Asymptotic Analysis for Optimal Investment and Consumption with

Transaction Costs

Karel Janeček Department of Mathematical Sciences Carnegie Mellon University Pittsburgh, PA 15213 kjanecek@andrew.cmu.edu

Steven E. Shreve^{1} Department of Mathematical Sciences and Center for Computational Finance Carnegie Mellon University Pittsburgh, PA 15213 shreve@cmu.edu

February 28, 2003

Abstract

We consider an agent who invests in a stock and a money market and consumes in order to maximize the utility of consumption over an infinite planning horizon in the presence of a proportional transaction cost λ >. The utility function is of the form U(c) = c^{1−p}/(1 − p) for p > 0, p ≠ 1. We provide a heuristic and a rigorous derivation of the asymptotic expansion of the value function in powers of λ^{1/3}, and we also obtain asymptotic results on the boundary of the “no-trade” region.

Short title: Asymptotic Transaction Costs JEL classification: G13 Mathematics Subject Classification (1991): 90A09, 60H30, 60G44

1Work supported by the National Science Foundation under grant DMS-01-01407.

1 Introduction

We consider the problem of an agent seeking to optimally invest and consume in the presence of proportional transaction costs. The agent can invest in a stock, modeled as a geometric Brownian motion, and in a money market with constant rate of interest. She may also consume and get utility U(c) = c^{1−p}/(1 − p), where p > 0, p ≠ 1. In addition, the agent must pay a proportional transaction cost λ > 0 for transferring capital between the stock and money market. All consumption is done from the money market. The agent wishes to maximize the expected discounted integral over [0, ∞) of the utility of consumption.

When the transaction cost λ is zero, the agent’s optimal policy is to keep a constant proportion of wealth, which we call the Merton proportion and denote θ_{p}, invested in the stock; see Merton [40]. When λ > 0, the optimal policy is to trade as soon as the position is sufficiently far from the Merton proportion. More specifically, the agent’s optimal policy is to maintain her position inside a wedge called the No Trade (NT) region. Trading occurs when the position hits the boundaries of the NT region and no trading occurs in the interior of NT. If the agent’s position is initially outside NT, she should immediately sell or buy stock in order to move to the boundary of NT. Except when the left boundary of NT is the positive y-axis, the set of trading times has zero Lebesgue measure, and the total amount of capital traded can be characterized by a possible initial jump plus local time on the boundary of NT. When the left boundary of NT is 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 under restrictive conditions by Davis & Norman [15], and thoroughly analyzed by Shreve & Soner [43]. Although the nature of the solution can be characterized, there is no explicit formula for this solution.

The Hamilton-Jacobi-Bellman (HJB) equation for the optimal control problem with transaction costs is a partial differential equation in two variables. For the power utility functions considered here, this can be reduced to an equation in one variable. Numerical results are provided by [1], [46], [47]. A useful and perhaps more informative approach for obtaining explicit results, the approach of this paper, is to develop a power series expansion for the value function and the boundaries of the NT region in powers of λ. For example, Constantinides [9] numerically computed the effect of transaction costs on the value function for our problem, and observed that transaction costs have a “first-order effect on assets’ demand” and a “second-order effect on equilibrium asset return.” In this paper, we make these statements precise by observing that the width of the NT region is large, of order λ^{1/3}, whereas the effect of transaction cost on the value function is smaller, of order λ^{2/3}.

The asymptotic expansion is valid for any Merton proportion θ_{p}, except for the case in which all wealth is invested in stock (θ_{p} = 1). In particular, we refute the conjecture in [43] that the Merton line (the set of stock/money market positions which are in the Merton proportion) is outside NT for θ_{p} > 1. This is the case for sufficiently high transaction costs. However, for sufficiently small transaction costs the Merton line is inside NT.

In the presence of transaction costs, contingent claim pricing by replication, or more generally by super-replication, has received considerable attention but often leads to a trivial result: the cheapest strategy is buy-and-hold. See [6], [14], [18], [20], [30], [31], [33], [34], [37], [44].

One alternative to the extremely conservative super-replication method for contingent claim prices, pioneered by Leland [36], is to strike a trade-off between transaction costs and “hedge slippage,” and this leads to a modified Black-Scholes equation; see, for example, [3], [4], [5], [7], [19], [25], [23], [45]. Another method, proposed by Hodges and Neuberger [24], is to price an option so that a utility maximizer is indifferent between either having a certain initial capital for investment or else holding the option but having initial capital reduced by the price of the option. This produces both a price and a hedge, the latter being the difference in the optimal trading strategies in the problem without the option and the problem with the option. This utility-based option pricing is examined in [10], [11], [16]. A formal asymptotic analysis of such an approach appears in [49]. Once again, the methodology developed in this paper suggests how to make this analysis rigorous. We note that in some of these papers the utility function is U(c) = c^{1−p}/(1 − p) with p restricted to be in (0, 1). We include p > 1 in our analysis because p ∈ (0, 1) leads to intolerably risky behavior. See Samuelson [42] for the argument in words of one syllable that this is the case even for logarithmic utility (p = 1).

The transaction cost problem with multiple stocks was studied by [1], [29]. For a jump diffusion model, see [21]. Transaction cost problems have dual formulations which can shed light on their solutions; see [12], [13], [17] [38]. Other papers which study super-martingales and conditions for no-arbitrage in these models are [12], [26], [27], [28], [35], [48].

In Section 2 we set out the model. Section 3 provides a heuristic expansion of the value function in powers of λ^{1/3}. The key results of Section 3 are proved in Section 4, using viscosity sub- and supersolutions to the HJB equation.

2 Model set-up and known results

The set-up of the model is similar to Shreve & Soner [43]. An agent is given an initial position of x dollars in the money market and y dollars in stock. The stock price is given by dS_{t} = αS_{t} dt + σS_{t} dW_{t}, where α and σ are positive constants and {W_{t}, t ≥ 0} is a standard Brownian motion on a filtered probability space (Ω, F, {F_{t}}_{t≥0}, ℙ). We assume a constant interest rate r < α. The agent must choose a policy consisting of three adapted processes C, L, and M. The consumption process C is nonnegative and integrable on each finite interval. The processes L and M are nondecreasing and right-continuous with left limits, and L_{0−} = M_{0−} = 0. L_{t} represents the cumulative dollar value of stock purchased up to time t, while M_{t} is the cumulative dollar value of stock sold.

Let X_{t} denote the wealth invested in the money market and Y_{t} the wealth invested in stock, with X_{0−} = x, Y_{0−} = y. The agent’s position evolves as

dX_{t} = (rX_{t} − C_{t}) dt − (1 + λ) dL_{t} + (1 − λ) dM_{t},  (2.1)
dY_{t} = αY_{t} dt + σY_{t} dW_{t} + dL_{t} − dM_{t}.  (2.2)

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

Define the solvency region

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

The policy (C, L, M) is admissible for (x, y) if (X_{t}, Y_{t}) given by (2.1), (2.2) is in \overline{S} for all t ≥ 0. We denote by A(x, y) the set of all such policies. We note that A(x, y) ≠ ∅ if and only if (x, y) ∈ \overline{S}; see [43], Remarks 2.1 and 2.2.

We introduce the agent’s utility function U_{p} defined for all c ≥ 0 by U_{p}(c) ≜ c^{1−p}/(1 − p). (An analysis along the lines of this paper is also possible for U_{0}(c) = log c, but we omit that in the interest of brevity.) Let β > 0 be a positive discount rate and define the value function

v(x, y) = sup_{(C,L,M)∈A(x,y)} 𝔼 ∫_{0}^{∞} e^{−βt}U_{p}(C_{t}) dt, (x, y) ∈ \overline{S}.

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

θ_{p} = \frac{1}{p} \frac{α − r}{σ^{2}} ,

in the stock. We call θ_{p} the Merton proportion. For λ = 0,

v(x, y) = \frac{1}{1 − p}A^{−p}(p)(x + y)^{1−p},

where

A(p) ≜ \frac{β − r(1 − p)}{p} − \frac{1}{2}σ^{2}θ_{p}^{2}(1 − p).

The optimal consumption in feedback form is C_{t} = A(p)(X_{t} + Y_{t}).

We assume throughout that A(p) > 0, which guarantees that the value function for the problem with zero transaction cost is finite.

We introduce the convex dual function Ũ_{p} : (0, ∞) ↦ ℝ defined by

Ũ_{p}(c̃) ≜ sup_{c>0} \{U_{p}(c) − cc̃\} = \frac{p}{1 − p} c̃^{−(1−p)/p}.  (2.3)

The supremum in (2.3) is attained by c = c̃^{−1/p}.

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

min {Lv − Ũ_{p}(v_{x}), −(1 − λ)v_{x} + v_{y}, (1 + λ)v_{x} − v_{y}} = 0,  (2.4)

where the second-order differential operator L is given by

(Lv)(x, y) = β v(x, y) − \frac{1}{2}σ^{2}y^{2} v_{yy}(x, y) − αy v_{y}(x, y) − rx v_{x}(x, y).

The optimal policy can be described in terms of two numbers 0 < z_{1} < z_{2} < 1/λ which define the no-trade region (see [43], Theorem 11.2)

NT ≜ \{(x, y) ∈ S; z_{1} < \frac{y}{x + y} < z_{2}\} .

If Y_{t}/(X_{t} + Y_{t}) < z_{1} one should buy stock in order to bring this ratio to the boundary y/(x + y) = z_{1} of NT. If Y_{t}/(X_{t} + Y_{t}) > z_{2} one should sell stock in order to bring this ratio to the other boundary y/(x + y) = z_{2} of NT. For θ_{p} < 1 we must have 0 < z_{1} < θ_{p} < z_{2} < 1, so that NT is in the first quadrant. For θ_{p} = 1, we have 0 < z_{1} < z_{2} = 1. In this paper we show that for θ_{p} > 1 and sufficiently small λ, 1 < z_{1} < θ_{p} < z_{2}, so NT is in the second quadrant.

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

v(γx, γy) = γ^{1−p} v(x, y), (x, y) ∈ S.  (2.5)

This is because (C, L, M) ∈ A(x, y) ⇔ (γC, γL, γM) ∈ A(γx, γy). Consequently, the problem reduces to that of a single variable. With T ≜ (−1/λ, 1/λ), we define

u(z) = v(1 − z, z),  z ∈ T .  (2.6)

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)^{1−p} u(\frac{y}{x + y} ), (x, y) ∈ S.  (2.7)

The HJB equation corresponding to (2.4) for the function u(z) is

\begin{matrix}min\{Du(z) − Ũ((1 − p) u(z)− & z u′(z)), λ(1 − p)u(z) + (1 − λz) u′(z), \\ λ(1 − p) u(z) − (1 + λz) u′(z)\} = 0, \end{matrix} (2.8)

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

\begin{aligned}Du(z) = & (pA(p) + \frac{1}{2}σ^{2}p(1 − p)(z − θ_{p})^{2}) u(z) \\ & + pσ^{2}z(1 − z)(z − θ_{p}) u′(z) − \frac{1}{2}σ^{2}z^{2}(1 − z)^{2} u′′(z). \end{aligned}

Because v(x, y) is continuous on \overline{S} and of class C^{2} in S \\{(x, y); y = 0\} ([43], Corollary 10.2 and Theorem 11.6), the function u is continuous on \overline{T} , twice continuously differentiable on T except possibly at z = 1, and satisfies the HJB equation (2.8). Moreover, [43] shows that

λ(1 − p)u(z) − (1 + λz)u′(z) = 0, −\frac{1}{λ} < z ≤ z_{1},  (2.9)
Du(z) − Ũ((1 − p)u(z) − zu′(z)) = 0, z_{1} ≤ z ≤ z_{2},  (2.10)
λ(1 − p)u(z) + (1 − λz)u′(z) = 0, z_{2} ≤ z < \frac{1}{λ}.  (2.11)

Since the function u is twice continuously differentiable except possibly at z = 1, at each of z_{1} ≠ 1 and z_{2} ≠ 1 two of the above three equations hold. Moreover, v_{yy} is continuous at x = 0, and v(0, y) satisfies the HJB equation (2.4) ([43], Corollary 10.3). It is also the case that (1−z)^{2}u′′(z) is continuous for z ∈ T , and lim_{z→1}(z − 1)^{2}u′′(z) = 0 ([43], (A.5)). Thus, two of the above three equations hold at each z_{i} if z_{i} = 1.

Equations (2.9) and (2.11) are consequences of the directional derivative of v(x, y) being zero in the directions of transaction in the regions in which it is optimal to buy stock and to sell stock, respectively. These equations imply

u(z) = u(z_{1}) (\frac{1 + λz}{1 + λz_{1}} )^{1−p} , −\frac{1}{λ} < z ≤ z_{1},  (2.12)

u(z) = u(z_{2}) (\frac{1 − λz}{1 − λz_{2}} )^{1−p} , z_{2} ≤ z < \frac{1}{λ}.  (2.13)

Lemma 2.1. Assume p > 0, p ≠ 1. For a > 0 and b < a we have

\begin{aligned}Ũ(a − b) & = \frac{p}{1 − p} (a − b) (a^{− \frac{1}{p}} + \frac{b}{p} a^{− \frac{1+p}{p}} + O(b^{2})) \\ & = \frac{p}{1 − p} a^{− \frac{1−p}{p}} + b a^{− \frac{1}{p}} + O(b^{2}). \end{aligned}  (2.14)

Proof. We write

Ũ(a − b) = \frac{p}{1 − p} (a − b)^{− \frac{1−p}{p}} = \frac{p}{1 − p}(a − b) (a − b)^{− \frac{1}{p}} .

A Taylor series expansion yields (a − b)^{−1/p} = a^{−1/p} + \frac{1}{p} b a^{−(1+p)/p} + O(b^{2}), and we get the desired result.

3 Heuristic derivation by Taylor series

In this section we derive several terms of a power series expansion of the value function by a heuristic method. One can get an idea on the size of the NT wedge by the following argument. When transaction costs are introduced, it is too expensive for an an agent to keep the proportion of capital in stock equal to θ_{p}. Suppose the agent decides to instead keep the proportion inside an interval centered at θ_{p} having width w. She then incurs an associated cost of transaction which is the product λℓ of the transaction cost λ and the amount of transacting (local time) ℓ accumulated by the state process at the boundaries of the NT wedge. Suppose now that the no-transaction interval has width ρw for some ρ > 0. One could multiply the stock volatility by ρ, which is equivalent to scaling the Brownian motion by ρ, and then the local time on the boundary of the NT wedge would also scale by ρ. If one subsequently scales time by 1/ρ^{2}, local time is also scaled by 1/ρ^{2}, and we have returned to the original volatility. The net effect of these two scalings is to scale local time by 1/ρ. In other words, the amount of transacting is inversely proportional to the width of the NT wedge.

On the other hand, by permitting the state process to lie in a wedge rather than at the optimal proportion θ_{p}, the agent loses utility due to displacement from the optimal proportion. This is proportional to the square of the displacement. To see that, one can consider the problem with zero transaction cost and wealth process V_{t} = X_{t} + Y_{t} given by

dV_{t} = rV_{t} dt + (α − r)θV_{t} dt − cV_{t} dt + σθV_{t} dW (t),

where θ is a constant proportion of wealth maintained in the stock at all times and c is a constant fraction of wealth being consumed at all times. It is convenient to take θ to be of the form θ_{p} + ϵ and c to be of the form (1 + δ)A(p)V_{t}, so that ϵ = 0 and δ = 0 provide the optimal solution to the zero transaction cost problem. One can then compute

𝔼V_{t}^{1−p} = V_{0}^{1−p} exp { (β − A(p) − B(ϵ) − (1 − p)A(p)δ) t } ,

where B(ϵ) = \frac{1}{2}p(1 − p)σ^{2}θ_{p}^{2}ϵ^{2}. This yields expected discounted utility of consumption

\frac{1}{1 − p}𝔼 ∫_{0}^{∞} e^{−βt}((1 + δ)A(p)V_{t})^{1−p} dt

= \frac{1}{1 − p}V_{0}^{1−p}A^{−p}(p)(1 + δ)^{1−p} (1 + \frac{B(ϵ)}{A(p)} + (1 − p)δ)^{−1} .

For fixed ϵ, this is maximized by taking δ = B(ϵ)/(pA(p)), and that value of δ results in expected utility

\frac{1}{1 − p}𝔼 ∫_{0}^{∞} e^{−βt}((1 + δ)A(p)V_{t})^{1−p} dt

= \frac{1}{1 − p}V_{0}^{1−p}A^{−p}(p) (1 + \frac{1}{2}(1 − p)σ^{2}θ_{p}^{2}A^{−1}(p)ϵ^{2})^{−p}
= \frac{1}{1 − p}V_{0}^{1−p}A^{−p}(p) − \frac{p}{2}V_{0}^{1−p}σ^{2}θ_{p}^{2}A^{−1−p}(p)ϵ^{2} + O(ϵ^{4}).

The first term in the last expression is the value function when there is zero transaction costs. The second order ϵ^{2} term is the loss due to displacement.

Suppose an agent faced with transaction costs chooses a no-transaction wedge whose width is order λ^{q} for some q > 0. The amount of transacting will be of order λ^{−q} and the marginal loss due to transacting will be of order λ^{1−q}. On the other hand, the marginal loss due to displacement will be of order λ^{2q}. The agent chooses q to balance these marginal losses, i.e., chooses q = \frac{1}{3} so that the width of the NT wedge is λ^{1/3} and the loss in the value function due to the presence of transaction cost λ is of order λ^{2/3}.

There is no explicit solution to (2.10) in the interval [z_{1}, z_{2}]. Guided by the above discussion, we thus assume that in this region u(z) has an expansion in powers of λ^{1/3}, and we expect the coefficient of λ^{1/3} to be zero. In order to work with this expansion, we need to also include the variable z, and we do that using powers of (z − θ_{p})^{1/3}:

\begin{aligned}u(z) & = γ_{0} − γ_{1} λ^{\frac{1}{3}} − γ_{2} λ^{\frac{2}{3}} − γ_{3} λ − γ_{40} λ^{\frac{4}{3}} − γ_{41} (z − θ_{p})λ \\ & − γ_{42} (z − θ_{p})^{2}λ^{\frac{2}{3}} − γ_{43} (z − θ_{p})^{3}λ^{\frac{1}{3}} − γ_{44} (z − θ_{p})^{4} + O(λ^{\frac{5}{3}} ). \end{aligned} (3.1)

We argued above that z − θ_{p} = O(λ^{1/3}) for z ∈ NT. This assumption together with (3.1) leads to the expansions

u′(z) = −γ_{41} λ − 2γ_{42} (z − θ_{p})λ^{\frac{2}{3}} − 3γ_{43} (z − θ_{p})^{2}λ^{\frac{1}{3}}
−4γ_{44} (z − θ_{p})^{3} + O(λ^{\frac{4}{3}} ),  (3.2)
u′′(z) = −2γ_{42} λ^{\frac{2}{3}} − 6γ_{43} (z − θ_{p})λ^{\frac{1}{3}} − 12γ_{44} (z − θ_{p})^{2} + O(λ). (3.3)

For z_{1} ≤ z ≤ z_{2} we have

Du(z) = pA(p)γ_{0} − pA(p)γ_{1} λ^{\frac{1}{3}} − pA(p)γ_{2} λ^{\frac{2}{3}} + \frac{1}{2}σ^{2}p(1 − p)(z − θ_{p})^{2}γ_{0} (3.4)

+σ^{2}z^{2}(1 − z)^{2} (γ_{42} λ^{\frac{2}{3}} + 3γ_{43} (z − θ_{p})λ^{\frac{1}{3}} + 6γ_{44} (z − θ_{p})^{2}) + O(λ).

Furthermore,

(1 − p) u(z) − z u′(z) = (1 − p)γ_{0} − (1 − p)γ_{1}λ^{\frac{1}{3}} − (1 − p)γ_{2}λ^{\frac{2}{3}} + O(λ). (3.5)

Setting a = (1 − p)γ_{0} and b = (1 − p)γ_{1} λ^{1/3} + O(λ^{2/3}) in Lemma 2.1, we obtain

Ũ((1−p) u(z)−z u′(z))=\frac{p}{1 − p}((1−p)γ_{0})^{− \frac{1−p}{p}}+(1−p)γ_{1}((1−p)γ_{0})^{− \frac{1}{p}} λ^{\frac{1}{3}} +O(λ^{\frac{2}{3}} ).

(3.6)

Equating first the O(1) terms and then the O(λ^{1/3}) terms in (3.4) and (3.6) (see (2.10)), we conclude that

γ_{0} = \frac{1}{1 − p}A^{−p}(p), γ_{1} = 0.

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

Since γ_{1} is zero, we can now set b = (1 − p)γ_{2} λ^{2/3} + O(λ) in Lemma 2.1, and obtain (after substituting for γ_{0})

Ũ((1 − p) u(z) − z u′(z)) = \frac{p}{1 − p}A^{1−p}(p) + (1 − p)γ_{2}A(p) λ^{\frac{2}{3}} + O(λ). (3.7)

From (2.12) and (2.13) we observe that u′′(z_{1}) = O(λ^{2}) and u′′(z_{2}) = O(λ^{2}). From the continuity of (1 − z)^{2}u′′(z) we get from (3.4)

Du(z_{i}) = \frac{p}{1 − p}A^{1−p}(p)−pA(p)γ_{2} λ^{\frac{2}{3}} + \frac{1}{2}σ^{2}p(z_{i} −θ_{p})^{2}A^{−p}(p)+O(λ), i = 1, 2,

where we have omitted the terms in (3.4) arising from u′′(z_{i}) = O(λ^{2}). Setting this equal to (3.7) at z_{i} results in (z_{i} − θ_{p})^{2} = \frac{1}{4}ν^{2} λ^{\frac{2}{3}} + O(λ), with

ν = \sqrt{\frac{8}{pσ^{2}} A^{1+p}(p)γ_{2}}.  (3.8)

We thus have

z_{1} − θ_{p} = −\frac{1}{2}ν λ^{\frac{1}{3}} + O(λ^{\frac{2}{3}} ), z_{2} − θ_{p} = \frac{1}{2}ν λ^{\frac{1}{3}} + O(λ^{\frac{2}{3}} ).  (3.9)

We may also equate the λ^{2/3} terms in (3.4) and (3.7) at z = θ_{p}, z = z_{1}, and z = z_{2}, and substitute for z_{i} − θ_{p} from (3.9), to obtain

\begin{aligned}A(p)γ_{2} & = σ^{2} θ_{p}^{2}(1 − θ_{p})^{2}γ_{42}, \\ A(p)γ_{2} & = \frac{1}{8}σ^{2}pA^{−p}(p)ν^{2} + σ^{2} θ_{p}^{2}(1 − θ_{p})^{2} (γ_{42} − \frac{3}{2}γ_{43}ν + \frac{3}{2}γ_{44}ν^{2}) , \\ A(p)γ_{2} & = \frac{1}{8}σ^{2}pA^{−p}(p)ν^{2} + σ^{2} θ_{p}^{2}(1 − θ_{p})^{2} (γ_{42} + \frac{3}{2}γ_{43}ν + \frac{3}{2}γ_{44}ν^{2}) , \end{aligned}

which implies

γ_{42} = \frac{A(p)γ_{2}}{σ^{2} θ_{p}^{2}(1 − θ_{p})^{2}} , γ_{43} = 0, γ_{44} = −\frac{pA^{−p}(p)}{12θ_{p}^{2}(1 − θ_{p})^{2}} .  (3.10)

Finally, we observe from (2.12) and from (3.1) that

u′(z_{1}) = \frac{λ(1 − p)}{1 + λz_{1}}u(z_{1}) = λ(1 − p)u(z_{1}) + O(λ^{2}) = A^{−p}(p) λ + O(λ^{\frac{5}{3}} ),

and similarly for u′(z_{2}). On the other hand, (3.2) and (3.9) imply that

\begin{aligned}u′(z_{1}) & = −γ_{41} λ + γ_{42}ν λ − \frac{3}{4}γ_{43}ν^{2} λ + \frac{1}{2}γ_{44}ν^{3} λ + O(λ^{\frac{4}{3}} ), \\ u′(z_{2}) & = γ_{41} λ + γ_{42}ν λ + \frac{3}{4}γ_{43}ν^{2} λ + \frac{1}{2}γ_{44}ν^{3} λ + O(λ^{\frac{4}{3}} ). \end{aligned}

It follows that

−γ_{41} + γ_{42}ν − \frac{3}{4}γ_{43}ν^{2} + \frac{1}{2}γ_{44}ν^{3} = A^{−p}(p) = γ_{41} + γ_{42}ν + \frac{3}{4}γ_{43}ν^{2} + \frac{1}{2}γ_{44}ν^{3}.

Since γ_{43} = 0 (see (3.10)), we conclude

γ_{42}ν + \frac{1}{2}γ_{44}ν^{3} = A^{−p}(p), γ_{41} = 0.  (3.11)

We have thus written the value function as

u(z) = \frac{1}{1 − p}A^{−p}(p) − γ_{2} λ^{\frac{2}{3}} + O(λ).

Furthermore, we can solve for γ_{2} and the width of the NT interval ν from equations (3.8), (3.10), and (3.11) to obtain

γ_{2} = (\frac{9}{32}pθ_{p}^{4}(1 − θ_{p})^{4})^{\frac{1}{3}} A^{−1−p}(p)σ^{2}, ν = (\frac{12}{p}θ_{p}^{2}(1 − θ_{p})^{2})^{\frac{1}{3}} . (3.12)

4 Rigorous asymptotic expansion

Definition 4.1. Let w: T → ℝ be continuous, taking values in (0, ∞) if 0 < p < 1 and taking values in (−∞, 0) if p > 1. Assume there are two points 0 < ζ_{1} < ζ_{2} < 1/λ such that

w(z) = w(ζ_{1}) (\frac{1 + λz}{1 + λζ_{1}} )^{1−p} , −\frac{1}{λ} < z ≤ ζ_{1},  (4.1)

w(z) = w(ζ_{2}) (\frac{1 − λz}{1 − λζ_{2}} )^{1−p} , ζ_{2} ≤ z < \frac{1}{λ}.  (4.2)

Assume further that w is of class C^{2} on (ζ_{1}, ζ_{2}) and the one-sided limits of w′′ exist at the endpoints of this interval. If

λ(1 − p)w(z) + (1 − λz)w′(z) ≥ 0, −\frac{1}{λ} < z < ζ_{2},  (4.3)
Dw(z) − Ũ((1 − p)w(z) − zw′(z)) ≥ 0, −\frac{1}{λ} < z < \frac{1}{λ},  (4.4)
λ(1 − p)w(z) − (1 + λz)w′(z) ≥ 0, ζ_{1} < z < \frac{1}{λ},  (4.5)

we say (w, ζ_{1}, ζ_{2}) is a supersolution triple. If (4.1), (4.2) and the reverse of inequality (4.4) for ζ_{1} < z < ζ_{2} hold, we say (w, ζ_{1}, ζ_{2}) is a subsolution triple.

If (w, ζ_{1}, ζ_{2}) is a supersolution (subsolution) triple, then

φ(x, y) ≜ (x + y)^{1−p}w(y/(x + y)), (x, y) ∈ S  (4.6)

is a supersolution (subsolution) of (2.4).

Lemma 4.2. Assume p > 0, p ≠ 1. If (w, ζ_{1}, ζ_{2}) is a supersolution (subsolution) triple, then w(z) ≥ u(z) (w(z) ≤ u(z)) for z ∈ T .

Proof. For 0 < p < 1, this follows from [43], Comparison Theorem A.3. We give the proof for p > 1.

Let (w, ζ_{1}, ζ_{2}) be a supersolution triple, and let φ be given by (4.6). In light of (2.7), is suffices to prove that φ(x_{0}, y_{0}) ≥ v(x_{0}, y_{0}) for fixed but arbitrary (x_{0}, y_{0}) ∈ S. Following the construction of Theorem 9.2 of [43], we let (C, L, M) be an optimal policy for this initial condition, i.e., with (X_{t}, Y_{t}) given by (2.1), (2.2), we have (X(0), Y (0)) ∈ ∂NT if (x_{0}, y_{0}) ∉ NT, (X_{t}, Y_{t}) ∈ \overline{NT} for all t ≥ 0, and

L_{t} = ∫_{0}^{t} 𝕀_{{Y_{s}/(X_{s}+Y_{s})=z_{1}}} dL_{s}, M_{t} = ∫_{0}^{t} 𝕀_{{Y_{s}/(X_{s}+Y_{s})=z_{2}}}dM_{s},
C_{t} = (v_{x}(X_{t}, Y_{t}))^{−1/p}, t ≥ 0.

Then v(x_{0}, y_{0}) = 𝔼 ∫_{0}^{∞} e^{−βt}U_{p}(C_{t}) dt.

The function φ is of class C^{2} in S except possibly on the lines y/(x+y) = ζ_{i}, i = 1, 2. We can mollify φ to obtain a C^{2} function, apply Itô’s rule to this C^{2} function, and then pass to the limit. On the two lines where φ may not be C^{2}, (X_{t}, Y_{t}) spends Lebesgue-measure zero time and we will obtain Itô’s rule for φ(X_{t}, Y_{t}). An exception would be if ζ_{2} = z_{2} = 1, because in this case (X_{t}, Y_{t}) remains on the y−axis once the y-axis is reached. A direct computation shows, however, that φ_{yy}(x, y) is continuous there, even if ζ_{2} = z_{2} = 1, and hence we will still obtain Itô’s rule. Therefore,

(e^{−βt}φ(X_{t}, Y_{t}))  (4.7)
= −e^{−βt}[Lφ(X_{t}, Y_{t}) dt+C_{t}φ_{x}(X_{t}, Y_{t})+((1 + λ)φ_{x}(X_{t}, Y_{t})−φ_{y}(X_{t}, Y_{t})) dL_{t}

+( − (1 − λ)φ_{x}(X_{t}, Y_{t}) + φ_{y}(X_{t}, Y_{t})) dM_{t}] + e^{−βt}φ_{y}(X_{t}, Y_{t}) σ dW_{t}

≤ −e^{−βt}U_{p}(C_{t}) dt + e^{−βt}φ_{y}(X_{t}, Y_{t}) σ dW_{t},

where we have used the supersolution property, the fact that Lφ = Du, and the fact (see(2.3)) that

Ũ_{p}(φ_{x}(X_{t}, Y_{t})) ≥ U_{p}(C_{t}) − C_{t}φ_{x}(X_{t}, Y_{t}).  (4.8)

We want to integrate (4.7) and eventually argue that the expected value of the Itô integral is zero. Define τ_{n} = inf{t ≥ 0; |X_{t} + Y_{t}| ≤ 1/n}, to obtain

∫_{0}^{n∧τ_{n}} e^{−βt}U_{p}(C_{t}) dt + e^{−β(n∧τ_{n})}φ(X_{n∧τ_{n}}, Y_{n∧τ_{n}})  (4.9)
≤ φ(x_{0}, y_{0}) + σ ∫_{0}^{n∧τ_{n}} e^{−βt}φ_{y}(X_{t}, Y_{t}) dW_{t}.

We can perform a similar analysis for the value function. The function v is C^{2} except on the y-axis. The x-axis is not in \overline{NT}, which contains (X_{t}, Y_{t}) for all t ≥ 0. At the y-axis, v_{yy} is continuous ([43], Corollary 10.3). This permits us to compute the differential of v(X_{t}, Y_{t}), and similarly to (4.9) we obtain

∫_{0}^{n∧τ_{n}} e^{−βt}U_{p}(C_{t}) dt + e^{−β(n∧τ_{n})}v(X_{n∧τ_{n}}, Y_{n∧τ_{n}})  (4.10)
= v(x_{0}, y_{0}) + σ ∫_{0}^{n∧τ_{n}} e^{−βt}v_{y}(X_{t}, Y_{t}) dW_{t}.

As n → ∞, we have either τ_{n} → ∞ or τ_{n} → τ_{0} ≜ inf{t ≥ 0 : X_{t} = Y_{t} = 0} < ∞. On the set {lim_{n→∞} τ_{n} = τ_{0} < ∞}, equation (4.10), the inequality U_{p} ≤ 0 and the fact that v(0, 0) = −∞ imply ∫_{0}^{τ_{0}} e^{−βt}v_{y}(X_{t}, Y_{t}) dW_{t} = −∞. This is impossible, because Itô integrals are either finite or have lim sup = − lim inf = ∞ (see [32], Chap. 3, Problem 4.11 and p. 232). We conclude that τ_{n} → ∞ almost surely. Taking expectations in (4.10) and letting n → ∞, we obtain

lim_{n→∞} 𝔼e^{−β(n∧τ_{n})}v(X_{n∧τ_{n}}, Y_{n∧τ_{n}}) = v(x_{0}, y_{0}) − 𝔼 ∫_{0}^{∞} e^{−βt}U_{p}(C_{t}) dt = 0.

(4.11)

Because of (2.7), (4.6) and the boundedness of u and w for z ∈ NT, either φ(x, y) = 0, or there are constants c_{1} and c_{2} such that

c_{1}φ(x, y) ≤ v(x, y) ≤ c_{2}φ(x, y) whenever z_{1} ≤ \frac{y}{x + y} ≤ z_{2}.

From (4.11) we conclude that lim_{n→∞} 𝔼e^{−β(n∧τ_{n})}φ(X_{n∧τ_{n}}, Y_{n∧τ_{n}}) = 0. Taking expectations and the limit in (4.9), we have v(x_{0}, y_{0}) = 𝔼 ∫_{0}^{∞} e^{−βt}U_{p}(C_{t}) dt ≤ φ(x_{0}, y_{0}), as desired.

Now let (w, ζ_{1}, ζ_{2}) be a subsolution triple. We show for fixed but arbitrary (x_{0}, y_{0}) ∈ S that φ(x_{0}, y_{0}) ≤ v(x_{0}, y_{0}). This time we construct a (suboptimal) policy (C, L, M) for which Y (0)/(X(0)+Y (0)) is either ζ_{1} or ζ_{2} if y_{0}/(x_{0}+y_{0}) ∉ (ζ_{1}, ζ_{2}), (X_{t}, Y_{t}) ∈ [ζ_{1}, ζ_{2}] for all t ≥ 0, and

L_{t} = ∫_{0}^{t} 𝕀_{{Y_{s}/(X_{s}+Y_{s})=ζ_{1}}} dL_{s}, M_{t} = ∫_{0}^{t} 𝕀_{{Y_{s}/(X_{s}+Y_{s})=ζ_{2}}} dM_{s},

C_{t} = (φ_{x}(X_{t}, Y_{t}))^{−1/p}, t ≥ 0.

For this policy, we have the reverse equality in (4.9).^{2} Using this inequality in place of (4.10), we argue as before that lim_{n→∞} τ_{n} = ∞ almost surely. φ ≤ 0, (4.9) with the inequality reversed implies 𝔼 ∫_{0}^{n∧τ_{n}} e^{−βt}U_{p}(C_{t}) dt ≥ φ(x_{0}, y_{0}) and letting n → ∞, we obtain v(x_{0}, y_{0}) ≥ 𝔼 ∫_{0}^{∞} e^{−βt}U_{p}(C_{t}) dt ≥ φ(x_{0}, y_{0}).

2We 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 trading policy such that the terms containing the integrals dL_{t} and dM_{t} in (4.7) are zero, and consumption policy so that we get equality in (4.8).

Theorem 4.3. Assume p > 0, p ≠ 1 and A(p) > 0. Then the value function u satisfies

u(θ_{p}) = \frac{1}{1 − p}A^{−p}(p) − (\frac{9}{32}p θ_{p}^{4}(1 − θ_{p})^{4})^{\frac{1}{3}} A^{−1−p}(p) σ^{2} λ^{\frac{2}{3}} + O(λ). (4.12)

Proof. We assume θ_{p} ≠ 1. For θ_{p} = 1 see Remark 4.6.

Step 1: Choice of constants and variables

We recall the constants γ_{2} and ν of (3.12). Set ξ = \sqrt{\frac{2}{3}(1 − p)γ_{2}A^{p}(p) + B}, where B is a constant chosen to make the expression under the square root positive. We next define

h(δ) = \frac{3}{2} δ^{2}λ^{\frac{2}{3}} − \frac{1}{ν^{2}} · δ^{4} + \frac{3}{2}B δ^{2}λ^{\frac{2}{3}}

and choose a positive constant M satisfying

M > θ_{p}A^{−p}(p) + \frac{1}{4}σ^{2}pν^{2}ξ^{2}A^{−p−1}(p).  (4.13)

We define functions

f_{1}^{±}(δ) = νλ^{\frac{1}{3}} − (1 − p)νγ_{2}A^{p}(p) λ ± (1 − p)νMA^{p}(p) λ^{\frac{4}{3}} − (1 − p)h(−δ) λ^{\frac{1}{3}}
+(λ^{− \frac{2}{3}} + (θ_{p} − δ)λ^{\frac{1}{3}} )h′(−δ),  (4.14)
f_{2}^{±}(δ) = ν λ^{\frac{1}{3}} − (1 − p)νγ_{2}A^{p}(p) λ ± (1 − p)νMA^{p}(p) λ^{\frac{4}{3}} − (1 − p)h(δ) λ^{\frac{1}{3}}
+(−λ^{− \frac{2}{3}} + (θ_{p} + δ)λ^{\frac{1}{3}} )h′(δ).  (4.15)

It is shown in Appendix A that there are numbers

δ_{1}^{±} = \frac{1}{2}νλ^{\frac{1}{3}} (1 − ξλ^{\frac{1}{3}} ) + o(λ^{\frac{2}{3}} ), δ_{2}^{±} = \frac{1}{2}νλ^{\frac{1}{3}} (1 − ξλ^{\frac{1}{3}} ) + o(λ^{\frac{2}{3}} )  (4.16)

satisfying f_{i}^{±}(δ_{i}^{±}) = 0, i = 1, 2.

Step 2: Construction of super/subsolutions.

Choose λ > 0 small enough that ζ_{1}^{±} ≜ θ_{p} − δ_{1}^{±} and ζ_{2}^{±} ≜ θ_{p} + δ_{2}^{±} all lie in (0, 1/λ). (We have θ_{p} > 0 since α > r.) Define

w_{±}(z) = \begin{cases}(\frac{A^{−p}(p)}{1−p} − γ_{2} λ^{\frac{2}{3}} ± Mλ − \frac{A^{−p}(p)}{ν}h(ζ_{1}^{±} − θ_{p}))(\frac{1+λz}{1+λζ_{1}^{±}} )^{1−p} , −\frac{1}{λ} ≤ z ≤ ζ_{1}^{±}, \\ \frac{1}{1−p}A^{−p}(p) − γ_{2} λ^{\frac{2}{3}} ± Mλ − \frac{A^{−p}(p)}{ν}h(z − θ_{p}), ζ_{1}^{±} ≤ z ≤ ζ_{2}^{±}, \\ (\frac{A^{−p}(p)}{1−p} − γ_{2} λ^{\frac{2}{3}} ± Mλ − \frac{A^{−p}(p)}{ν}h(ζ_{2}^{±} − θ_{p})) (\frac{1−λz}{1−λζ_{2}^{±}} )^{1−p} , ζ_{2}^{±} ≤ z ≤ \frac{1}{λ}. \end{cases}

The reader can verify that if M were zero, then in the region [ζ_{1}^{±}, ζ_{2}^{±}] the formula for w_{±}(z) agrees with the power series expansion

γ_{0} − γ_{2}λ^{\frac{2}{3}} − γ_{42}(z−θ_{p})^{2}λ^{\frac{2}{3}} − γ_{44}(z−θ_{p})^{4},

where the coefficients γ_{0}, γ_{2}, γ_{42}, and γ_{44} are those worked out in the previous section. The term ±Mλ in the definition of w_{±} will be used to create supersolution and subsolution triples. We have the derivative formula

w′_{±}(z) = \begin{cases}\frac{λ}{1+λz} (1 − p)w_{±}(z), & −\frac{1}{λ} < z ≤ ζ_{1}^{±}, \\ −ν^{−1}A^{−p}(p) h′(z − θ_{p}), & ζ_{1}^{±} ≤ z ≤ ζ_{2}^{±}, \\ −\frac{λ}{1−λz} (1 − p)w_{±}(z), & ζ_{2}^{±} ≤ z < \frac{1}{λ}. \end{cases}

The equations f_{1}^{±}(δ_{1}) = 0 and f_{2}^{±}(δ_{2}) = 0 guarantee that w′_{±} is defined and continuous at ζ_{1}^{±} and ζ_{2}^{±}. We also have

w′′_{±}(z) = \begin{cases}−\frac{λ^{2}}{(1+λz)^{2}} p(1 − p)w_{±}(z), & −\frac{1}{λ} < z < ζ_{1}^{±}, \\ −ν^{−1}A^{−p}(p) h′′(z − θ_{p}), & ζ_{1}^{±} < z < ζ_{2}^{±}, \\ −\frac{λ^{2}}{(1−λz)^{2}} p(1 − p)w_{±}(z), & ζ_{2}^{±} < z < \frac{1}{λ}. \end{cases}

The function w_{±}(z) is C^{2} except at ζ_{1}^{±} and ζ_{2}^{±}, and at these two points, the one-sided second derivatives exist and equal the respective one-sided limits of the second derivatives.

Step 3: Verification that (w_{−}, ζ_{1}^{−}, ζ_{2}^{−}) is a subsolution triple.

It suffices to verify

Dw_{−}(z) − Ũ((1 − p)w_{−}(z) − zw′_{−}(z)) ≤ 0, ζ_{1}^{−} < z < ζ_{2}^{−}.  (4.17)

To do this, we simultaneously work with both w_{−} and w_{+}. We thereby develop an inequality for w_{+} needed in the subsequent supersolution verification. We use the facts that z − θ_{p} = O(λ^{1/3}), so h(z − θ_{p}) = O(λ^{4/3}), h′(z − θ_{p}) = O(λ).

For ζ_{1}^{±} < z < ζ_{2}^{±} we have (1 − p)w_{±}(z) − zw′_{±}(z) = a − b, where a = A^{−p}(p) and b = (1−p)γ_{2}λ^{2/3} ∓(1−p)Mλ−A^{−p}(p)ν^{−1}zh′(z−θ_{p})+O(λ^{4/3}). Lemma 2.1 implies

Ũ((1 − p)w_{±}(z) − zw′_{±}(z))  (4.18)
= \frac{p}{1 − p}A^{1−p}(p) + A(p) [(1 − p)γ_{2}λ^{\frac{2}{3}} ∓ (1 − p)Mλ − \frac{A^{−p}(p)}{ν}zh′(z − θ_{p})]
+O(λ^{\frac{4}{3}} )
= pA(p)w(z) + γ_{2}A(p)λ^{\frac{2}{3}} ∓ Mλ − \frac{A^{1−p}(p)}{ν}zh′(z − θ_{p}) + O(λ^{\frac{4}{3}} ).

Therefore,

Dw_{±}(z) − Ũ((1 − p)w_{±}(z) − zw′_{±}(z))
= \frac{1}{2}σ^{2}p(1 − p)(z − θ_{p})^{2}w_{±}(z) − \frac{1}{2}σ^{2}z^{2}(1 − z)^{2}w′′_{±}(z)

−γ_{2}A(p)λ^{\frac{2}{3}} ± Mλ + \frac{1}{ν} A^{1−p}(p)zh′(z − θ_{p}) + O(λ^{\frac{4}{3}} )

= \frac{1}{2}σ^{2}pA^{−p}(p)(z − θ_{p})^{2} + \frac{1}{2ν} σ^{2}z^{2}(1 − z)^{2}A^{−p}(p)h′′(z − θ_{p})

−γ_{2}A(p)λ^{\frac{2}{3}} ± Mλ + \frac{1}{ν} A^{1−p}(p)zh′(z − θ_{p}) + O(λ^{\frac{4}{3}} ).

Writing z = θ_{p} + (z − θ_{p}) and 1 − z = 1 − θ_{p} − (z − θ_{p}), we derive the relation

z^{2}(1 − z)^{2} = θ_{p}^{2}(1 − θ_{p})^{2} + 2θ_{p}(1 − θ_{p})(1 − 2θ_{p})(z − θ_{p}) + O(λ^{\frac{2}{3}} ).

Using this and the formulas h′(δ) = 3δλ^{2/3} −4δ^{3}/ν^{2} +3Bδλ^{2/3}, h′′(δ) = 3λ^{2/3} − 12δ^{2}/ν^{2} + 3Bλ^{2/3}, we obtain

Dw_{±}(z) − Ũ((1 − p)w_{±}(z) − zw′_{±}(z))
= [\frac{1}{2}σ^{2}pA^{−p}(p) − \frac{6}{ν^{3}} σ^{2}θ_{p}^{2}(1 − θ_{p})^{2}A^{−p}(p)] (z − θ_{p})^{2}

+ [\frac{3}{2ν} σ^{2}θ_{p}^{2}(1 − θ_{p})^{2}A^{−p}(p) − γ_{2}A(p)] λ^{\frac{2}{3}}

+σ^{2}θ_{p}(1 − θ_{p})(1 − 2θ_{p})A^{−p}(p) [\frac{3(z − θ_{p})λ^{\frac{2}{3}}}{ν} − \frac{12(z − θ_{p})^{3}}{ν^{3}} ]

+θ_{p}A^{−p}(p) [\frac{3(z − θ_{p})λ^{\frac{2}{3}}}{ν} − \frac{4(z − θ_{p})^{3}}{ν^{3}} ] ± Mλ + O(λ^{\frac{4}{3}} ).

The definitions of ν and γ_{2} imply that the first two terms on the right-hand side are zero. Because

\frac{z − θ_{p}}{ν} = \frac{1}{2}λ^{\frac{1}{3}} + O(λ^{\frac{2}{3}} ),

the third term is O(λ^{4/3}), and we can simplify the fourth term to obtain

Dw_{±}(z) − Ũ((1 − p)w_{±}(z) − zw′_{±}(z)) = θ_{p}A^{−p}(p)λ ± Mλ + O(λ^{\frac{4}{3}} ) (4.19)

for z_{1}^{±} < z < z_{2}^{±}. From (4.13) we have (4.17) for all sufficiently small λ > 0. This completes the verification that (w_{−}, ζ_{1}^{−}, ζ_{2}^{−}) is a subsolution triple.

Step 4: Verification that (w_{+}, ζ_{1}^{+}, ζ_{2}^{+}) is a supersolution triple \underline{Step 4a:} Interval (−1/λ, ζ_{1}^{+}).

We must show that that (4.3) and (4.4) hold in this interval. Since (1 − p)w_{+}(z) > 0 for sufficiently small λ > 0, so is w′_{+}(z), and thus (4.3) holds.

It remains to verify that for sufficiently small λ

Dw_{+}(z) − Ũ((1 − p)w_{+}(z) − zw′_{+}(z)) ≥ 0, −\frac{1}{λ} < z < ζ_{1}^{+}.  (4.20)

In this interval,

(1 − p)w_{+}(z) − zw′_{+}(z) = \frac{1 − p}{1 + λz} w_{+}(z)

= \frac{1}{1 + λζ_{1}^{+}} (\frac{1 + λz_{1}^{+}}{1 + λz} )^{p} (A^{−p}(p) − (1 − p)γ_{2}λ^{\frac{2}{3}}

+ (1 − p)Mλ − \frac{1}{ν} (1 − p)A^{−p}(p)h(ζ_{1}^{+} − θ_{p})).

Using the first equality in Lemma 2.1 and the equality Ũ(αc̃) = α^{−(1−p)/p} Ũ(c̃), we get

Ũ((1 − p)w_{+}(z) − zw′_{+}(z))

= (1 + λz_{1}^{+})^{\frac{1−p}{p}} (\frac{1 + λz}{1 + λζ_{1}^{+}} )^{1−p} \frac{p}{1 − p}

× (A^{−p}(p) − (1 − p)γ_{2}λ^{\frac{2}{3}} + (1 − p)Mλ − \frac{1 − p}{ν}A^{−p}(p)h(ζ_{1}^{+} − θ_{p}))

× (A(p) + \frac{1 − p}{p}γ_{2}A^{p+1}(p)λ^{\frac{2}{3}} − \frac{1 − p}{p}MA^{p+1}(p)λ + O(λ^{\frac{4}{3}} ))

= (1 + λζ_{1}^{+})^{\frac{1−p}{p}} w_{+}(z)

× (pA(p) + (1 − p)γ_{2}A^{p+1}(p)λ^{\frac{2}{3}} − (1 − p)MA^{p+1}(p)λ + O(λ^{\frac{4}{3}} )) .

But (1 + λζ_{1}^{+})^{(1−p)/p} = 1 + \frac{1−p}{p}λζ_{1}^{+} + O(λ^{2}) = 1 + \frac{1−p}{p}λθ_{p} + O(λ^{4/3}), and thus

Ũ((1 − p)w_{+}(z) − zw′_{+}(z))
= w_{+}(z)(pA(p) + (1 − p)γ_{2}A^{p+1}(p)λ^{\frac{2}{3}} − (1 − p)MA^{p+1}(p)λ
+(1 − p)θ_{p}A(p)λ + O(λ^{\frac{4}{3}} )).  (4.21)

It is easy to verify that for λ > 0 sufficiently small, the function k(z) ≜ (z − θ_{p}) + λz(1 − z)/(1 + λz) attains its maximum over (−1/λ, ζ_{1}^{+}] at ζ_{1}^{+} and k(ζ_{1}^{+}) < 0. Therefore

k^{2}(z) ≥ k^{2}(ζ_{1}^{+}) = (δ_{1}^{+})^{2} + δ_{1}^{+} O(λ) = \frac{1}{4}ν^{2}λ^{\frac{2}{3}} − \frac{1}{2}ν^{2}ξλ + o(λ), −\frac{1}{λ} < z ≤ ζ_{1}^{+}.

It follows that for sufficiently small λ > 0

Dw_{+}(z) − Ũ((1 − p)w_{+}(z) − zw′_{+}(z))

= \{\frac{1}{2}p σ^{2}((z − θ_{p}) + z(1 − z)\frac{λ}{1 + λz} )^{2} − γ_{2}A^{p+1}(p) λ^{\frac{2}{3}} + MA^{p+1}(p)λ

− θ_{p}A(p)λ + O(λ^{\frac{4}{3}} )\}(1 − p)w_{+}(z)

≥ \{\frac{1}{2}p σ^{2}k(ζ_{1}^{+}) − γ_{2}A^{p+1}(p) λ^{\frac{2}{3}} + MA^{p+1}(p)λ − θ_{p}A(p)λ + O(λ^{\frac{4}{3}} )\}

× (1 − p)w_{+}(z)

= \{(\frac{1}{8}pσ^{2}ν^{2} − γ_{2}A^{p+1}(p)) λ^{\frac{2}{3}} + (MA^{p+1}(p) − \frac{1}{4}pσ^{2}ν^{2}ξ − θ_{p}A(p)) λ

+ o(λ)}(1 − p)w_{+}(z)

= \{(MA^{p+1}(p) − \frac{1}{4}pσ^{2}ν^{2}ξ − θ_{p}A(p)) λ + o(λ)\}(1 − p)w_{+}(z) ≥ 0,

where we have used (4.13) in the last step. \underline{Step 4b:} Interval [ζ_{2}^{+}, 1/λ). This is analogous to Step 4a. \underline{Step 4c:} Interval (ζ_{1}^{+}, ζ_{2}^{+}). From (4.19) and (4.13) we have

Dw_{+}(z) − Ũ((1 − p)w_{+}(z) − zw_{+}(z)) = (θ_{p}A^{−p}(p) + M) λ + O(λ^{\frac{4}{3}} ) ≥ 0. We must also show that

g_{1}(z) ≜ λ(1 − p)w_{+}(z) − (1 + λz)w′_{+}(z) ≥ 0, ζ_{1}^{+} < z < ζ_{2}^{+}.  (4.22)

For z ∈ (ζ_{1}^{+}, ζ_{2}^{+}), we have z − θ_{p} = O(λ^{1/3}). Using this fact, we compute

g_{1}(z) = A^{−p}(p)λ − (1 − p)γ_{2}λ^{\frac{5}{3}} + (1 − p)Mλ^{2}

−\frac{3(1 − p)}{2ν}A^{−p}(p)(z − θ_{p})^{2}λ^{\frac{5}{3}} + \frac{1 − p}{ν^{3}}A^{−p}(p)(z − θ_{p})^{4}λ

+\frac{3(1 + λz)}{ν}A^{−p}(p)(z − θ_{p})λ^{\frac{2}{3}} − \frac{4(1 + λz)}{ν^{3}}A^{−p}(p)(z − θ_{p})^{3},

g′_{1}(z) = \frac{12}{ν}A^{−p}(p)λ^{\frac{2}{3}} [\frac{1}{4} − (\frac{z − θ_{p}}{νλ^{\frac{1}{3}}} )^{2} + O(λ)] .

We know that g_{1}(ζ_{1}^{+}) = 0 and thus, to prove (4.22), it suffices to show that g′_{1} is positive on [ζ_{1}^{+}, ζ_{2}^{+}]. Because −(z − θ_{p})^{2} is a concave function of z, it suffices to check the endpoints. We have for i = 1, 2 that

(\frac{ζ_{i}^{+} − θ_{p}}{νλ^{\frac{1}{3}}} )^{2} = (\frac{1}{2}(1 − ξλ^{\frac{1}{3}} ) + o(λ^{\frac{1}{3}} ))^{2} = \frac{1}{4} − \frac{1}{2}ξλ^{\frac{1}{3}} + o(λ^{\frac{1}{3}} ).

Therefore,

g′_{1}(ζ_{i}^{+}) = \frac{12}{ν}A^{−p}(p)λ^{\frac{2}{3}} [\frac{1}{2}ξλ^{\frac{1}{3}} + o(λ^{\frac{1}{3}} )] > 0

for sufficiently small λ > 0. The proof that g_{2}(z) = λ(1 − p)w_{+}(z) + (1 − λz)w′_{+}(z) is positive for z ∈ [ζ_{1}^{+}, ζ_{2}^{+}] is analogous. This completes the proof that (w_{+}, ζ_{1}^{+}, ζ_{2}^{+}) is a supersolution triple.

\underline{Conclusion:} We note that w_{±}(θ_{p}) = \frac{1}{1−p}A^{−p}(p)−γ_{2} λ^{\frac{2}{3}} ±Mλ, and so Lemma 4.2 implies

\frac{1}{1 − p}A^{−p}(p) − γ_{2}λ^{\frac{2}{3}} − Mλ ≤ u(θ_{p}) ≤ \frac{1}{1 − p}A^{−p}(p) − γ_{2}λ^{\frac{2}{3}} + Mλ.

Corollary 4.4. Assume p > 0, p ≠ 1 and A(p) > 0. For fixed z ∈ T , the value function satisfies

u(z) = \frac{1}{1 − p}A^{−p}(p) − (\frac{9}{32}p θ_{p}^{4}(1 − θ_{p})^{4})^{\frac{1}{3}} A^{−1−p}(p) σ^{2} λ^{\frac{2}{3}} + O(λ). (4.23)

Proof. In the proof of Theorem 4.3 we constructed a supersolution w_{+} and a subsolution w_{−} such that w_{+}(z)−w_{−}(z) = O(λ), w_{±}(z) = w_{±}(θ)+O(λ) for fixed z ∈ T . It follows that u(z) = w_{±}(z) + O(λ) = w_{±}(θ) + O(λ) = u(θ) + O(λ).

Theorem 4.5. Assume p > 0, p ≠ 1, A(p) > 0 and θ_{p} ≠ 1. Then with ν given by (3.12), we have

z_{1} = θ_{p} − \frac{1}{2}νλ^{\frac{1}{3}} + O(λ^{\frac{2}{3}} ), z_{2} = θ_{p} + \frac{1}{2}νλ^{\frac{1}{3}} + O(λ^{\frac{2}{3}} ).

Proof. The value function u is concave, so u′ is monotone. By taking the derivative of u in the BS and SS regions (see (2.12) and (2.13)) we see that

u′(z) = O(λ) ∀z ∈ [z_{1}, z_{2}].  (4.24)

It follows that for z = z_{1} and z = z_{2}, and hence for all z ∈ [z_{1}, z_{2}],

Ũ((1 − p)u(z) − zu′(z)) = \frac{p}{1 − p}A^{1−p}(p) + (1 − p)γ_{2}A(p)λ^{\frac{2}{3}} + O(λ).

We also know that u′′(z) is continuous for z ∈ T \{1}. From equations (2.12) and (2.13) it follows that u′′(z_{i}) = O(λ^{2}) for z_{i} ≠ 1. We can thus write

\begin{aligned}Du(z_{i}) & = (pA(p) + \frac{1}{2}σ^{2}p(1 − p)(z_{i} − θ_{p})^{2})u(z_{i}) + O(λ) \\ & = \frac{p}{1 − p}A^{1−p}(p) − pγ_{2}A(p)λ^{\frac{2}{3}} + \frac{1}{2}σ^{2}pA^{−p}(p)(z_{i} − θ_{p})^{2} + O(λ). \end{aligned}

If z_{i} = 1, the term (1 − z_{i})^{2}u′′(z_{i}) is set equal to zero (see [43], equation (A.5)), and the above equation still holds. In order to satisfy Equation (2.8) we must have

\begin{aligned}0 & = Du(z_{i}) − Ũ((1 − p)u(z_{i}) − z_{i}u′(z_{i})) \\ & = −γ_{2}A(p) λ^{\frac{2}{3}} + \frac{1}{2}σ^{2}pA^{−p}(p)(z_{i} − θ_{p})^{2} + O(λ). \end{aligned}  (4.25)

z_{i} = θ_{p} ± λ^{\frac{1}{3}} \sqrt{\frac{2}{pσ^{2}} A^{p+1}(p)γ_{2}} + O(λ^{\frac{2}{3}} ) = θ_{p} ± (\frac{3}{2p}θ_{p}^{2}(1 − θ_{p})^{2})^{\frac{1}{3}} λ^{\frac{1}{3}} + O(λ^{\frac{2}{3}} ),

where the − sign is for z_{1} and the + sign is for z_{2}.

Remark 4.6. The proof of Theorem 4.3 is valid so long as θ_{p} ≠ 1. The case of θ_{p} = 1 can be considered a singular case for which the parameter ν appearing in a denominator in the definition of h(δ) is zero. The intuition is that if the optimal proportion in the risky asset is 100% of wealth, then the investor does not incur any transaction costs due to adjusting the position size, as soon as the agent’s position equals the optimum of 100% in stock. However, in order to consume the agent must transfer money from stock to money market and pay the transaction cost. We could regard this transaction cost as a consumption tax. It follows that the loss in value function is of order of λ rather than λ^{2/3}. In [43], Theorems 11.2 and 11.6 show that in this case z_{2} = θ_{p} = 1 > z_{1} > 0, and [43], Corollary 9.10 asserts that

v(x, y) = \frac{1}{1 − p}A^{−p}(p)(x + (1 − λ)y)^{1−p}, (x, y) ∈ S, x ≤ 0,

or equivalently,

u(z) = v(1 − z, z) = \frac{1}{1 − p}A^{−p}(p)(1 − λz)^{1−p}, 1 ≤ z < \frac{1}{λ}.

We see then that

u(θ_{p}) = \frac{1}{1 − p}A^{−p}(p) − A^{−p}(p)λ + O(λ^{2}),  (4.26)

which is consistent with (4.12).

The proof of Theorem 4.5 is valid for θ_{p} = 1, except the last step. When θ_{p} = 1, we have γ_{2} = 0 and (4.25) reduces to

z_{1} − θ_{p} = z_{1} − 1 = O(λ^{\frac{1}{2}} ).

Remark 4.7. The optimal consumption policy c is given by v_{x}^{−1/p} (see (2.4) and (2.3)). Since v(x, y) = (x + y)^{1−p}u(y/(x + y)) (see (2.7)), we have

v_{x}(x, y) = (1 − p)(x + y)^{−p}u (\frac{y}{x + y} ) − y(x + y)^{−1−p}u′ (\frac{y}{x + y} )
= (1 − p)(x + y)^{−p}u (\frac{y}{x + y} ) + O(λ),

where we have used (4.24). Therefore, Theorem 4.3 implies that the optimal consumption in feedback form is

c = (A^{−p}(p)−(1−p)γ_{2}λ^{\frac{2}{3}} )^{− \frac{1}{p}} (x+y)+O(λ) = (A(p)+ \frac{1 − p}{p}γ_{2}λ^{\frac{2}{3}} )(x+y)+O(λ).

Observe that for sufficiently small λ the existence of transaction costs increases the size of consumption for p ∈ (0, 1), while the consumption is decreased for p > 1. The intuition explaining this observation is related to the fact that the index of intertemporal substitution, which is equal to 1/p, is high for small p. An agent with index of intertemporal substitution higher than 1 will optimally avoid some transaction costs resulting from trading by consuming faster.

Remark 4.8. There is considerable evidence that

u(θ_{p}) = \frac{1}{1 − p}A^{−p}(p) − γ_{2}λ^{\frac{2}{3}} − θ_{p}A^{−p}(p) λ + O(λ^{\frac{4}{3}} ).  (4.27)

We have just seen in (4.26) that this is the case when θ_{p} = 1 (and consequently γ_{2} = 0). In the proof of Theorem 4.3, any choice of M > θ_{p}A^{−p}(p) gives us a subsolution of the form

w_{−}(θ_{p}) = \frac{1}{1 − p}A^{−p}(p) − γ_{2}λ^{\frac{2}{3}} − Mλ + O(λ^{\frac{4}{3}} ).  (4.28)

(The second term on the right-hand side of (4.13) is needed only for the supersolution argument.) Finally, the coefficient −θ_{p}A^{−p}(p) on λ can be obtained by a tedious heuristic analysis along the lines of Section 3.

In Remark 4.6 we introduced the concept of consumption tax as an interpretation of the transaction cost an agent must pay in order to first move capital from stock to money market before consuming it. In that remark, the agent was ideally 100% invested in stock. If instead the agent seeks to hold a proportion θ_{p} ≠ 1 in stock, then consuming proportionally from stock and money market, the agent would pay a consumption tax λθ_{p} times the total consumption. To highest order, the optimal consumption level is thus (1 − λθ_{p}) of what it would be if there were no transaction cost, and this multiplies the value function by

(1 − λθ_{p})^{1−p} = 1 − (1 − p)θ_{p} λ + O(λ^{2}).

The value function for zero transaction cost when wealth is 1 is A^{−p}(p)/(1 − p), and so after this multiplication, the value function has been reduced by

θ_{p}A^{−p}(p) λ,

which is the order λ term we see in (4.27). We thus expect the value function for the problem with transaction cost λ to be reduced from the zero-transaction cost value function A^{−p}(p)/(1 − p) by at least θ_{p}A^{−p}(p)λ, this reduction being due solely to the cost of moving capital from stock to money market in order to consume. There is also a cost of trading to stay in the NT wedge, which reduces the value function by γ_{2}λ^{\frac{2}{3}} , but cannot further reduce the value function by an order λ 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 4.5. We already have from Theorem 4.5 and its proof that z_{i} − θ_{p} = O(λ^{1/3}) and that u′(z) = O(λ) for z ∈ [z_{1}, z_{2}]. The mean-value theorem gives

u(z_{i}) = u(θ_{p}) + O(λ^{\frac{4}{3}} ),

and

(1 − p)u(z_{1}) − z_{1}u′(z_{1}) = \frac{(1 − p)u(z_{1})}{1 + λz_{1}}
= A^{−p}(p) − (1 − p)γ_{2} λ^{\frac{2}{3}} − (2 − p)θ_{p}A^{−p}(p) λ + O(λ^{\frac{4}{3}} ),
(1 − p)u(z_{2}) − z_{2}u′(z_{2}) = \frac{(1 − p)u(z_{2})}{1 − λz_{2}}
= A^{−p}(p) − (1 − p)γ_{2} λ^{\frac{2}{3}} + pθ_{p}A^{−p}(p) λ + O(λ^{\frac{4}{3}} ).

Using Lemma 2.1, we obtain

Ũ((1 − p)u(z_{2}) − z_{2}u′(z_{2}))

= \frac{p}{1 − p}A^{1−p}(p) + (1 − p)A(p)γ_{2} λ^{\frac{2}{3}} + (2 − p)θ_{p}A^{1−p}(p) λ + O(λ^{\frac{4}{3}} ),

Ũ((1 − p)u(z_{1}) − z_{1}u′(z_{1}))

= \frac{p}{1 − p}A^{1−p}(p) + (1 − p)A(p)γ_{2} λ^{\frac{2}{3}} − pθ_{p}A^{1−p}(p) λ + O(λ^{\frac{4}{3}} ).

We write for z_{1} and z_{2} similarly as in Theorem 4.5

Du(z_{i}) = \frac{p}{1 − p}A^{1−p}(p)−pA(p)γ_{2} λ^{\frac{2}{3}} −pθ_{p}A^{1−p}(p) λ+\frac{1}{2}σ^{2}pA^{−p}(p)(z_{i}−θ_{p})^{2}+O(λ^{\frac{4}{3}} ).

From Du(z_{i}) = Ũ((1 − p)u(z_{i}) − z_{i}u′(z_{i})), we now see that

(z_{1} − θ_{p})^{2} = \frac{2}{pσ^{2}} (A^{1+p}(p)γ_{2} λ^{\frac{2}{3}} + 2θ_{p}A(p) λ) + O(λ^{\frac{4}{3}} ), (4.29)
(z_{2} − θ_{p})^{2} = \frac{2}{pσ^{2}} A^{1+p}(p)γ_{2} λ^{\frac{2}{3}} + O(λ^{\frac{4}{3}} ),  (4.30)

which implies

\begin{aligned}z_{1} − θ_{p} & = −\frac{1}{2}ν λ^{\frac{1}{3}} − \frac{4θ_{p}A(p)}{σ^{2}p ν} λ^{\frac{2}{3}} + O(λ), \\ z_{2} − θ_{p} & = \frac{1}{2}ν λ^{\frac{1}{3}} + O(λ). \end{aligned}

This suggests that the optimal policy is to keep a wider wedge on the right side of the Merton proportion θ_{p}. This extra width makes sense because consumption reduces the money market position.

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

z_{1} − θ_{p} = z_{1} − 1 = −\sqrt{\frac{4θ_{p}A(p)}{σ^{2}p}} λ^{\frac{1}{2}} + O(λ^{\frac{5}{6}} ), z_{2} − θ_{p} = z_{2} − 1 = O(λ).

In fact, when θ_{p} = 1, we have z_{2} = θ_{p}.

Remark 4.9. In the key formulas derived in this paper, the transaction cost parameter λ appears in combination with Γ ≜ θ_{p}(1−θ_{p}). According to Theorem 4.3, the highest order loss in the value function due to transaction costs is

(\frac{9p}{32})^{\frac{1}{3}} A^{−1−p}σ^{2} (Γ^{2}λ)^{\frac{2}{3}}  (4.31)

From Theorem 4.5, we see that

z_{i} − θ_{p} = (−1)^{i} (\frac{3}{2p})^{\frac{1}{3}} (Γ^{2}λ)^{\frac{1}{3}} .  (4.32)

One way to see the intrinsic nature of the quantity Γ^{2}λ is to define the proportion of capital in stock, θ_{t} = Y_{t}/(X_{t} + Y_{t}), and apply Itô’s formula when (X_{t}, Y_{t}) is generated by the optimal triple (C, L, M), for which L and M are continuous, to derive the equation

dθ_{t} = θ_{t}(1 − θ_{t})\frac{dS_{t}}{S_{t}} − rθ_{t}(1 − θ_{t}) dt − σ^{2}θ_{t}^{2}(1 − θ_{t}) dt + \frac{θ_{t}C_{t}}{X_{t} + Y_{t}} dt

+\frac{λθ_{t} + 1}{X_{t} + Y_{t}} dL_{t} + \frac{λθ_{t} − 1}{X_{t} + Y_{t}} dM_{t}.

We see that the response of θ_{t} to relative changes in the stock price is θ_{t}(1−θ_{t}). Our desire is to keep θ_{t} in the interval [z_{1}, z_{2}]. If the leading term in the above equation were scaled to be ρθ_{t}(1 − θ_{t})\frac{dS_{t}}{S_{t}}, then because the dt terms have a lower order effect on the dynamics, this would be approximately equivalent to scaling time by ρ^{2}, and so the local time of θ_{t} at the endpoints of [z_{1}, z_{2}] would approximately be scaled by ρ^{2}. If at the same time the transaction cost were replaced by λ/ρ^{2}, then the total amount of transacting would be approximately unaffected. The quantity θ_{t}^{2}(1 − θ_{t})^{2}λ is invariant under this scaling. Because the optimal policy keeps θ_{t} near θ_{p}, the quantity Γ^{2}λ appears to be intrinsic.

When replicating an option by trading, the position held by the hedging portfolio, denominated in shares of stock, is call the delta of the option, and the sensitivity of the delta to changes in the stock price is the gamma. We have here a similar situation, except that θ_{t} is the proportion of capital held in stock, rather than the number of shares of stock, and θ_{t}(1 − θ_{t}) is the sensitivity of this proportion to relative changes in the stock price.

It is interesting to note that the quantity Γ^{2}λ also plays a fundamental role in the formal asymptotic expansions of Whalley and Wilmott [49]. In fact, even the constants (\frac{9}{32})^{\frac{1}{3}} and (\frac{3}{2})^{\frac{1}{3}} 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 δ of the form O(λ^{1/3}). For such δ, we may write the terms of order λ and lower in f_{1}^{±}(δ) of (4.14) as

f_{1}^{±}(δ)

= ν λ^{\frac{1}{3}} − (1 − p)γ_{2}νA^{p}(p) λ − h′(δ) λ^{− \frac{2}{3}} − θ_{p}h′(δ) λ^{\frac{1}{3}} + h′(δ) δλ^{\frac{1}{3}} + O(λ^{\frac{4}{3}} )
= ν λ^{\frac{1}{3}} − (1 − p)γ_{2}νA^{p}(p) λ − 3δ + \frac{4}{ν^{2}} · \frac{δ^{3}}{λ^{\frac{2}{3}}} − 3Bδλ^{\frac{2}{3}} + O(λ^{\frac{4}{3}} ).

Consider δ_{0} ≜ \frac{1}{2}νλ^{1/3} (1 − ξ_{0}λ^{1/3}). Then

f_{1}^{±}(δ_{0}) = ν λ^{\frac{1}{3}} − (1 − p)γ_{2}νA^{p}(p) λ − \frac{3}{2}ν λ^{\frac{1}{3}} + \frac{3}{2}νξ_{0} λ^{\frac{2}{3}}

+ \frac{1}{2}ν λ^{\frac{1}{3}} (1 − 3ξ_{0}λ^{\frac{1}{3}} + 3ξ_{0}^{2}λ^{\frac{2}{3}} ) − \frac{3}{2}Bν λ + O(λ^{\frac{4}{3}} )

= ν(\frac{3}{2}ξ_{0}^{2} − (1 − p)γ_{2}A^{p}(p) − \frac{3}{2}B) λ + O(λ^{\frac{4}{3}} ).

With ξ = \sqrt{\frac{2}{3}(1 − p)γ_{2}A^{p}(p) + B} > 0, we take ξ_{0} = \sqrt{ξ^{2} + η}, where |η| <

ξ^{2}. Then f_{1}^{±}(δ_{0}) = \frac{3}{2}νη λ + O(λ^{\frac{4}{3}} ). Thus, for η > 0 we have f_{1}^{±}(δ_{0}) > 0 for sufficiently small λ > 0, and for η < 0 we have f_{1}^{±}(δ_{0}) < 0 for sufficiently small λ > 0. Therefore, for every η ∈ (0, ξ^{2}) and sufficiently small λ > 0, there exists

δ_{1}^{±} ∈ (\frac{1}{2}ν λ^{\frac{1}{3}} (1 − λ^{\frac{1}{3}} \sqrt{ξ^{2} − η}), \frac{1}{2}ν λ^{\frac{1}{3}} (1 − λ^{\frac{1}{3}} \sqrt{ξ^{2} + η}))

satisfying f_{1}^{±}(δ_{1}^{±}) = 0. In other words, δ_{1}^{±} = \frac{1}{2}νλ^{\frac{1}{3}} (1 − ξ λ^{\frac{1}{3}} ) + o(λ^{\frac{2}{3}} ). The

proof of the existence of δ_{2}^{±} is analogous.

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