }\forall\hat{\theta}^{t-1};$ $None$

and

$\int_{\hat{\theta}_{t}}^{\theta_{t}}D_{t}(\tilde{\theta}_{t};\hat{\theta}^{t-1})d\tilde{\theta}_{t}\geq\int_{\hat{\theta}_{t}}^{\theta_{t}}D_{t}(\hat{\theta}_{t},\tilde{\theta}_{t};\hat{\theta}^{t-1})d\tilde{\theta}_{t}\
\ \ \forall t,\forall\theta_{t},\forall\hat{\theta}_{t}{\rm\ and\
}\forall\hat{\theta}^{t-1}.$ $None$

Proof: ($\Leftarrow$ part) Observe first that, by (4), the left-hand side of
(5) is equal to
$U_{t}(\theta_{t};\hat{\theta}^{t-1})-U_{t}(\hat{\theta}_{t};\hat{\theta}^{t-1})$.
Observe next that the right-hand side of (5) is equal to

$\eqalign{&v(\theta_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))-v(\hat{\theta}_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))\cr-&\delta\int_{\hat{\theta}_{t}}^{\theta_{t}}\int_{\underline{\theta}}^{\overline{\theta}}D_{t+1}(\tilde{\theta}_{t+1};\hat{\theta}^{t})\
{\partial
F_{t+1}(\tilde{\theta}_{t+1}|\tilde{\theta}_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))\over\partial\theta_{t}}d\tilde{\theta}_{t+1}d\tilde{\theta}_{t}\cr=&v(\theta_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))-v(\hat{\theta}_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))\cr-&\delta\int_{\hat{\theta}_{t}}^{\theta_{t}}\int_{\underline{\theta}}^{\overline{\theta}}{dU_{t+1}(\tilde{\theta}_{t+1};\hat{\theta}^{t})\over
d\theta_{t+1}}\ {\partial
F_{t+1}(\tilde{\theta}_{t+1}|\tilde{\theta}_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))\over\partial\theta_{t}}d\tilde{\theta}_{t+1}d\tilde{\theta}_{t}\cr=&v(\theta_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))-v(\hat{\theta}_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))\cr-&\delta\int_{\underline{\theta}}^{\overline{\theta}}{dU_{t+1}(\tilde{\theta}_{t+1};\hat{\theta}^{t})\over
d\theta_{t+1}}\int_{\hat{\theta}_{t}}^{\theta_{t}}{\partial
F_{t+1}(\tilde{\theta}_{t+1}|\tilde{\theta}_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))\over\partial\theta_{t}}d\tilde{\theta}_{t}d\tilde{\theta}_{t+1}\cr=&v(\theta_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))-v(\hat{\theta}_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))\cr-&\delta\int_{\underline{\theta}}^{\overline{\theta}}{dU_{t+1}(\tilde{\theta}_{t+1};\hat{\theta}^{t})\over
d\theta_{t+1}}\Bigl{(}F_{t+1}(\tilde{\theta}_{t+1}|\theta_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))-F_{t+1}(\tilde{\theta}_{t+1}|\hat{\theta}_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))\Bigr{)}d\tilde{\theta}_{t+1}\cr=&v(\theta_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))-v(\hat{\theta}_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))\cr-&\delta\Bigl{[}U_{t+1}(\tilde{\theta}_{t+1};\hat{\theta}^{t})\Bigl{(}F_{t+1}(\tilde{\theta}_{t+1}|\theta_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))-F_{t+1}(\tilde{\theta}_{t+1}|\hat{\theta}_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))\Bigr{)}\Bigr{]}_{\underline{\theta}}^{\overline{\theta}}\cr+&\delta\int_{\underline{\theta}}^{\overline{\theta}}U_{t+1}(\tilde{\theta}_{t+1};\hat{\theta}^{t})d\Bigl{(}F_{t+1}(\tilde{\theta}_{t+1}|\theta_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))-F_{t+1}(\tilde{\theta}_{t+1}|\hat{\theta}_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))\Bigr{)}\cr=&v(\theta_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))-v(\hat{\theta}_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))\cr+&\delta\int_{\underline{\theta}}^{\overline{\theta}}U_{t+1}(\tilde{\theta}_{t+1};\hat{\theta}^{t})d\Bigl{(}F_{t+1}(\tilde{\theta}_{t+1}|\theta_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))-F_{t+1}(\tilde{\theta}_{t+1}|\hat{\theta}_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))\Bigr{)}.}$

The first expression follows from (3), the first equality follows from the
differential form of (4) for $U_{t+1}(\tilde{\theta}_{t+1}|\hat{\theta}^{t})$,
the second equality follows from the change in the order of integration, the
third equality follows from integrating out the inner integral, the fourth
equality follows from integration by parts, and the last inequality follows
from the fact that
$F_{t+1}(\underline{\theta}|\theta_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))=F_{t+1}(\underline{\theta}|\hat{\theta}_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))=0$
and
$F_{t+1}(\overline{\theta}|\theta_{t},\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))=\
F_{t+1}(\overline{\theta}|\hat{\theta}_{t},$
$\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t}))=1$. Putting together, this is
nothing but $(IC^{\prime}_{t})$.

($\Rightarrow$ part) Recall from the discussion preceding this theorem that
$(IC_{t})$ implies (4). It is also straightforward to see that $(IC_{t})$
implies (5): Follow the reverse steps of the previous argument. Q.E.D.

Pavan et al. (2014) call the inequality (5) as integral monotonicity. To
examine inequality (5) in some detail, let us consider the following
assumption.

Assumption 1. (The single-crossing condition) We have

(i) $\forall a:v(\hat{\theta},a)\geq v(\theta,a)$ for
$\hat{\theta}\geq\theta$, and

(ii) $v(\hat{\theta},\hat{a})-v(\theta,\hat{a})\geq
v(\hat{\theta},a)-v(\theta,a)$ for all $\hat{\theta}\geq\theta$ and
$\hat{a}\geq a$.

Observe that, for the last period $T$ when $T$ is finite, we have
$D_{T}(\hat{\theta}_{T},\theta_{T};\hat{\theta}^{T-1})=v_{\theta}(\theta_{T},\alpha_{T}(\hat{h}^{T},\hat{\theta}_{T}))$.
Thus, (5) is equivalent to

$\int_{\hat{\theta}_{T}}^{\theta_{T}}\bigl{(}v_{\theta}(\tilde{\theta}_{T},\alpha_{T}(\hat{h}^{T},\tilde{\theta}_{T}))-v_{\theta}(\tilde{\theta}_{T},\alpha_{T}(\hat{h}^{T},\hat{\theta}_{T}))\bigr{)}d\tilde{\theta}_{T}\geq
0$ $None$

for all $\theta_{T},\hat{\theta}_{T}$ and $\hat{h}^{T}$. Note that, if

$\forall\hat{h}^{T}:\
\alpha_{T}(\hat{h}^{T},\theta_{T})\geq\alpha_{T}(\hat{h}^{T},\hat{\theta}_{T})$
for $\theta_{T}\geq\hat{\theta}_{T}$ (7)

holds, then $(6)$ holds. Conversely, $(6)$ implies by $(4)$ that

$U_{T}(\theta_{T};\hat{\theta}^{T-1})-U_{T}(\hat{\theta}_{T};\hat{\theta}^{T-1})\geq
v(\theta_{T},\alpha_{T}(\hat{h}^{T},\hat{\theta}_{T}))-v(\hat{\theta}_{T},\alpha_{T}(\hat{h}^{T},\hat{\theta}_{T})),$

which is nothing but $(IC^{\prime}_{T})$. Interchanging the roles of
$\theta_{T}$ and $\hat{\theta}_{T}$, we have

$U_{T}(\hat{\theta}_{T};\hat{\theta}^{T-1})-U_{T}(\theta_{T};\hat{\theta}^{T-1})\geq
v(\hat{\theta}_{T},\alpha_{T}(\hat{h}^{T},\theta_{T}))-v(\theta_{T},\alpha_{T}(\hat{h}^{T},\theta_{T})).$

Combining these inequalities, we get

$\eqalign{&v(\hat{\theta}_{T},\alpha_{T}(\hat{h}^{T},\hat{\theta}_{T}))-v(\theta_{T},\alpha_{T}(\hat{h}^{T},\hat{\theta}_{T}))\geq
U_{T}(\hat{\theta}_{T};\hat{\theta}^{T-1})-U_{T}(\theta_{T};\hat{\theta}^{T-1})\cr\geq&v(\hat{\theta}_{T},\alpha_{T}(\hat{h}^{T},\theta_{T}))-v(\theta_{T},\alpha_{T}(\hat{h}^{T},\theta_{T})).}$

This implies (7) under Assumption 1. Hence, $(6)$ implies (7). The preceding
argument and Theorem 1 then imply the following result.

Corollary 1. Under Assumption 1 of single-crossing condition, $(IC_{T})$ holds
if and only if (4) and (7) hold.

The inequality (7) is the monotonicity condition. Note in particular that this
characterization applies to the static model, i.e, the one-period model in
which $T=1$.

For general $t$, we see that (5) holds if
$D_{t}(\hat{\theta}_{t},\theta_{t};\hat{\theta}^{t-1})$ is increasing in
$\hat{\theta}_{t}$.7 7 We use the term ‘increasing’ in the weak sense.8 8
Pavan et al. (2014) call this condition as average monotonicity. Note that
$D_{t}(\hat{\theta}_{t},\theta_{t};\hat{\theta}^{t-1})$ is increasing in
$\hat{\theta}_{t}$ if each individual term in (2) is increasing in
$\hat{\theta}_{t}$: Pavan et al. (2014) in turn call this condition as ex-post
monotonicity (with the additional assumption that
$F_{t+1}(\theta_{t+1}|\theta_{t},a_{t})=F_{t+1}(\theta_{t+1}|\theta_{t})$,
i.e., the transition kernel $F_{t+1}$ does not depend on the public action).
Moreover, we have the following sufficient condition for incentive
compatibility.

Corollary 2. Under Assumption 1 of single-crossing condition, the four
conditions below together with (4) imply incentive compatibility.

(i)
$\alpha_{t}(\hat{h}^{t},\theta_{t})\geq\alpha_{t}(\hat{h}^{t},\hat{\theta}_{t})$
for all $t$, $\hat{h}^{t}$ and $\theta_{t}\geq\hat{\theta}_{t}$;

(ii) $\partial F_{t+1}(\theta_{t+1}|\theta_{t},a_{t})/\partial\theta_{t}\leq
0$ for all $t$, $\theta_{t}$, $a_{t}$ and $\theta_{t+1}$;

(iii) $|\partial F_{t+1}(\theta_{t+1}|\theta_{t},a_{t})/\partial\theta_{t}|$
is increasing in $a_{t}$ for all $t$, $\theta_{t}$, and $\theta_{t+1}$; and

(iv) $\alpha_{s}(h^{s}_{+}(\hat{h}^{t},\hat{\theta}_{t}))$ is increasing in
$\hat{\theta}_{t}$ for all $\hat{h}^{t}$ and $s=t+1,\ldots,T$.

Proof: It is straightforward to see by an inspection of (2) that
$D_{t}(\hat{\theta}_{t},\theta_{t};\hat{\theta}^{t-1})$ is increasing in
$\hat{\theta}_{t}$ if the conditions (i) to (iv) hold. Theorem 1 then implies
incentive compatibility. Q.E.D.

Note that condition (i) is the monotonicity condition and condition (ii) is
the first-order stochastic dominance condition. Note also that condition (iii)
holds trivially when $F_{t+1}$ $(\theta_{t+1}|\theta_{t},a_{t})$ does not
depend on $a_{t}$, and that condition (iv) holds trivially when the current
decision does not depend on past reports.

By definition of $U_{1}(\theta_{1})$, the total expected payment the player
makes, i.e., the expected value of
$\sum_{t=1}^{T}\tau_{t}(\tilde{h}^{t},\tilde{\theta}_{t})$, is

$\eqalign{&\int_{\underline{\theta}}^{\overline{\theta}}v(\tilde{\theta}_{1},\alpha_{1}(\tilde{\theta}_{1}))f_{1}(\tilde{\theta}_{1})d\tilde{\theta}_{1}+\sum_{t=2}^{T}\delta^{t-1}\underbrace{\int_{\underline{\theta}}^{\overline{\theta}}\cdots\int_{\underline{\theta}}^{\overline{\theta}}}_{t-{\rm
times}}v(\tilde{\theta}_{t},\alpha_{t}(\tilde{h}^{t},\tilde{\theta}_{t}))\cr&\
\times
f_{t}(\tilde{\theta}_{t}|\tilde{\theta}_{t-1},\alpha_{t-1}(\tilde{h}^{t-1},\tilde{\theta}_{t-1}))\cdots
f_{2}(\tilde{\theta}_{2}|\tilde{\theta}_{1},\alpha_{1}(\tilde{\theta}_{1}))f_{1}(\tilde{\theta}_{1})d\tilde{\theta}_{t}\cdots
d\tilde{\theta}_{1}}$

$\hskip-227.62204pt-\int_{\underline{\theta}}^{\overline{\theta}}U_{1}(\tilde{\theta}_{1})f_{1}(\tilde{\theta}_{1})d\tilde{\theta}_{1}.$

Since

$\eqalign{&\int_{\underline{\theta}}^{\overline{\theta}}U_{1}(\tilde{\theta}_{1})f_{1}(\tilde{\theta}_{1})d\tilde{\theta}_{1}=\
\Bigl{[}-U^{1}(\tilde{\theta}_{1})(1-F_{1}(\tilde{\theta}_{1}))\Bigr{]}_{\underline{\theta}}^{\overline{\theta}}+\int_{\underline{\theta}}^{\overline{\theta}}{dU_{1}(\tilde{\theta}_{1})\over
d\theta_{1}}(1-F_{1}(\tilde{\theta}_{1}))d\tilde{\theta}_{1}\cr=&\
U_{1}(\underline{\theta})+\int_{\underline{\theta}}^{\overline{\theta}}D_{1}(\tilde{\theta}_{1})(1-F_{1}(\tilde{\theta}_{1}))d\tilde{\theta}_{1}}$

where the second equality holds by $(4)$, the total expected payment is equal
to

$\eqalign{&\int_{\underline{\theta}}^{\overline{\theta}}\Bigl{[}v(\tilde{\theta}_{1},\alpha_{1}(\tilde{\theta}_{1}))-v_{\theta}(\tilde{\theta}_{1},\alpha_{1}(\tilde{\theta}_{1}))\
{1-F_{1}(\tilde{\theta}_{1})\over
f_{1}(\tilde{\theta}_{1})}\Bigr{]}f_{1}(\tilde{\theta}_{1})d\tilde{\theta}_{1}\cr+&\sum_{t=2}^{T}\delta^{t-1}\int_{\underline{\theta}}^{\overline{\theta}}\cdots\int_{\underline{\theta}}^{\overline{\theta}}\Bigl{[}v(\tilde{\theta}_{t},\alpha_{t}(\tilde{h}^{t},\tilde{\theta}_{t}))+(-1)^{t}\
v_{\theta}(\tilde{\theta}_{t},\alpha_{t}(\tilde{h}^{t},\tilde{\theta}_{t})){1-F_{1}(\tilde{\theta}_{1})\over
f_{1}(\tilde{\theta}_{1})}\cr&\ \ \times{\partial
F_{2}(\tilde{\theta}_{2}|\tilde{\theta}_{1},\alpha_{1}(\tilde{\theta}_{1}))/\partial\theta_{1}\over
f_{2}(\tilde{\theta}_{2}|\tilde{\theta}_{1},\alpha_{1}(\tilde{\theta}_{1}))}\cdots{\partial
F_{t}(\tilde{\theta}_{t}|\tilde{\theta}_{t-1},\alpha_{t-1}(\tilde{h}^{t-1},\tilde{\theta}_{t-1}))/\partial\theta_{t-1}\over
f_{t}(\tilde{\theta}_{t}|\tilde{\theta}_{t-1},\alpha_{t-1}(\tilde{h}^{t-1},\tilde{\theta}_{t-1}))}\Bigr{]}\cr&\
\ \times
f_{t}(\tilde{\theta}_{t}|\tilde{\theta}_{t-1},\alpha_{t-1}(\tilde{h}^{t-1},\tilde{\theta}_{t-1}))\cdots
f_{1}(\tilde{\theta}_{1})d\tilde{\theta}_{t}\cdots d\tilde{\theta}_{1}\cr-&\
U_{1}(\underline{\theta})}$ $None$

by (2).

3 The optimal selling time

3.1. The optimal solution

We turn to the monopolistic seller’s problem. The seller owns an indivisible
object and decides when and at what price to sell it to maximize his revenue.
This is an instance of the dynamic mechanism design of the previous section,
where the mechanism designer is the seller and the players are the buyers.
Thus, the public action $a_{t}\in A$ is the probability of sales
$q_{t}\in[0,1]$ and the monetary transfer $z_{t}\in I\\!\\!R$ is the price of
the object. Assume without loss of generality that (i) the seller does not
attach a value to the object, i.e., the reservation value of the seller is
zero, and that (ii) there is a single buyer.9 9 We assume without loss of
generality that the seller’s cost is equal to zero. It is straightforward to
incorporate positive, and possibly time-varying, costs. Let us consider a
canonical monopolistic pricing in which the buyer’s valuation function is
given by $v(\theta,a)=\theta a=\theta q$. Then, we have
$v(\theta_{t},\alpha_{t}(h^{t},\theta_{t}))=\theta_{t}q_{t}(h^{t},\theta_{t})$
and
$v_{\theta}(\theta_{t},\alpha_{t}(h^{t},\theta_{t}))=q_{t}(h^{t},\theta_{t})$.

If we define10 10 Note that $\alpha_{1}(\theta_{1})=q_{1}(\theta_{1})$ and
$\alpha_{t}(h^{t},\theta_{t})=q_{t}(h^{t},\theta_{t})$ for $t\geq 2$.

$\eqalign{&\psi_{1}(\theta_{1})=\theta_{1}-{1-F_{1}(\theta_{1})\over
f_{1}(\theta_{1})};{\rm\ and\ for\
}t=2,\ldots,T,\cr&\psi_{t}(\theta_{1},\cdots,\theta_{t})=\theta_{t}+(-1)^{t}\
{1-F_{1}(\theta_{1})\over f_{1}(\theta_{1})}\cr&\ \ \times{\partial
F_{2}(\theta_{2}|\theta_{1},q_{1}(\theta_{1}))/\partial\theta_{1}\over
f_{2}(\theta_{2}|\theta_{1},q_{1}(\theta_{1}))}\cdots{\partial
F_{t}(\theta_{t}|\theta_{t-1},q_{t-1}(h^{t-1},\theta_{t-1}))/\partial\theta_{t-1}\over
f_{t}(\theta_{t}|\theta_{t-1},q_{t-1}(h^{t-1},\theta_{t-1}))},}$ $None$

then the buyer’s (total expected) payment (8) of the previous section, which
is the seller’s (total expected) revenue of the present problem, becomes

$\eqalign{&\int_{\underline{\theta}}^{\overline{\theta}}\psi_{1}(\tilde{\theta}_{1})q_{1}(\tilde{\theta}_{1})f_{1}(\tilde{\theta}_{1})d\tilde{\theta}_{1}+\sum_{t=2}^{T}\delta^{t-1}\int_{\underline{\theta}}^{\overline{\theta}}\cdots\int_{\underline{\theta}}^{\overline{\theta}}\psi_{t}(\tilde{\theta}_{1},\cdots,\tilde{\theta}_{t})q_{t}(\tilde{h}^{t},\tilde{\theta}_{t})\cr&\
\times
f_{t}(\tilde{\theta}_{t}|\tilde{\theta}_{t-1},q_{t-1}(\tilde{h}^{t-1},\tilde{\theta}_{t-1}))\cdots
f_{1}(\tilde{\theta}_{1})d\tilde{\theta}_{t}\cdots
d\tilde{\theta}_{1}-U_{1}(\underline{\theta}).}$ $None$

Observe that $\psi_{t}(\theta_{1},\cdots,\theta_{t})$ corresponds to the
virtual valuation of Myerson (1981) for the static mechanism design. Observe
also that the seller’s revenue does not depend on the transfer rule and thus
the revenue equivalence principle applies. The seller’s problem is to choose
the probability of sales $\\{q_{1}(\cdot),\cdots,q_{T}(\cdot)\\}$ for the
maximization of this revenue subject to the constraints

$\eqalign{&(i)\ 0\leq\sum_{t=1}^{T}q_{t}(h^{t},\theta_{t})\leq 1{\rm\ \ for\
all\ }h^{t}{\rm\ and\ }\theta_{t};\cr&(ii)\ U_{1}(\underline{\theta})\geq
0;{\rm\
and}\cr&(iii)\int_{\hat{\theta}_{t}}^{\theta_{t}}D_{t}(\tilde{\theta}_{t};\hat{\theta}^{t-1})d\tilde{\theta}_{t}\geq\int_{\hat{\theta}_{t}}^{\theta_{t}}D_{t}(\hat{\theta}_{t},\tilde{\theta}_{t};\hat{\theta}^{t-1})d\tilde{\theta}_{t}\
\ \ \forall t,\forall\theta_{t},\forall\hat{\theta}_{t}{\rm\ and\
}\forall\hat{\theta}^{t-1},{\rm\ where}\cr&\ \ \ \ \ \ \
D_{t}(\hat{\theta}_{t},\theta_{t};\hat{\theta}^{t-1})=\
q_{t}(\hat{h}^{t},\hat{\theta}_{t})-\delta\int_{\underline{\theta}}^{\overline{\theta}}D_{t+1}(\tilde{\theta}_{t+1};\hat{\theta}^{t})\
{\partial
F_{t+1}(\tilde{\theta}_{t+1}|\theta_{t},q_{t}(\hat{h}^{t},\hat{\theta}_{t}))\over\partial\theta_{t}}d\tilde{\theta}_{t+1}.}$

Note that the first constraint is the feasibility condition that the seller
can sell at most one indivisible object over the entire span of time, the
second constraint is the individual rationality condition (or the
participation constraint of the buyer), and the third constraint is inequality
(5) of Theorem 1, which concerns incentive compatibility.11 11 Note well that
inequality (4) of Theorem 1 is incorporated in the process of deriving the
seller’s revenue.

As is standard in the mechanism design literature, we solve for the relaxed
problem as ignoring the last constraint, and then check whether the solution
satisfies the ignored constraint. First, it is obvious for revenue
maximization that the second constraint is satisfied with equality, i.e,
$U_{1}(\underline{\theta})=0$. Next, the optimal solution for the relaxed
problem can be found by the following procedure. Let
$V_{t}(\theta_{1},\cdots,\theta_{t})$ be the continuation value in period $t$,
that is, the seller’s revenue evaluated in period $t$, given the realized
types of $\theta^{t}=(\theta_{1},\cdots,\theta_{t})$ and under the optimal
decisions in subsequent periods. Observe that
$V_{t}(\theta_{1},\cdots,\theta_{t})\geq 0$ for all $t$ and $\theta^{t}$.

The procedure for finding the optimal solution:12 12 We assume that, when the
seller is indifferent between selling in this period and selling in a future
period (or not selling), he chooses the latter. Any alternative assumption
regarding the seller’s behavior would be fine.

Start with period 1 and for $t=1,\ldots,T$:

[1] If $\psi_{t}(\theta_{1},\cdots,\theta_{t})\leq 0$, then do not sell in
this period and move to the next period.

[2] If $\psi_{t}(\theta_{1},\cdots,\theta_{t})>0$, then compare
$\psi_{t}(\theta_{1},\cdots,\theta_{t})$ with

$M\buildrel\rm
def\over{=}\delta\int_{\underline{\theta}}^{\overline{\theta}}V_{t+1}(\theta_{1},\cdots,\theta_{t},\tilde{\theta}_{t+1})f_{t+1}(\tilde{\theta}_{t+1}|\theta_{t},q_{t}(\hat{h}^{t},\theta_{t}))d\tilde{\theta}_{t+1}.$

[2-1] If $\psi_{t}(\theta_{1},\cdots,\theta_{t})\leq M$, then do not sell in
this period and move to the next period.

[2-2] If $\psi_{t}(\theta_{1},\cdots,\theta_{t})>M$, then sell in this period.

Moreover, if $\psi_{s}(\theta_{1},\cdots,\theta_{s})>0$ for all
$s=t+1,\ldots,T$ and $\theta^{s}=(\theta_{1},\cdots,\theta_{s})$ given
$\theta^{t}$, then $M$ is equal to

$\eqalign{M^{\prime}\buildrel\rm
def\over{=}\max_{s=t+1,\cdots,T}&\delta^{s-t}\int_{\underline{\theta}}^{\overline{\theta}}\cdots\int_{\underline{\theta}}^{\overline{\theta}}\psi_{s}(\theta_{1},\cdots,\theta_{t},\tilde{\theta}_{t+1},\cdots,\tilde{\theta}_{s})\cr\times&f_{s}(\tilde{\theta}_{s}|\tilde{\theta}_{s-1},q_{s-1}(h^{s-1}_{+}(\hat{h}^{t},\theta_{t}))\cdots
f_{t+1}(\tilde{\theta}_{t+1}|\theta_{t},q_{t}(\hat{h}^{t},\theta_{t}))d\tilde{\theta}_{s}\cdots
d\tilde{\theta}_{t+1}.}$

Note: We have $M=M^{\prime}=0$ when $t=T$ by convention. Hence, it is optimal
to sell the object in the last period if and only if
$\psi_{T}(\theta_{1},\cdots,\theta_{T})>0$. Observe that the object may not be
sold eventually for some realizations of private information.

Some papers in the literature, including Kakade et al. (2013) and Pavan et al.
(2014), have examined the repeated sales in which a monopolistic seller sells
a (perishable) good in each period. The feasibility constraint in this
scenario is $0\leq q_{t}(h^{t},\theta_{t})\leq 1$ for all $t,h^{t}$ and
$\theta_{t}$, and the optimal solution for the relaxed problem is
straightforward: Sell in this period if and only if
$\psi_{t}(\theta_{1},\cdots,\theta_{t})>0$. In comparison, the seller in our
setup can sell only one object over the entire span of time and hence the
seller’s problem is similar to the optimal stopping time problem.

3.2. Two-period examples

We first investigate some interesting examples with $T=2$. Observe that
inequality (5) is equal to

$\int_{\hat{\theta}_{2}}^{\theta_{2}}q_{2}(\hat{\theta}_{1},q_{1}(\hat{\theta}_{1}),\tilde{\theta}_{2})d\tilde{\theta}_{2}\geq\int_{\hat{\theta}_{2}}^{\theta_{2}}q_{2}(\hat{\theta}_{1},q_{1}(\hat{\theta}_{1}),\hat{\theta}_{2})d\tilde{\theta}_{2}$
$None$

for $t=2$ and

$\eqalign{&\int_{\hat{\theta}_{1}}^{\theta_{1}}q_{1}(\tilde{\theta}_{1})d\tilde{\theta}_{1}-\delta\int_{\hat{\theta}_{1}}^{\theta_{1}}\int_{\underline{\theta}}^{\overline{\theta}}q_{2}(\tilde{\theta}_{1},q_{1}(\tilde{\theta}_{1}),\tilde{\theta}_{2}){\partial
F_{2}(\tilde{\theta}_{2}|\tilde{\theta}_{1},q_{1}(\tilde{\theta}_{1}))\over\partial\theta_{1}}d\tilde{\theta}_{2}d\tilde{\theta}_{1}\cr\geq&\int_{\hat{\theta}_{1}}^{\theta_{1}}q_{1}(\hat{\theta}_{1})d\tilde{\theta}_{1}-\delta\int_{\hat{\theta}_{1}}^{\theta_{1}}\int_{\underline{\theta}}^{\overline{\theta}}q_{2}(\hat{\theta}_{1},q_{1}(\hat{\theta}_{1}),\tilde{\theta}_{2}){\partial
F_{2}(\tilde{\theta}_{2}|\tilde{\theta}_{1},q_{1}(\hat{\theta}_{1}))\over\partial\theta_{1}}d\tilde{\theta}_{2}d\tilde{\theta}_{1}}$
$None$

for $t=1$. Observe also that $(11)$ holds when
$q_{2}(\hat{\theta}_{1},q_{1}(\hat{\theta}_{1}),\theta_{2})$ is increasing in
$\theta_{2}$. To see whether the optimal solution for the relaxed problem is
indeed the optimal solution for the original problem, we will check below
whether these inequalities are satisfied. For comparison, we will solve for
both the case when $0\leq q_{t}\leq 1$ for $t=1,2$ (the repeated sales
scenario) and the case when $0\leq q_{1}+q_{2}\leq 1$.

Example 1. Let $F_{1}(\theta_{1})=\theta_{1}$ for $\theta_{1}\in[0,1]$ and
$F_{2}(\theta_{2}|\theta_{1})=\theta_{2}/\theta_{1}$ for
$\theta_{2}\in[0,\theta_{1}]$. We have $\partial
F_{2}/\partial\theta_{1}=-\theta_{2}/\theta_{1}^{2}\leq 0$,

$\psi_{1}(\theta_{1})=2\theta_{1}-1{\rm\ and\
\,}\psi_{2}(\theta_{1},\theta_{2})={\theta_{2}\over\theta_{1}}(2\theta_{1}-1).$

The seller’s revenue is

$\int_{0}^{1}(2\theta_{1}-1)q_{1}(\theta_{1})d\theta_{1}+\delta\int_{0}^{1}\int_{0}^{\theta_{1}}{\theta_{2}\over\theta_{1}}(2\theta_{1}-1)q_{2}(\theta_{1},q_{1}(\theta_{1}),\theta_{2}){1\over\theta_{1}}d\theta_{2}d\theta_{1}.$

(i) When $0\leq q_{t}\leq 1$ for $t=1,2$: It is straightforward to see that
the optimal solution is

$q_{1}(\theta_{1})=\cases{1&if $\theta_{1}>{1\over 2}$;\cr 0&otherwise\cr}$

and

$q_{2}(\theta_{1},q_{1}(\theta_{1}),\theta_{2})=\cases{1&if
$\theta_{1}>{1\over 2}$ and $\theta_{2}>0$;\cr 0&otherwise.\cr}$

It is also straightforward to see that $(11)$ and $(12)$ are satisfied since
$q_{1}$ is increasing in $\theta_{1}$ and $q_{2}$ is increasing in
$\theta_{1}$ and $\theta_{2}$. We may alternatively apply Corollary 2 for
incentive compatibility.

(ii) When $0\leq q_{1}+q_{2}\leq 1$: If $\theta_{1}\leq 1/2$, then the optimal
solution is to have $q_{1}=q_{2}=0$. If $\theta_{1}>1/2$, then the seller
compares $2\theta_{1}-1$ with

$\delta\int_{0}^{\theta_{1}}{\theta_{2}\over\theta_{1}}(2\theta_{1}-1){1\over\theta_{1}}d\theta_{2}={\delta(2\theta_{1}-1)\over
2}.$

Observe that the former is bigger than the latter. In summary, the optimal
solution is

$q_{1}(\theta_{1})=\cases{1&if $\theta_{1}>{1\over 2}$;\cr 0&otherwise\cr}$

and $q_{2}(\theta_{1},q_{1}(\theta_{1}),\theta_{2})=0$ for all $\theta_{1}$
and $\theta_{2}$. It is straightforward to see that $(11)$ and $(12)$ are
satisfied since $q_{1}$ is increasing in $\theta_{1}$ and $q_{2}=0$. We may
alternatively apply Corollary 2 for incentive compatibility.

Example 2. Let $F_{1}(\theta_{1})=\theta_{1}$ for $\theta_{1}\in[0,1]$ and
$F_{2}(\theta_{2}|\theta_{1})=\theta_{2}^{\theta_{1}}$ for
$\theta_{2}\in[0,1]$. We have $\partial
F_{2}/\partial\theta_{1}=\theta_{2}^{\theta_{1}}{\rm ln\,}\theta_{2}\leq 0$,

$\psi_{1}(\theta_{1})=2\theta_{1}-1{\rm\ and\
\,}\psi_{2}(\theta_{1},\theta_{2})=\theta_{2}+{1-\theta_{1}\over\theta_{1}}\theta_{2}{\rm
ln\,}\theta_{2}.$

The seller’s revenue is

$\int_{0}^{1}(2\theta_{1}-1)q_{1}(\theta_{1})d\theta_{1}+\delta\int_{0}^{1}\int_{0}^{1}\Bigl{(}\theta_{2}+{1-\theta_{1}\over\theta_{1}}\theta_{2}{\rm
ln\,}\theta_{2}\Bigr{)}q_{2}(\theta_{1},q_{1}(\theta_{1}),\theta_{2})\theta_{1}\theta_{2}^{\theta_{1}-1}d\theta_{2}d\theta_{1}.$

(i) When $0\leq q_{t}\leq 1$ for $t=1,2$: It is straightforward to see that
the optimal solution is

$q_{1}(\theta_{1})=\cases{1&if $\theta_{1}>{1\over 2}$;\cr 0&otherwise\cr}$

and

$q_{2}(\theta_{1},q_{1}(\theta_{1}),\theta_{2})=\cases{1&if
$\theta_{2}>e^{-\theta_{1}/(1-\theta_{1})}$;\cr 0&otherwise.\cr}$

Note that $\theta_{2}>e^{-\theta_{1}/(1-\theta_{1})}$ if and only if
$\theta_{1}>-{\rm ln\,}\theta_{2}/(1-{\rm ln\,}\theta_{2})$. It is also
straightforward to see that $(11)$ and $(12)$ are satisfied since $q_{1}$ is
increasing in $\theta_{1}$ and $q_{2}$ is increasing in $\theta_{1}$ and
$\theta_{2}$. We may alternatively apply Corollary 2 for incentive
compatibility.

(ii) When $0\leq q_{1}+q_{2}\leq 1$: If $\theta_{1}\leq 1/2$ and
$\theta_{1}\leq-{\rm ln\,}\theta_{2}/(1-{\rm ln\,}\theta_{2})$, then the
optimal solution is to have $q_{1}=q_{2}=0$. If $\theta_{1}\leq 1/2$ and
$\theta_{1}>-{\rm ln\,}\theta_{2}/(1-{\rm ln\,}\theta_{2})$, then the optimal
solution is to have $q_{1}=0$ and $q_{2}=1$. If $\theta_{1}>1/2$, then the
seller compares $2\theta_{1}-1$ with

$\delta\int_{0}^{1}\max\\{\theta_{2}+{1-\theta_{1}\over\theta_{1}}\theta_{2}{\rm
ln\,}\theta_{2},0\\}\theta_{1}\theta_{2}^{\theta_{1}-1}d\theta_{2}.$

Since there are only two periods, let us assume $\delta=1$ to avoid
unnecessary complications. Observe that, in period 2, we have $q_{2}=1$ if
$\theta_{1}>-{\rm ln\,}\theta_{2}/(1-{\rm ln\,}\theta_{2})$, i.e.,
$\theta_{2}>e^{-\theta_{1}/(1-\theta_{1})}$. Hence,

$\eqalign{&\int_{0}^{1}\max\\{\theta_{2}+{1-\theta_{1}\over\theta_{1}}\theta_{2}{\rm
ln\,}\theta_{2},0\\}\theta_{1}\theta_{2}^{\theta_{1}-1}d\theta_{2}=\int_{e^{-\theta_{1}/(1-\theta_{1})}}^{1}\bigl{(}\theta_{2}+{1-\theta_{1}\over\theta_{1}}\theta_{2}{\rm
ln\,}\theta_{2}\bigr{)}\theta_{1}\theta_{2}^{\theta_{1}-1}d\theta_{2}\cr=&\Bigl{(}1-{2\theta_{1}\over
1-\theta_{1}}e^{-\theta_{1}(1+\theta_{1})/(1-\theta_{1})}\Bigr{)}/\bigl{(}1+\theta_{1}\bigr{)}^{2}.}$

We find that $2\theta_{1}-1$ is greater than the last term if and only if
$\theta_{1}>0.652$. In summary, the optimal solution is

$q_{1}(\theta_{1})=\cases{1&if $\theta_{1}>0.652$;\cr 0&otherwise\cr}$

and

$q_{2}(\theta_{1},q_{1}(\theta_{1}),\theta_{2})=\cases{1&if $\theta_{1}\leq
0.652$ and $\theta_{2}>e^{-\theta_{1}/(1-\theta_{1})}$;\cr 0&otherwise.\cr}$

For incentive compatibility, we cannot apply Corollary 2 since $q_{2}$ is not
increasing in $\theta_{1}$. We have to check inequalities $(11)$ and $(12)$
directly. Since $q_{2}$ is increasing in $\theta_{2}$ for all $\theta_{1}$,
$(11)$ is satisfied. As for $(12)$, we divide cases.

[1] When $\hat{\theta}_{1}<\theta_{1}\leq 0.652$: The left-hand side (LHS
henceforth) of $(12)$ is equal to
$-\int_{\hat{\theta}_{1}}^{\theta_{1}}\int_{c(\tilde{\theta}_{1})}^{1}\tilde{\theta}_{2}^{\tilde{\theta}_{1}}{\rm
ln\,}\tilde{\theta}_{2}d\tilde{\theta}_{2}d\tilde{\theta}_{1}$ where
$c(\tilde{\theta}_{1})=e^{-\tilde{\theta}_{1}/(1-\tilde{\theta}_{1})}$ and the
right-hand side (RHS henceforth) of $(12)$ is equal to
$-\int_{\hat{\theta}_{1}}^{\theta_{1}}\int_{d}^{1}\tilde{\theta}_{2}^{\tilde{\theta}_{1}}{\rm
ln\,}\tilde{\theta}_{2}d\tilde{\theta}_{2}d\tilde{\theta}_{1}$ where
$d=e^{-\hat{\theta}_{1}/(1-\hat{\theta}_{1})}$. Since

$LHS-
RHS=-\int_{\hat{\theta}_{1}}^{\theta_{1}}\int_{c(\tilde{\theta}_{1})}^{d}\tilde{\theta}_{2}^{\tilde{\theta}_{1}}{\rm
ln\,}\tilde{\theta}_{2}d\tilde{\theta}_{2}d\tilde{\theta}_{1}\geq 0$

by the fact that $c(\tilde{\theta}_{1})\leq d$ and
$\tilde{\theta}_{2}^{\tilde{\theta}_{1}}{\rm ln\,}\tilde{\theta}_{2}\leq 0$,
$(12)$ is satisfied.

[2] When $\hat{\theta}_{1}\leq 0.652<\theta_{1}$: We have
$LHS=\int_{0.652}^{\theta_{1}}d\tilde{\theta}_{1}-\int_{\hat{\theta}_{1}}^{0.652}\int_{c(\tilde{\theta}_{1})}^{1}\tilde{\theta}_{2}^{\tilde{\theta}_{1}}{\rm
ln\,}\tilde{\theta}_{2}$ $d\tilde{\theta}_{2}d\tilde{\theta}_{1}$ and
$RHS=-\int_{\hat{\theta}_{1}}^{\theta_{1}}\int_{d}^{1}\tilde{\theta}_{2}^{\tilde{\theta}_{1}}{\rm
ln\,}\tilde{\theta}_{2}d\tilde{\theta}_{2}d\tilde{\theta}_{1}$. Thus,

$\eqalign{LHS-
RHS=&\int_{0.652}^{\theta_{1}}d\tilde{\theta}_{1}-\int_{\hat{\theta}_{1}}^{0.652}\int_{c(\tilde{\theta}_{1})}^{d}\tilde{\theta}_{2}^{\tilde{\theta}_{1}}{\rm
ln\,}\tilde{\theta}_{2}d\tilde{\theta}_{2}d\tilde{\theta}_{1}\cr+&\int_{0.652}^{\theta_{1}}\int_{d}^{1}\tilde{\theta}_{2}^{\tilde{\theta}_{1}}{\rm
ln\,}\tilde{\theta}_{2}d\tilde{\theta}_{2}d\tilde{\theta}_{1}.}$

Note that
$-\int_{\hat{\theta}_{1}}^{0.652}\int_{c(\tilde{\theta}_{1})}^{d}\tilde{\theta}_{2}^{\tilde{\theta}_{1}}{\rm
ln\,}\tilde{\theta}_{2}d\tilde{\theta}_{2}d\tilde{\theta}_{1}\geq 0$ by the
fact that $c(\tilde{\theta}_{1})\leq d$ and
$\tilde{\theta}_{2}^{\tilde{\theta}_{1}}{\rm ln\,}\tilde{\theta}_{2}$ $\leq
0$. Note also that

$\eqalign{&\int_{0.652}^{\theta_{1}}\Bigl{(}1+\int_{d}^{1}\tilde{\theta}_{2}^{\tilde{\theta}_{1}}{\rm
ln\,}\tilde{\theta}_{2}d\tilde{\theta}_{2}\Bigr{)}d\tilde{\theta}_{1}\geq\int_{0.652}^{\theta_{1}}\Bigl{(}1+\int_{0}^{1}\tilde{\theta}_{2}^{\tilde{\theta}_{1}}{\rm
ln\,}\tilde{\theta}_{2}d\tilde{\theta}_{2}\Bigr{)}d\tilde{\theta}_{1}\cr=&\int_{0.652}^{\theta_{1}}\Bigl{(}1-{1\over(1+\tilde{\theta}_{1})^{2}}\Bigr{)}d\tilde{\theta}_{1}\geq
0.}$

Hence, $LHS-RHS\geq 0$ and $(12)$ is satisfied.

[3] When $0.652<\hat{\theta}_{1}<\theta_{1}$: We have $q_{1}=1$ and $q_{2}=0$,
so both LHS and RHS of $(12)$ are equal to $\theta_{1}-\hat{\theta}_{1}$.
Hence, $(12)$ is satisfied.

Example 3. Let $F_{1}(\theta_{1})=\theta_{1}$ for $\theta_{1}\in[0,1]$ and
$F_{2}(\theta_{2}|\theta_{1})=\theta_{2}-2(\theta_{1}-{1\over
2})\theta_{2}(1-\theta_{2})$ for $\theta_{2}\in[0,1]$.13 13 This is the
distribution in Example 6 of Pavan et al. (2014). We have $\partial
F_{2}/\partial\theta_{1}=-2\theta_{2}(1-\theta_{2})\leq 0$,

$\psi_{1}(\theta_{1})=2\theta_{1}-1{\rm\ and\
\,}\psi_{2}(\theta_{1},\theta_{2})={\theta_{1}\theta_{2}^{2}\over
1-\theta_{1}-\theta_{2}+2\theta_{1}\theta_{2}}.$

The seller’s revenue is

$\int_{0}^{1}(2\theta_{1}-1)q_{1}(\theta_{1})d\theta_{1}+\delta\int_{0}^{1}\int_{0}^{1}2\theta_{1}\theta_{2}^{2}q_{2}(\theta_{1},q_{1}(\theta_{1}),\theta_{2})d\theta_{2}d\theta_{1}.$

(i) When $0\leq q_{t}\leq 1$ for $t=1,2$: It is straightforward to see that
the optimal solution is

$q_{1}(\theta_{1})=\cases{1&if $\theta_{1}>{1\over 2}$;\cr 0&otherwise\cr}$

and

$q_{2}(\theta_{1},q_{1}(\theta_{1}),\theta_{2})=1{\rm\ for\ all\
}\theta_{1}>0{\rm\ and\ }\theta_{2}>0.$

It is also straightforward to see that $(11)$ and $(12)$ are satisfied since
$q_{1}$ is increasing in $\theta_{1}$ and $q_{2}$ is increasing in
$\theta_{1}$ and $\theta_{2}$. We may alternatively apply Corollary 2 for
incentive compatibility.

(ii) When $0\leq q_{1}+q_{2}\leq 1$: If $\theta_{1}\leq 1/2$, then the optimal
solution is to have $q_{1}=0$. If $\theta_{1}>1/2$, then the seller compares
$2\theta_{1}-1$ with

$\delta\int_{0}^{1}2\theta_{1}\theta_{2}^{2}d\theta_{2}={2\delta\theta_{1}\over
3}.$

If $\theta_{1}>3/(6-2\delta)$, then we have $q_{1}=1$ and $q_{2}=0$.
Otherwise, move to period 2 and see whether $\theta_{2}>0$. In summary, the
optimal solution is

$q_{1}(\theta_{1})=\cases{1&if $\theta_{1}>{3\over 6-2\delta}$;\cr
0&otherwise\cr}$

and

$q_{2}(\theta_{1},q_{1}(\theta_{1}),\theta_{2})=\cases{1&if
$0<\theta_{1}\leq{3\over 6-2\delta}$ and $\theta_{2}>0$;\cr 0&otherwise.\cr}$

For incentive compatibility, we cannot apply Corollary 2 since $q_{2}$ is not
increasing in $\theta_{1}$. We have to check inequalities $(11)$ and $(12)$
directly. Since $q_{2}$ is increasing in $\theta_{2}$ for all $\theta_{1}$,
$(11)$ is satisfied. As for $(12)$, we divide cases.

[1] When $\hat{\theta}_{1}<\theta_{1}\leq 3/(6-2\delta)$: We have $q_{1}=0$
and $q_{2}=1$, so both LHS and RHS of $(12)$ are equal to
$\delta\int_{\hat{\theta}_{1}}^{\theta_{1}}\int_{0}^{1}2\tilde{\theta}_{2}(1-\tilde{\theta}_{2})d\tilde{\theta}_{2}d\tilde{\theta}_{1}=\delta(\theta_{1}-\hat{\theta}_{1})/3$.
Hence, $(12)$ is satisfied.

[2] When $\hat{\theta}_{1}\leq 3/(6-2\delta)<\theta_{1}$: We have

$LHS=\int_{{3\over
6-2\delta}}^{\theta_{1}}d\tilde{\theta}_{1}+\delta\int_{\hat{\theta}_{1}}^{{3\over
6-2\delta}}\int_{0}^{1}2\tilde{\theta}_{2}(1-\tilde{\theta}_{2})d\tilde{\theta}_{2}d\tilde{\theta}_{1}=\theta_{1}-{3\over
6-2\delta}+{\delta\over 3}\Bigl{(}{3\over 6-2\delta}-\hat{\theta}_{1}\Bigr{)}$

and

$RHS=\delta\int_{\hat{\theta}_{1}}^{\theta_{1}}\int_{0}^{1}2\tilde{\theta}_{2}(1-\tilde{\theta}_{2})d\tilde{\theta}_{2}d\tilde{\theta}_{1}={\delta\over
3}(\theta_{1}-\hat{\theta}_{1}).$

Since

$LHS-RHS=\bigl{(}1-{\delta\over 3}\Bigr{)}\Bigl{(}\theta_{1}-{3\over
6-2\delta}\Bigr{)}>0,$

$(12)$ is satisfied.

[3] $3/(6-2\delta)<\hat{\theta}_{1}<\theta_{1}$: We have $q_{1}=1$ and
$q_{2}=0$, so both LHS and RHS of $(12)$ are equal to
$\theta_{1}-\hat{\theta}_{1}$. Hence, $(12)$ is satisfied.

3.3. The general analysis

We resume back to the case with an arbitrary number of periods. We begin with
some special cases.

Example 4. (Independent private information) Let the private information be
independent across periods so that $\partial F_{t+1}/\partial\theta_{t}=0$ for
all $t$. Then, we have $\psi_{t}(\theta_{1},\cdots,\theta_{t})=\theta_{t}$ for
$t=2,\ldots,T$. The optimal solution can be found by the procedure given above
using $M^{\prime}$. (See the paragraph after Proposition 1 below for a further
discussion on the optimal solution.) In addition,
$D_{t}(\hat{\theta}_{t},\theta_{t};\hat{\theta}^{t-1})=q_{t}(\hat{h}^{t},\hat{\theta}_{t})$
and thus the inequality $(5)$ becomes

$\int_{\hat{\theta}_{t}}^{\theta_{t}}q_{t}(\hat{h}^{t},\tilde{\theta}_{t})d\tilde{\theta}_{t}\geq\int_{\hat{\theta}_{t}}^{\theta_{t}}q_{t}(\hat{h}^{t},\hat{\theta}_{t})d\tilde{\theta}_{t}.$

This is satisfied when $q_{t}(\hat{h}^{t},\theta_{t})$ is increasing in
$\theta_{t}$. Observe that we have an increasing optimal
$q_{t}(\hat{h}^{t},\theta_{t})$ for $t=2,\ldots,T$ since
$\psi_{t}(\theta_{1},\cdots,\theta_{t})=\theta_{t}$. We also have an
increasing optimal $q_{1}(\theta_{1})$ when
$\psi_{1}(\theta_{1})=\theta_{1}-(1-F_{1}(\theta_{1}))/f(\theta_{1})$ is
increasing in $\theta_{1}$, i.e., when the virtual valuation is regular as
Myerson (1981) has named it. A sufficient condition for a regular virtual
valuation is that the distribution $F_{1}$ is IFR, i.e,. the hazard rate (or
failure rate) $f_{1}(\theta_{1})/(1-F_{1}(\theta_{1}))$ is increasing.14 14
Examples of IFR distributions are exponential, uniform, normal, logistic,
power (for $c\geq 1$), Weibull (for $c\geq 1$), and gamma (for $c\geq 1$)
where $c$ is the shape parameter.

We note as comparison that, for the repeated sales scenario in which $0\leq
q_{t}(h^{t},\theta_{t})\leq 1$ for all $t=1,\ldots,T$, the optimal solution
achieves the first-best outcome, that is, $q_{t}(\hat{h}^{t},\theta_{t})=1$,
except for the first period.15 15 Hence, there is no (downward) distortion. As
observed by the authors of previous works, the reason is that the seller and
the buyer face symmetric information except for period 1 since period-$t$
information for $t\geq 2$ is independent of period-1 information.

Example 5. (First-order autoregressive process) Suppose for $t\geq 2$ that
$\tilde{\theta}_{t}=\gamma\theta_{t-1}+(1-\gamma)\tilde{\epsilon}_{t}$ where
$\tilde{\epsilon}_{t}$’s are independently and identically distributed random
variables over $\Theta=[\underline{\theta},\overline{\theta}]$ and
$\gamma\in[0,1)$. Then, we have

$\psi_{t}(\theta_{1},\ldots,\theta_{t})=\theta_{t}-\gamma^{t-1}{1-F_{1}(\theta_{1})\over
f_{1}(\theta_{1})}$

for $t=1,\ldots,T$. Observe that, when
$\psi_{t}(\theta_{1},\cdots,\theta_{t})>0$, we have

$\psi_{t+1}(\theta_{1},\cdots,\theta_{t+1})=\gamma\theta_{t}+(1-\gamma)\epsilon_{t+1}-\gamma^{t}{1-F_{1}(\theta_{1})\over
f_{1}(\theta_{1})}=\gamma\psi_{t}(\theta_{1},\cdots,\theta_{t})+(1-\gamma)\epsilon_{t+1}>0$

and thus $\psi_{s}(\theta_{1},\cdots,\theta_{s})>0$ for all $s=t+1,\ldots,T$
and $\theta^{s}=(\theta_{1},\cdots,\theta_{s})$. The optimal solution can be
found by the procedure given above using $M^{\prime}$. (See the paragraph
after Proposition 1 below for a further discussion on the optimal solution.)
In addition, when $F_{1}$ is IFR, the four conditions of Corollary 2 are
satisfied and thus incentive compatibility holds.

We note as comparison that, for the repeated sales scenario in which $0\leq
q_{t}(h^{t},\theta_{t})\leq 1$ for all $t=1,\ldots,T$, the optimal solution is
such that $q_{t}(h^{t},\theta_{t})=1$ if and only if
$\theta_{t}\geq\gamma^{t-1}(1-F_{1}(\theta_{1}))/f_{1}(\theta_{1})$. Hence,
$q_{t}(h^{t},\theta_{t})$ approaches the first-best outcome as $t$ increases.
This property was first observed by Besanko (1985) in a model of multi-period
contracts.

Let us next extend some examples of the previous subsection.

Example 6. Let $F_{1}(\theta_{1})=\theta_{1}$ for $\theta_{1}\in[0,1]$ and
$F_{t}(\theta_{t}|\theta_{t-1})=\theta_{t}/\theta_{t-1}$ for
$\theta_{t}\in[0,\theta_{t-1}]$ and $t\geq 2$. We have $\partial
F_{t}/\partial\theta_{t-1}=-\theta_{t}/\theta_{t-1}^{2}\leq 0$ and

$\psi_{t}(\theta_{1},\cdots,\theta_{t})={\theta_{t}\over\theta_{1}}(2\theta_{1}-1).$

The seller’s revenue is

$\int_{0}^{1}(2\theta_{1}-1)q_{1}(\theta_{1})d\theta_{1}+\sum_{t=2}^{T}\delta^{t-1}\int_{0}^{1}\cdots\int_{0}^{\theta_{t-1}}{\theta_{t}\over\theta_{1}}(2\theta_{1}-1)q_{t}(h^{t},\theta_{t})\prod_{s=2}^{t}{1\over\theta_{s-1}}d\theta_{t}\cdots
d\theta_{1}.$

When $\theta_{1}\leq 1/2$, the optimal solution is to have
$q_{1}=\cdots=q_{T}=0$. For the case when $\theta_{1}>1/2$, let us assume that
$T$ is finite and use backward induction. In period $T$, it is optimal to have
$q_{T}=1$ if and only if $\theta_{T}>0$. In period $T-1$, the seller compares
$\psi_{T-1}(\theta_{1},\cdots,\theta_{T-1})=\theta_{T-1}(2\theta_{1}-1)/\theta_{1}$
with

$\delta\int_{0}^{\theta_{T-1}}{\theta_{T}\over\theta_{1}}(2\theta_{1}-1){1\over\theta_{T-1}}d\theta_{T}=\delta{\theta_{T-1}\over
2\theta_{1}}(2\theta_{1}-1).$

Hence, it is optimal to have $q_{T-1}=1$ if and only if $\theta_{T-1}>0$.
Proceeding backwards, the optimal solution is

$q_{1}(\theta_{1})=\cases{1&if $\theta_{1}>{1\over 2}$;\cr 0&otherwise\cr}$

and $q_{t}(h^{t},\theta_{t})=0$ for all $t\geq 2$, $h^{t}$ and $\theta_{t}$.
It is straightforward to see that $(5)$ is satisfied for all $t=1,\ldots,T$.
We may alternatively apply Corollary 2 for incentive compatibility.

Example 7. Let $F_{1}(\theta_{1})=\theta_{1}$ for $\theta_{1}\in[0,1]$ and
$F_{t}(\theta_{t}|\theta_{t-1})=\theta_{t}^{\theta_{t-1}}$ for
$\theta_{t}\in[0,1]$ and $t\geq 2$. We have $\partial
F_{t}/\partial\theta_{t-1}=\theta_{t}^{\theta_{t-1}}{\rm ln\,}\theta_{t}\leq
0$ and

$\psi_{t}(\theta_{1},\cdots,\theta_{t})=\theta_{t}+(-1)^{t}{1-\theta_{1}\over\theta_{1}}\theta_{t}{\rm
ln\,}\theta_{2}\cdots{\rm ln\,}\theta_{t}.$

The seller’s revenue is

$\eqalign{&\int_{0}^{1}(2\theta_{1}-1)q_{1}(\theta_{1})d\theta_{1}\cr+&\sum_{t=2}^{T}\delta^{t-1}\int_{0}^{1}\cdots\int_{0}^{1}\Bigl{(}\theta_{t}+(-1)^{t}{1-\theta_{1}\over\theta_{1}}\theta_{t}\prod_{s=2}^{t}{\rm
ln\,}\theta_{s}\Bigr{)}q_{t}(h^{t},\theta_{t})\bigl{(}\prod_{s=2}^{t}\theta_{s-1}\theta_{s}^{\theta_{s-1}-1}\bigr{)}d\theta_{t}\cdots
d\theta_{1}.}$

Let us assume that $T$ is finite and $\delta=1$. In period $T$, it is optimal
to have $q_{T}=1$ if and only if

$\theta_{T}>exp\Bigl{(}(-1)^{T-1}{\theta_{1}\over
1-\theta_{1}}\prod_{s=2}^{T-1}\bigl{(}{\rm
ln\,}\theta_{s}\bigr{)}^{-1}\Bigr{)}\buildrel\rm def\over{=}k_{T}.$

Let $V_{T}(\theta_{1},\cdots,\theta_{T})$ be the continuation value function
from the optimal allocation in period $T$. In period $T-1$, the seller
compares $\psi_{T-1}(\theta_{1},\cdots,\theta_{T-1})$ with

$\eqalign{&\int_{0}^{1}V_{T}(\theta_{1},\cdots,\theta_{T})f_{T}(\theta_{T}|\theta_{T-1})d\theta_{T}=\int_{0}^{1}\max\\{\psi_{T}(\theta_{1},\cdots,\theta_{T}),0\\}f_{T}(\theta_{T}|\theta_{T-1})d\theta_{T}\cr=&\int_{k_{T}}^{1}\Bigl{(}\theta_{T}+(-1)^{T}{1-\theta_{1}\over\theta_{1}}\theta_{T}\prod_{s=2}^{T}{\rm
ln\,}\theta_{s}\Bigr{)}\theta_{T-1}\theta_{T}^{\theta_{T-1}-1}d\theta_{T}.}$

We can find $k_{T-1}$ such that it is optimal to have $q_{T-1}=1$ if and only
if $\theta_{T-1}>k_{T-1}$. Proceeding backwards, in period $t$, the seller
compares $\psi_{t}(\theta_{1},\cdots,\theta_{t})$ with

$\int_{0}^{1}V_{t+1}(\theta_{1},\cdots,\theta_{t+1})f_{t+1}(\theta_{t+1}|\theta_{t})d\theta_{t+1},$

where $V_{t+1}(\theta_{1},\cdots,\theta_{t+1})$ is the continuation value
function from the optimal allocation in period $t+1$. In summary, the optimal
allocation is, for $t=1,\ldots,T$,

$q_{t}(h^{t},\theta_{t})=\cases{1&if $\theta_{s}\leq k_{s}$ for
$s=1,\cdots,t-1$ and $\theta_{t}>k_{t}$;\cr 0&otherwise.\cr}$

Let us take the case when $T=3$ as an illustration. We have

$\psi_{3}(\theta_{1},\theta_{2},\theta_{3})=\theta_{3}-{1-\theta_{1}\over\theta_{1}}\theta_{3}{\rm
ln\,}\theta_{2}{\rm ln\,}\theta_{3}$

and it is optimal to have $q_{3}=1$ if and only if
$\theta_{3}>k_{3}=exp(\theta_{1}/((1-\theta_{1}){\rm ln\,}\theta_{2}))$. In
period 2, the seller compares

$\psi_{2}(\theta_{1},\theta_{2})=\theta_{2}+{1-\theta_{1}\over\theta_{1}}\theta_{2}{\rm
ln\,}\theta_{2}$

with

$\eqalign{&\int_{k_{3}}^{1}\Bigl{(}\theta_{3}-{1-\theta_{1}\over\theta_{1}}\theta_{3}{\rm
ln\,}\theta_{2}{\rm
ln\,}\theta_{3}\Bigr{)}\theta_{2}\theta_{3}^{\theta_{2}-1}d\theta_{3}\cr=&{\theta_{2}\over\theta_{2}+1}(1-k_{3}^{\theta_{2}+1})-{1-\theta_{1}\over\theta_{1}}\theta_{2}{\rm
ln\,}\theta_{2}{-1+k_{3}^{\theta_{2}+1}\bigl{(}1-(\theta_{2}+1){\rm
ln\,}k_{3}\bigr{)}\over(\theta_{2}+1)^{2}}.}$

We find that there exists $k_{2}$ such that it is optimal to have $q_{2}=1$ if
and only if $\theta_{2}>k_{2}$. Let $V_{2}(\theta_{1},\theta_{2})$ be the
continuation value function from the optimal allocation in period 2. In period
1, the seller compares $\psi_{1}(\theta_{1})=2\theta_{1}-1$ with

$\eqalign{&\int_{0}^{1}V_{2}(\theta_{1},\theta_{2})f_{2}(\theta_{2}|\theta_{1})d\theta_{2}\cr=&\int_{k_{2}}^{1}\psi_{2}(\theta_{1},\theta_{2})f_{2}(\theta_{2}|\theta_{1})d\theta_{2}+\int_{0}^{k_{2}}\int_{k_{3}}^{1}\psi_{3}(\theta_{1},\theta_{2},\theta_{3})f_{3}(\theta_{3}|\theta_{2})f_{2}(\theta_{2}|\theta_{1})d\theta_{3}d\theta_{2}.}$

We were not able to algebraically find the explicit solution for $k_{1}$ and
$k_{2}$ nor to check incentive compatibility.

As the previous example demonstrates, it may sometimes be a difficult task to
find an optimal allocation as well as to check incentive compatibility. Even
when $\psi_{s}(\theta_{1},\cdots,\theta_{s})>0$ for all $s=t+1,\ldots,T$ and
$\theta^{s}=(\theta_{1},\cdots,\theta_{s})$ so that $M^{\prime}$ of the
procedure for finding the optimal solution can be used, the seller has to
compare $\psi_{t}(\theta_{1},\cdots,\theta_{t})$ with

$\eqalign{&\delta^{s-t}\int_{\underline{\theta}}^{\overline{\theta}}\cdots\int_{\underline{\theta}}^{\overline{\theta}}\psi_{s}(\theta_{1},\cdots,\theta_{t},\tilde{\theta}_{t+1},\cdots,\tilde{\theta}_{s})\cr\times&f_{s}(\tilde{\theta}_{s}|\tilde{\theta}_{s-1},q_{s-1}(h^{s-1}_{+}(\hat{h}^{t},\theta_{t}))\cdots
f_{t+1}(\tilde{\theta}_{t+1}|\theta_{t},q_{t}(\hat{h}^{t},\theta_{t}))d\tilde{\theta}_{s}\cdots
d\tilde{\theta}_{t+1}}$

for all $s=t+1,\cdots,T$. This may be cumbersome. In contrast, it is in
general not optimal to use the myopic stopping rule in which the seller
compares $\psi_{t}(\theta_{1},\cdots,\theta_{t})$ only with

$\delta\int_{\underline{\theta}}^{\overline{\theta}}\psi_{t+1}(\theta_{1},\cdots,\theta_{t},\tilde{\theta}_{t+1})f_{t+1}(\tilde{\theta}_{t+1}|\theta_{t},q_{t}(\hat{h}^{t},\theta_{t}))d\tilde{\theta}_{t+1},$

i.e., the discounted expected value of
$\psi_{t+1}(\theta_{1},\cdots,\theta_{t+1})$. However, it is straightforward
to present a necessary and sufficient condition for the myopic stopping rule
to be optimal.

Proposition 1. Suppose $\psi_{s}(\theta_{1},\cdots,\theta_{s})>0$ for all
$s=t+1,\ldots,T$ and $\theta^{s}=(\theta_{1},\cdots,\theta_{s})$ at
$\theta^{t}$. The myopic stopping rule is optimal if and only if

$\eqalign{&\int_{\underline{\theta}}^{\overline{\theta}}\psi_{s+1}(\theta_{1},\cdots,\theta_{s},\tilde{\theta}_{s+1})f_{s+1}(\tilde{\theta}_{s+1}|\theta_{s},q_{s}(\hat{h}^{s},\theta_{s}))d\tilde{\theta}_{s+1}\cr>&\
\delta\int_{\underline{\theta}}^{\overline{\theta}}\int_{\underline{\theta}}^{\overline{\theta}}\psi_{s+2}(\theta_{1},\cdots,\theta_{s},\tilde{\theta}_{s+1},\tilde{\theta}_{s+2})f_{s+2}(\tilde{\theta}_{s+2}|\tilde{\theta}_{s+1},q_{s+1}(\hat{h}^{s},\theta_{s},q_{s}(\hat{h}^{s},\theta_{s}),\tilde{\theta}_{s+1}))\cr&\times
f_{s+1}(\tilde{\theta}_{s+1}|\theta_{s},q_{s}(\hat{h}^{s},\theta_{s}))d\tilde{\theta}_{s+2}d\tilde{\theta}_{s+1}}$

holds for all $s=t,\cdots,T-1$ and
$\theta^{s}=(\theta_{1},\cdots,\theta_{s})$. Proof: Note that LHS is the
expected value of $\psi_{s+1}(\theta_{1},\cdots,\theta_{s+1})$ and RHS is the
discounted expected value of $\psi_{s+2}(\theta_{1},\cdots,\theta_{s+2})$. If
LHS is greater than RHS for all $s=t,\cdots,T-1$, then we have a decreasing
sequence of values and thus the maximum $M^{\prime}$ defined in the procedure
for finding the optimal solution is obtained at $s=t+1$. Hence, the myopic
rule is optimal. It is easy to see that the converse also holds. Q.E.D.

Observe that, for Example 4 above, the myopic stopping rule is optimal if and
only if $\mu_{t+1}>\delta\mu_{t+2}$ where $\mu_{t}$ is the expected value of
$\tilde{\theta}_{t}$, i.e.,
$\mu_{t}=\int_{\underline{\theta}}^{\overline{\theta}}\tilde{\theta}_{t}f_{t}(\tilde{\theta}_{t})d\tilde{\theta}_{t}$.
In particular, the myopic rule is optimal when $\mu_{t}=\mu$ for all
$t=2,\cdots,T$ and $\delta<1$. Hence, it is optimal to sell in the first
period $t$ with $\theta_{t}>\delta\mu_{t+1}$. Observe next that, for Example 5
above, the myopic stopping rule is optimal if and only if

$\gamma\psi_{t}+(1-\gamma)\mu_{\epsilon}>\delta\Bigl{(}\gamma^{2}\psi_{t}+(1-\gamma^{2})\mu_{\epsilon}\Bigr{)}$

where $\mu_{\epsilon}$ is the expected value of $\tilde{\epsilon}_{t}$ for all
$t$. This inequality holds when $\gamma=0$ and $\delta<1$. For
$\gamma\in(0,1)$, the inequality is equivalent to

$\psi_{t}>{\bigl{(}\delta(1-\gamma^{2})-(1-\gamma)\bigr{)}\mu_{\epsilon}\over\gamma(1-\delta\gamma)}.$

When $\delta=1$, the myopic stopping rule is optimal if and only if
$\theta_{t}>\gamma^{t-1}(1-F_{1}(\theta_{1})/f_{1}(\theta_{1})+\mu_{\epsilon}$.
It is also easy to observe that the myopic stopping rule is optimal for
Example 6.

We next examine ex-post individual rationality. In an optimal allocation, we
have $q_{t}=1$ for at most one $t\in\\{1,\ldots,T\\}$ and $q_{s}=0$ for all
other $s$’s. If $q_{t}=1$, then the seller’s revenue (or the price of the
object) in period $t$ is $\psi_{t}(\theta_{1},\cdots,\theta_{t})$. Observe
also that the seller’s revenue is zero whenever $q_{s}=0$ for
$s\in\\{1,\ldots,T\\}$. The buyer’s ex-post payoff in period $t$ with
$q_{t}=1$ is thus

$\eqalign{&\theta_{t}-\psi_{t}(\theta_{1},\cdots,\theta_{t})\cr=&{1-F_{1}(\theta_{1})\over
f_{1}(\theta_{1})}\Bigl{(}-{\partial
F_{2}(\theta_{2}|\theta_{1},q_{1}(\theta_{1}))/\partial\theta_{1}\over
f_{2}(\theta_{2}|\theta_{1},q_{1}(\theta_{1}))}\Bigr{)}\cdots\Bigl{(}-{\partial
F_{t}(\theta_{t}|\theta_{t-1},q_{t-1}(h^{t-1},\theta_{t-1}))/\partial\theta_{t-1}\over
f_{t}(\theta_{t}|\theta_{t-1},q_{t-1}(h^{t-1},\theta_{t-1}))}\Bigr{)},}$

and that in period $t$ with $q_{t}=0$ is zero. Hence, we have:

Proposition 2. Ex-post individual rationality holds in an optimal allocation
under the first-order stochastic dominance condition of

$\partial F_{t+1}(\theta_{t+1}|\theta_{t},a_{t})/\partial\theta_{t}\leq 0$ for
all $t,\theta_{t},a_{t}$ and $\theta_{t+1}$.

Proof: Obvious since the buyer’s ex-post payoff is equal to
$\theta_{t}-\psi_{t}(\theta_{1},\cdots,\theta_{t})\geq 0$ when $q_{t}=1$, and
zero when $q_{t}=0$. Q.E.D.

3.4. Comparative statics

Let us perform some comparative static analysis regarding the seller’s revenue
and the selling time. Observe first that the seller’s revenue

$\psi_{t}(\theta_{1},\cdots,\theta_{t})=\theta_{t}-{1-F_{1}(\theta_{1})\over
f_{1}(\theta_{1})}\prod_{s=2}^{t}\Bigl{(}-{\partial
F_{s}(\theta_{s}|\theta_{s-1},q_{s-1}(h^{s-1},\theta_{s-1}))/\partial\theta_{s-1}\over
f_{s}(\theta_{s}|\theta_{s-1},q_{s-1}(h^{s-1},\theta_{s-1}))}\Bigr{)}$

in period $t$ increases as (i) $f_{1}(\theta_{1})/(1-F_{1}(\theta_{1}))$
increases and (ii) $-{\partial F_{s}/\partial\theta_{s-1}\over f_{s}}$
decreases.16 16 It also increases in $\theta_{t}$ obviously. Hence, the
revenue in period $t$ increases as the private information becomes less
persistent in the sense of (ii). This is due to the fact that the information
rent becomes smaller. In particular, the revenue is highest when the private
information is independent across periods (i.e., $\partial
F_{t+1}/\partial\theta_{t}=0$ for all $t$) and thus there exists no private
information except for period 1.

Observe however that this does not necessarily imply that the seller’s total
expected revenue increases as (i) or (ii) holds, since the (conditional)
density function also changes. Note that the total expected revenue of $(10)$
can be written as

$\eqalign{&\int_{\underline{\theta}}^{\overline{\theta}}\bigl{[}\tilde{\theta}_{1}f_{1}(\tilde{\theta}_{1})-(1-F_{1}(\tilde{\theta}_{1}))\bigr{]}q_{1}(\tilde{\theta}_{1})d\tilde{\theta}_{1}\cr&+\sum_{t=2}^{T}\delta^{t-1}\int_{\underline{\theta}}^{\overline{\theta}}\cdots\int_{\underline{\theta}}^{\overline{\theta}}\Bigl{[}\tilde{\theta}_{t}f_{1}(\tilde{\theta}_{1})\prod_{s=2}^{t}f_{s}(\tilde{\theta}_{s}|\tilde{\theta}_{s-1},q_{s-1}(\tilde{h}^{s-1},\tilde{\theta}_{s-1}))\cr&-(1-F_{1}(\tilde{\theta}_{1}))\prod_{s=2}^{t}\bigl{(}-\partial
F_{s}(\tilde{\theta}_{s}|\tilde{\theta}_{s-1},q_{s-1}(\tilde{h}^{s-1},\tilde{\theta}_{s-1}))/\partial\theta_{s-1}\bigr{)}\Bigr{]}q_{t}(\tilde{h}^{t},\tilde{\theta}_{t})d\tilde{\theta}_{t}\cdots
d\tilde{\theta}_{1}}$

with $U_{1}(\underline{\theta})=0$. We thus have:

Proposition 3. The seller’s (total expected) revenue increases as

(i)
$\int_{\underline{\theta}}^{\overline{\theta}}\bigl{[}\tilde{\theta}_{1}f_{1}(\tilde{\theta}_{1})-(1-F_{1}(\tilde{\theta}_{1}))\bigr{]}d\tilde{\theta}_{1}$
increases;

(ii) for all $t=2,\ldots,T$:

$\eqalign{&\int_{\underline{\theta}}^{\overline{\theta}}\cdots\int_{\underline{\theta}}^{\overline{\theta}}\Bigl{[}\tilde{\theta}_{t}f_{1}(\tilde{\theta}_{1})\prod_{s=2}^{t}f_{s}(\tilde{\theta}_{s}|\tilde{\theta}_{s-1},q_{s-1}(\tilde{h}^{s-1},\tilde{\theta}_{s-1}))\cr&-(1-F_{1}(\tilde{\theta}_{1}))\prod_{s=2}^{t}\bigl{(}-\partial
F_{s}(\tilde{\theta}_{s}|\tilde{\theta}_{s-1},q_{s-1}(\tilde{h}^{s-1},\tilde{\theta}_{s-1}))/\partial\theta_{s-1}\bigr{)}\Bigr{]}d\tilde{\theta}_{t}\cdots
d\tilde{\theta}_{1}}$

increases; and

(iii) $\delta$ increases.

Proof: Obvious. Q.E.D.

As for the selling time, we have:

Proposition 4. Assume $\partial
F_{t+1}(\theta_{t+1}|\theta_{t},a_{t})/\partial\theta_{t}\leq 0$ for all
$t,\theta_{t},a_{t}$ and $\theta_{t+1}$. Assume also that the myopic stopping
rule is optimal. The probability that the seller sells the object earlier,
that is, sells in period $t$ rather than period $t+1$, increases as

(i) $\theta_{t}$ increases and/or
$E[\theta_{t+1}|\theta_{t},q_{t}(h^{t},\theta_{t})]$ decreases;

(ii) $\int_{\underline{\theta}}^{\overline{\theta}}\bigl{(}-\partial
F_{t+1}(\tilde{\theta}_{t+1}|\theta_{t},q_{t}(h^{t},\theta_{t}))/\partial\theta_{t}\bigr{)}\
d\tilde{\theta}_{t+1}$ increases;

(iii) $f_{1}(\theta_{1})/(1-F_{1}(\theta_{1}))$ increases (decreases, resp.)
when

$1+\delta\int_{\underline{\theta}}^{\overline{\theta}}\partial
F_{t+1}(\tilde{\theta}_{t+1}|\theta_{t},q_{t}(h^{t},\theta_{t}))/\partial\theta_{t}\,d\tilde{\theta}_{t+1}>(<,resp.)\
0$;

(iv) for all $s=2,\cdots,t$:

$-{\partial
F_{s}(\theta_{s}|\theta_{s-1},q_{s-1}(h^{s-1},\theta_{s-1}))/\partial\theta_{s-1}\over
f_{s}(\theta_{s}|\theta_{s-1},q_{s-1}(h^{s-1},\theta_{s-1}))}$

decreases (increases, resp.) when

$1+\delta\int_{\underline{\theta}}^{\overline{\theta}}\partial
F_{t+1}(\tilde{\theta}_{t+1}|\theta_{t},q_{t}(h^{t},\theta_{t}))/\partial\theta_{t}\,d\tilde{\theta}_{t+1}>(<,resp.)\
0$; and

(v) $\delta$ decreases.

Proof: In period $t$, the seller compares

$\theta_{t}-{1-F_{1}(\theta_{1})\over
f_{1}(\theta_{1})}\prod_{s=2}^{t}\Bigl{(}-{\partial
F_{s}(\theta_{s}|\theta_{s-1},q_{s-1}(h^{s-1},\theta_{s-1}))/\partial\theta_{s-1}\over
f_{s}(\theta_{s}|\theta_{s-1},q_{s-1}(h^{s-1},\theta_{s-1}))}\Bigr{)}$

with

$\eqalign{&\delta\int_{\underline{\theta}}^{\overline{\theta}}\Bigl{[}\tilde{\theta}_{t+1}-{1-F_{1}(\theta_{1})\over
f_{1}(\theta_{1})}\prod_{s=2}^{t+1}\Bigl{(}-{\partial
F_{s}(\theta_{s}|\theta_{s-1},q_{s-1}(h^{s-1},\theta_{s-1}))/\partial\theta_{s-1}\over
f_{s}(\theta_{s}|\theta_{s-1},q_{s-1}(h^{s-1},\theta_{s-1}))}\Bigr{)}\Bigr{]}\cr&\times
f_{t+1}(\tilde{\theta}_{t+1}|\theta_{t},q_{t}(h^{t},\theta_{t}))d\tilde{\theta}_{t+1}.}$

If the first term is non-positive, then the seller does not sell in period $t$
and moves to period $t+1$. So, assume that the first term is positive. If the
second term is non-positive, then the seller sells in period $t$. So, assume
that the second term is positive. It is obvious that the probability that the
seller sells the object earlier increases as $\delta$ decreases. Next,
subtracting the second term from the first term, we get the inequality

$\theta_{t}-\delta
E[\theta_{t+1}|\theta_{t},q_{t}(h^{t},\theta_{t})]>{1-F_{1}(\theta_{1})\over
f_{1}(\theta_{1})}\prod_{s=2}^{t}\Bigl{(}-{\partial
F_{s}/\partial\theta_{s-1}\over
f_{s}}\Bigr{)}\Bigl{(}1-\delta\int_{\underline{\theta}}^{\overline{\theta}}-{\partial
F_{t+1}\over\partial\theta_{t}}d\tilde{\theta}_{t+1}\Bigr{)},$

from which the results follow straightforwardly. Q.E.D.

4 Discussion

We have studied the problem of when to sell an indivisible object to the set
of Markovian buyers. We have formulated the seller’s problem as a dynamic
mechanism design problem. We have provided the procedure for finding the
optimal solution and shown how to check incentive compatibility. We have also
examined sufficient conditions for the optimality of myopic stopping rule and
ex-post individual rationality. In addition, we have presented some
comparative static results regarding the seller’s revenue and the selling
time.

It is straightforward to extend the analysis to the case of many buyers. Let
$I=\\{1,\ldots,n\\}$ be the set of buyers and let
$\theta_{i,t}\in\Theta_{i}=[\underline{\theta}_{i},\overline{\theta}_{i}]$ be
player $i$’s type in period $t$. As usual, let
$\theta_{t}=(\theta_{1,t},\ldots,\theta_{n,t})$ denote the type profile in
period $t$ and let $\theta^{t}=(\theta_{1},\ldots,\theta_{t})$ and
$a^{t}=(a_{1},\ldots,a_{t})$. Let $F_{i,t}$ be the distribution of
$\theta_{i,t}$ and assume that the transition kernel is independent across
buyers so that

$F_{t}(\theta_{t}|\theta_{t-1},a_{t-1})=\prod_{i\in
I}F_{i,t}(\theta_{i,t}|\theta_{i,t-1},a_{t-1}).$

Then, essentially the same results obtain by defining
$\psi_{i,t}(\theta_{1},\cdots,\theta_{t})$ similar to $(9)$ and utilizing

$\psi_{t}(\theta_{1},\cdots,\theta_{t})=\max_{i\in
I}\psi_{i,t}(\theta_{1},\cdots,\theta_{t}).$

Observe next that it is almost trivial to incorporate the seller’s cost of
producing the object: Just introduce $c_{t}$ as the cost in period $t$ and
subtract it from the revenue.

We have analyzed the optimal selling time of an indivisible object under the
Markovian assumption. It is an interesting research agenda to extend the
analysis to (i) many heterogeneous objects, and/or (ii) non-Markovian
environments.

REFERENCES

Bergemann, D. and Pavan, A. (2015), “Introduction to symposium on dynamic
contracts and mechanism design,” Journal of Economic Theory 159(B), 679-701.

Bergemann, D. and Said, M. (2011), “Dynamic auctions,” Cochran, J., Cox, L.,
Keskinocak, P., Kharoufeh, J., and Smith, C. (Eds.), Wiley Encyclopedia of
Operations Research and Management Science, Wiley, 1511-1522.

Bergemann, D. and Välimäki, J. (2019), “Dynamic mechanism design: An
introduction,” Journal of Economic Literature 57, 235-274.

Besanko, D. (1985), “Multi-period contracts between principal and agent with
adverse selection,” Economics Letters 17, 33-37.

Board, S. and Skrzypacz, A. (2016), “Revenue management with forward-looking
buyers,” Journal of Political Economy 124, 1046-1087.

Chow, Y.S., Robbins, H. and Siegmund, D. (1971), Great Expectations: The
Theory of Optimal Stopping, Houghton Mifflin.

Furguson, T. (2006), “Optimal Stopping and Applications,” Electronic text
available at http://www.math.ucla.edu/~tom/Stopping/Contents.html.

Gallien, J. (2006), “Dynamic mechanism design for online commerce,” Operations
Research 54, 291-300.

Gershkov, A., Moldovanu, B. and Strack, P. (2018), “Revenue-maximizing
mechanisms with strategic customers and unknown, Markovian demand,” Management
Science 64, 2031-2046.

Kakade, S., Lobel, I. and Nazerzadeh, H. (2013), “Optimal dynamic mechanism
design and the virtual-pivot mechanism,” Operations Research 61, 837-854.

Lippman, S. and McCall, J. (1976), “The economics of job search: A survey,”
Economic Inquiry 14, 155-189.

McCall, J. (1965), “The economics of information and optimal stopping rules,”
Journal of Business 38, 300-317.

Mierendorff, K. (2016), “Optimal dynamic mechanism design with deadlines,”
Journal of Economic Theory 161, 190-222.

Myerson, R. (1981), “Optimal auction design,” Mathematics of Operations
Research 6, 58-73.

Pai, M. and Vohra, R. (2013), “Optimal dynamic auctions and simple index
rules,” Mathematics of Operations Research 38, 682-697.

Pavan, A. (2017), “Dynamic mechanism design: Robustness and endogenous types,”
Honoré, B, Pakes, A., Piazzesi, M., and Samuelson, L. (Eds.), Advances in
Economics and Econometrics: Eleventh World Congress, Cambridge University
Press, 1-62.

Pavan, A., Segal, I. and Toikka, J. (2014), “Dynamic mechanism design: A
Myersonian approach,” Econometrica 82, 601-653.

Shiryaev, A.N. (1978), Optimal Stopping Rules, Springer.

Stigler, G. (1961), “The economics of information,” Journal of Political
Economy 69, 213-225.

Vohra, R. (2012), “Dynamic mechanism design,” Surveys in Operations Research
and Management Science 17, 60-68.