Asymptotic Analysis by Karel Janecek

8/7/2019 Asymptotic Analysis by Karel Janecek

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Asymptotic Analysis for OptimalInvestment and Consumption with

Transaction Costs

Karel JanecekDepartment of Mathematical Sciences

Carnegie Mellon UniversityPittsburgh, PA 15213

[email protected]

Steven E. Shreve1

Department of Mathematical Sciences andCenter for Computational Finance

Carnegie Mellon UniversityPittsburgh, PA 15213

[email protected]

July 7, 2003

Abstract

We consider an agent who invests in a stock and a money market andconsumes in order to maximize the utility of consumption over aninfinite planning horizon in the presence of a proportional transactioncost > 0. The utility function is of the form U(c) = c1p/(1p) for

p > 0, p = 1. We provide a heuristic and a rigorous derivation of theasymptotic expansion of the value function in powers of 1/3, andwe also obtain asymptotic results on the boundary of the no-traderegion.

Short title: Asymptotic Transaction CostsJEL classification: G13Mathematics Subject Classification (1991): 90A09, 60H30, 60G44

1 Work supported by the National Science Foundation under grants DMS-0103814 and DMS-0139911.


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1 Introduction

Consider an agent seeking to optimally invest and consume in the presenceof proportional transaction costs. The agent can invest in a stock, modelled

as a geometric Brownian motion, and in a money market with constant rateof interest. She may also consume and get utility U(c) = c1p/(1p), wherep > 0, p = 1. In addition, the agent must pay a proportional transactioncost > 0 for transferring capital between the stock and money market. Allconsumption is done from the money market. The agent wishes to maximizethe expected discounted integral over [0, ) of the utility of consumption.

When the transaction cost is zero, the agents optimal policy is to keepa constant proportion of wealth, which we call the Merton proportion anddenote p, invested in the stock; see Merton [40]. When > 0, the optimalpolicy is to maintain a position inside a wedge called the No Trade (N T)region. Trading occurs when the position hits the boundaries of the N Tregion and no trading occurs in the interior of N T. If the agents positionis initially outside N T, she should immediately sell or buy stock in order

to move to the boundary of N T. Except when the left boundary of N T isthe positive y-axis, the set of trading times has zero Lebesgue measure, andthe total amount of capital traded can be characterized by a possible initial

jump plus local time on the boundary ofN T. When the left boundary ofN Tis the positive y-axis, the agent will stay on the y-axis once she arrives there.This problem was formulated by Magill & Constantinides [39], solved underrestrictive conditions by Davis & Norman [14], and thoroughly analyzed byShreve & Soner [43]. An explicit formula for the solution is not known.

The Hamilton-Jacobi-Bellman (HJB) equation for this problem is a par-tial differential equation in two variables. For the power utility functionsconsidered here, this can be reduced to an equation in one variable. Nu-merical results are provided by [1], [46], [47]. A useful and perhaps moreinformative approach for obtaining explicit results, the approach of this pa-

per, is to develop a power series expansion for the value function and theboundaries of the N T region in powers of . For example, Constantinides[8] numerically computed the effect of transaction costs on the value func-tion for our problem, and observed that transaction costs have a first-ordereffect on assets demand and a second-order effect on equilibrium assetreturn. This effect has been made precise by formal power series expan-sions in a variety of models. Atkinson & Wilmott [2] accomplished this for amodel with fixed transaction costs studied by Morton & Pliska [41]. Whal-ley & Wilmott [49] applied this approach to utility-based option pricing,described below. Korn [33] developed it for maximization of exponentialutility in the presence of fixed and proportional transaction costs. Korn &Laue [34] carried this analysis farther, including the case of two stocks anda money market. In this paper, we treat only the case of proportional trans-action costs, first obtaining a formal power series expansion in the spiritof the previous papers, but then providing a rigorous justification for theleading terms in the expansion. In particular, we observe that the width of

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the NT region is large, of order 1/3, whereas the effect of transaction coston the value function is smaller, of order 2/3.

The asymptotic expansion of this paper is valid for any Merton propor-tion p, except for the case in which all wealth is invested in stock (p = 1).In particular, we refute the conjecture in [43] that the Merton line (the setof stock/money market positions which are in the Merton proportion) isoutside N T for p > 1. This is the case for sufficiently high transactioncosts. However, for small transaction costs the Merton line is inside N T.

In the presence of transaction costs, contingent claim pricing by repli-cation, or more generally by super-replication, has received considerableattention but often leads to a trivial result: the cheapest strategy is buy-and-hold. See [6], [13], [17], [19], [28], [29], [31], [32], [37], [44].

One alternative to the conservative super-replication method for contin-gent claim prices, pioneered by Leland [36], is to strike a balance betweentransaction costs and hedge slippage, and this leads to a modified Black-Scholes equation; see, e.g., [3], [4], [5], [7], [18], [21], [23], [45]. Anothermethod, proposed by Hodges and Neuberger [22], is to price an option so

that a utility maximizer is indifferent between having a certain initial capitalfor investment or holding the option but having initial capital reduced bythe price of the option. This produces both a price and a hedge, the latterbeing the difference in the optimal trading strategies in the problem with-out the option and the problem with the option. This utility-based optionpricing is examined in [9], [10], [15]. A formal asymptotic analysis of suchan approach appears in [49]. Once again, the methodology developed in thispaper suggests how to make this analysis rigorous. We note that in someof these papers the utility function is U(c) = c1p/(1 p) with p restrictedto be in (0, 1). We include p > 1 in our analysis because p (0, 1) leads tointolerably risky behavior. See Samuelson [42] for the argument in words ofone syllable that this is the case even for logarithmic utility (p = 1).

The transaction cost problem with multiple assets was studied by [1], [6],

[27], [28]. For a jump diffusion model, see [20]. Transaction cost problemshave dual formulations which can shed light on their solutions; see [11],[12], [16] [38]. Other papers which study super-martingales and conditionsfor no-arbitrage in these models are [24], [25], [26], [35], [48].

In Section 2 we set out the model. Section 3 provides a heuristic expan-sion of the value function in powers of 1/3. The key results of Section 3 areproved in Section 4, using viscosity sub- and supersolutions.

2 Model set-up and known results

The model is similar to Shreve & Soner [43]. An agent is given an initialposition of x dollars in money market and y dollars in stock. The stockprice is given by dSt = St dt + St dWt, where and are positiveconstants and {Wt, t 0} is a standard Brownian motion on a filteredprobability space

, F, {Ft}t0,P

. We assume a constant interest rate

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r < . The agent chooses a policy of three adapted processes C, L, andM. The consumption process C is nonnegative and integrable on each finiteinterval. The processes L and M are nondecreasing and right-continuouswith left limits, and L0 = M0 = 0. Lt represents the cumulative value ofstock purchased up to time t, and M

tis the cumulative value of stock sold.

Let Xt denote the wealth invested in the money market and Yt the wealthinvested in stock, with X0 = x, Y0 = y. The agents position evolves as

dXt = (rXt Ct) dt (1 + ) dLt + (1 ) dMt, (2.1)

dYt = Yt dt + Yt dWt + dLt dMt. (2.2)

The constant (0, 1) appearing in these equations accounts for propor-tional transaction costs, which are paid from the money market account.

Define the solvency region

S {(x, y); x + (1 + ) y > 0, x + (1 ) y > 0} .

The policy (C,L,M) is admissible for (x, y) if (Xt, Yt) given by (2.1), (2.2)

is in S for all t 0. We denote by A(x, y) the set of all such policies. Wenote that A(x, y) = if and only if (x, y) S; see [43], Remarks 2.1, 2.2.We introduce the agents utility function Up defined for all c 0 by

Up(c) c1p/(1 p). (An analysis is also possible for U0(c) = log c.) Let

> 0 be a positive discount rate and define the value function

v(x, y) = sup(C,L,M)A(x,y)

E

0

etUp(Ct) dt, (x, y) S.

This problem when = 0 was solved by Merton [40], who determinedthat the optimal policy always keeps a wealth proportion

p =1

p

r

2, (2.3)

in the stock. We call p the Merton proportion. For = 0,

v(x, y) =1

1 pAp(p)(x + y)1p, (2.4)

where

A(p) r(1 p)

p

1

222p(1 p). (2.5)

The optimal consumption in feedback form is Ct = A(p)(Xt + Yt).We assume throughout that A(p) > 0, which is necessary and sufficient

for finiteness of the value function for the zero transaction cost problem.We introduce the convex dual function

Up : (0, ) R defined by

Up(c) supc>0

{Up(c) cc} = p1 p c(1p)/p. (2.6)

The supremum in (2.6) is attained by c = c1/p.

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x

Money Market

y

Stock

y

x+y= p

y

x+y= z1

y

x+y= z2

x + (1 + )y = 0

x + (1 )y = 0

Buy

Stock

Region

Sell

Stock

Region

No Transaction Region N T

z = 1

z = 1

z = z2

z = p

z = z1z = 0

z = 1

Fig. 1 The solvency region

Shreve and Soner [43] show that the value function is a smooth solutionof the Hamilton-Jacobi-Bellman (HJB) equation

min

Lv Up(vx), (1 )vx + vy, (1 + )vx vy = 0, (2.7)where the second-order differential operator L is given by

(Lv)(x, y) = v(x, y) 1

2

2y2 vyy(x, y) y vy(x, y) rx vx(x, y).

The optimal policy can be described in terms of two numbers 0 < z1 < z2 1 and sufficientlysmall , 1 < z1 < p < z2, so N T is in the second quadrant.

Power utility functions lead to homotheticity of the value function: for > 0,

v(x,y) = 1p v(x, y), (x, y) S. (2.8)

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This is because (C,L,M) A(x, y) (C,L,M) A(x,y). Con-sequently, the problem reduces to that of a single variable. With T (1/, 1/), we define

u(z) = v(1 z, z), z T. (2.9)

In other words, we make the change of variables z = y/(x + y), 1 z =x/(x + y), which maps the solvency region S onto the interval T. Then

v(x, y) = (x + y)1p u

y

x + y

, (x, y) S. (2.10)

The HJB equation corresponding to (2.7) for the function u(z) is

min

Du(z) U(1 p) u(z) zu(z), (1 p)u(z) + (1 z)u(z),(1 p) u(z) (1 + z) u(z)

= 0, (2.11)

where (see [43] p. 681, substituting for , 1 + for 1/(1 ), 1 p for p)

Du(z) = pA(p) + 12 2p(1 p)(z p)2u(z)+ p2z(1 z)(z p) u

(z) 1

22z2(1 z)2 u(z).

Because v(x, y) is continuous on S and of class C2 in S \ {(x, y); x = 0}([43], Corollary 10.2 and Theorem 11.6), the function u is continuous on T,C2 on T except possibly at z = 1, and satisfies (2.11). Moreover, [43] shows

(1 p)u(z) (1 + z)u(z)=0, 1

< z z1, (2.12)

Du(z) U(1 p)u(z) zu(z)=0, z1 z z2, (2.13)(1 p)u(z) + (1 z)u(z)=0, z2 z 0 and b < a we have

U(a b) = p1 p

(a b)

a1p +

b

pa

1+pp + O

b2

=p

1 pa

1pp + b a

1p + Ob2.

(2.17)

Proof We write

U(a b) = p1 p

a b1p

p =p

1 p(a b) (a b)

1p .

A Taylor series expansion of the function f(x) = (a x)1/p yields (a b)1/p = a1/p + 1p b a

(1+p)/p + O

b2

, and we get the desired result.

3 Heuristic derivation by Taylor series

In this section we heuristically derive several terms of a power series ex-pansion of the value function. One can get an idea on the size of the N T

wedge by the following argument. When transaction costs are introduced,it is too expensive for an agent to keep the proportion of capital in stockequal to p. Suppose the agent decides to instead keep the proportion insidean interval centered at p having width w. She then incurs an associatedcost of transaction which is the product of the transaction cost and theamount of transacting (local time) accumulated by the state process at theboundaries of the N T wedge. Suppose now that the no-transaction intervalhas width w for some > 0. One could multiply the stock volatility by, which is equivalent to scaling the Brownian motion by , and then thelocal time on the boundary of the N T wedge would also scale by . If onesubsequently scales time by 1/2, local time is also scaled by 1/2, and wehave returned to the original volatility. The net effect of these two scalingsis to scale local time by 1/. In other words, the amount of transacting isinversely proportional to the width of the N T wedge.

On the other hand, by permitting the state process to lie in a wedgerather than at the optimal proportion p, the agent loses utility due todisplacement from the optimal proportion. This is proportional to the squareof the displacement. To see that, one can consider the problem with zerotransaction cost and wealth process Xt + Yt given by

d(Xt + Yt) = r(Xt + Yt) dt + ( r)(Xt + Yt) dt c(Xt + Yt) dt

+ (Xt + Yt) dW(t),

where is a constant proportion of wealth maintained in the stock at alltimes and c is a constant fraction of wealth being consumed at all times.It is convenient to take to be of the form p + and c to be of the form

(1 + )A(p)(Xt + Yt), so that = 0 and = 0 provide the optimal solutionto the zero transaction cost problem. One can then compute

E(Xt + Yt)1p = (X0 + Y0)

1p exp

A(p) B() (1 p)A(p)

t

,

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where B() = 12p(1 p)22. This yields expected utility

1

1 pE

0

et

(1 + )A(p)(Xt + Yt)

1p

dt

= 11 p

(X0 + Y0)1pAp(p)(1 + )1p 1 + B()

A(p)+ (1 p)1 .

For fixed , this is maximized by taking = B

)/(pA(p)

, and that valueof results in expected utility

1

1 pE

0

et

(1 + )A(p)(Xt + Yt)1p

dt

=1

1 p(X0 + Y0)

1pAp(p)

1 +

1

2(1 p)2A1(p)2p

(3.1)

=1

1 p(X0 + Y0)

1pAp(p) p

2(X0 + Y0)

1p2A1p(p)2 + O(4).

The first term above is the value function when there is zero transactioncosts. The order 2 term is the lowest order loss due to displacement.

Suppose an agent faced with transaction costs chooses a no-transactionwedge whose width is order q for some q > 0. The amount of transactingwill be of order q and the marginal loss due to transacting will be oforder 1q. On the other hand, the marginal loss due to displacement willbe of order 2q. The agent chooses q to balance these marginal losses, i.e.,chooses q = 13 so that the width of the N T wedge is

1/3 and the loss in

the value function due to the presence of transaction cost is of order 2/3.In Remark 4.9 we show how to carry this probabilistic discussion farther toidentify the coefficients multiplying the terms

13 in the N T width and

23

in the value funciton loss. In this section, we identify these coefficients by

an analysis based on asymptotic expansion.There is no explicit solution to (2.13) in the interval [z1, z2]. Guided

by the above discussion, we thus assume that in this region u(z) has anexpansion in powers of 1/3, and we expect the coefficient of 1/3 to bezero. In order to work with this expansion, we need to also include thevariable z, and we do that using powers of z p:

u(z) = 0 1 13 2

23 3 40

43 41 (z p)

42 (z p)2

23 43 (z p)

313 44 (z p)

4 + O

53

.

(3.2)

We argued above that z p = O

1/3

for z [z1, z2]. Because we

can always move from z [z1, z2] to p by paying a transaction cost of |z p| = O4/3, we must have |u(z) u(p)| = O4/3, and so thelarger terms (z p), (z p)

2, (z p)1/3, (z p)

3, (z p)21/3,

(z p)2/3 do not appear in (3.2).

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Using z p = O

1/3

together with (3.2) leads to the expansions

u(z)=41 242 (z p)23 343 (z p)

213

444 (z p)3 + O

43

, (3.3)u(z)=242 23 643 (z p) 13 1244 (z p)2 + O(). (3.4)For z1 z z2 we have

Du(z)=pA(p)0 pA(p)1 13 pA(p)2

23 +

1

22p(1 p)(z p)

20 (3.5)

+2z2(1 z)2

42 23 + 343 (z p)

13 + 644 (z p)

2

+ O().

Furthermore,

(1 p) u(z) z u(z) = (1 p)0 (1 p)113 (1 p)2

23 + O(). (3.6)

Setting a = (1 p)0 and b = (1 p)1 1/3 + O2/3 in Lemma 1, weobtain

U(1 p) u(z) z u(z) = p1 p

(1 p)0 1p

p

+ (1 p)1

(1 p)0 1p 13 + O 23 . (3.7)

Equating first the O(1) terms and then the O

1/3

terms in (3.5) and (3.7)(see (2.13)), we conclude that

0 =1

1 pAp(p), 1 = 0.

Observe that 0 is the value v(1 z, z) for zero transaction costs.

Since 1 is zero, we can now set b = (1 p)2 2/3 + O() in Lemma 1,and obtain (after substituting for 0)

U(1 p) u(z) z u(z) = p1 p

A1p(p) +( 1 p)2A(p) 23 + O(). (3.8)

From (2.15) and (2.16) we observe that u(z1) = O

2

and u(z2) =

O

2

. From the continuity of (1 z)2u(z) we get from (3.5)

Du(zi) =p

1 pA1p(p)pA(p)2

23 +

1

22p(zip)

2Ap(p)+O(), i = 1, 2,

where we have omitted the terms in (3.5) arising from u(zi) = O

2

.

Setting this equal to (3.8) at zi results in (zi p)2

=

1

4

2

23

+ O(), with

=

8

p2A1+p(p)2. (3.9)

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We thus have

z1 p = 1

2

13 + O

23

, z2 p =

1

2

13 + O

23

. (3.10)

We may also equate the

2/3

terms in (3.5) and (3.8) at z = p, z = z1,and z = z2, and substitute for zi p from (3.10), to obtain

A(p)2 = 2 2p(1 p)

242,

A(p)2 =1

82pAp(p)2 + 2 2p(1 p)

2

42

3

243+

3

244

2

,

A(p)2 =1

82pAp(p)2 + 2 2p(1 p)

2

42 +

3

243+

3

244

2

,

which implies

42 =A(p)2

2 2p(1 p)2

, 43 = 0, 44 = pAp(p)

122p(1 p)2

. (3.11)

Finally, we observe from (2.15) and from (3.2) that

u(z1) =(1 p)

1 + z1u(z1) = (1 p)u(z1) + O

2

= Ap(p) + O

53

,

and similarly for u(z2). On the other hand, (3.3) and (3.10) imply that

u(z1) = 41 + 42 3

443

2 +1

244

3 + O

43

,

u(z2) = 41 + 42 +3

443

2 +1

244

3 + O

43

.

It follows that

41 + 42 34

432 + 1

244

3 = Ap(p) = 41 + 42+ 34

432 + 1

244

3.

Since 43 = 0 (see (3.11)), we conclude

42+1

244

3 = Ap(p), 41 = 0. (3.12)

We have thus written the value function as

u(z) =1

1 pAp(p) 2

23 + O().

Furthermore, we can solve for 2 and the width of the N T interval fromequations (3.9), (3.11), and (3.12) to obtain

2 =

9

32p4p(1 p)

4

13

A1p(p)2, =

12

p2p(1 p)

2

13

. (3.13)

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4 Rigorous asymptotic expansion

Definition 1 Let w : T R be continuous, taking values in (0, ) if p (0, 1) and in (, 0) if p > 1. Assume 0 < 1 < 2 < 1/ satisfy

w(z) = w(1) 1 + z

1 + 1

1p , 1

< z 1, (4.1)

w(z) = w(2)

1 z

1 2

1p, 2 z 0. Then the value functionu satisfies

u(p) =1

1 pAp(p)

9

32p 4p(1 p)

413

A1p(p) 2 23 +O(). (4.12)

Proof We assume p = 1. For p = 1 see Remark 1.

Step 1: Choice of constants and variables

We recall the constants 2 and of (3.13). Set =

23(1 p)2A

p(p) + B,

where B is a constant chosen to make the expression under the square rootpositive. We next define

h() =3

22

23

1

2 4 +

3

2B 2

23

3 We used the optimal policy for the supersolution argument to get equality in

(4.10). For the subsolution argument we get the appropriate inequality in (4.10)by using a suboptimal policy. However, we need to pick the trading policy sothat the terms containing the integrals dLt and dMt in (4.7) are zero, and theconsumption policy so that we get equality in (4.8).

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and choose a positive constant M satisfying

M > pAp(p) +

1

42p22Ap1(p). (4.13)

We define functions

f1 ()=13 (1 p)2A

p(p) (1 p)M Ap(p) 43 (1 p)h()

13

+

23 + (p )

13

h(), (4.14)

f2 ()= 13 (1 p)2A

p(p) (1 p)M Ap(p) 43 (1 p)h()

13

+

23 + (p + )

13

h(). (4.15)

It is shown in Appendix A that there are numbers

1 =1

2

13 (1

13 ) + o

23

, 2 =

1

2

13 (1

13 ) + o

23

(4.16)

satisfying fi (i ) = 0, i = 1, 2.

Step 2: Construction of super/subsolutions.

Choose > 0 small enough that 1 p 1 and

2 p +

2 all lie

in (0, 1/). (We have p > 0 since > r.) Define

w(z) =

Ap(p)1p 2

23 M A

p(p) h(

1 p)

1+z1+

1

1p,

11p

Ap(p) 2 23 M A

p(p) h(z p),

Ap(p)1p 2

23 M A

p(p) h(

2 p)

1z12

1p,

for 1/ z 1 , 1 z

2 ,

2 z 1/ respectively. The reader

can verify that if M were zero, then in the region [1 , 2 ] the formula for

w(z) agrees with the power series expansion

0 223 42(zp)

223 44(zp)

4,

where the coefficients 0, 2, 42, and 44 are those worked out in the pre-vious section. The term M in the definition of w will be used to createsupersolution and subsolution triples. We have the derivative formula

w(z) =

1+z (1 p)w(z), 1 < z

1 ,

1Ap(p) h(z p), 1 z

2 ,

1z (1 p)w(z), 2 z 0 for all small > 0, so is w+(z), and thus (4.3) holds.

It remains to verify that for sufficiently small

Dw+(z) U(1 p)w+(z) zw+(z) 0, 1 < z < +1 . (4.20)In this interval,

(1 p)w+(z) zw+(z) =

1 p

1 + zw+(z)

=1

1 + +

1 1 + z+1

1 + z p

Ap(p) (1 p)2

23

+ (1 p)M 1

(1 p)Ap(p)h(+1 p)

.

We use the equality U(c) = (1p)/p U(c) and the first equality in Lemma 1:U(1 p)w+(z) zw+(z)= (1 + z+1 )

1pp

1 + z

1 + +1

1p p1 p

Ap(p) (1 p)2

23 + (1 p)M

1 p

Ap(p)h(+1 p)

A(p) +

1 p

p2A

p+1(p)23

1 p

pMAp+1(p) + O

43

= (1 + +1 )

1pp w+(z)

pA(p) + (1 p)2A

p+1(p)23 (1 p)MAp+1(p) + O

43

.

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But (1 + +1 )(1p)/p = 1 + 1pp

+1 + O

2

= 1 + 1pp p + O

4/3

; thus

U

(1 p)w+(z) zw+(z)

= w+(z)pA(p) + (1 p)2Ap+1(p) 23 (1 p)MAp+1(p)+(1 p)pA(p) + O

43

. (4.21)

It is easy to verify that for > 0 sufficiently small, the function k(z) (z p) + z(1 z)/(1 + z) attains its maximum over (1/,

+1 ] at

+1

and k(+1 ) < 0. Therefore

k2(z) k2(+1 ) = (+1 )

2++1 O() =1

42

23

1

22 +o

, 1

< z +1 .

It follows that for sufficiently small > 0

Dw+(z) U(1 p)w+(z) zw+(z)=

1

2p 2

(z p) + z(1 z)

1 + z

2 2A

p+1(p) 23 + MAp+1(p)

pA(p) + O

43

(1 p)w+(z)

1

2p 2k(+1 ) 2A

p+1(p) 23 + MAp+1(p) pA(p) + O

43

(1 p)w+(z)

=

1

8p22 2A

p+1(p)

23 +

MAp+1(p)

1

4p22 pA(p)

+o()(1 p)w+(z)=

MAp+1(p)

1

4p22 pA(p)

+ o()

(1 p)w+(z) 0,

where we have used (4.13) in the last step.

Step 4b: Interval [+2 , 1/). This is analogous to Step 4a.

Step 4c: Interval (+1 , +2 ). From (4.19) and (4.13) we have

Dw+(z)

U

(1 p)w+(z) zw+(z)

=

pA

p(p) + M

+ O

43

0.

We must also show that

g1(z) (1 p)w+(z) (1 + z)w+(z) 0,

+1 < z <

+2 . (4.22)

16


18/27

For z (+1 , +2 ), we have z p = O

1/3

. Using this fact, we compute

g1(z) = Ap(p) (1 p)2

53 + (1 p)M 2

3(1 p)

2

Ap(p)(z p)2

53 +

1 p

3

Ap(p)(z p)4

+3(1 + z)

Ap(p)(z p)

23

4(1 + z)

3Ap(p)(z p)

3,

g1(z) =12

Ap(p)

23

1

4

z p

13

2+ O()

.

We know that g1(+1 ) = 0 and thus, to prove (4.22), it suffices to show that

g1 is positive on [+1 ,

+2 ]. Because (z p)

2 is a concave function of z, itsuffices to check the endpoints. We have for i = 1, 2 that

+i p

13

2=

1

2(1

13 ) + o

13

2=

1

4

1

2

13 + o

13

.

Therefore,

g1(+i ) =

12

Ap(p)

231

2

13 + o

13

> 0

for sufficiently small > 0. The proof that

g2(z) = (1 p)w+(z) + (1 z)w+(z)

is positive for z [+1 , +2 ] is analogous. This completes the proof that

(w+, +1 ,

+2 ) is a supersolution triple.

Conclusion:We note that w(p) =

11p

Ap(p) 2 23 M, and so Lemma 2

implies

1

1 p Ap(p) 22

3 M u(p) 1

1 p Ap(p) 22

3 + M .

Corollary 1 Assume p > 0, p = 1 andA(p) > 0. For fixed z T, the value function satisfies

u(z) =1

1 pAp(p) 9

32p 4p(1 p)

4 13

A1p(p) 2 23 + O(). (4.23)

Proof In the proof of Theorem 1 we constructed a supersolution w+ and asubsolution w such that w+(z) w(z) = O(), w(z) = w() + O()for fixed z T. It follows that u(z) = w(z) + O() = w() + O() =u() + O().

Theorem 2 Assume p > 0, p = 1, A(p) > 0 and p = 1. Then with given

by (3.13), we have

z1 = p 1

2

13 + O

23

, z2 = p +

1

2

13 + O

23

.

17


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Proof The value function u is concave, so u is monotone. By taking thederivative ofu in the BS and SS regions (see (2.15) and (2.16)) we see that

u(z) = O() z [z1, z2]. (4.24)

It follows that for z = z1 and z = z2, and hence for all z [z1, z2],U(1 p)u(z) zu(z) = p1 p

A1p(p) + (1 p)2A(p)23 + O().

We also know that u(z) is continuous for z T \ {1}. From equa-tions (2.15), (2.16) it follows that u(zi) = O

2

, zi = 1. We thus have

Du(zi) =pA(p) +

1

22p(1 p)(zi p)

2

u(zi) + O()

=p

1 pA1p(p) p2A(p)

23 +

1

22pAp(p)(zi p)

2 + O().

If zi = 1, the term (1 zi)2u(zi) is set equal to zero (see [43], equation

(A.5)), and the above equation still holds. To satisfy (2.11) we must have

0 = Du(zi) U(1 p)u(zi) ziu(zi)= 2A(p)

23 +

1

22pAp(p)(zi p)

2 + O().(4.25)

It follows that

zi = p 13

2

p2Ap+1(p)2 + O

23

= p

3

2p2p(1 p)

2

13

13 + O

23

,

where the sign is for z1 and the + sign is for z2.

Remark 1 The proof of Theorem 1 is valid so long as p = 1. The caseof p = 1 can be considered a singular case for which the parameter appearing in a denominator in h() is zero. The intuition is that if theoptimal proportion in the risky asset is 100% of wealth, then as soon as theagents position reachs the optimal 100% in stock, the investor no longerincurs transaction costs from adjusting it. However, to consume the agentmust transfer money from stock to money market and pay the transactioncost. We could regard this cost as a consumption tax. It follows that theloss in value function is of order rather than 2/3. In [43], Theorems 11.2,11.6 show that in this case z2 = p = 1 > z1 > 0, and Corollary 9.10 asserts

v(x, y) =1

1 pAp(p)

x + (1 )y

1p

, (x, y) S, x 0,

or equivalently,

u(z) = v(1 z, z) =1

1 pAp(p)(1 z)1p, 1 z 1.This is explained by the fact that the index of intertemporal substitution,1/p, is high for small p. An agent with index of intertemporal substitutiongreater than 1 will avoid some transaction cost by consuming faster.

Remark 3There is considerable evidence that

u(p) =1

1 pAp(p) 2

23 pA

p(p) + O

43

. (4.27)

We have just seen in (4.26) that this is the case when p = 1 (and conse-quently 2 = 0). In the proof of Theorem 1, any choice of M > pA

p(p)gives us a subsolution of the form

w(p) =1

1 pAp(p) 2

23 M + O

43

. (4.28)

(The second term on the right-hand side of (4.13) is needed only for thesupersolution argument.) Finally, the coefficient pA

p(p) on can beobtained by a tedious heuristic analysis along the lines of Section 3.

In Remark 1 we introduced the concept of consumption tax as an in-terpretation of the transaction cost an agent must pay in order to movecapital from stock to money market before consuming it. In that remark,

19


21/27

the agent was100% invested in stock. If instead the agent seeks to hold aproportion p = 1 in stock, then consuming proportionally from stock andmoney market, the agent would pay a consumption tax p times the to-tal consumption. To highest order, the optimal consumption level is thus(1

p) of what it would be if there were no transaction cost, and this

multiplies the value function by

(1 p)1p = 1 (1 p)p + O(

2).

The value function for zero transaction cost when wealth is 1 is Ap(p)/(1p), and so after this multiplication, the value function has been reduced bypA

p(p) , which is the order term we see in (4.27). We thus expectthe value function for the problem with transaction cost to be reducedfrom the zero-transaction cost value function Ap(p)/(1 p) by at leastpA

p(p), this reduction being due solely to the cost of moving capitalfrom stock to money market in order to consume. There is also a cost oftrading to stay in the NT wedge, which reduces the value function by 2

23 ,

but cannot further reduce the value function by an order term because

then the value function would fall below the lower bound (4.28).Under the assumption that (4.27) holds, one can improve the calculations

in Theorem 2. We already have from Theorem 2 and its proof that zi p =O(1/3) and that u(z) = O() for z [z1, z2]. The mean-value theoremgives

u(zi) = u(p) + O

43

,

and

(1 p)u(z1) z1u(z1) =

(1 p)u(z1)

1 + z1

= Ap(p)(1 p)2 23 (2p)pA

p(p) + O

43

,

(1 p)u(z2) z2u(z2) =

(1 p)u(z2)

1 z2= Ap(p) (1 p)2

23 + ppA

p(p) + O

43

.

Using Lemma 1, we obtainU(1 p)u(z2) z2u(z2)=

p

1 pA1p(p) + (1 p)A(p)2

23 + (2 p)pA

1p(p) + O

43

,

U(1 p)u(z1) z1u(z1)=

p

1 pA1p(p) + (1 p)A(p)2

23 ppA

1p(p) + O

43

.

We write for z1 and z2 similarly as in Theorem 2,

Du(zi) = p1 p A1p(p) pA(p)2

2

3 ppA1p(p)

+1

22pAp(p)(zi p)

2 + O

43

.

20


22/27

From Du(zi) = U(1 p)u(zi) ziu(zi), we now see that(z1 p)

2 =2

p2

A1+p(p)2

23 + 2pA(p)

+ O

43

, (4.29)

(z2 p)2

=

2

p2 A1+p

(p)2

23

+ O 43 , (4.30)which implies

z1 p = 1

2

13

4pA(p)

2p 23 + O(),

z2 p =1

2

13 + O().

This suggests that the optimal policy is to keep a wider wedge on the rightside of the Merton proportion p. This extra width makes sense becauseconsumption reduces the money market position.

We can do the same calculation for p = 1, in which case 2 = = 0.Taking the square roots in (4.29), (4.30), we now have

z1 p = z1 1 =

4pA(p)

2p12 + O

56

, z2 p = z2 1 = O().

In fact, when p = 1, we have z2 = p, as we saw in Remark 1.

Remark 4 In the key formulas derived in this paper, the transaction costparameter appears in combination with p(1 p). According toTheorem 1, the highest order loss in the value function due to transactioncosts is

9p

32

13

A1p(p)2

2 23 . (4.31)

From Theorem 2, we see that

zi p = (1)i

3

2p

13

2 13 + O

23

. (4.32)

One way to see the intrinsic nature of the quantity 2 is to define theproportion of capital in stock, t = Yt/(Xt + Yt), and apply Itos formulawhen (Xt, Yt) is generated by the optimal triple (C,L,M), for which L andM are continuous, to derive the equation

dt = t(1 t)dStSt

rt(1 t) dt 22t (1 t) dt +

tCtXt + Yt

dt

+t + 1

Xt + YtdLt +

t 1

Xt + YtdMt. (4.33)

We see that the response of t to relative changes in the stock price ist(1 t). When replicating an option by trading, the position held by thehedging portfolio, denominated in shares of stock, is called the deltaof the

21


23/27

option, and the sensitivity of the delta to changes in the stock price is thegamma. We have here a similar situation, except that t is the proportionof capital held in stock, rather than the number of shares of stock, andt(1 t) is the sensitivity of this proportion to relative changes in the stockprice.

We can now obtain the quantities in (4.31) and (4.32) by the followingheuristic argument. Suppose when the transaction cost is > 0 we choose tokeep t in an interval [p w, p + w], trading in order to create reflection atthe endpoints of this interval. As compared to the optimal expected utilitycorresponding to zero transaction costs, this entails two losses: the loss dueto displacement (not always having t = p) and the loss due to transactioncosts. The first increases with w while the second decreases. We wish tochoose w to balance these losses, i.e., to minimize their sum.

In (3.1) we computed the loss due to displacement to be

L() =p

2(X0 + Y0)

1p2A1p(p)2.

In the present case, the displacement is a random, time-varying quantitytaking values in [w, w]. However, for small w > 0, for each positive t thedistribution of t in [w, w] is approximately uniform. Therefore, the lossdue to displacement in the present model is approximately

DL(w) =1

2w

ww

L() d =p

6(X0 + Y0)

1p2A1p(p)w2.

Consider a Brownian motion doubly reflected in order to stay in [w, w],i.e., Zt = Wt + Lt Mt, where Zt [w, w] for all t, Lt grows onlywhen Zt = w and Mt grows only when Zt = w. Then limtELt/t =limtEMt/t = 1/(4w). (Use Itos formula to write

Ef(Zt) = f(Z0) +1

2 t

0

Ef(Zu) du + f(w)ELt f

(w)EMt,

divide by t, and let t to conclude

1

4w

ww

f(x) dx + f(w) limt

1

tELt f

(w) limt

1

tEMt = 0

for any C2 function f.) The amount of transacting done at both bound-aries grows at rate 1/(2w) per unit time. Because the dW(t) term in (4.33)is multiplied by t(1 t), and scaling Brownian motion by a constant islike scaling time by the square of the constant, the amount of transactingdone to keep t in [w, w] grows at rate

22/(2w) per unit time. Thiscauses a relative loss of capital

=

22

2w

per unit time. Incurring this loss is equivalent to having zero transactioncost but with the mean rate of return of both the stock and the money

22


24/27

market reduced by . According to (2.3)(2.5), the optimal expected utilityin such a problem is

1

1 p A(p) +

(1 p)

p p

(X0 + Y0)1p

=1

1 pAp(p)(X0 + Y0)

1p Ap1(p)(X0 + Y0)1p + O(2).

Therefore, the loss due to transaction costs is approximately

T C(w) = Ap1(p)(X0 + Y0)1p =

1

2w22Ap1(p)(X0 + Y0)

1p.

The total loss is approximately T L(w) = DL(w) + T C(w). SettingT L(w) = 0 and solving for w, we obtain

w =

3

2p

13

(2)13 ,

the leading term in (4.32). With this value of w, we have

T L(w) =

9p

32

13

A1p(p)2(2)23 (X0 + Y0)

1p,

thereby obtaining the term in (4.31).It is interesting to note that the quantity 2 also plays a fundamental

role in the formal asymptotic expansions of Whalley and Wilmott [49]. In

fact, even the constants932

13 and32

13 in (4.31) and (4.32) appear in [49],

the first at the end of Section 3.3 and the second in equation (3.10).

A Width of the NT interval

We shall only consider of the form O1/3. For such , we may write theterms of order and lower in f1 () of (4.14) as

f1 ()

= 13 (1 p)2A

p(p) h() 23 ph

() 13 + h()

13 + O

43

=

13 (1 p)2A

p(p) 3 +4

2

3

23

3B23 + O

43

.

Consider 0 12

1/3

1 01/3

. Then

f1 (0) = 13 (1 p)2A

p(p) 3

2

13 +

3

20

23

+ 12

131 30 13 + 320 23 32 B + O 43 = 3

220 (1 p)2A

p(p) 3

2B

+ O

43

.

23


25/27

With =

23 (1 p)2A

p(p) + B > 0, we take 0 =

2 + , where

|| < 2. Then f1 (0) =32 +O

43

. Thus, for > 0 we have f1 (0) > 0,

and for < 0 we have f1 (0) < 0 for sufficiently small > 0. Therefore,for every 0, 2 and sufficiently small > 0, there exists

1

1

2

13

1

13

2

,1

2

13

1

13

2 +

satisfying f1 (1 ) = 0. In other words,

1 =

12

13

1

13

+ o

23

. The

proof of the existence of 2 is analogous.

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Asymptotic Analysis by Karel Janecek

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