Optimizing Venture Capital Investments in a Jump Diffusion Model

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1. Optimizing Venture Capital Investments in a Jump Diffusion ModelErhan Bayraktar Masahiko Egami AbstractWe study two practical optimization problems in relation to venture capital investments and/or Re- search and Development (R&D) investments. In the rst problem, given the amount of the initial investment and the cash ow structure at the initial public offering (IPO), the venture capitalist wants to maximize overall discounted cash ows after subtracting subsequent investments, which keep the invested company solvent. We describe this problem as a mixture of singular stochastic control and optimal stopping problems. The singular control corresponds to nding an optimal subsequent investment policy so that the value of the investee company stays solvent. The optimal stopping corresponds to nding an optimal timing of making the company public. The second problem is concerned with optimal dividend policy. Rather than selling the company at an IPO, the investor may want to harvest technological achievements in the form of dividend when it is appropriate. The optimal control policy in this problem is a mixture of singular and impulse controls.Key words: Venture capital investments, R&D, IPO, stochastic control, optimal stopping, singular control, impulse control, jump diffusions. JEL Classication: O32, O33, C61. Mathematics Subject Classication (2000) : Primary: 49N25, Secondary: 60G40. 1 IntroductionIn accordance with the recent theoretical and practical development in the area of real options, modeling venture capital investment and R&D has become increasingly an important topic. Davis et al. [3] reviews the literature of this area. One of the most important issues is modeling the dynamics of the value process of start-up companies and/or R&D projects. Among many approaches, one approach is to use jump models with Poisson arrivals. For example, Willner [11] uses a deterministic drift component and stochastic jumps whose size follows a gamma distribution. A similar model is presented by Pennings and Lint [8], who also model with a deterministic drift and E. Bayraktar was supported in part by the National Science Foundation, under grant DMS-0604491. E. Bayraktar and M. Egami are with Department of Mathematics, University of Michigan, Ann Arbor, MI, 48109-1043, USA. Email:{erhan, egami}@umich.edu. We thank Savas Dayanik for valuable comments and are most grateful to Anthea Au Yeung, Ikumi Koseki, T ru Masuda, and Akihiko Yasuda for giving insights and good memories through helpful and enjoyableo conversations.1 2. a jump part whose size follows a Weibull distribution with scale parameter two. In the spirit of these papers, we will model the value of the process with a jump diffusion. More specically, we are assuming that the company or the R&D project has (unproven) innovative technologies and hence the appreciation of the company value occurs when there is a technological breakthrough or discovery of innovative methods. Let {, F , P} be a probability space hosting a Poisson random measure N (dt, dy) on [0, ) R and Brownian motion W = (Wt )t0 , adapted to some ltration F = {Ft }t0 satisfying the usual conditions. The mean measure of N is (dt, dy) dtF (dy), where > 0 is constant, and F (dy) is the common distribution of the jump sizes. The (uncontrolled) value process X 0 of the invested company is described as follows: 000 dXt = (Xt )dt + (Xt )dWt + yN (dt, dy), t (0) ,(1.1)0in which (0) 0 inf{t 0; Xt < 0}.(1.2)In this set up the jumps of X 0 come from a compound Poisson process whose jump size distribution is F . To obtain explicit results we will take this dsitribution to be exponential. In practice, investments in promising start-up com- panies are made through venture capital funds (often called private equity fund as well) that raise capital from institutional investors such as banks, insurance companies, university endowment funds, and pension/retirement funds. Venture capital funds screen out those start-up companies and select several companies to invest in. In many cases, the venture capital funds actively help and advise them by taking a seat on the board of the invested compa- nies. When there is a technological breakthrough, this jump is materialized in the following way: the company and the venture capital fund reevaluate the value of the company stock by using expected cash ow methods provided that the company is successful in manufacturing real products or by using comparable transactions in the past. As a result, some new investors may become willing to invest in the company at the re-evaluated price in this second round funding. Hence the appreciation of the stock value can be modelled by an arrival of jumps. The size and arrival rate of jumps can be estimated by track records of the venture capital funds. The nal objective of these venture capital investments is, in many cases, to make the company public through initial public offerings (IPOs) or to sell to a third party at a premium. However, in due course, there are times when the start-up company faces the necessities to solicit new (additional) capital. In turn, the venture capitalist has to make decisions on whether to make additional investments. We refer to this type of problem as the IPO problem. Let us mention some advantages of using a jump diffusion model rather than a piecewise deterministic Markov model as in other works in the literature. As we discussed in the previous paragraph, until going public in the IPO market, the start-up company evolves while proving the merits and applicabilities of their technology. At an early stage, the companys growth mostly depends on the timing and magnitude of jump part in (1.1). At the time when the company invites second and third round investors, it is often the case that they have generated some cash ows from their operation while jumps of great magnitude are not necessarily expected. At these stages, the diffusion part of (1.1) is becoming increasingly inuential. Hence the jump diffusion model can represent start-up companies of various stages by appropriately modifying the parameters of the model. To address the issue of subsequent investments in the IPO problem, we rst solve an optimal stopping problem of a reected jump diffusion. In this problem, the venture capitalist does not allow the companys value to go below a xed level, say a, and attempts to nd an optimal stopping rule for the IPO (Section 2.2). Based on the results obtained for this problem, we proceed to a more general case in which the venture capitalist can also choose a (Section 2.3). This problem is a mixture of singular stochastic control and optimal stopping. The singular control is how the venture capitalist exercises controls or interventions in terms of additional capital infusions. The 2 3. optimal stopping is, given a certain reward function at the IPO market, to nd an optimal timing of making the company public. In summary, while making decisions with respect to additional investments, the venture capitalist seeks to nd an optimal stopping rule in order to maximizing her return, after subtracting the present value of her intermediate investments.Another problem of interest is the following: Rather than selling outright the interest in the start-up company or R&D investments, the investor may want to extract values out of the company or project in the form of dividend until the time when the value becomes zero. This situation may be more suitable in considering R&D investments because one wants to harvest technological achievements when appropriate, while one keeps the project running. We refer to this type of problem as the harvesting problem (Section 3). We prove that the following policy is optimal: One sets a threshold, say b > 0, and every time the company value becomes equal to (due to the diffusion part) or beyond the threshold (due to Poisson jump part), one takes the value down to level b in the form of dividend. The optimal strategy is a mixture of singular and impulse control.The rest of the paper is structured as follows: In section 2, we solve the IPO problem, rst by setting the lower threshold level a xed and later by allowing this level vary. In section 3, we solve the harvesting problem. For each problem, we rst derive a verication lemma. Next, we construct a candidate solution and verify the optimality of this candidate by showing that the conditions prescribed in the verication lemma are all satised. We also provide some static sensitivity analysis to the model parameters. In section 4, we give our concluding remarks and compare the values of IPO and harvesting problems.2 The IPO Problem2.1 Problem DescriptionLet us consider the problem of the venture capitalist now. The dynamics of the value of the start-up company is described as (1.1). After making the initial investment in the amount of x, the venture capitalist can make interventions in the form of additional investments. Hence the controlled process X is written as follows:dXt = (Xt )dt + (Xt )dWt +yN (dt, dy) + dZt , X0 = x,(2.1)0in which Z = (Zt )t0 is a continuous, non-decreasing (except at t = 0) Ft -adapted process. Such a control Z is said to be admissible. We will write Z V, if Z is admissible. The venture capitalists purpose is to nd the best F-stopping time to make the company public through an IPO to maximize the present value of the discounted future cash-ow. We will denote the set of all F stopping times by S. The discounted future cash-ow of the venture capitalist if he applies the control Z, and makes an initial public offering at time is 0J ,Z (x) Ex e( 0) h(X 0 ) es dZs , S, Z V,(2.2)0in which 0 inf{t 0; Xt < 0}, (2.3)is the time of bankruptcy of the company, et dZt = Z0 + et dZt ,(2.4) 0 (0,) 3 4. and h : R+ R+ has the following form: h(x) rx,(2.5) with r > 1. The parameter r is determined by the IPO market. Let us discuss some rationale of our model specication. First of all, valuation of IPOs is itself a very challenging subject and, to our knowledge, no complete solutions have yet been obtained. The pricing mechanism at the IPO market is complex, involving uncertainties with respect to the future of the newly publicized companies. A widely observed and recommended procedure both by academics and market practitioners is using comparable rm multiples: The subject companys operational and nancial information is compared with those of publicly- owned comparable companies, especially with ones newly made public. For example, the price-earning (P/E) ratio and/or market-to-book (M/B) ratio are multiplied by a certain number called multiples (that may vary from industry to industry) to calculate the IPO value. These numbers of the comparable rms serve as benchmarks. Kim and Ritter [7] found that P/E multiples using forecasted earnings result in much more accurate valuations than using historical earnings. In this pricing process, the role of investment banks (they often serve underwriters as well) is critical. They, together with the rm, evaluate the current operational performance, analyze the comparable rm multiples, project future earnings, assess the market demand for IPO stocks and set the timing of IPO. This procedure inherently involves signicant degree of variation on prices and introduces discontinuity of the post- IPO value from the pre-IPO value, since post-IPO value is, to a certain extent, market driven, while pre-IPO value is mostly company specic. This justies our choice of reward function h(). The post IPO value rx is strictly greater than the pre IPO value x. Further evidence of this discontinuity can be obtained by the literature about initial stock returns on IPO markets. Johnston and Madura [6] examined the initial (i.e. rst day trading) returns of IPOs (both Internet rms and non- Internet rms) during January 1, 1996 to December 31, 2000. They found that the average initial returns of Internet rms IPOs was 78.50%. They also reviewed the papers by Ibbotson and Jaffe [4], Reilly [9] and Ritter [10] and tabulated those authors ndings. Ritter [10], for example, found average initial returns as high as 48.4% for IPOs that occurred during 1980-1981. Thus, it is widely observed that the IPO companies are priced at a premium and that these premia are the most important sources of income to the venture capitalist. This phenomenon of abnormal initial returns is incorporated in our model with r > 1. 2.2 Optimal Stopping of a Reected Jump DiffusionIn this section we will assume that Z in (2.1) has the following forma Zt = (a x)1{x h(x) for all x (a, b(a)), (v) (A )v(x) < 0 for all x (b(a), ), v(x) = h(x) for all x [b(a), ),then v(x) = V (x; a), x R+ , and b(a)inf{t 0; Xt b(a)} is optimal.Proof. Let us dene (n) inf{t 0; Xt n}. Let S. When we apply It s formula to the semimartingaleo X (see e.g. Jacod and Shiryaev [5]), we obtain (n)0e( (n)0) v(X (n)0 ) = v(x) + es (A )v(Xs )ds + 1{x 0 and (x) = . Then the action of the innitesimal generator of X 0 on a test function f is given by 1Af (x) = f (x) + 2 f (x) + (f (x + y) f (x))ey dy. (2.13)20Let us dene1 2 2 G() + . (2.14)2 Note that 0 Ex eXt = exp (G()t) .(2.15)Lemma 2.2. The equation G() = has two positive roots 1 , 2 and one negative root 3 satisfying0 < 1 < < 2 , and 3 > 0.(2.16) Proof. Let us denote 1 2 2A() ( + ),B() .(2.17) 2 It follows that lim B() = , lim B() = , lim B() = 0, (2.18) lim A() = lim A() = ,(2.19) and that A(0) = ( + ) < B(0) = . (2.20)Moreover, A() is strictly decreasing on (, / 2 ) and strictly increasing on (/ 2 , ); B() is strictly increas- ing on (, ) and strictly decreasing on (, ). The claim is a direct consequence of these observations. Let us dene v0 (x; a) A1 e1 x + A2 e2 x + A3 e3 x ,(2.21)for some A1 , A2 , A3 R and b > a, which are tobe determined. We set the candidate value function as x a + v0 (a; a), x [0, a], v(x; a) v (x; a),x [a, b], (2.22) 0 rx,x [b, ), Our aim is to determine these constants so that v(; a) satises the conditions of the verication lemma. 6 7. We will choose A1 , A2 , A3 R and b > a to satisfy A1 e1 b + A2 e2 b + A3 e3 b = rb,(2.23a) A1 1 bA2 2 b A3 3 b11 e + 2 e +3 e+r b+ = 0,(2.23b)1 A1 e1 a + 2 A2 e2 a 3 A3 e3 a = 1,(2.23c) 1 b2 b 3 b1 A1 e + 2 A2 e 3 A3 e = r. (2.23d)For the function v in (2.22) to be well-dened, we need to verify that this set of equations have a unique solution. But before let us point how we came up with these equations. The expressions (2.23a), (2.23c) and (2.23d) come from continuous pasting at b, rst-order smooth pasting at a and rst order smooth pasting at b, respectively. Equation (2.23b) on the other hand comes from evaluatingbx 1 (A)v(x; a) = v0 (x)+ 2 v0 (x)+v0 (x + y)ey dy + r (x + y)ey dy (+)v0 (x) = 0. 20bx(2.24)Lemma 2.3. For any given a, there is a unique (A1 , A2 , A3 , b) R3 (a, ) that solves the system of equations (2.23a)-(2.23d). Moreover, b > max{a, b }, in which1 b (2.25) and A1 > 0, A2 > 0. Proof. Using (2.23a), (2.23a) and (2.23a) we can determine A1 , A2 and A3 as functions of b: r ( 1 )[2 3 (b + 1) + (2 3 )] 1 b A1 (b) =e=: D1 (b)e1 b , 2 (3 + 1 )(2 1 )r (2 )[1 3 (b + 1) + (1 3 )] 2 b A2 (b) = 2e=: D2 (b)e1 b , (2.26) (3 + 2 )(2 1 )r ( + 3 )[2 1 (b + 1) (2 + 1 )] 3 b A3 (b) = 2e =: D3 (b)e3 b . (3 + 1 )(2 + 3 )Let us deneR(b)1 A1 (b)e1 a + 2 A2 (b)e2 a 3 A3 (b)e3 a . (2.27)To verify our claim we only need to show that there is one and only one root of the equation R(b) = 1. Observe that R(a) = 1 A1 (a)e1 a + 2 A2 (a)e2 a 3 A3 (a)e3 a = 1 D1 (b) + 2 D2 (b) 3 D3 (b)(2.28) = 1 A1 e1 b + 2 A2 e2 b 3 A3 e3 b = r > 1,and thatlim R(b) = .(2.29) b The derivative of b R(b) isR (b) = 1 A1 (b)e1 a + 2 A2 (b)e2 a 3 A3 (b)e3 a(2.30)= 1 C1 e1 (ba) + 2 C2 e2 (ba) + 3 C3 e3 (ba) (1 2 3 b + Y ), 7 8. in whichr 1 r2 r + 3C1> 0, C2 > 0, and C3 > 0, 2 (3 + 1 )(2 1 ) 2 (3 + 2 )(2 1 ) 2 (3 + 1 )(3 + 2 )(2.31)and Y1 2 3 + (1 2 + 2 3 + 1 3 ).(2.32)Observe that Y= b .(2.33)1 2 3 From (2.30) it follows that on (, b ] the function b R(b) is increasing, and on [b , ) it is decreasing. If b a, then it follows directly from R(a) = r > 1 and limb R(b) = that there exists a unique b > a such that R(b) = 1. On the other hand, if b > a, then R(x) > 1 on x [a, b ]. Again, since limb R(b) = , there exists a unique b > b such that R(b) = 1. Let us show that A1 (b) > 0 for the unique root of R(b) = 1. Observe that A1 (b ) = 0 and A1 (b ) > 0. Moreover, b is the only local extremum of the function b A1 (b), and limb A1 (b) = 0. Since this function is decreasing on [b , ), A1 (b) > 0. Similarly, A2 (b) > 0.Remark 2.1. It follows from (2.23a),(2.23c) and (2.23d) that v(b; a) = rb,v (a; a) = 1 < v (b(a); a) = r.(2.34)Lemma 2.4. Let A1 , A2 , A3 , and b be as in Lemma 2.3 and v0 (; a) be as in (2.21). Then if A3 0, then v0 (; a) is convex for all a 0. Otherwise, there exists a unique point x < b such that, v0 (; a) is concave on [0, x) and convex on (, ). x Proof. The rst and the second derivative of v0 (; a) (dened in (2.22)) arev0 (x; a) = A1 1 e1 x + A2 2 e2 x 3 A3 e3 x ,v0 (x; a) = A1 1 e1 x + A2 2 e2 x + 3 A3 e3 x . 2 22(2.35) From Lemma 2.3 we have thatA1 > 0 and A2 > 0.(2.36)If A3 0, then (2.35) and (2.34) imply that v (x; a) > 0, x [a, b(a)], i.e., v(; a) is convex on [a, b(a)].22Let us analyze the case when A3 < 0. In this case the functions x A1 1 e1 x + A2 2 e2 x and x 2 3 x 3 A3 e intersect at a unique point x > 0. The function v0 (; a) (dened in (2.22)) decreases on [0, x) and increases on [, ). Now from (2.34) it follows that x < b(a).x 2.2.3 Verication of OptimalityProposition 2.1. Let us denote the unique b in Lemma 2.3 by b(a) to emphasize its dependence on a. Then v(; a) dened in (2.22) is equal to V (; a) of (2.8). Proof. The function v in (2.22) already satises (A )v(x; a) = 0, x (a, b(a)),v(x; a) = rx,x [b(a), ), v (a; a) = 1. (2.37)8 9. Therefore, we only need to show that(A )v(x; a) < 0,x (b(a), ),and that v(x; a) > rx,x (a, b(a)), (2.38)Let us prove the rst inequality.r (A )v(x; a) = r + rx,x > b(a).(2.39) So, (A )v(x; a) < 0, for x > b(a) if and only if1 b(a) > = b . (2.40) However, we already know from Lemma 2.3 that (2.40) holds.Let us prove the second inequality in (2.38). If A3 0, then Lemma 2.4 imply that v (x; a) > 0, x [a, b(a)], i.e., v(; a) is convex on [a, b(a)]. Therefore v (; a) is increasing on [a, b(a)] and v (x; a) [1, r) on [a, b). Since x v(x; a) intersects the function x rx at b(a), v(x; a) > rx, x [a, b(a)). Otherwise there would exist a point x [a, b) such that v (x ; a) > r. If A3 < 0, then the function v0 (; a) (dened in (2.22)) decreases on [0, x) and increases on [, ), in which x x < b(a), by Lemma 2.4. If x a, then v (x; a) < r for x [a, b(a)) since v (a; a) = 1, v (x; a) < 1 for x (a, x] and v (x; a) < r for x (, b(a)). On the other hand if x < a, then v (x; a) [1, r) for x x [a, b(a)) since v (; a) is increasing on this interval and v (b; a) = r. So in any case v (x; a) < r on [a, b(a)). Since v(b; a) = rb, then v(x) > rx, x [a, b). Otherwise there would exist a point x [a, b) such that v (x ; a) > r. Figure 1 shows the value function and its derivatives. As expected the value function v(; 1) is concave at rst and becomes convex before it coincides with the line h(). It can be seen that v(; a) satises the conditions of the verication lemma.v xv x1.2 71 60.8 50.6 40.4 30.2 xx12 3 4 5 61 2 345 6(a)(b) Figure 1: The IPO problem with parameters (, , , , , r) = (0.05, 0.75, 1.5, 0.25, 0.1, 1.25) and a = 1: (a) The value function v(x) with b(a) = 4.7641. (b) v (x) is also continuous in x R+ . 2.3 A Mixed Optimal Stopping and Singular Control ProblemIn this section, we will analyze a scenario in which the venture capitalist controls the value of the company, by pumping in extra cash, besides deciding when the company should go public. We will assume that the set of 9 10. controls available to the venture capitalist are of local time type. The venture capitalist will maximize the expected value of her discounted future income. The goal of the venture capitalist is to nd an a 0 and S such that a ,Z aU (x)sup sup J ,Z (x) = sup V (x; a) = J (x), x 0,(2.41) a[0,B] Sa[0,B] if (a , ) [0, B] S exists. Here, for any point a the control Z a is given by (2.6). In this optimization problem, we impose a budget constraint a B, which is due to the niteness of the cash reserves of the venture capitalist. The main result of this section is Proposition 2.2. We will show that V (x; a), for all x 0, is maximized at either a = 0 or a = B. We will start with analyzing the local extremums of the function a v0 (x; a), x 0. We will derive the second order smooth t condition at a, from a rst order derivative condition.Lemma 2.5. Recall the denition of the function v0 (; a) from (2.22). If a 0 is a local extremum of the function a v0 (x; a), for any x 0, then v0 (; a) = 0.a Proof. Let us denote A1 (a) A1 (b(a)), A2 (a) A2 (b(a)), and A3 (a) A3 (b(a)), (2.42)in which the functions A1 (), A2 () and A3 () are given by (2.26). The derivativedv0 a a a (x; a) = A1 ()e1 x + A2 ()e2 x + A3 ()e3 x = 0 (2.43) da for all x 0 if and only if a a aA1 () = A2 () = A3 () = 0, (2.44)since the functions x e1 x , x e2 x and x e2 x , x 0, are linearly independent.It follows from (2.23c) that for any a 0 1 A1 (a)e1 a + 2 A2 (a)e2 a 3 A3 (a)e3 a = 1. (2.45)Taking the derivative with respect to a we get2 2 2 (1 A1 (a) + 1 A1 (a))e1 a + (2 A2 (a) + A2 (a))e2 a + (3 A3 (a) 3 A3 (a))e3 a = 0.(2.46)Evaluating this last expression at a = a we obtain 2 2 2 1 A1 ()e1 a + 2 A2 ()e2 a + 3 A3 ()e3 a = v0 (; a) = 0,a a a a (2.47)where we used (2.44).Lemma 2.6. Let a be as in Lemma 2.5 and a b(a), a 0, be as in Proposition 2.1. Then b () = 0. Thea point a is a unique local extremum of a v0 (x; a), for all x 0, if and only if a is the unique local extremum of a b(a). If a is the unique local extremum of b(), then b () > 0. Moreover, a = argmina0 (b(a)). a Proof. Let A1 () be as in (2.42). Since a dA1 () = A1 (b())b () = 0, a a(2.48)db 10 11. and for any a, b(a) > b , in which b is the unique local extremum of the function b A1 (b), it follows that b () = 0.a Assume that a is the unique local extremum of b(). Then A1 (a) = A2 (a) = A3 (a) = 0,(2.49)if and only if a = a. Using (2.43), it is readily seen that a v0 (x; a) has a unique local extremum and that this local extremum is equal to a. On the other hand we know from Lemma 2.3 that b(a) > a for all a. Therefore, if b() has a unique local extremum at a, it can not be a local maximum. On the other hand, if there were an a = a such that b(a) b(),a then there would be a local maximum in (min{a, a}, max{a, a}), which yields a contradiction. Lemma 2.7. Recall the denition of v(; a), a 0, from (2.22). For any a1 0, a2 0 if b(a1 ) > b(a2 ), then v(x; a1 ) > v(x; a2 ), x 0. Proof. We will rst show that v0 (x; a1 ) = A1 (a1 )e1 x + A2 (a1 )e3 x + A3 (a1 )e3 x v0 (x; a2 ) = A1 (a2 )e1 x + A2 (a2 )e3 x + A3 (a2 )e3 x , (2.50) for x [0, b(a2 )], in which A1 , A2 , A3 are dened in (2.42). Since b(a2 ) b(a1 ),A1 (b(a1 )) < A1 (b(a2 )),A2 (b(a1 )) < A2 (b(a2 )),A3 (b(a1 )) > A3 (b(a2 )).(2.51)This follows from the fact that the functions A1 (), A2 () are increasing and A3 () is decreasing on [b , ) and that b(a) > b , for any a 0. See (2.26) and Lemma 2.3. Let us deneW (x) v0 (x; a1 ) v0 (x; a2 ), x R.(2.52) The derivativeW (x) < 0, x R,and (2.53) lim W (x) = , lim W (x) = .(2.54)x xTherefore, W () has a unique root. We will show that this root, which we will denote by k, satises b(a2 ) < k < b(a1 ). It follows from Lemma 2.4 that v0 (; a) is convex on [b(a), ), for any a. Moreover, for any a 0, v0 (; a) smoothly touches the function h() (see (2.23a) and (2.23d)), and stays above h() since v0 (; a) is convex. Now, since b(a2 ) < b(a1 ), for the function W () to have a unique root, that unique root has to satisfy b(a2 ) < k < b(a1 ). This proves (2.50).From (2.50) it follows that v(x; a1 ) v(x; a2 ), for any x [min{a1 , a2 }, b(a2 )]. But v(x; a2 ) = rx for x b(a2 ) and v(x; a) = v0 (x; a1 ) > rx, x [b(a2 ), b(a1 )] and v(x; a) = rx, x b(a1 ). Therefore, we have v(x; a1 ) v(x; a2 ),x min{a1 , a2 }. (2.55)In what follows we will show that the inequality in (2.55) also holds on x min{a1 , a2 }. Let us assume that a1 < a2 . It follows from (2.53) and v0 (a2 ; a2 ) = 1 (see (2.23c)) that v0 (a2 ; a1 ) < 1. (2.56)11 12. Therefore, v0 (; a1 ) does not intersect x xa2 +v0 (a2 , a2 ) x [a1 , a2 ]. Otherwise, at the point of intersection, say x, v0 (, a1 ) > 1, which together with (2.56) contradicts Lemma 2.4. This implies that v(x; a1 ) > v(x; a2 ), x x [a1 , a2 ]. Let us assume that a1 > a2 and that b(a2 ) a1 . Then, v0 (x; a2 ) < x a1 + v0 (a1 ; a1 ). Otherwise, v0 (; a2 ) intersects x x a1 + v0 (a1 ; a1 ) at x0 (a2 , a1 ). Then v0 (x0 ; a2 ) > 1. Since v0 (a; a) = 1 for any a 0, using Lemma 2.4, it follows that v0 (a1 ; a2 ) > v0 (a1 ; a1 ). This yields a contradiction since v0 (; a2 ) and v0 (; a1 ) do not intersect for any x < k, in which k > b(a2 ) a1 . Therefore, v(x; a1 ) > v(x; a2 ), x [a2 , a1 ].Finally, let us assume that a1 > a2 and that b(a2 ) < a1 . Since v(x; a2 ) = rx and v0 (x; a1 ) > rx for x [b(a2 ), a1 ] it follows that v(x; a1 ) > v(x; a2 ), x [a2 , a1 ]. Now, the proof is complete.Corollary 2.1. Recall the denition of v(; a) from (2.22). Let a be the unique local extremum of a b(a), a 0. Then v(x; a) v(x; a), x 0, for all a 0. Proof. The proof follows from Lemmas 2.6 and 2.7.Corollary 2.2. Let a be the unique local extremum of a b(a), a 0. Then function v(; a) is convex. Moreover, v (x; a) > 1, x > a. a a Proof. Let A3 () be as in (2.42). If A3 () 0 then v0 (; a) is convex by Lemma 2.4. aIf A3 () < 0, then the function a 3 a 3 3 v0 (x; a) = A1 ()1 e1 x + A2 ()2 e2 x 3 A3 ()e3 x > 0. a(2.57)Since v0 ( , a) = 0, then (2.57) implies that v0 (x; a) > 0 for x > a. The convexity of v(; a) follows, since it is a equal to v0 (; a) on [, b()] and is linear everywhere else. a a Since v (; a) = 1 (see Remark 2.1), it follows from the convexity of v(; a) that v (x; a) > 1 for x > a. a Note that the second order smooth t condition v (; a) = 0 yields a solution that minimizes V (x; a), x 0,a a 0, as a result of Corollary 2.1. In the next proposition we nd the maximizer.Proposition 2.2. Assume that a b(a), a 0 has a unique local extremum. ThenU (x) = max v(x; a),(2.58) a{0,B}in which U is given by (2.41). Proof. It follows from Lemma 2.6 that a b(a), has a unique local extremum, and this the local extremum is a minimum. Therefore, a b(a), a [0, B] is maximized at either of the boundaries. The result follows from Lemma 2.7.Our results in this section hinge on the existence and uniqueness of a in Lemma 2.5. We were not able to obtain this result and leave this as an open problem. 12 13. Conjecture 2.1. Let the functions A1 (), A2 () and A3 () be dened by (2.42). And let a b(a) denote the unique b (for a given a) in Lemma 2.3. Then there exist a unique a 0 that satises 2 2 2 v0 (; a) = 1 A1 ()e1 a + 2 A2 ()e2 a + 3 A3 ()e3 a = 0.a a a a (2.59)Using the same parameters as in Figure 1, we solve (2.23a)-(2.23d) and (2.59) numerically and nd a. We observe in Figure 2 that (a) a is the minimizer of b(a), and (b) v(x; 0) v(x; a) for x R+ .b av x;a v x;0 v x;a tilde7 86.5 1660.85.5 0.6 45 0.44.520.2a 24 6 8 xx1 2 3 43.5 1 2 3 4 5 6 (a) (b)(c) Figure 2: Using the parameters (, , , , ) = (0.05, 0.75, 1.5, 0.25, 0.1): (a) a = 3.884 minimizes the function b(a) with b() = 4.741. (b) The corresponding value function v(x; a) (solid line) is below v(x; 0)a (dashed line). (c) v(x; 0) v(x; a). Before ending this section, we provide sensitivity analysis of the optimal stopping barrier to the parameters of the problem. We use the parameter sets (, , , , ) = (0.05, 0.75, 1.5, 0.25, 0.1) with r = 1.25 and a = 0 and vary one parameter with the others xed at the base case. Figure 3 shows the results. In fact, all the graphs show monotone relationship between b(a) and the parameters, which is intuitive. Larger (that means smaller 1/) leads to a smaller threshold value since the mean jump size is small (Graph (a)). Similarly, larger leads to a larger threshold value since the frequency of jumps is greater and the investor can expects higher revenue. (Graph (b)). In the same token, if the absolute value of is greater, the process inclines to return to zero more frequently. Hence the investor cannot expect high revenue due to the time value of money. (Graph (c)). The result of the l! arger volatility expands the continuation region since the process X has a greater probability to reach further out within a xed amount of time. Hence the investor can expect the process to reach a higher return level (Graph (d)). 3 The Harvesting Problem3.1 Problem DescriptionIn this section, the investor wants to extract the value out of the company intermittently (i.e., receives dividends from the company) when there are opportunities to do so. This problem might t better the case of R&D invest- ments rather than the venture capital investments. Namely, the company or R&D project has a large technology platform, based on which applications are made and products are materialized from time to time. Each time it occurs, the investor tries to sell these products or applications and in turn receives dividends. There are many papers about dividend payout problems that consider continuous diffusion processes. See, for example, Bayraktar and Egami [1] and the references therein. One of the exceptions, to our knowledge, is Dassios and Embrechts [2] that analyze, using the Laplace transform method, the downward jump case. The absolute value of the jumps are 13 14. bb 17.512 1510 12.58 1067.55 42.5 2 0.5 1 1.5 22.50.5 1 1.5 2(a) (b)b b 55.6 45.4 35.2 2 0.20.4 0.6 0.8 1 14.8 4.6-0.6 -0.5 -0.4 -0.3 -0.2 -0.1(c) (d) Figure 3: Sensitivity analysis of the harvesting (dividend payout) problem to the parameters. The basis parameters are (, , , , ) = (0.05, 0.75, 1.5, 0.25, 0.1): (a) jump size parameter , (b) arrival rate , (c) drift rate and (d) volatility .exponentially distributed. In [2], the investor extracts dividends every time when a piecewise deterministic Markov process hits a certain boundary (i.e., singular control). In what follows, the dividend payments are triggered by sporadic jumps of the process as well as the diffusion part. Whenever the value of the company exceed a certain value, which may occur continuously or via jumps, dividends are paid out. So the investor applies a mixture of singular and impulse controls. Again, we consider the jump diffusion model (1.1) for the intrinsic value of the company. Accounting for the dividend payments the value of the company follows: dXt = (Xt )dt + (Xt )dWt + yN (dt, dy) dZt(3.1) 0in which Z = (Zt )t0 is a continuous non-decreasing (expect at t = 0) F-adapted process, i.e., Z V, is the dividend payment policy.The investor wants to maximize the discounted expected value of the payments she receives, which is given by 0 J Z (x) Ex et dZt (3.2) 0in which 0 is dened as in (2.3) denotes the time of insolvency. The investor wants to determine the optimal dividend policy Z that satisesV (x) sup J Z (x) = J Z (x),(3.3) ZVif such a Z V exists.14 15. 3.2 A Mixed Singular and Impulse Control Problem3.2.1 Verication LemmaLemma 3.1. Let us assume that () is bounded. If non-negative function v C1 (R+ ) is also twice continuously differentiable except at countably many points and satises (i) (A )v(x) 0, x 0,(ii) v (x) 1, x 0, (iii) v (x) 0 (i.e. v is concave),thenv(x) V (x),x 0. (3.4)Moreover, if there exists point b R+ such that v C1 (R+ ) C2 (R+ {b}) such that(iv) (A )v(x) = 0 , v (x) > 1, for all x [0, b),(v) (A )v(x) < 0, v(x) = x b + v(b), x > b,in which the integro-differential operator A is dened by (2.9), thenv(x) = V (x)x R+ ,and, (3.5)Zt = (Xt b)1{Xt >b} + Lb ,t t 0,(3.6)in which t Lt =1{Xs =b} dLb ,s t 0,(3.7) 0 is optimal. Proof. Let (n) be as in the proof of Lemma 2.1. Using It s formula for semimartingales (see e.g. Jacod and o Shiryaev [5]) (n)0 (n)0 e( (n)0 ) v(X (n)0 ) = v(x) + es (A )v(Xs )ds es v (Xs )dZs(c)00 (n)0 (n)0 +es (Xs )dWs +es (v(Xs + y) v(Xs ))(N (ds, dy) (ds, dy))0 00(3.8)+ ek v(Xk ) v(Xk ) , 0k (n)0 (n)0+ es (v(Xs ) v(Xs + y)) N (ds, dy)0 0in which {k }kN is an increasing sequence of F stopping times that are the times the process X jump due to jumps in Z that do not occur at the time of Poisson arrivals. Z (c) is the continuous part of Z, i.e., (c) ZtZt (Zs Zs ). (3.9)0st15 16. The controller is allowed to choose the jump times of Z to coincide with the jump times of N . But this is taken into account in (3.10) in the last line. Observe that the expression on this line is zero if the jump times of Z never coincide with those of the Poisson random measure.Equation (3.10) can be written as (n)0 (n)0 e( (n)0) v(X (n)0 ) = v(x) +es (A )v(Xs )ds es dZs00 (n)0 (n)0 +es (1 v (Xs ))dZs + es (v(Xs + y) v(Xs ))(N (ds, dy) (ds, dy)) 0 00 (n)0 + ek v(Xk ) v(Xk ) + (Xk Xk )v (Xk ) +es (Xs )dWs 0k (n)00 (n)0 + es (v(Xs ) v(Xs + y) + (y + Xs Xs )v (Xs + y)) N (ds, dy) 0 0 (3.10)After taking expectations the stochastic integral terms vanish. Also, the concavity of v implies thatv(y) v(x) v (x)(y x) 0, for any y > x.(3.11)Now together with the, Assumptions (i), (ii), (iii) we obtain (n)0 v(x) Ex e( (n)0) v(X (n)0 ) + es dZs .0Equation (3.4) follows from the bounded and monotone convergence theorems. When the control Z dened in (3.6) is applied, the third line (3.10) is equal to (x b)1x>b , since the jump times of Zt coincide with that of the Poisson random measure N except at time zero if X0 = x > b. The fourth line is also zero, because v() is linear on [b, ). After taking expectations and then using assumptions (iv) and (v), monotone and bounded convergence theorems we obtain0 v(x) = Ex es dZs ,(3.12)0which proves the optimality of Z and v() = V (). 3.2.2 Construction of a Candidate SolutionAs in Section 2.2.2 we will assume that the mean measure of the Poisson random measure N is given by (dt, dy) = dtey dy, (x) = where > 0 and (x) = .Let us dene v0 (x) B1 e1 x + B2 e2 x + B3 e3 x , x 0, (3.13) for B1 , B2 , B3 R that are to be determined. We set our candidate function to be v (x) x [0, b),0v(x) (3.14) x b + v0 (b), x [b, ).16 17. We will choose B1 , B2 , B3 and b to satisfy B1 1 bB2 2 bB3 3 b 1 e +ee+ B1 e1 b + B2 e2 b + B3 e3 b + = 0, (3.15)1 2 3 + 1 B1 e1 b + 2 B2 e2 b 3 B3 e3 b = 1 (3.16)1 B1 e1 b + 2 B2 e2 b + 3 B3 e3 b = 0, 2 2 2 (3.17) B1 + B2 + B3 = 0. (3.18)Equation (3.15) by explicitly evaluating bx (A )v(x) = v (x) + v(x + y)F (dy) +(v(b) + (x + y b))ey dy ( + )v(x) 0 bxand setting it to zero. Equations (3.16) and (3.17) are there to enforce rst and second order smooth t at point b. The last equation imposes the function v to be equal to zero at point zero. The vale function, V satises this condition since whenever the value process X hits level zero bankruptcy is declared.Lemma 3.2. There exists unique solution B1 , B2 , B3 and b to the system of equations (3.15), (3.16), (3.17), and (3.18) if and only if the quantity + / > 0. Moreover, B1 > 0, B2 > 0 and B3 < 0. Proof. Using (3.15), (3.16), and (3.17), we can express B1 , B2 and B3 as functions of b: For all b > 0, we havee1 b 2 3 ( 1 ) B1 (b) = > 0, 1 (2 1 )(1 + 3 )e2 b3 1 (2 ) B2 (b) = > 0, (3.19) 2 (2 + 3 )(2 1 )e 3 b 1 2 ( + 3 ) B3 (b) = < 0. 3 (1 + 3 )(2 + 3 ) Let us dene Q(b) B1 (b) + B2 (b) + B3 (b), b 0. (3.20) Our claim follows once we show that the function b Q(b), b 0 has a unique root. The derivative of Q()Q (b) = B1 (b) + B2 (b) + B3 (b) < 0, (3.21)therefore Q() is decreasing. Explicitly computing Q(0) in (3.19), we obtain1 + / Q(0) > 0 if and only if 1 2 3 + (1 2 + 2 3 + 3 1 ) => 0. 1 2 3 Sincelim Q(b) = (3.22) b the claim follows. 3.2.3 Verication of OptimalityLemma 3.3. Let B1 , B2 , B3 and b be as in Lemma 3.2. Then v dened in (3.14) satises(i) (A )v(x) < 0 for x (b, ), (ii) v (x) > 1 on x [0, b), (iii) v (x) < 0 on x [0, b).17 18. Proof. (i): On x (b, ), v(x) = (x b) + v0 (b), we compute (A )v(x) = + / (x b) v0 (b ) < + / v0 (b) = lim(A )v(x) = lim(A )v(x) = 0. xbxbHere we used the continuity of v(x), v (x) and v (x) at x = b. (ii) and (iii): Since B1 , B2 > 0 and B3 < 0,v0 (x) = B1 1 e1 x + B2 1 e2 x B3 1 e3 x > 0,3 2 3(3.23)i.e., v0 () is monotonically increasing in x. It follows from (3.17) that v0 (b) = 0, therefore v0 (x) < 0 on x [0, b). This proves (iii). Since v0 (x) < 0, x R+ , v0 () is decreasing on R+ . It follows from (3.17) that v0 (b) = 1. Therefore, v0 (x) > 1 on x [0, b). This proves (ii). Proposition 3.1. Suppose that + > 0. Let B1 , B2 , B3 and b be as in Lemma 3.2. Then the function v() dened in (3.14) satisesv(x) = V (x) = sup J Z (x). (3.24)ZVand Z dened in (3.6) is optimal. Proof. Note that (A )v(x) = 0, x [0, b) as a result of (3.15). The function v() is linear on [b, ). It follows from Lemma 3.3 that the function v() satises all the conditions in the verication lemma. Figure 4 shows the value function and its derivatives. As expected the value function is concave and is twice continuously differentiable. Finally, we perform some sensitivity analysis of the optimal barrier b with respectv xv x v x 14 x6 0.5 1 1.5 2 2.5 3 125 -10 104 -2083 6-302 41 2 -40 xx0.5 1 1.52 2.5 30.5 1 1.5 2 2.5 3 -50 (a)(b) (c) Figure 4:The harvesting (dividend payout) problem with parameters (, , , , )= (0.05, 0.75, 1.5, 0.25, 0.1): (a) The value function v(x) with b = 1.276. (b) v (x) and v (x) to show that the optimality conditions are satised. to the parameters of the problem. Figure 5 shows the results. Graph (a): The rst graph shows that as the expected value of jump size 1/ decreases, so does the threshold level b, as one would expect. Graph (b): It is interesting to observe that b increases rst and start decreasing when reaches a certain level, say max . A possible interpretation is as follows: In the range of (0, max ), i.e. for small , one wants to extract a large amount of cash whenever jumps occur since the opportunities are limited. As gets larger, one starts to be willing to let the process live longer by extracting smaller amounts each time. On the other hand, after max , one becomes comfortable with receiving more dividends, causing the declining trend of b . Graph (c): The small in the 18 19. absolute value sense implies that it takes more time to hit the absorbing state. Accordingly, it is safe to extract a large amount of dividend. However, when the cost increases up to a certain level, say , it becomes risky to extract and hence b increases. It is observed that after the cost level is beyond , one would become more desperate to take a large dividend at one time in the fear of imminent insolvency caused by a large (in the absolute value sense). This is the downward trend of b on the left side of . Graph (d): As the volatility goes up, then the process tends to spend more time away from zero in both the positive and negative real line. Accordingly, the threshold level increases to follow the process. bb1.2 1.5 1 0.81 0.6 0.4 0.5 0.2 24 6 8101 234 5 6 7 (a)(b)bb 1.5 3 1.4 2.5 1.3 2 1.2 1.5 1.1 10.5-0.4 -0.3 -0.2-0.1 0.9 0.2 0.4 0.6 0.8 1 (c)(d) Figure 5: Sensitivity analysis of the harvesting (dividend payout) problem to the parameters. The basis parameters are (, , , , ) = (0.05, 0.75, 1.5, 0.25, 0.1): (a) jump size parameter , (b) arrival rate , (c) drift rate and (d) volatility . 4 Concluding RemarksBefore concluding, we compare two value functions, one for the IPO problem and the other for the harvesting problem. We set parameters , , , , and equal and vary the level of r > 1, the expected return at the IPO market. Figure 6 exhibits the two value functions: v(x; 0) (solid line) for the IPO problem with a = 0 and v(x) (dashed line) for the harvesting problem. We consider three different values of r here; (a) r = 1.25, (b) r = 1.5 and (c) r = 2. It can be observed that as r increases, the value function for the IPO problem shifts upward for all the points of x R+ . This jump diffusion model, although simple, gives a quick indication as to which strategy (IPO or harvesting) is more advantageous given the initial investment amount x. Moreover, as we discussed, this model has both diffusion and jump components, allowing us to model different stage of the start-up company by modifying the relative size of diffusion parameter and jump parameter /.19 20. V x , V x;0 V x , V x;0 V x , V x;0 128 8106 6 864 442 22x x x1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6(a) (b) (c) Figure 6: The comparison of two value functions with (, , , , ) = (0.05, 0.75, 1.5, 0.25, 0.1): (a) r = 1.25. (b) r = 1.5 and (c) r = 2 where the value function for the IPO problem with a = 0 is shown in solid line and the value function for the harvesting problem is shown in dashed line.References[1] E. Bayraktar and M. Egami. 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