The Silk Road:
Tax competition among nation states*

Kaz Miyagiwa
Osaka University and Emory University
Abstract
All international trade involves the shipment of commodities from one nation to another.
Many commodities, before reaching their final destinations, are transshipped throughseveral nations, each having independent authorities to tax commodities in transit. When
trade is repeated over time, such “middle” nations may lose monopoly power overcommodities in transit.
JEL Classification Codes: D4, F1, H2. L13
Keywords: international tax competition, international trade, economic geography
Correspondence: Kaz Miyagiwa, Institute for Social and Economic Research, Osaka
University, 6-1 Mihogaoka, Ibaraki, Osaka 567-0047, JAPAN. e-mail:kmiyagiwa@email.com*

I thank Yuka Ohno, and seminar participants at Kobe University, for their helpful comments andsuggestions on an early version of this paper. Errors are my responsiblity.

1. Introduction
The ancient civilizations of China and the Mediterranean were connected by thelegendary Silk Road, the nearly 5, 000-mile-long trade route that meandered throughmany areas of influence, in which the independent nation states exacted tolls fromtraveling merchants. Analyzing a game-theoretic model of such trade routes, Karni and
Chakrabarti (1997) have found that the nation states collectively suffer from the double
marginalization problem so that unification of them into one empire, such as achieved
under the Mongols, leads to more trade and tax revenues. Gardner, Gaston and Masson(2002) have reached similar conclusions in their theoretical analysis of cargo shipments
down the medieval Rhine when local barons constructed castles along the river to exact
tolls for the right of passage.
As these examples vividly illustrate, in market economies trade between
producers and final users of the products involve successive layers of intermediaries.
Manufactures and farm produce are sold to wholesalers and then to retailers before
reaching their final consumers. Oil and natural gas can reach overseas consumers throughpipelines across foreign countries, which demand user fees. Other examples include
telephone calls involving local and long-distance or international telephone service
providers, and vacation trips consisting of hub-to-hub flights by one carrier and local
flights by another. In each of these cases, the double marginalization problem can arise
when more than one intermediary have monopoly power over the commodities in transit.
As is well known, however, the problem can be mitigated if two or more in a chain ofmonopolies are integrated to coordinate prices. Furthermore, vertical integration is known
to be socially desirable as the price is lowered and output increased.

2
As Feinberg and Kamien (2001) have argued, however, the double
marginalization problem arises from the implicit assumption that prices and tolls are precommitted to before transactions take place. This indeed is the assumption made in theworks of Karni and Chakrabarti (1997) and Gardner et al. (2002). If trade instead occurs
sequentially, the lack of commitment leads to the disappearance of the double
marginalization problem, but as shown by Feinberg and Kamien (2001) the economy
instead suffers from the holdup problem. To demonstrate this result, suppose that amerchant must deliver his ware to the market over the road that goes through two
domains. Assuming that the merchandise loses its value unless delivered to the market,
suppose that the baron of the second domain, when the merchant has arrived there,
demands the entire market value of the merchandise in exchange for the right of passage
through his domain. Then, the merchant is indifferent between acceding to this demandand discontinuing the journey, for either way he will lose the value of his merchandise.
Therefore the baron of the second domain can capture (almost) the entire value of themerchandise with a slightly lower tax to induce the merchant to complete his journey.
However, a forward-looking merchant anticipates the taxing strategy of thesecond baron. Since he loses all his value once having reached the second domain, themerchant will not embark on a journey unless he can cross the first domain tax-free. The
conclusion: when trade occurs the baron of the second domain captures the entire valueof the merchandise. Since only the second baron exercises monopoly power, there is no
double marginalization. Furthermore, social welfare is greater than when the two baronscommit to the taxes before the merchant embarks on a journey.

3
In this paper I argue that the hold-up problem in the Feinberg-Kamien analysis isa consequence of the assumption that the game is played only once. Although one canthink of real-world examples in which one-shot games are appropriate, many transactions
are rarely one-shot affairs. Therefore, in this paper I analyze a model of infinitely
repeated transactions. I show that trade thrives despite trade costs, and that theequilibrium price is lower than when the barons contract on prices in a one-shot game,but higher than when all the domains are integrated into a single monopoly. The intuition
underlying these results is as follows. In a repeated-game setting, the baron of the domain
which merchants last cross can always act myopically, exacting the value of thecommodity as in the one-shot game. However, if it is assumed that no merchant will
embark on a journey in the future once he suffers financial losses in the past, then the lastbaron may want to exact a lower tax to guarantee the continuation of trade. If this longrun strategy yields greater revenues than the one-time gain from acting myopically, thentrade can continue occur. The objective of this paper is to investigate the nature of suchan equilibrium outcome.
When there are more than two monopolies in a succession, we show that theentire value of the merchandise is shared between the first and the last baron while thebarons of all the “middle” domains fail to capture any rent despite their monopoly
positions over commodities in transit. The following is the reason. What makes the baronof the last domain act myopically or patiently is the sum of taxes the merchants paid
before getting there. If this sum is high, the last baron has to lower his tax for thecontinuation of trade in the future, which makes it more attractive to act myopically. I
show that there is a threshold level of sum of taxes collected by all the barons before the4last. If the actual sum of taxes the merchant paid exceeds this threshold the last baron acts
myopically, thereby ending future trade. Thus, it is in the best interest of all the otherbarons to keep this sum below the threshold to ensure future trade. But then the baron of
the first domain exercises a first-mover advantage, setting his tax equal to that threshold,leaving no room for maneuver for the barons of the middle domains. Thus, the first andthe last monopolists exact the entire value of the merchandise.
The remainder of the paper is organized in four sections. The next section revisits
the Feinberg-Kamien model. Section 3 extends the model to cases of repeated interplays.
Section 4 extends the model further to a case of multiple road segments. Section 5
concludes.

2. The one-shot game
Suppose that merchants produce a homogeneous product in one country and sell itin another country. Merchants must deliver the product to the market. Choose units sothat each merchant delivers one unit of merchandise to the market. Let x denote thequantity of the product delivered to the market. Treat x as a real number to ease
exposition. Market (inverse) demand is given by p(x), a differentiable function withrespect to x > 0, with first and second derivatives denoted by p’(x) < 0 and p”(x) ≤ 0.
To reach the market merchants must go through two nation states and pay taxes
there. For example, the first nation may be their home country, which imposes an export
tax while the second nation may be where the market is, imposing an import tax.
Consider the following four-stage game.
Stage 1: The first state announces a tax rate t1 ≥ 0.

5Stage 2: Observing t1, all the merchants simultaneously decide whether to cross thefirst state or not.
Stage 3: After the merchants have crossed the first state, the second posts a tax rate t2≥ 0.
Stage 4: Observing t2, the merchants decide whether to cross the second state.
Nation state i incurs costs ci (i = 1, 2) per merchant crossing its territory (ci ≥ 0). Eachstate’s payoff is the net tax revenue it collects from the merchants (the difference between
the total taxes collected and the total cost incurred). A merchant’s payoff is the difference
between the market price of the commodity (zero in case of non-delivery) and the sum oftaxes he pays to get to the market. There are no other markets, and all the taxes are sunkafter they are paid. We normalize each merchant’s default payoff to zero, and adopt the
tie-breaking rule that a merchant sets out on a journey as long as he expects a nonnegative payoff. Finally, we assume that p(0) > c1 + c2, so delivery of the commodity to
the market is socially desirable.
We solve the game backwards. Let x1 be the number of merchants who has paid
t1 and crossed the first state. Now they observe t2, and each merchant reasons as follows.
If he crosses the second state and expects x2 (≤ x1) other merchants to do the same, his
payoff will bep(x2) – t2 – t1.
On the other hand, if he does not, his payoff will be – t1, since the first tax payment issunk. Therefore, a merchant goes forward if and only ifp(x2) – t2 – t1 ≥ – t1

6
or
(1)

p(x2) – t2 ≥ 0.
Moving backward to the third stage of the game, state 2 chooses t2 to maximize

the net tax revenue (t2 – c2)x2 subject to the constraint (1) and x2 ≤ x1. Ignoring thesecond inequality for the moment, maximization of the Lagrangian(t2 – c2)x2 + λ[p(x2) – t2]
yields the optimality conditions
t2 – c2 + λp’(x2) = 0; and x2 – λ = 0,
where λ is the Lagrangian multiplier. The two equations above combine to yield(2)p(x2) + p’(x2)x2 – c2 = 0.

The first two terms constitute marginal revenue. Thus, (2) is the standard first-ordercondition for a monopolist facing demand p(x) and constant marginal cost c2. Given themassumption on p(x), (2) has a unique solution, denoted by x2 . The (maximum) monopoly
rent to state 2 equalsmmm

π2 ≡ [p(x2 ) – c2]x2 .
Substituting into the constraint equation from (1), we obtain state 2’s optimal tax rate:mm

t2 = p(x2 ).
m

Now return to the constraint x2 ≤ x1 that we have so far ignored. Suppose that x2mm< x1. Then, if all x1 merchants paid the tax t2 = p(x2 ) and crossed the second state, thempayoff to each merchant will be p(x1) – p(x2 ) – t1 which is less than – t1. Therefore, not7all merchants will cross the second state to reach the market. In equilibrium, themmerchants have rational expectations so that exactly x2 merchants cross the second statemwhile x1 – x2 do not. However, each merchant has the same payoff of – t1.
m

Suppose alternatively that x1 ≤ x2 . Then, the left-hand side of (2) is strictlympositive at x1 < x2 . Therefore, the optimal tax is t2 = p(x1), implying all the merchants x1
cross the second segment, but again they all earn – t1. To sum, state 2’s optimal strategymm

m

is t2 = p(x1) if x1 ≤ x2 and t2 = p(x2 ) if x1 > x2 .
We now move back to the second stage of the game. Given state 2’s optimal
strategy, the merchants, once having crossed the first state, expect to earn the negative
profit, – t1, whether they would cross the second or not. Therefore, the merchants will
embark on journeys only if t1 ≤ 0. But state 1 is willing to charge t1 = 0 only if c1 = 0.
Alternatively, suppose that that c1 = 0 but that each merchant incurs some trade
cost c > 0, which can also be his opportunity cost. Since state 2 exacts the entire sales
revenue from the merchants that have crossed the first state, the merchants cannot recover
their trade costs. Thus, the merchants never travel. Thus, the presence of trade cost borneby the merchants or by the first nation state, no matter how small, results in the absenceof trade in the one-shot game, even though trade is socially desirable. Thus,

Proposition 1: In the one-shot game with two nation states, trade takes place if and only
if the first nation state incurs no trade cost (c1 = 0) and the merchants’ opportunity costs

8
are zero. When trade occurs, the second state captures the entire value of themerchandise.

3. Repeated interactions
In this section we reconsider the above problem in a repeated-game setting. There
are an infinite number of periods. Each merchant makes one delivery per period. Whileno trade is still a possible equilibrium outcome under the conditions of the previous
section, there is an alternative equilibrium outcome in which trade occurs. In this section
we examine the properties of such an equilibrium.
Assume that each merchant adopts the following strategy. In the first period hedecides whether to embark on a journey. In any subsequent period, he sets out on a newjourney if and only if he has never suffered a loss in the past. The merchants who stayedhome in the first period never travel in later periods.
I next specify the equilibrium strategies of two nation states that result in trade. I
look for a stationary subgame-perfect equilibrium outcome with the following
characteristics. In each period, the constant number x* of merchants embark on newjourneys and the states demand the taxes at constant rates, t1* and t2*, respectively. I
massume that x* ≤ x2 , and justify this assumption shortly.
I first show that the second state’s equilibrium tax is given by t2 = p(x*) – t1. To
see this, suppose that the equilibrium tax satisfies t2 < p(x*) – t1. Then, the merchants
would earn positive profits, so the second nation can raise the tax to increase its tax
revenue. If p(x*) > t2 > p(x*) – t1 all the x* merchants will cross the second state but all

9will suffer losses, p(x*) – t1 – t2 < 0, ending all future trade. Finally, if t2 > p(x*), then as
I explained in the previous section, only a fraction, say, x2, of the x* merchants will crossthe second state to deliver the merchandise, where x2 is given by t2 = p(x), while theremainder, i.e., x* - x2, will end their journeys prematurely. In this case, however, all thex* merchants suffer the negative profit – t1 and will never come back in the future. Thus,
t2 = p(x*) – t1 is the only tax consistent with the equilibrium outcome in which there istrade in every period.
I have shown that the equilibrium net revenue per period to the second nation
state is [p(x*) – t1 - c2]x*. Adding up over periods leads to the discounted sum of profits:
v2 = [p(x*) – t1 – c2]x*/(1 - δ)
where δ ∈(0, 1) is the (common) discount factor. Now, consider a deviation by state 2.
Deviating from the equilibrium, state would set the tax rate equal to p(x*) to earn theentire value, [p(x*) – c2]x*, of the merchandise for one period and no future revenues.
Thus the second state has no incentive to deviate if
v2 = [p(x*) – t1 – c2]x*/(1 – δ) ≥ [p(x*) – c2]x*,
which simplifies to:t1 ≤ δ[p(x*) – c2].
The above condition then yields state 2’s best responses to t1:(3)

t2 = p(x*) – t1

if t1 ≤ δ[p(x*) – c2]

t2 = p(x*)

if t1 ≥ δ[p(x*) – c2]

10
I next show that the first nation’s optimal tax is(4)

t1 = δ[p(x*) – c2].

To prove this, suppose contrarily the first state posts a lower tax, t1 < δ[p(x*) – c2]. The
merchants observe this and expect state 2 to set the tax equal to t2 = p(x*) – t1, once they
have gotten there. If a small group of additional merchants (of size ε > 0) decide also to
cross the first state, state 2 will adjust the tax rate to t2 = p(x* + ε) – t1 for ε small enoughto satisfy the inequality: t1 ≤ δ[p(x* + ε) – c2]. Then, all x* + ε merchants journey to themarket. This shows that x* is not the equilibrium number of merchants. On the other
hand, if the first state sets a higher tax, t1 > δ[p(x*) – c2], state 2 will act myopically,
thereby disturbing the equilibrium. Thus, in the equilibrium (4) must hold. (4) thus
defines a mapping from t1 to x*.
Now, state 1 sets t1 to maximize the net tax revenue per period, (t1 – c1)x*, where
x* is determined by t1 via (4). It is more convenient to invert the above mapping andrestate state 1’s problem as:
Maxx* {δ[p(x*) – c2] – c1}x*.
The first-order condition(5)

δ[p(x*) – c2] – c1 + δx*p’(x*) = 0

determines the equilibrium number of traveling merchants x*. Finally, when evaluated atmm

x2 (see Eq. 2), the left-hand side of (5) equals – c1 < 0, implying x* < x2 . This justifiesmmy focus on the case in which x* ≤ x2 .

11
Now that the equilibrium value of x* is found, the first state’s equilibrium tax rate
is given by(6)

t1* = δ[p(x*) – c2].

(5) and (6) lead tot1* – c1 = δ[p(x*) – c2] – c1 = – δx*p’(x*) > 0,
indicating that state 1 earns a strictly positive profit. The optimal tax for state 2, t2*,obtains from substituting for t1* from (6) into (3). I summarize the findings of thissection in the next proposition.

Proposition 2: In the stationary equilibrium with two nation states there is an equilibriumoutcome with trade with the following features.
m(A) The number of merchants who journey is x* given by (5), and x* < x2 .
(B) The optimal tax rates of the two nation states are
t1* = δ[p(x*) – c2],t2* = (1 - δ)p(x*) + δc2,(C) Both states 1 and 2 earn positive net tax revenues.

The intuition for Result 2.C is as follows. If the tax rate the merchants paid tostate 1 is not too high, state 2 can collect sufficiently large tax revenues. If the sum ofsuch revenues over a long haul exceeds the payoff from acting myopically, state 2 will
prefer that trade continues; otherwise it will act myopically. Then, state 1 can capitalize

12
on its first-mover advantage to raise its tax high enough to make state 2 indifferent
between those two options, thereby collecting some positive net tax revenues.
In sum, in contrast to the case in which all agents interact just once, repeatedinterplays result in trade despite trade costs, and yield strictly positive payoffs for bothnation states. Suppose that there are only x0 potential merchants (x0 < x*). In this case,the proposition still holds, as can easily be confirmed, with x0 replacing x* in Results 2.A
and 2.B.

4. Many nations states
This section extends the above analysis to the case in which the trade route goes
through multiple (more than two) nation states. A case of three nation states is sufficient
to capture the essential features of such an extension. Now merchants go through nation
states 1, 2 and 3 in that order.
Look again for a set of stationary equilibrium strategies that induce the same
number of merchants to embark on journeys every period. Then, a procedure similar tothe one employed in the previous section establishes the following best responses forstate 3:(7)

t3 = p(x*) – t1 – t2

if t1 + t2 ≤ δ[px*) – c3]

t3 = p(x*)

if t1 + t2 > δ[p(x*) – c3],

where x* again denotes the (yet undetermined) number of merchants who travel theentire trade route in the equilibrium. The conditions in (7) indicate that the sum of thefirst two taxes holds the key to whether state 3 acts myopically or not.

13
I next show that the equilibrium tax rate for state 2 ist2* = δ[p(x*) – c3] – t1.

(8)

To prove this, observe first that state 2 will never set a tax lower than t2*, because raising
the tax rate up to δ[p(x*) – c3] – t1 will not trigger state 3 to act myopically, and hence
can increase its tax revenue. Consider next a higher tax rate, t2* > δ[p(x*) – c3] – t1.
Then, the merchants, having crossed the first state, would infer from the third state’s best
responses in (7) that, once having traversed the second state, the third state will act
myopically. The merchant’s net income then would be – (t1 + t2), if he crosses the second
state, whereas he can earn the income – t1 by choosing not to cross it. Thus, no merchants
would cross the second state. This proves my claim. Additionally, to be optimal, t2* must
give state 2 a non-negative payoff; i.e., we need this additional condition:
δ[p(x*) – c3] – t1 ≥ c2.

(9)Now, turning to state 1, I show that, if all the x* merchants pay t1 and embark onjourneys, x* must satisfy (9) with strict equality: i.e.,t1 = δ[p(x*) – c3] – c2.

(10)

If t1 is strictly less than the right-hand side of (10), more merchants are willing to travel,
disturbing the equilibrium number x* of merchants. Eq. (10) thus defines a mapping fromt1 to x*.
State 1’s problem is now stated: choose t1 to maximize the net income (t1 – c1)x*
subject to (10). Using (10), the first-order condition is written:

14
(11)

δ[p(x*) + x*p’(x*) – c3] – c2 – c1 = 0,

which defines x*. Now, define bym

x3 ≡ argmax [p(x) – c3]x,
the optimal output for the third state when it acts myopically. Then the left-hand side ofmm(11) is negative at x3 , implying x* < x3 .
By (10) state 1’s optimal tax rate is given by(12)

t1* = δ[p(x*) – c3] – c2.

(11) and (12) imply
t1* – c1 = δ[p(x*) – c3] – c2 – c1 = - δx*p’(x*) > 0
so state 1’s payoff is strictly positive. Substituting from (12) into (8) shows, however,that t2* = c2. Thus, net tax revenue is zero for state 2. Further substitution shows apositive payoff for state 3 as given in the next proposition.

Proposition 3: The model with three nation states has a stationary equilibrium with thefollowing properties.
(A) The equilibrium number x* of merchants set out on journeys every period isdetermined by (11),(B) The optimal tax rates for state i = 1, 2, 3 are
t1* = δ[p(x*) – c3] – c2,
t2* = c2t3* = (1 - δ)p(x*) + δc3,15(C) Only states 1 and 3 earn strictly positive rents every period.

What is most surprising about this proposition is the fact that nation state 2 breaks
even despite its monopoly position over the trade route. This result has the follow
explanation. State 3 chooses the myopic action over the long-term equilibrium action
unless the sum of two taxes the merchants pay before getting there are sufficiently low.
Then, state 1 can take advantage of its first-mover position to set the tax just high enoughto make state 3 indifferent between the myopic and the long-term strategy. Therefore, if
state 2 sets the tax rate higher than its cost to earn positive tax revenues, no merchants
will travel.
The next proposition generalizes Proposition 3 (the proof is similar and omitted).

Proposition 4: The model of N (> 2) segmented roads has a stationary equilibrium, inwhich the optimal tax rates are
t1* = δ[p(x*) – cN] – ∑ k = 2 ck ,N −1

tk* = ck;k = 2, …., N – 1,

tN* = (1 - δ)p(x*) + δcN,
where x*, the equilibrium number of merchants, is the solution toδ[p(x) + xp’(x) – cN] – ∑ k =1 ck = 0.
N −1

5. Concluding remarks

16
All international trade involves the shipment of commodities from one nation toanother. Many commodities, before reaching their final destinations, are transshipped
through several nations, each with independent tax authorities. We find that, if trade
continues over time, only the nation states occupying the first and the last segment of the
trade route can extract the monopoly rent through taxation.
Our analysis leads to the following speculation. The Silk Road benefited only the
Chinese and the Roman Empire as they controlled the beginning and the end of the traderoute. To prosper along the Silk Road, the other nation states needed more than themonopoly power to tax commodities in transit. Similar fates may haunt the present-day
nations in like positions. For example, Egypt and Panama, despite their unique positions
to control the bulk of world trade, seem unable to exploit their monopoly power.
Similarly, countries, though which the pipelines carry oil and natural gas to the finaldestinations, seem unable to capture much of the monopoly rent.

17
Appendix
We show that the model of Section 3 has no stationary equilibrium in which the numberof merchants who travel is greater than x2m. Suppose that such an equilibrium exists somthat x* ≥ x2 . Then, the optimal tax rate for state 2 is t2 = p(x*) – t1. If it behavesmmyopically, state 2 would set the tax rate equal to p(x2 ) rather than at p(x*) to earn themmonopoly rent π2 defined in Section 2. State 2 has no incentive to behave myopically if[p(x*) – t1 – c2]x*/(1 – δ) ≥ π2m.
This condition simplifies tomt1 ≤ p(x*) – c2 – (1 – δ)π2 /x*
In the equilibrium we havemt1 = p(x*) – c2 – (1 – δ)π2 /x*.
This equation maps from t1 to the equilibrium x*. State 1 chooses t1 to maximize the net
tax collection. Equivalently, it chooses x* to maximize(t1 – c1)x* = [p(x*) – c2 – c1]x* – (1 – δ)π2m
The x* therefore fulfils the first-order condition:p(x*) – c2 – c1 + x*p’(x*) = 0.
m

Evaluated at x2 , the left-hand side of the above equation is negative, implying x* < x2m.
This contradicts our initial assumption.❏

18
References
Feinberg, Y., and Kamien, M. I., 2001, Highway robbery: complementary monopoly andthe hold-up problem, International Journal of Industrial Organization 19, 1603-1621.
Gardner, R., Gaston, N., and Masson, R. T., 2002. Tolling the Rhine in 1254:
complementary monopoly revisited, unpublished paper
Karni, E., and Chakrabarti, S. K., 1997, Political structure, taxes, and trade, Journal of
Public Economics 64, 241-258.