数理解析研究所講究録
第 1557 巻 2007 年 152-161

152

An Interpretation of the “Greatest Hits”-like
Behavior
Yuhki Hosoya
Graduate School of Economics
Keio University
2-15-45 Mita Minato-ku
Tokyo 108-8345
e-mail:ukki@jcom.home.ne.jp
Abstract

We try to explain the strange behavior for suppliers of books orsoftwares which release the cheap(paperback) edition of their products
only when this product has a good sale, though its cost is nearly zero.
We construct a differential equation like the replicator dynamics ofa strategic form game, and show that if supplier’s decision is biased,
then his payoff may be better than not biased.
Keywords: cheap edition, replicator dynamics, biased decision.
JEL classification numbers: C73, D21.

1

Introduction

The purpose of this article is to explain the strange behavior for suppliersof books or softwares which release the cheap(or, paperback) edition of theirproducts only when this product has a good sale1. The cost to release thecheap(paperback) edition is usually very cheaper than to create their new
product2, so they can get new profit even if this cheap edition sells only athausand copies. Therefore, this behavior is strange and seems to $co$ ntradict
1The term “greatest hits” means the series of the cheap edition of the Sony PlayStation’s
softwares.

2So we can assume its cost is approximately .
$0$

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the profit-maximization behavior of suppliers. If they are rational, then theymust release the cheap edition of “every” products3.
This paper tries to explain its behavior and to show that this strange
behavior of suppliers does not necessarily mean their irrational choice. The
basic idea is here; if they always release the cheap edition, then most consumers wait for the release of cheap edition and tends to not buy it. So
that, to control such waiting behavior, supplier does not release the cheap
edition when his product does not have a good sale. In other words, suppliers
exchange the long-run profit for short-run profit.
To express this idea, we use a dynamic system similar to the replicator
dynamics of a strategic form game represents this situation, except the supplier’s behavior is biased so that he tends to not release the cheap edition
when many consumers wait for it. We show that if this bias is sufficiently
high, then supplier’s payoff may be improved than his bias is .
Compared to other alternative explanations(especially using the folk theorem), this explanation has some virtues. First, this explanation need not
require the clever action of consumers. Second, supplier’s decision process
to realize the long-run equilibrium emerges specifically, and matches actual
behavior. So this explanation is better than the others.
In section 2, we display our model and derive the result of this model.
Section 3 is the concluding remarks.
$0$2Model and result

We constracts a strategic form game below;

and a dynamic system;$\dot{x}/x=(1-x)[(u-p_{H})-y(d-p_{L})])$$\dot{y}/y=(1-y)[(1-x)p_{L}+D]-b(1-x)$ .

3Moreover, we can argue that the motivation to release the cheap edition of a productwhich has a good sale is smaller than which does not have a good sale, because in the lattercase this product has missed relatively large potential demand. Although this argument
seems to have some persuasive power, it completely contradicts the actual behavior of
suppliers.

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If $b=0$ , then this system is equal to the replicator dynamics of above game.
We assume $p_{H}>p_{L}>0,$ $d-p_{L}>u-p_{H}>0,$ $D>0$ .
The interpretation of these models is here: player 1 represents a consumer
and player 2 represents a supplier. Player 1 considers whether he buys aproduct immediately. If he gets it immediately, then he gains . If not, hewaits for the release of the cheap edition of this product. If player 2 releases
the cheap edition, player 1 can get it and gain . If he does not release it,then player 1 waits until this product becomes old-fashioned and the worthof it tends to .
denotes the original retail price of this product, anddenotes the price of cheap edition. So we as sume $p_{H}>p_{L}>0$ .
If $u-p_{H}\geq d-p_{L}$ , then consumer should buy immediately. Such consumer
must not relate the determination of release of cheap edition, so we assume$d-p_{L}>u-p_{H}$ . Moreover, if $u-p_{H}\leq 0$ , then he should wait for cheap edition.
Such consumer’s behavior is so simple that we treat it only implicitly andassume if player 2 does not release the cheap edition, then he automatically$loses-D<0$ , the sales of such consumer. Therefore, we treat only the case$d-p_{L}>u-p_{H}>0$ .
The replicator dynamics of this game represents the short-run behavior ofplayers. At first, both players blindly believe their current action is the best.
As time goes by, however, they may observe their action is not the best andchange it. Changing action occurs in proportion to the difference between
the payoff of current action and of social average. So both the growth rateof the proportion of the people buying immediately(denoted by x) and of
the probability of releasing cheap edition(denoted by y) are equal to above
difference4, i.e.,
$u$

$d$

$0$

$\dot{x}/x$$p_{L}$$=u-p_{H}-[x(u-p_{H})-(1-x)y(d-p_{L})]$$=$$\dot{y}/y$$p_{H}$$(1-x)[(u-p_{H})-y(d-p_{L})]$ ,$=xp_{H}+(1-x)p_{L}-[y(xp_{H}+(1-x)p_{L})+(1-y)(xp_{H}-D)]$$=$$(1-y)[(1-x)p_{L}+D]$ .

Finally, the term $-b(1-x)$ represents supplier’s decision is biased in thesense that if is large and his product does not have a good sale, then hedoes not release the cheap edition.
Then we get a result below;$x$

proposition 1 (1)If $b \leq\frac{(p_{L}+D)[(d-p_{L})-(u-p_{H})]}{d-p_{L}}$ then any solution
4We interpret as the proportion of the consumer choosing

and as the probability
interpretation of mixed
model,
in
the
cheap
So
our
edition.
supplier
releasetochoosesthatstrategy of each player is different.
$x$

$B$

$y$

155
$(x, y)(t)$ of this dynamic system passing through the interior of $[0,1]\cross[0,1]$

converges to $(x^{*}, y^{*})=(0, \frac{p_{L}-b+D}{p_{L}+D})$ .
(2) $Ifb>\frac{(p_{L}+D)[(d-p_{L})-(u-p_{H})]}{d-p_{L}}$ then this dynamic system has an

asymptotically stable equilibrium point such that,$(x^{*}, y^{*})=( \ovalbox{\tt\small REJECT}\frac{u-p_{H}}{d-p_{L}})$.

The implication of this proposition is here. (1) means if supplier’s choice
is not biased(that is, $b=0$ )) then he tends always to release the cheap
edition and consumers tend always to wait for release of it. We call thisresult the short-run equilibrium5. In contrast, (2) means if supplier’s choice
is sufficiently biased, then it is possible that a certain number of consumers
buy immediately. Moreover, this equilibrium point converges to $(1,$as $barrow+\infty$ , and the payoff of supplier converges to$\frac{u-p_{H}}{d-p_{L}})$$p_{H}- \frac{u-p_{H}}{d-p_{L}}D$. If

is sufficiently small, its payoff is larger than , the payoff of short-run
equilibrium. So we can conclude that if $D$ is sufficiently small, supplier’s
biased decision may improve long-run payoff.
$D$$p_{L}$

proof Suppose $(x, y)(t)$ is the solution and $(x, y)(t^{*})$ is in the interior of$[0,1]\cross[0,1]$ . At first, we confirm $(x, y)(t)$ is extendable on [[ and$t^{*},$$+\infty$

for every.
Clearly, $(x, y)(t)$ is the solution of the dynamic system below;$(x, y)(t)\in[0,1]\cross[0,1]$$\dot{x}$$\dot{y}$$t\geq t^{*}$$=$$x(1-x)[(u-p_{H})-y(d-p_{L})]$ ,$=$$y[(1-y)[(1-x)p_{L}+D]-b(1-x)]$ ,where the right-hand side is defined on. As is well known, if $(x, y)(t)$is not extendable on [[, there has to besuch that $(x, y)(t)\not\in$$[0,1]\cross[0,1]$ . Therefore, it suffices to show that $(x, y)(t)\in[0,1]\cross[0,1]$ forall.
Suppose $x(t)\leq 0$ for some $t>t^{*}$ . Then we have $x(t’)=0$ for some$t’\in[t^{*}, t]$ and thus $x\equiv 0$ , implying $x(t^{*})=0$ , a contradiction. So we have$x(t)>0$ for all $t>t^{*}$ . Similarly, we get $x(t)<1$ and $y(t)>0$ for all $t>t^{*}$ .
Next, suppose $y(t)>1$ for some $t>t^{*}$ . Let is the maximal element of the$\mathbb{R}^{2}$$t^{*},$$t\geq t^{*}$$+\infty$$t\geq t^{*}$$\overline{t}$5In fact, the point $(0,1)$ represents the only Nash equilibrium of this game.

156
nonempty set $\{t’\in[t^{*})t]|y(t)=1\}$ . If $b>0$ , we have ] $=-b(1-x(t])<0$ ,so that, there exists] [ such that $y(\hat{t})<1$ . Hence we have there exists[ such that $y(t’)=1$ , contradicting the definition of . If $b=0$ , wehave
and thus $y(t)=1$ , a contradiction. So we must have $(x, y)(t)$ is.
extendable on$\dot{y}(t$$\hat{t}\in$$t$$\overline{t}$)$t’\in]\hat{t},$$t$$\overline{t}$$y\equiv 1$$[t^{*},$$+\infty[$(1) We separates this situation into three cases; I) $b=0$ , II) $0<b<$$\frac{(p_{L}+D)[(d-p_{L})-(u-p_{H})]}{d-p_{L}}$1II) $b= \frac{(p_{L}+D)[(d-p_{L})-(u-p_{H})]}{d-p_{L}}$ .

Figure 1: $b=0$ .

Case I: Assume $b=0$ . Then, we have $y(t)$ increases whenever $y(t)\in$ ] $0,1[$ ,and thus $y(t)$ converges to $y^{*}= \sup y([t^{*}, +\infty[)\in[0,1]\cross[0,1]$ . If $y^{*}<1$ ,$\dot{y}(t)=y(t)(1-y(t))[(1-x(t))p_{L}+D]\geq y(t^{*})(1-y^{*})D>0$ ,, so that, $y(t)arrow+\infty$ as $tarrow\infty$ , a contradiction. So $y^{*}=1$ .
such that $0<x(t5<1$ and
Therefore, we must have there exists$y(t$]and thus $x(t)$ converges toHence we have $\dot{x}(t)<0$ for all

for all$t\geq t^{*}$$\overline{t}\geq t^{*}$$> \frac{u}{d}-pg-p_{L}$$t\geq\overline{t}$157$x^{*}= \inf x([\overline{t}, +\infty[)$$\dot{x}(t)$. If $x^{*}>0$ ,$x(t)(1-x(t))[(u-p_{H})-y(t)(d-p_{L})]$$=$$x^{*}(1-x(t\gamma)[u-p_{H}-y(t](d-p_{L})]<0$ ,$\leq$, so that, $x(t)arrow-\infty$ as $tarrow\infty$ , a contradiction. So $x^{*}=0$ .
for all
This completes the proof of case I.
$t\geq\overline{t}$Figure 2: $0<b< \frac{(p_{L}+D)((d-p_{L})-(u-p_{H}))}{d-p_{L}}$Case II: Assume $0<b< \frac{(p_{L}+D)[(d-p_{L})-(u-p_{H})]}{d-p_{L}}$See figure 2. We separates $[0,1]\cross[0,1]$ into three areas. Area A isline, and area is separated fromseparated from area by theby thecurve. We can easily check that
$B$

$C$

area.

$B$

$\dot{x}=0$$\dot{y}=0$$\dot{x}=0\Leftrightarrow y=\frac{u-p_{H}}{d-p_{L}}$,$\dot{y}=0\Leftrightarrow y=\frac{(1-x)(p_{L}-b)+D}{(1-x)p_{L}+D}\equiv g(x)$.

158
By easy computation, we have$g’(x)= \frac{bD}{((1-x)p_{L}+D)^{2}}>0$,$g”(x)= \frac{2bDp_{L}}{((1-x)p_{L}+D)^{3}}>0$,

whenever $x<1$ . Note that $g( O)=\frac{p_{L}-b+D}{p_{L}+D}$ .
Now, we will show that every solution passing through area A must enterarea B. In fact,$\dot{y}(t)$$\geq$$y(t^{*})[ \frac{(d-p_{L})-(u-p_{H})}{(d-p_{L})}[(1-x(t))p_{L}+D]-b(1-x(t))]$$>$$y(t^{*})x(t)D \frac{(d-p_{L})-(u-p_{H})}{d-p_{L}}$$\geq$$y(t^{*})x(t^{*})D \frac{(d-p_{L})-(u-p_{H})}{d-p_{L}}>0$,

for some
So we must nave.
Next, we will show that every solution passing through area must notreenter area A. Suppose $(x, y)(t^{*})$ is in area and $(x,y)(t)$ is in area A forsome $t>t^{*}$ . Then $y(t^{*})> \frac{u-p_{H}}{d-p_{L}}>y(t)$ . Define $\overline{t}=\sup\{t’\in[t^{*}, t]|y(t’)=$whenever$y(t) \leq\frac{u-p_{H}}{d-p_{L}}$$t\geq t^{*}$$y(t)> \frac{u-p_{H}}{d-p_{L}}$

$B$

$B$

$\frac{u-p_{H}}{d-p_{L}}\}$. Then$y(t”)= \frac{u-p_{H}}{d-p_{L}}$$\dot{y}(\overline{t}>0$, implying

for some $t”\in$ ]$\overline{t},$$t$$y(t’)> \frac{u-p_{H}}{d-p_{L}}$for some$t’\in[\overline{t},t]$

and thus], a contradiction.

Therefore, we have there exists such that $\dot{x}(t)<0$ for all
and thus$x(t)$ converges to.
$(x,y)(t)$
Supposebelongs to neither area nor the curve $\{(x, g(x))|x\in$$[0,1]\}$ for all. Then $\dot{y}(t)>0$ for all
and thus $y(t)$ converges to. Suppose $x^{*}>0$ . Then,$t\geq\overline{t}$$\overline{t}$$x^{*}= \inf x([\overline{t}, +\infty[)$$C$$t\geq\overline{t}$$t\geq\overline{t}$$y^{*}= \sup y([\overline{t}, +\infty[)$$\dot{x}(t)\leq x^{*}(1-x(t\gamma)[(u-p_{H})-y(t=)(d-p_{L})]<0$,. Hence we must have $x(t)arrow-\infty$ .as $tarrow\infty$ , a contradiction.
for all, we
So we have $x^{*}=0$ . Since $(x, y)(t)$ never belongs to area for all
must have $y^{*}\leq g(O)$ . Suppose $y^{*}<g(O)$ . Define$t\geq\overline{t}$$C$$t\geq\overline{t}$$s^{*}= \inf\{(1-y)((1-x)p_{L}+D)-b(1-x)|x\in[0,1],$ $y \in[\frac{u-p_{H}}{d-p_{L}},$ $y^{*]}\}$ .

159

Figure 3:$b= \frac{(p_{L}+D)((d-p_{L})-(u-p_{H}))}{d-p_{L}}$.

Since $y^{*}<g(O)$ and $g’(x)>0$ for all $x<1$ , we must have $s^{*}>0$ andthus] $s^{*}>0$ for all, implying $y(t)arrow+\infty$ as $tarrow\infty$ , acontradiction.
Lastly, we treat the case there exists such that $(x, y)(t)$ is in area C.
We will show that $(x, y)(t)$ stays at area whenever. Suppose on the$\dot{y}(t)\geq y(t$$t\geq\overline{t}$$\hat{t}$$t\geq\hat{t}$$C$

contrary $(x, y)(t)$ is in area for some
. Let be the greatest elementof $\{t’\in[\hat{t}, t]|\dot{y}(t’)=0\}$ . Then we must haveand thus there exists$t’\in]tt]\sim$, such that $\dot{y}(t’)<0$ . Since $\dot{y}(t)>0$ , there exists が’ ] ] such that, contradicting the definition of .
So we have $(x, y)(t)$ is in area for all. Therefore, we must have$\dot{y}(t)<0$$\dot{x}(t)<0$ and, and thus $(x, y)(t)arrow(x^{*}, y^{*})$ , where
whenever$x^{*}= \inf x([\hat{t}, +\infty[)$ and $y^{*}= \inf y([\hat{t}, +\infty[)$. We can easily check that $x^{*}=0$
and $y^{*}\geq g(O)$ . Suppose $y^{*}>g(O)$ . Then there exists
such that$B$$t\geq\hat{t}$$t\sim$$\dot{x}(t)\sim<0$$\sim$$\in$$\dot{y}(t^{u})=0$

$t,$

$t$

$t\sim$$C$$t\geq\hat{t}$$t\geq\hat{t}$$t\sim\geq\hat{t}$160$y^{*}\geq x(t)\sim$. Define$s^{*}= \sup\{(1-y)((1-x)p_{L}+D)-b(1-x)|x\in[0, x(t)], y\in\sim[y^{*}, 1]\}$ .

Then $s^{*}<0$ and thus $\dot{y}(t)\leq y^{*}s^{*}<0$ for all, a contradiction.

, implying $y(t)arrow-\infty$ as$t\geq t\sim$$tarrow\infty$

Case III: Assume $b= \frac{(p_{L}+D)[(d-p_{L})-(u-p_{H})]}{d-p_{L}}$ .
line
curve closses the
The difference from case II is where theat $x=0$ . By the similar argument on case II, we can show that there exists
such that $(x, y)(t)$ is in area C. The rest of the proof is the same as
the last part of the proof of case II.
$\dot{x}=0$$\dot{y}=0$$\hat{t}>t^{*}$(2) Since$(x^{*})y^{*})$

is a solution of$g(x)= \frac{u-p_{H}}{d-p_{L}}$, we must have$\dot{x}=\dot{y}=0$at

is asymptotically stable. To dothis, it suffices to show that for every eigenvalue of the Jacobian matrix, its$(x^{*}, y^{*})$. So it suffices to show that$(x^{*}, y^{*})$$\lambda$

161
real part is negative. Since $\dot{x}=0$ andatis$\dot{y}=0$at$(x^{*}, y^{*})$, its Jacobian matrix$(x^{*}, y^{*})$$(y^{*}[b-(1-y^{*})p_{L}]0$$-y^{*}((1-x^{*})p_{L}+D)-x^{*}(1-x^{*})(d-p_{L}))$and thus, its characteristic equation is$\lambda^{2}+y^{*}((1-x^{*})p_{L}+D)\lambda+x^{*}(1-x^{*})y^{*}(d-p_{L})[b-(1-y^{*})p_{L}]=0$Let$\lambda_{1},$$\lambda_{2}$be the solution of this equation. Then,$\lambda_{1}+\lambda_{2}=-y^{*}((1-x^{*})p_{L}+D)<0$,$\lambda_{1}\lambda_{2}=x^{*}(1-x^{*})y^{*}(d-p_{L})[b-(1-y^{*})p_{L}]$andis not a real number, thenmust be less than . So we can suppose
loss of generality. Then,

Note that, if
real part of$\overline{\lambda}_{1}=\lambda_{2}$$\lambda_{1}$$\lambda_{1},$$\lambda_{2}$

$0$

.

3

$\lambda_{1},$$\lambda_{2}<0$. It completes the proof.

, so thewithout$\lambda_{1}+\lambda_{2}<0$$\lambda_{2}\in \mathbb{R}$$\lambda_{1},$$b-(1-y^{*})p_{L}> \frac{D[(d-p_{L})-(u-p_{H})]}{d-p_{L}}>0$

implying.

,$\blacksquare$

Concluding remarks

We try to explain the strange behavior for suppliers of books or softwares
which release the cheap edition of their products only when this product has
a good sale, and show that if supplier’s short-run decision is biased, then his
long-run payoff may improve. So suppliers need not be irrational, and his
strange behavior may improve his payoff.
To conclude above, we used a dynamic system similar to the replicator
dynamics of a strategic form game represents this situation, except the supplier’s behavior is biased. Compared to alternative explanations(especially,using the folk theorem), this explanation has some virtues. First, this explanation need not require the clever action of consumers. Second, supplier’s
decision process to realize the long-run equilibrium emerges specifically, andmatches actual behavior. So this explanation is better than the others.

References[1] Hofbauer, J. and Sigmund, K. (1998) Evolutionary Games and Populational Dynamics, Cambridge University Press.

[2] Weibull, J. (1995) Evolutionary Game Theory, MIT Press.