Optimal Investment and Consumption for A Portfolio with Stochastic Dividends

Transcription

1 Optimal Investment and Consumption for A Portfolio with Stochastic Dividends Tao Pang 1 and Katherine Varga 2 November 12, 215 Abstract: In this paper, we consider a portfolio optimization model in which an investor can invest in a risky asset and a riskless asset. For the risky asset, not only does its price fluctuate from time to time, but its dividend payment is also unstable. This model can be used to model a stock with stochastic dividends, or a farmland with stochastic productivity rate, or a commercial real estate with unstable rental income. The goal is to choose the optimal investment and consumption (or withdrawal) to achieve the best satisfaction level measured by utility functions. We use the dynamic programming principle to derive the Hamilton-Jacobi-Bellman equation, which we proceed to prove existence of a classical solution. Our value function is chosen to maximize the expected total discounted HARA utility of consumption. The optimal investment and consumption strategies are derived and verified. Keywords: Portfolio Optimization; Utility Function; Stochastic Dividend. 1 Introduction In the classical Merton portfolio optimization problem, an investor distributes his wealth between a risky asset (e.g., stock) and a riskless asset. The price 1 Tao Pang, Department of Mathematics, North Carolina State University, Raleigh, NC , USA. tpang@ncsu.edu. 2 Katherine Varga, CoBank, 634 S. Fiddlers Green Circle, Greenwood Village, CO 8111, USA. kvarga@cobank.com. 1

2 P t of the risky asset is governed by geometric Brownian motion: dp t P t = µdt + σdb t, (1.1) where µ and σ are the positive constant drift and volatility, respectively, and B t is a one-dimensional standard Brownian motion. The controls are investment and consumption, that is, at any time t the investor chooses how much of his wealth to invest and how much to consume. The goal is to maximize his expected discounted utility of consumption. There are many literatures that consider portfolio optimization of Merton s type. Fleming and Pang, in 5], assume stock price is governed by geometric Brownian motion and the risk-free interest rate is stochastic. Some related models were considered in Pang 19, 2]. On the other hand, some researchers have considered the stochastic volatility model (see Fleming- Hernández-Hernández 3, 4], Hata-Sheu (11, 12]), Nakai 15], and Noh- Kim16]). More extensions of Merton s model can be found in Bielecki-Pliska 1], Fleming-Sheu 7], Zariphopoulou 22] and the references therein. Here we consider an economic unit with productive capital and liability in the form of debt. The capital can be regarded as a risky asset and the liability bears a risk-free interest rate. In the classical Merton s model, the price of the risky asset is assumed to be a geometric Brownian motion (see (1.1)), which assumes the expected return rate for the risk asset is a constant (µ in (1.1)). However, as we know, the expected return rate may not be a constant all the time. For example, when the stock price has been well above its 2 day moving average, it is expected that the future expected return rate will be much lower. So here we consider a stochastic model for the price of the capital in which the expected return rate depends on the current market price. In addition, typically, a stock pays some dividends, but the dividend payments may not be stable. So in our model, we also consider the stochastic dividend payment. For capital with productivity, the productive output can be treated as dividend payment. Examples of capital with risky market price and unstable productivity rate include but are not limited to: stock with dividend, farm land with farm output (rice, corns, fruits, etc.), commercial real estate with rental income and retail stores with sale profits. As an example in the Merton problem context, the economic unit may be interpreted as an investor who receives dividends on his investments. The worth of the investor depends on particular sources of uncertainty. Dividends and stock prices fluctuate over time, so we allow them to be stochastic. This 2

3 model can be regarded as an optimal investment and withdrawal (consumption) problem for an investment unit, such as an private equity investor, an investment company, or a real estate company, a company with chain retail stores, etc. Our work is motivated by similar models for optimal investment and consumption that were considered in Fleming and Pang 6] in which they consider stochastic productivity. Some related models have been considered in Chang-Pang-Yang 2], Pang-Hussain 21], in which some portfolio optimization models for stochastic systems with delays are considered. The rest of this paper is organized as the following. In Section 2, the problem is formulated as a stochastic control problem. The dynamic programming equation (Hamilton-Jacobi-Bellman (HJB) equation) for the value function is derived in Section 3. By virtue of the sub-super solution method, we establish the existence results for the HJB equation in Section 4. Finally, in Section 5, we prove that the solution of the HJB equation is equal to our value function and the optimal consumption and investment strategies are derived. 2 Problem Formulation Consider an investor who, at time t, owns N t shares of stock at price per share P t. The total worth of investments is given by K t = N t P t. Then we can write dp t dk t = N t dp t + P t dn t = K t + I t dt, P t where I t is the investment rate at time t. For example, if dt represents one month and he decides at time t to invest an additional $1 per month at a $1 price per share, then the number of shares purchased will be dn t = I tdt P t = 1. The investor s debt, L t, increases with interest payments, investment, and consumption, and decreases with income. Thus our equation for the change in debt is given by dl t = (rl t + I t + C t D t )dt, (2.1) where r is a constant interest rate, C t is the consumption rate or the withdrawal rate, and D t is the rate of income from production. In other 3

4 words, D t is the total earned in dividends per unit time at time t. The total in dividends is proportional to the productivity of capital, or dividend rate, b t : D t dt = b t N t dt. (2.2) The investor s net worth is given by X t = K t L t. (2.3) We require X t >. It is worth noting that the model applies to any economic unit with productive capital and liabilities. For another example, consider a farm on which produce is grown and then sold for profit. N t may represent the number of acres of the farm, with P t being the property value per acre. Debt increases with property taxes, the purchase of new land, and consumption, and decreases with income from selling the produce. Another example is the commercial real estate for an investment company. Not only the market price for the commercial real estate will change from time to time, the rental income, which can be treated as the dividend, is unstable, too. So this model can be used to describe a company with risky investment on commercial real estate. From here on, we continue our explanations with the example of the investor. We consider the dividend rate and stock price to be stochastic. In particular, we allow the dividend rate to be governed by the following equation: b t dt = bdt + σdb t, (2.4) where b, σ > are constants. The general equation for stock price is dp t P t = µ t dt + σ t d B t. (2.5) Here B t and B t are one-dimensional standard Brownian motions, which we allow to be correlated with correlation constant ρ ( 1, 1]. That is, Ed B t db t ] = ρdt. We restrict ρ 1 for technical reasons, which will be apparent later. In addition, it is reasonable to assume that the dividend process and the stock 4

5 price process are not perfectly negatively correlated. The change in net worth of the investor is given by dx t = K t dp t P t + I t dt (rl t + I t + C t D t )dt. (2.6) The controls k t = K t and c t = C t are the proportion of wealth invested and X t X t the proportion of wealth consumed, respectively. Substituting L t = K t X t, (2.2), (2.5), and our controls into (2.6), we simplify to obtain ( ) ] b dx t = X t + µ t r k t + (r c t ) dt + σk tx t db t + σ t k t X t d P t P B t. (2.7) t In this paper we specify certain conditions on the the stock price drift and volatility. Here we allow the drift to be a bounded function of price, and take the volatility to be constant. We use the dynamic programming principle to derive the Hamilon-Jacobi-Bellman (HJB) equation, from which we proceed to prove existence of a classical solution using a subsolution/supersolution method, and then verify that the solution obtained is equal to our value function, the maximum expected discounted utility of consumption. Assume that the unit price of capital, P t, satisfies the stochastic differential equation dp t P t = µ(p t )dt + σd B t, (2.8) where the volatility σ > is constant, and the drift parameter µ is a bounded function price. That is, we suppose there exists a constant M 1 > such that Let µ(p t ) M 1. Y t = log P t. For notational purposes, we will now write µ(y t ) rather than µ(p t ). Using Ito s rule with (2.8), we have dy t = 1 dp t 1 1 (dp P t 2 Pt 2 t ) 2 = µ(y t )dt + σd B t 1 2 σ2 dt = (µ(y t ) 1 ) 2 σ2 dt + σd B t, 5

6 and so dy t = µ(y t )dt + σd B t, (2.9) where µ(y t ) = µ(y t ) 1 2 σ2. Note that µ is also bounded: µ(y) µ(y) σ2 M σ2 M 2. (2.1) Now we can rewrite (2.7) as (be Y dx t = X t t +µ(y t ) r ) k t +(r c t ) ]dt+x t σk t e Yt db t + σk t d B ] t. (2.11) Note that we have two state variables, X t and Y t. Let their initial values be given by X = x and y = y. Recall that our controls k t = K t /X t and c t = C t /X t are the proportions of wealth invested and consumed, respectively. We require that they belong to the admissible control space, Π, as defined below. Definition 2.1. (Admissible Control Space) The pair (k t, c t ) is said to be in the admissible control space Π if (k t, c t ) is an R 2 -process which is progressively measurable with respect to a (B t, B t )-adapted family of σ- algebras (F t, t ). Moreover, we require that k t, c t, and ( ) ( ) P r kt 2 dt < = 1, P r c t dt < = 1 for all T >. Our goal is to maximize the expected total discounted HARA utility of consumption. The objective function is given by J(x, y, k., c.) = E x,y e δt 1 ] γ (c tx t ) γ dt, (2.12) which we wish to maximize subject to the constraints (k t, c t ) Π and x >, while y R. The parameter δ is a positive discount factor. The corresponding value function is V (x, y) = sup (k t,c t) Π E x,y e δt 1 ] γ (c tx t ) γ dt. (2.13) 6

7 3 Dynamic Programming Equations Using the dynamic programming principle, we can get the following dynamic programming equation for V (x, y) : δv = σ2 2 V yy + µ(y)v y + rxv x + max cxv x + 1γ ] c (cx)γ (3.1) ] + max (be y + µ(y) r)kxv x + ( σ 2 + ρσ σe y )kxv xy + k2 x 2 q(y) V xx, k 2 where q(y) σ 2 e 2y + 2ρσ σe y + σ 2. (3.2) Equation (3.1) is the Hamilton-Jacobi-Bellman (HJB) equation for V (x, y) as defined in (2.13). We assume ρ 1. To be more precise, we define a constant 1 < ρ < (that may be arbitrarily close to 1), and require ρ ρ, 1]. Then q(y) is bounded below by a positive constant. Actually, we have q(y) σ 2 e 2y + 2ρ σ σe y + σ 2 = (σe y + ρ σ) 2 + (1 ρ 2 ) σ 2 (1 ρ 2 ) σ 2 >. (3.3) For the non-log HARA utility function, we can show that V (x, y) is homogeneous in x with order y. Using equation (2.11) for dx t, and for any constant λ >, we have d(λx t ) = λ dx t, λx = λx, and J(λx, y, k., c.) = E x,y e δt 1 ] γ (c tλx t ) γ dt = λ γ E x,y e δt 1 ] γ (c tx t ) γ dt = λ γ J(x, y, k., c.). Then V (x, y) = sup J(x, y, k., c.) = sup x γ J(1, y, k., c.) = x γ V (1, y). (k.,c.) Π (k.,c.) Π 7

8 We can then write V in the form V (x, y) = 1 γ xγ W (y). (3.4) Using equation (3.4), we get a reduced DPE for W : δw = σ2 2 W yy + µ(y)w y + γrw + max ] γcw + c γ c + γ max k (be y + µ(y) r ) kw + ( σ 2 + ρσ σe y )kw y k2 (1 γ)q(y)w 2 Let Q(y) = log W (y). Then W y W = Q y and W yy W = Q yy + Q 2 y. (3.5) Then we can get the equation for Q(y) as the following δ = σ2 2 (Q yy + Q 2 y) + µ(y)q y + γr + max γce Q + c γ] + γ max G(y, Q y, k), (3.6) c k where G(y, p, k) be y + µ(y) r + ( σ 2 + ρσ σe y )p ]k k2 (1 γ)q(y). (3.7) 2 The candidates for the optimal controls are c = e Q(y) γ 1, k = ] be y + µ(y) r + ( σ 2 + ρσ σe y + )Q y, (3.8) (1 γ)q(y) where a + max(a, ). Plug in the above controls, then we can get the equation for Q: δ = σ2 2 (Q yy + Q 2 y) + µ(y)q y + γr + (1 γ)e Q γ 1 + γg(y, Qy, k ). (3.9) From the definition of k and G, it is easy to verify that G(y, p, k ), G(y,, k ) = Ψ(y), (3.1) 8 ].

9 where (be y + µ(y) r) 2 if be y + µ(y) r >, Ψ(y) 2q(y)(1 γ) otherwise. (3.11) Define H(y, z, p) σ2 2 p2 µ(y)p γg(y, p) γr + δ (1 γ)e z γ 1. (3.12) Note that H is a continuous function in each variable. Now we can rewrite Equation (3.9) as σ 2 2 Q yy = H(y, Q, Q y ). (3.13) Equation (3.13) is the Hamilton-Jacobi-Bellman equation for Q(y) corresponding to the value function (2.13), where V (x, y) = 1 γ xγ e Q(y). 4 Existence Results To prove existence of solution to (3.13), we use the method of subsolution and supersolution. Definition 4.1. Q(y) is a subsolution (supersolution) of (3.13) provided that σ 2 2 Q yy ( ) H(y, Q, Q y ). (4.1) In addition, if ˆQ is a subsolution, Q is a supersolution, and ˆQ Q, then we say that ˆQ, Q is an ordered pair of subsolution/supersolution. We wish to identify a particular subsolution and supersolution. In doing so, we will make use of the following lemma. Lemma 4.1. Ψ(y), as defined in (3.11), is bounded. Proof. We first consider the case ρ 1. Using the identity (a + b) 2 9

10 2a 2 + 2b 2, and the fact that σ, σ, e y >, we have Ψ(y) = (be y + µ(y) r) 2 2(1 γ)(σ 2 e 2y + 2ρσ σe y + σ 2 ) (be y + µ(y) r) 2 2(1 γ)(σ 2 e 2y + σ 2 ) 2b2 e 2y + 2(µ(y) r) 2 2(1 γ)(σ 2 e 2y + σ 2 ). By performing some simplifications and applying our upper bound on µ(y), we get 2b 2 e 2y + 2(µ(y) r) 2 2(1 γ)(σ 2 e 2y + σ 2 ) = b2 e 2y σ 2 + σ 2 (µ(y) 2 r) 2 σ 2 (1 γ)(σ 2 e 2y + σ 2 ) = b2 (σ 2 e 2y + σ 2 ) + σ 2 (µ(y) 2 r) 2 b 2 σ 2 σ 2 (1 γ)(σ 2 e y + σ 2 ) = b 2 σ 2 (1 γ) + (µ(y) r)2 b2 σ2 σ 2 (1 γ)(σ 2 e 2y + σ 2 ) b 2 σ 2 (1 γ) + (M 1 + r) 2 b2 σ2 σ 2 (1 γ)(σ 2 e 2y + σ 2 ). The last step is true because µ(y) r µ(y) + r M 1 + r, (M 1 + r) 2 b 2 σ 2 /σ 2, then r. If Ψ(y) b 2 σ 2 (1 γ). (4.2) Otherwise, we have Ψ(y) b 2 σ 2 (1 γ) + (M 1 + r) 2 b2 σ2 σ 2 = (M 1 + r) 2 (1 γ) σ 2 σ 2 (1 γ). Thus, for ρ 1, { Ψ(y) max b 2 σ 2 (1 γ), (M 1 + r) 2 σ 2 (1 γ) }. 1

11 We can get a similar bound if ρ is negative. If ρ ρ <, then ρ(σe y σ) 2, which implies 2ρσ σe y ρσ 2 e 2y + ρ σ 2. Then we have Ψ(y) = = (be y + µ(y) r) 2 2(1 γ)(σ 2 e 2y + 2ρσ σe y + σ 2 ) (be y + µ(y) r) 2 2(1 γ)(σ 2 e 2y + ρσ 2 e 2y + ρ σ 2 + σ 2 ) (be y + µ(y) r) 2 2(1 γ)(1 + ρ)(σ 2 e 2y + σ 2 ) 2b 2 e 2y + 2(µ(y) r) 2 2(1 γ)(1 + ρ)(σ 2 e 2y + σ 2 ) { 1 b 2 max 1 + ρ σ 2 (1 γ), (M } 1 + r) 2. σ 2 (1 γ) Let Then { Ψ(y) max b 2 σ 2 (1 γ), (M 1 + r) 2 σ 2 (1 γ) and so we see that Ψ(y) is bounded. }, Ψ max { Ψ, Ψ 1 + ρ }. (4.3) Ψ(y) Ψ <, (4.4) We now identify an ordered pair of subsolution/supersolution to (3.13). Lemma 4.2. Suppose < γ < 1 and In addition, define K 1 (γ 1) log Then any constant K K 1 is a subsolution of (3.13). δ > γr. (4.5) ] δ γr. (4.6) 1 γ Proof. Since K is a constant, its derivatives are all. So we only need to show that γψ(y) γr + δ (1 γ)e K γ 1. 11

12 Since K K 1 = (γ 1) log e K γ 1 e K 1 γ 1 = exp log ( ) δ γr, and (γ 1) <, we have 1 γ Therefore, using Ψ(y) and γ >, we can get ( )] δ γr = δ γr 1 γ 1 γ. γψ(y) γr + δ (1 γ)e K γ 1 γr + δ + γr δ. This completes the proof. Lemma 4.3. Suppose < γ < 1 and δ > γ(r + Ψ), (4.7) where Ψ is defined by (4.3). In addition, define ] δ γ(r + Ψ) K 2 (γ 1) log. (4.8) 1 γ Then any constant K K 2 is a supersolution of (3.13). Proof. The proof is similar to the proof of Lemma 4.2, but instead we want to show that γψ(y) γr + δ (1 γ)e K γ 1. ( ) δ γ(r + Ψ) Since K K 2 = (γ 1) log and (γ 1) <, we have 1 γ ( )] e K γ 1 δ γ(r + Ψ) δ γ(r + Ψ) exp log =. 1 γ 1 γ Therefore, using that Ψ(y) Ψ, we can get γψ(y) γr + δ (1 γ)e K γ 1 γψ(y) γr + δ δ + γ(r + Ψ) This completes the proof. = γ( Ψ Ψ(y)). Lemma 4.4. Suppose K 1 and K 2 are defined by (4.6) and (4.8) respectively, (4.7) holds, and < γ < 1. Then K 1, K 2 is an ordered subsolution/supersolution pair of (3.13). 12

13 Proof. Since Ψ, δ > γ(r + Ψ) implies δ > γr. Using γ >, we can easily get K 1 K 2. Thus K 1, K 2 is an ordered subsolution/supersolution pair of (3.13). The following theorem gives existence of a classical solution to (3.13). The result follows from Theroem 3.8 of 5]. Theorem 4.5. Suppose K 1 and K 2 are defined by (4.6) and (4.8) respectively, (4.7) holds, and < γ < 1. Then (3.13) has a bounded classical solution Q(y) which satisfies K 1 Q(y) K 2. (4.9) Proof. By Lemma 4.4, we have that K 1, K 2 is an ordered subsolution/supersolution pair of (3.13). Note that H is strictly increasing with respect to z. In order to use Theorem 3.8 from 5], we need to show that, for H defined by (3.12) H(y, z, p) C 1 (p 2 + C 2 ), (4.1) for some constants C 1 > and C 2 >. We start by obtaining such a bound on G(y, p, k) defined by (3.7): G(y, p, k) = be y + µ(y) r + ( σ 2 + ρσ σe y )p ]k k2 (1 γ)q(y) 2 be y + µ(y) r + ( σ 2 + ρσ σe y )p ] 2 2(1 γ)q(y) 2(be y + µ(y) r) 2 + 2( σ 2 + ρσ σe y ) 2 p 2 2(1 γ)q(y) = 2Ψ(y) + 2( σ2 + ρσ σe y ) 2 p 2, (4.11) (1 γ)q(y) where q(y) = σ 2 e 2y + 2ρσ σe y + σ 2. Consider the coefficient of p 2 in the second term of (4.13), and note that it is positive. If we let u = e y, we see that this coefficient is a rational expression involving a quadratic in u divided by another quadratic in u, where u >. The limit as u tends to infinity (u ) of such an expression is a constant. In addition, we have 13

14 q(y) q >, so the denominator does not approach zero. Therefore this coefficient is bounded above by some positive constant C 1 : 2( σ 2 + ρσ σe y ) 2 (1 γ)q(y) Using (4.12) along with (4.4), we have that C 1. (4.12) G(y, p, k) C 1 p Ψ. (4.13) Now using (3.12), (2.1), (4.13), and (4.9), we get the following bound for K 1 z K 2 : H(y, z, p) σ2 z 2 p2 + µ(y)p + γ max G(y, p, k) + γr + δ + (1 γ)e γ 1 k σ2 2 p2 + M 2 p + γ( C 1 p Ψ) + γr + δ + (1 γ)e K1 C 1 (p 2 + C 2 ) where C 1 and C 2 are positive constants. Therefore by 5], since H is strictly increasing with respect to z and we can bound H as in (4.1), (3.13) has a classical solution Q(y) such that K 1 Q(y) K 2. γ 1 Now that we have existence of a classical solution Q(y) to the HJB Equation (3.13). It is easy to verify that Ṽ (x, y) = 1 γ xγ e Q(y) (4.14) is a classical solution of the HJB equations (3.1). Next, we will show that Ṽ (x, y) is equal to the value function defined by (2.13). 5 Verification Theorem Before stating the Verification Theorem, we will need a few results. First, let Ṽ be given by (5.9). Now the candidate for the optimal controls are c (Y t ) = e Q(Y t ) γ 1, k be Yt + µ(y t ) r + ( σ 2 + ρσ σe Yt ) (Y t ) = Q ] + y, (1 γ)q(y t ) (5.1) 14

15 Lemma 5.1. Suppose < γ < 1 and (4.7) holds. If Q(y) is a classical solution of (3.13) which satisfies (4.9), then Q y (y) is bounded. That is, there exists a constant A > such that Q y (y) A. (5.2) Proof. We know Q(y) is bounded by constants K 1 and K 2. Since Q is a classical solution of (3.13), Q C 2. So its first and second derivatives are defined everywhere. Thus to prove that Q y is bounded, it is sufficient to prove that it is bounded at its local extrema. Suppose Q y has a local max or min at y. Then Q yy (y ) =. Therefore we have = σ2 2 Q 2 y(y ) + µ(y ) Q y (y ) + γr δ + (1 γ)e Q(y) γ 1 + γ max G(y, Q y (y ), k). k Since max k G(y, p, k) and γr + (1 γ)e Q γ 1, we have σ 2 2 Q 2 y(y ) + µ(y ) Q y (y ) δ. (5.3) The left-hand side of (5.3) is a quadratic expression in Q y (y ). Since the quadratic is concave up, the only Q y -values that satisfy (5.3) are those bounded between the two roots of the quadratic. The roots are given by r 1 µ(y ) µ 2 (y ) + 2 σ 2 δ and r σ 2 2 µ(y ) + µ2 (y ) + 2 σ 2 δ, σ 2 where r 1 r 2. We need bounds for Q y that are independent of y. Using (2.1), we have r 1 M 2 M σ 2 δ r σ 2 1 and r 2 M 2 + M σ 2 δ r σ 2 2, so that r 1 Q y (y ) r 2. (5.4) Thus Q y is bounded at y. Since y was an arbitrary maximum or minimum point of Qy, Qy is bounded at its maxima and minima. Therefore, Qy is bounded. 15

16 Lemma 5.2. Suppose < γ < 1 and (4.7) holds. Let Q(y) denote a classical solution of (3.13) which satisfies (4.9). Define k (y) as in (5.1). Then there exist positive constants A 1 and A 2 such that k (Y t ) A 1, e Yt k (Y t ) A 2. (5.5) Proof. Note that k, so we do not need to bound it in absolute value. We bound k (Y t ) and e Yt k (Y t ) as follows: k (Y t ) = be Yt + µ(y t ) r + ( σ 2 + ρσ σe Yt ) Q ] + y (Y t ) (1 γ)q(y t ) be Y t + µ(y t ) r + ( σ 2 + ρσ σe Yt ) Q y (Y t ) (1 γ)q(y t ) be Yt + M 1 + r + ( σ 2 + ρ σ σe Yt )A (1 γ)q(y t ) A 1, (5.6) and e Yt k (Y t ) e Yt be Y t + M 1 + r + ( σ 2 + ρ σ σe Yt )A ] (1 γ)q(y t ) A 2. (5.7) To justify (5.6) and (5.7), we apply the same reasoning we used for the bound (4.12) in the proof of Theorem 4.5. Let Z t = e Yt. Note that (5.6) is the quotient of a linear expression in Z t and a quadratic expression in Z t, where Z t >. Since q(y t ) is bounded below by a positive constant, the limits of (5.6) as Z t approach positive and negative infinity both exist. Therefore the quotient is bounded. The expression in (5.7) is the quotient of a quadratic in Y t by another quadratic in Y t, and it s limits as Y t approaches both positive and negative infinity exist as well. Therefore (5.7) holds. Lemma 5.3. Suppose < γ < 1 and (4.7) holds. Let Q(y) denote a classical solution of (3.13) which satisfies (4.9). For the process X t defined in (2.11), and k (y) and c (y) defined by (5.1), we have for any fixed m >. EX m t ] < (5.8) 16

17 Proof. The proof is very similar to the proof of 18] Chapter 1, Lemma 1.1. So we omit it here. We now state our main result. Theorem 5.4. (Verification Theorem) Suppose < γ < 1 and (4.7) holds. Let Q(y) denote a classical solution of (3.13) which satisfies (4.9). Denote Ṽ (x, y) 1 γ xγ e Q(y). (5.9) Then we have Ṽ (x, y) V (x, y), (5.1) where V (x, y) is the value function defined by (2.13). Moreover, the optimal control policy is k (y) = be y + µ(y) r + ( σ 2 + ρσ σe y ) Q ] + y (y), c (y) = e Q(y) γ 1. (1 γ)q(y) (5.11) Proof. Since Q is a classical solution of (3.13), we can show that Ṽ is a classical solution of (3.1). For any admissible control (k t, c t ) Π, using Ito s rule for f(t, Ṽ ) = e δt Ṽ, we get de δt Ṽ (X t, Y t )] = δe δt Ṽ (X t, Y t )dt + e δt dṽ (X t, Y t ). (5.12) Applying Ito s rule to Ṽ, we have dṽ (X t, Y t ) = ṼxdX t + ṼydY t + 1 2Ṽxx(dX t ) 2 + Ṽxy(dX t )(dy t ) + 1 2Ṽyy(dY t ) 2 = σ 2 2 Ṽyy + µ(y t )Ṽy + rx t Ṽ x cx t Ṽ x + (be Yt + µ(y t ) r)k t X t Ṽ x + k t X t ( σ 2 + ρ σσe Yt )Ṽxy + k2 t Xt 2 q(y t ) Ṽ xx ]dt 2 +σe Yt k t X t Ṽ x db t + ( σk t X t Ṽ x + σṽy) d Bt. 17

18 The integral form of (5.12) is e δt Ṽ (X T, y T ) Ṽ (x, y) = = + e δt dṽ (X t, Y t ) δe δt Ṽ (X t, Y t )dt σ e δt 2 2 Ṽyy + µ(y t )Ṽy + (r c + (be Yt + µ(y t ) r)k)x t Ṽ x + kx t ( σ 2 + ρ σσe Yt )Ṽxy + k2 Xt 2 q(y t ) ] Ṽ xx δṽ dt 2 ) σe Yt kx t Ṽ x db t + ( σkxt Ṽ x + σṽy d Bt. (5.13) For any admissible control k t, c t, we have the following inequality: σ 2 2 Ṽyy + µ(y t )Ṽy + rx t Ṽ x c t X t Ṽ x + (be Yt + µ(y t ) r)k t X t Ṽ x + ( σ 2 + ρσ σe Yt )k t X t Ṽ xy + k2 t Xt 2 q(y t ) Ṽ xx 2 δṽ (x, y) 1 γ (cx t) γ. (5.14) Combining (5.13) and (5.14), we have where e δt Ṽ (X T, y T ) Ṽ (x, y) m T = + e δt δṽ (X t, Y t ) 1 γ (c tx t ) γ ]dt σe Yt k t X t Ṽ x db t + δe δt Ṽ (X t, Y t )dt ( ) σk t X t Ṽ x + σṽy d B t = e δt 1 γ (c tx t ) γ dt + m T + m T, (5.15) σe Yt k t X t Ṽ x db t and m T = It is not hard to show that m t, m t are local martingales. ( ) σk t X t Ṽ x + σṽy d B t. 18

19 Rearranging equation (5.15) to obtain a bound on Ṽ, we have Ṽ (x, y) e δt 1 γ (c tx t ) γ dt + e δt Ṽ (X T, y T ) m T m T. (5.16) Next we replace T with T τ R, where τ R = inf{t > ; Xt 2 + Yt 2 we can get Ṽ (x, y) τr = R 2 }, then e δt 1 γ (c tx t ) γ dt+e δt τ R Ṽ (X T τr, y T τr ) m T τr m T τr. Since m T and m T are local martingales, Em T τr ] = and E m T τr ] =. So taking the expectation of both sides gives us τr Ṽ (x, y) E x,y e δt 1 γ (c tx t ) γ dt + E x,y e δt τ R Ṽ (X T τr, y T τr ) τr E x,y e δt 1 γ (c tx t ) γ dt. (5.17) The second inequality is true because Ṽ >. Now we take the limit as R tends to infinity and use Fatou s lemma, which states ] E lim inf X n lim inf EX n]. n n So we have τr Ṽ (x, y) lim E x,y e δt 1 ] R γ (c tx t ) γ dt E x,y e δt 1 ] γ (c tx t ) γ dt. Now, letting T tend to infinity and using Fatou s lemma again, Ṽ (x, y) lim E x,y e δt 1 ] T γ (c tx t ) γ dt E x,y e δt 1 ] γ (c tx t ) γ dt (5.18). Since (5.18) holds for any admissible controls c t and k t, we have Ṽ (x, y) sup (k t,c t) Π E x,y e δt 1 γ (c tx t ) γ dt = V (x, y). (5.19) 19

20 Next we must show the reverse inequality, Ṽ (x, y) V (x, y). We start by showing that (kt, c t ) Π. First, we note that k be Yt + µ(y t ) r + ( σ 2 + ρσ σe Yt ) (Y t ) = Q ] + y (Y t ) and c (Y t ) = e Q(Yt ) γ 1 (1 γ)q(y t ) are progressively measurable. Second, we must show that ( ) ( ) P r (kt ) 2 dt < = 1, P r c t dt < = 1. (5.2) By Lemma 5.2, we have that kt A 1 and noting that K 1 Q(y) K 2, we can easily get the above inequalities. Therefore (kt, c t ) Π. Using kt and c t instead of arbitrary k t, c t >, we have equality in (5.16): Ṽ (x, y) = e δt 1 γ (c t X t ) γ dt + e δt Ṽ (X T, y T ) m T m T, (5.21) where m T = σe Yt kt X t Ṽ x db t and m T = ( ) T σk t X t Ṽ x + σṽy d B t. We wish to show that m T and m T are martingales, so that these terms vanish when taking the expectation of both sides in (5.21). Using Lemmas 5.1, 5.2, and 5.3, we have ] ( E x,y σe Y t ) 2dt kt X t Ṽ x and = E x,y σ2 γ 2 y2 2e 2K 2 ( σe Yt kt X t 1 ) 2 γ Xγ t e dt] Q E x,y X 2(1+γ) t ] ) 2dt E x,y ( σk t X t Ṽ x + σṽy ( = E x,y σk t X t 1 γ Xγ t e Q + σ 1 ) 2 γ Xγ t e Q Qy dt] E x,y 2 σ2 γ 2 (k t ) 2 X 2(1+γ) t e 2 Qdt ] + E x,y 2 σ2 ] 2 σ2 γ 2 y2 1e 2K 2 E x,y X 2(1+γ) t dt + 2 σ2 γ 2 e2k 2 A 2 <. γ 2 X2γ t ] dt <, ] e 2 Q Q2 y dt ] E x,y X 2γ t dt 2

21 Therefore m T and m T are martingales. Taking the expectation in (5.21), we obtain Ṽ (x, y) = E x,y e δt 1 ] ] γ (c t X t ) γ dt + e δt E x,y Ṽ (XT, y T ). (5.22) Since V (x, y) Ṽ (x, y), as shown above, we see that E x,y e δt 1 ] γ (c t X t )dt Ṽ (x, y) <. (5.23) This implies that lim inf T E x,y e δt 1γ ] (c T X T ) γ =, (5.24) which is easily seen with a proof by contradiction. Note that c t is bounded below by a positive constant: Then we have = lim inf T which implies E x,y e δt 1γ (c T X T ) γ ] lim inf T c t e K 2 γ 1 c >. E x,y c γ lim inf T E x,y e δt 1 ] γ Xγ T, e δt 1 ] γ Xγ T =. (5.25) Using (5.25) and the fact that e Q e K 2, we have ] lim inf E x,y e δt Ṽ (X T, y T ) =. (5.26) T Now taking the lim inf of (5.22) as T approaches infinity, and using the monotone convergence theorem, we get Ṽ (x, y) = lim inf E x,y e δt 1 ] T γ (c t X t ) γ dt = E x,y e δt 1 ] γ (c t X t ) γ dt. (5.27) Finally, by (5.27) and the definition of V, we have Ṽ (x, y) V (x, y). (5.28) Combining (5.19) and (5.28), we get Ṽ (x, y) = V (x, y). 21

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