Uniform approximation of the Cox-Ingersoll-Ross process

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1 Uniform approximation of the Cox-Ingersoll-Ross process G.N. Milstein J.G.M. Schoenmakers arxiv: v1 [math.pr] 3 Dec 013 December 4, 013 Abstract The Doss-Sussmann DS) approach is used for uniform simulation of the Cox- Ingersoll-Ross CIR) process. The DS formalism allows to express trajectories of the CIR process through solutions of some ordinary differential equation ODE) depending on realizations of a Wiener process involved. By simulating the first-passage times of the increments of the Wiener process to the boundary of an interval and solving the ODE, we uniformly approximate the trajectories of the CIR process. In this respect special attention is payed to simulation of trajectories near zero. From a conceptual point of view the proposed method gives a better quality of approximation from a path-wise point of view) than standard, or even exact simulation of the SDE at some discrete time grid. AMS 000 subject classification. Primary 65C30; secondary 60H35. Keywords. Cox-Ingersoll-Ross process, Doss-Sussmann formalism, Bessel functions, confluent hypergeometric equation. 1 Introduction The Cox-Ingersoll-Ross process V t) is determined by the following stochastic differential equation SDE) dv t) = kλ V t))dt + σ V dwt), V t 0 ) = V 0, 1) where k, λ, σ are positive constants, and w is a scalar Brownian motion. Due to [6] this process has become very popular in financial mathematical applications. The CIR process is used in particular as volatility process in the Heston model [13]. It is known [14], [15]) that for V 0 > 0 there exists a unique strong solution V t0,v 0 t) of 1) for all t t 0 0. The CIR process V t) = V t0,v 0 t) is positive in the case kλ σ and nonnegative in the case kλ < σ. Moreover, in the last case the origin is a reflecting boundary. As a matter of fact, 1) does not satisfy the global Lipschitz assumption. The difficulties arising in a simulation method for 1) are connected with this fact and with the natural requirement of preserving nonnegative approximations. A lot of approximation Ural Federal University, Lenin Str. 51, Ekaterinburg, Russia; Grigori.Milstein@usu.ru Weierstrass-Institut für Angewandte Analysis und Stochastik, Mohrenstrasse 39, Berlin, Germany; schoenma@wias-berlin.de 1

2 methods for the CIR processes are proposed. For an extensive list of articles on this subject we refer to [3] and [7]. Besides [3] and [7] we also refer to [1,, 11, 1], where a number of discretization schemes for the CIR process can be found. Further we note that in [17] a weakly convergent fully implicit method is implemented for the Heston model. Exact simulation of 1) is considered in [5, 9] see [3] as well). In this paper, we consider uniform pathwise approximation of V t) on an interval [t 0, t 0 + T ] using the Doss-Sussmann transformation [8], [0], [19]) which allows for expressing any trajectory of V t) by the solution of some ordinary differential equation that depends on the realization of wt). The approximation V t) will be uniform in the sense that the path-wise error will be uniformly bounded, i.e. sup V t) V t) r almost surely, ) t 0 t t 0 +T where r > 0 is fixed in advance. In fact, by simulating the first-passage times of the increments of the Wiener process to the boundary of an interval and solving this ODE, we approximately construct a generic trajectory of V t). Such kind of simulation is more simple than the one proposed in [5] and moreover has the advantage of uniform nature. Let us consider the simulation of a standard Brownian motion W on a fixed time grid t 0, t i,..., t n = T. Although W may be even exactly simulated at the grid points, the usual piecewise linear interpolation W t) = t i+1 t W t i ) + t t i W t i+1 ) t i+1 t i t i+1 t i is not uniform in the sense of ). Put differently, for any large) positive number A, there is always a positive probability though possibly small) that sup W t) W t) > A. t 0 t t 0 +T Therefore, for path dependent applications for instance, such a standard, even exact, simulation method may be not desirable and a uniform method preserving ) may be preferred. We note that the original DS results rely on a global Lipschitz assumption that is not fulfilled for 1). We therefore have introduced the DS formalism that yields a corresponding ODE which solutions are defined on random time intervals. If V gets close to zero however, the ODE becomes intractable for numerical integration and so, for the parts of a trajectory V t), that are close to zero, we are forced to use some other not DS) approach. For such parts we here propose a different uniform simulation method. Another restriction is connected with the condition := 4kλ σ ) /8 > 0. We note that the case > 0 is more general than the case kλ σ that ensures positivity of V t), and stress that in the literature virtually all convergence proofs for methods for numerical integration of 1) are based on the assumption kλ σ. We expect that the results here obtained for > 0 can be extended to the case where 0, however in a highly nontrivial way. Therefore, the case 0 will be considered in a subsequent work. The next two sections are devoted to DS formalism in connection with 1) and to some auxiliary propositions. In Sections 4 and 5 we deal with the one-step approximation and the convergence of the proposed method, respectively. Section 6 is dedicated to the uniform construction of trajectories close to zero.

3 The Doss-Sussmann transformation.1 Due to the Doss-Sussmann approach [8], [14], [19], [0]), the solution of 1) may be expressed in the form V t) = F Xt), wt)), 3) where F = F x, y) is some deterministic function and Xt) is the solution of some ordinary differential equation depending on the part ws), 0 s t, of the realization w ) of the Wiener process wt). Let us recall the Doss-Sussmann formalism according to [19], V.8. In [19] one consideres the Stratonovich SDE The function F = F x, y) is found from the equation dv t) = bv )dt + γv ) dwt). 4) F y = γf ), F x, 0) = x, 5) and Xt) is found from the ODE dx dt = 1 bf Xt), wt)), X0) = V 0). 6) F/ xxt), wt)) It turns out that application of the DS formalism after the Lamperti transformation Ut) = V t) see [7]) leads to more simple equations. The Lamperti transformation yields the following SDE with additive noise du = U k U)dt + σ dw, U0) = V 0) > 0, where 7) = Let us seek the solution of 7) in the form 4kλ σ. 8) 8 Ut) = GY t), wt)) 9) in accordance with 3)-6). Because the Ito and Stratonovich forrms of equation 7) coincide, we have bu) = U k U, γu) = σ. The function G = Gy, z) is found from the equation i.e., G z = σ, Gy, 0) = y, Gy, z) = y + σ z, 10) and Y t) is found from the ODE dy dt = Y + σ wt) k Y + σ wt)), Y 0) = U0) = V 0) > 0. 11) 3

4 From 9), 10), and solution of 11), we formally obtain the solution Ut) of 7): Ut) = Y t) + σ wt). 1) Hence V t) = U t) = Y t) + σ wt)). 13). Since the Doss-Sussmann results rely on a global Lipschitz assumption that is not fulfilled for 1), solution 13) has to be considered only formally. In this section we therefore give a direct proof of the following more precise result. Proposition 1 Let Y 0) = U0) = V 0) > 0. Let τ be the following stopping time: τ := inf{t : V t) = 0}. Then equation 11) has a unique solution Y t) on the interval [0, τ), the solution Ut) of 7) is expressed by formula 1) on this interval, and V t) is expressed by 13). Proof. Let wt), V t)) be the solution of the SDE system dw = dwt), dv = kλ V )dt + σ V t)dw t), which satisfies the initial conditions w0) = 0, V 0) > 0. Then Ut) = V t) > 0 is a solution of 7) on the interval [0, τ). Consider the function Y t) = Ut) σ wt), 0 t < τ. Clearly, Y t) + σ wt) > 0 on [0, τ). Due to Ito s formula, we get dy t) = dut) σ dwt) = dt Y + σ wt) k Y + σ wt))dt, i.e., the function Ut) σ wt) is a solution of 11). The uniqueness of Y t) follows from the uniqueness of V t)..3 So far we were starting at the moment t = 0. It is useful to consider the Doss- Sussmann transformation with an arbitrary initial time t 0 > 0 which even may be a stopping time, for example, 0 t 0 < τ). In this case, we obtain instead of 11) for Y = Y t; t 0 ) = Ut) σ wt) wt 0)) = V t) σ wt) wt 0)), t 0 t < t 0 + τ, the equation dy dt = Y + σ wt) wt 0)) k Y + σ wt) wt 0))), 14) Y t 0 ; t 0 ) = V t 0 ), t 0 t < t 0 + τ, with given by 8). Clearly, V t) = Y t; t 0 ) + σ wt) wt 0))), t 0 t < t 0 + τ. 15) 4

5 3 Auxiliary propositions 3.1 Let us consider in view of 14) solutions of the ordinary differential equations which are given by dy 0 dt = y 0 k y0, y 0 t 0 ) = y 0 > 0, t t 0 0, 16) y 0 t) = y 0 t 0,y 0 t) = [y 0e kt t 0) + k 1 e kt t 0) )] 1/, t t 0. 17) In the case > 0, i.e., 4kλ > σ, we have: if y 0 > /k then y 0 t 0,y 0 t) /k as t and if 0 < y 0 < /k then y 0 t 0,y 0 t) /k as t. Further y 0 t) = /k is a solution of 16). In the case = 0, the solution y 0 t 0,y 0 t) 0 under t for any y 0 > 0. We note that the case 0 is more general than the case kλ σ we recall that in the latter case V t) > 0, t t 0 ). In the case < 0, the solution y 0 t 0,y 0 t) is convexly decreasing under not too large y 0. It attains zero at the moment t given by t = t k ln y 0 /k /k and yt 0 0,y 0 t) =. In what follows we deal with the case 4kλ σ = 0. 19) Our next goal is to obtain estimates for solutions of the equation dy dt = 18) y + σ ϕt) k y + σ ϕt)), yt 0) = y 0, t 0 t t 0 + θ, 0) cf. 14) ) for a given continuous function ϕt). Lemma Let 0. Let y i t), i = 1,, be two solutions of 0) such that y i t)+ σ ϕt) > 0 on [t 0, t 0 + θ], for some θ with 0 θ T. Then y t) y 1 t) y t 0 ) y 1 t 0 ), t0 t t 0 + θ. 1) Proof. We have From here dy t) y 1 t)) = y t) y 1 t)) ) y 1 t) + σϕt) + k y1 t) + σ ) ϕt)) dt. y t) + σ ϕt) k y t) + σ ϕt)) whence 1) follows. y t) y 1 t)) = y t 0 ) y 1 t 0 )) t y s) y 1 s)) + [ t 0 y 1 s) + σ ϕs))y s) + σϕs)) k y s) y 1 s)) ]ds y t 0 ) y 1 t 0 )), 5

6 Remark 3 It is known that for δ > 1 the Bessel process BES δ has the representation Zt) = Z0) + δ 1 t 0 1 ds + W t), 0 t <, Zs) where W is standard Brownian Motion, Zt) 0 a.s., and that in particular E t 0 1 ds < Zs). See [18]; for δ 1 the representation of BES δ is less simple and involves the concept of local time.) From this fact it is not difficult to show that for > 0 the solution of 7) may be represented as Ut) = Ut 0 ) + t t 0 t 0 Us) k Us))ds + σ wt) wt 0)), U0) > 0, t 0 t <. Thus, with Y t) = Ut) σ wt) wt 0)), it holds that t Y t) = Y t 0 ) + Y s) + σ ws) wt 0)) k Y s) + σ 0))) ) ws) wt ds, for Y 0) = U0) > 0, 0 t <, and that in particular Y is continuous and of bounded variation. From this it follows that ) holds for t 0 t t 0 + T when > 0 and ϕt) = wt) wt 0 ) is an arbitrary Brownian trajectory, and then inequality 1) in Lemma goes through for θ = T. 3.3 Now consider 0) for a continuous function ϕ satisfying ϕt) r, t 0 t t 0 + θ t 0 + T, 3) for some r > 0 and 0 θ T. Along with 16), 0) with 3), we further consider the equations dy dt = y + σr k y + σ r), yt 0) = y 0, 4) dy dt = y σr k y σ r), yt 0) = y 0. 5) Let us assume that y 0 σr > 0, and consider an η > 0, to be specified below, that satisfies y 0 η σr > 0. 6) The solutions of 16), 0) with 3), 4), and 5) are denoted by y 0 t), yt), y t), and y + t), respectively, where y 0 t) is given by 17). By using 17) we derive straightforwardly that y t) = [y 0 + σ r) e kt t 0) + k 1 e kt t 0) )] 1/ σ r, t 0 t t 0 + θ, 7) y + t) = [y 0 σ r) e kt t 0) + k 1 e kt t 0) )] 1/ + σ r, t 0 t t 0 + θ. 8) Note that y t) + σr/ > 0 and y + t) > σr/, t 0 t t 0 + θ. Due to the comparison theorem for ODEs see, e.g., [10], Ch. 3), the inequality y + σ r k y + σ r) y + σ ϕt) k y + σ ϕt)) 6 y σ r k y σ r),

7 which is fulfilled in view of 3) for y > σr/, then implies that y t) yt) y + t), t 0 t t 0 + θ. 9) The same inequality holds for yt) replaced by y 0 t). We thus get yt) y 0 t) y + t) y t), t 0 t t 0 + θ. 30) 4kλ σ Proposition 4 Let = 0, the inequalities 3) and 6) be fulfilled for a 8 fixed η > 0, and let θ T. We then have yt) y 0 t) Crt t0 ) Crθ, t 0 t t 0 + θ, with 31) C = σk + 4σ 3η e k T. In particular, C is independent of t 0, y 0, and r provided 6) holds). Proof. We estimate the difference y + t) y t). It holds y + t) = z t) + σ r, y t) = z + t) σ r, y + t) y t) = σr z + t) z t)), 3) where z ± t) = [y 0 ± σ r) e kt t 0) + k 1 e kt t 0) )] 1/. Further, z + t) z t) = z+ t)) z t)) z + t) + z t) = y 0σre kt t 0) z + t) + z t). 33) Using the inequality a + b) 1/ a + b/a for any a > 0 and b 0, we get z + t) y 0 + σ r)e k t t 0) + k z t) y 0 σ r)e k t t 0) + k 1 e kt t 0) ) y 0 + σ r)e k t t 0), 1 e kt t 0) ) y 0 σ r)e k t t 0), whence Therefore z + t) + z t) y 0 e k t t0) + 1 e kt t0) ) k 1 z + t) + z t) 1 y 0 e k t t 0) 1 e k t t 0) ky 0 σ 4 r ) y 0 y 0 σ 4 r ) e kt t0) 1).. 7

8 From 33) we have that and so due to 3) we get z + t) z t) σre k t t 0) 1 0 y + t) y t) σr1 e k t t 0) ) + ky 0 σ 4 r ) e kt t0) 1) σr e k t t0) e k t t0) ). ky0 σ 4 r ) Since 1 e qϑ qϑ for any q 0, ϑ 0, and y 0 σ 4 r 3 4 η under 6), we obtain 0 y + t) y t) σrk t t 0) + 4σr 3kη e k t t 0) kt t 0 ). From this and 30), 31) follows with C = σk + 4σ 3η e k T. Corollary 5 Under the assumptions of Proposition 4, we get by taking η = y 0, yt) y 0 t) σk + 4σ ) e k 3y0 T rθ, := D 1 + D ) rθ, t y0 0 t t 0 + θ, where D 1 and D only depend on the parameters of the CIR process under consideration and the time horizon T. 4 One-step approximation Let us suppose that for t m, t 0 t m < t 0 + T, V t m ) is known exactly. In fact, t m may be considered as a realization of a certain stopping time. Consider Y = Y t; t m ) on some interval [t m, t m + θ m ] with y m := Y t m ; t m ) = V t m ), given by the ODE cf. 14)), Assume that dy dt = Y + σ wt) wt m)) k Y + σ wt) wt m))), 34) Y t m ; t m ) = V t m ), t m t t m + θ m. y m = V t m ) σr. 35) Due to 15), the solution V t) of 1) on [t m, t m + θ m ] is obtained via V t) = Y t; tm ) + σ wt) wt m)), t m t t m + θ m. 36) Though equation 34) is just) an ODE, it is not easy to solve it numerically in a straightforward way because of the non-smoothness of wt). We are here going to construct an 8

9 approximation y m t) of Y t; t m ) via Proposition 4. To this end we simulate the point t m + θ m, wt m + θ m ) wt m )) by simulating θ m as being the first-passage stopping) time of the Wiener process wt) wt m ), t t m, to the boundary of the interval [ r, r]. So, wt) wt m ) r for t m t t m + θ m and, moreover, the random variable wt m + θ m ) wt m ), which equals either r or +r with probability 1/, is independent of the stopping time θ m. A method for simulating the stopping time θ m is given in Subsection 4.1 below. Proposition 4 and Corollary 5 then yield, Y t; t m ) y m t) D 1 + D ym t m+1 := mint m + θ m, t 0 + T ), where y m t) is the solution of the problem ) r t m+1 t m ), t m t t m+1 with 37) dy m dt = y m k ym, y m t m ) = Y t m ; t m ) = V t m ) that is given by 17) with t m, y m ) = t m, V t m )). We so have, V t) = Y t; tm ) + σ wt) wt m)) = y m t) + σ wt) wt m)) + ρ m t), where due to 37), ρ m t) D 1 + D ) r t ym m+1 t m ), t m t t m+1. 38) We next introduce the one-step approximation V t) of V t) on [t m, t m+1 ] by V t) := y m t) + σ wt) wt m)), t m t t m+1. 39) Since wt m+1 ) wt m ) = r if t m+1 = t m + θ m < t 0 + T, and wt m+1 ) wt m ) r if t m+1 = t 0 + T, the one-step approximation 39) for t = t m+1 is given by V t m+1 ) := y m t m+1 ) + σ wt m+1) wt m )) = 40) y m t m+1 ) + σ { rξm with P ξ m = ±1) = 1/, if t m+1 = t m + θ m < t 0 + T, ζ m if t m+1 = t 0 + T, with ζ m = wt 0 + T ) wt m ) being drawn from the distribution of W t0 +T t m conditional on max 0 s t 0 +T t m W s r, 41) where W is an independent standard Brownian motion. For details see Subsection 4.1 below. We so have the following theorem. Theorem 6 For the one-step approximation V t m+1 ) due to the exact starting value V t m ) = V t m ) = ym, we have the one step error V tm+1 ) V t m+1 ) D 1 + D ) r t m+1 t m ). 4) V t m ) 9

10 4.1 Simulation of θ m and ζ m For simulating θ m we utilize the distribution function Pt) := P τ < t), where τ is the first-passage time of the Wiener process W t) to the boundary of the interval [ 1, 1]. A very accurate approximation Pt) of Pt) is the following one: Pt) Pt) = t 0 P s)ds with πt 3 e t 3e t + 5e t ), 0 < t π, P t) = π t 9π t 5π t π e 8 3e 8 + 5e 8 ), t > π, and it holds P t) P t) , and sup Pt) Pt) , sup t 0 see for details [16], Ch. 5, Sect. 3 and Appendix A3 ). Now simulate a random variable U uniformly distributed on [0, 1], Then compute τ = P 1 U) which is distributed according to P. That is, we have to solve the equation Pτ) = U, for instance by Newton s method or any other efficient solving routine. Next set θ m = r τ m. For simulating ζ m in 40) we observe that 41) is equivalent with rw r t 0 +T t m) conditional on max W u 1. 0 u r t 0 +T t m) We next sample ϑ from the distribution function Qx; r t 0 + T t m )), where Qx; t) is the known conditional distribution function see [16], Ch. 5, Sect. 3) t 0 Qx; t) := P W t) < x max 0 s t W s) < 1), 1 x 1, 43) and set ζ m = rϑ. The simulation of the last step looks rather complicated and may be computationally expensive. However it is possible to take for wt 0 +T ) wt ν ) simply any value between r and r, e.g. zero. This may enlarge the one-step error on the last step but does not influence the convergence order of the elaborated method. Indeed, if we set wt 0 +T ) wt ν ) to be zero, for instance, on the last step, we get V t 0 + T ) = y ν t 0 +T ) instead of 40), and ν V t0 + T ) V t 0 + T ) r D 1 + D ) t m+1 t m ) + σr, 44) V t m ) m=0 Remark 7 We have in any step Eθ n = r, the random number of steps before reaching t 0 + T, say ν + 1, is finite with probability one, and Eν = O1/r ). For details see [16], Ch. 5, Lemma 1.5. In a heuristic sense this means that, if we have convergence of order Or), we obtain accuracy O h), for an expected) number of steps O1/h) similar to the standard Euler scheme. 10

11 5 Convergence theorem In this section we develop a scheme that generates approximations V t 0 ) = V t 0 ), V t 1 ),..., V t n+1 ), where n = 0, 1,,..., and t 1,..., t n+1 are realizations of a sequence of stopping times, and show that the global error in approximation V t n+1 ) is in fact an aggregated sum of local errors, i.e., r n m=0 D 1 + D ) t m+1 t m ) rt D 1 + D ), V t m ) ηn with y m = V t m ), provided that y m σr for m = 0,..., n, and so η n := min 0 m n y m σr. Let us now describe an algorithm for the solution of 1) on the interval [t 0, t 0 + T ] in the case 0. Suppose we are given V t 0 ) and r such that V t0 ) σr. For the initial step we use the one-step approximation according to the previous section and thus obtain see 40) and 4)) V t 1 ) = y 0 t 1 ) + σ wt 1) wt 0 )), V t1 ) = V t 1 ) + ρ 0 t 1 ), where Suppose that ρ 0 t 1 ) D 1 + D ) r t 1 t 0 ) =: C 0 rt 1 t 0 ). 45) V t 0 ) V t 1 ) σr. We then go to the next step and consider the expression V t) = Y t; t1 ) + σ wt) wt 1)), 46) where Y t; t 1 ) is the solution of the problem see 34)) dy dt = Y + σ wt) wt 1)) k Y + σ wt) wt 1))), 47) Y t 1 ; t 1 ) = V t 1 ), t 1 t t 1 + θ 1. Now, in contrast to the initial step, the value V t 1 ) is unknown and we are forced to use V t 1 ) instead. Therefore we introduce Y t; t 1 ) as the solution of the equation 47) with initial value Y t 1 ; t 1 ) = V t 1 ). From the previous step we have that 11

12 Y t1 ; t 1 ) Y t 1 ; t 1 ) = V t1 ) V t 1 ) = ρ0 t 1 ) C 0 rt 1 t 0 ). Hence, due to Lemma, Y t; t1 ) Y t; t 1 ) ρ 0 t 1 ) C 0 rt 1 t 0 ), t 1 t t 1 + θ 1. 48) Let θ 1 be the first-passage time of the Wiener process wt 1 + ) wt 1 ) to the boundary of the interval [ r, r]. If t 1 + θ 1 < t 0 + T then set t := t 1 + θ 1, else set t := t 0 + T. In order to approximate Y t; t 1 ) for t 1 t t let us consider along with equation 47) the equation dy 1 dt = y k 1 y1, y 1 t 1 ) = Y t 1 ; t 1 ) = V t 1 ). Due to Proposition 3 and Corollary 5 it holds that Y t; t1 ) y 1 t) D 1 + D ) r t t 1 ) =: C 1 r t t 1 ), V t 1 ) t 1 t t. 49) and so by 48) we have Y t; t1 ) y 1 t) rc0 t 1 t 0 ) + C 1 t t 1 )), t 1 t t. 50) We also have see 46)) V t) = Y t; t1 ) + σ wt) wt 1)) = y 1 t) + σ wt) wt 1)) + R 1 t), 51) where R 1 t) rc0 t 1 t 0 ) + C 1 t t 1 )), t 1 t t. 5) We so define the approximation V t) := y 1 t) + σ wt) wt 1)), that satisfies 53) V t) = V t) + R 1 t), t 1 t t. 54) and then set V t ) = y 1 t ) + σ wt ) wt 1 )) = 55) y 1 t ) + σ { rξ1 with P ξ 1 = ±1) = 1/, if t = t 1 + θ 1 < t 0 + T,, ζ 1 if t = t 0 + T, cf. 40) and 41). We thus end up with a next approximation V t ) such that V t ) V t ) = R 1 t ) rc0 t 1 t 0 ) + C 1 t t 1 )). 56) From the above description it is obvious how to proceed analogously given a generic approximation sequence of approximations V t m ), m = 0, 1,,..., n, with V t 0 ) = V t 0 ), 1

13 that satisfies by assumption V t m ) σr, for m = 0,..., n, and 57) n 1 V tn ) V t n ) r D 1 + D ) t m+1 t m ) V t m ) 58) Indeed, consider the expression m=0 n 1 =: r C m t m+1 t m ). m=0 V t) = Y t; tn ) + σ wt) wt n)), where Y t; t n ) is the solution of the problem dy dt = Y + σ wt) wt n)) k Y + σ wt) wt n))), 59) Y t n ; t n ) = V t n ), t n t t n + θ n, for a θ n > 0 to be determined. Since V t n ) is unknown we consider Y t; t n ) as the solution of the equation 59) with initial value Y t n ; t n ) = V t n ). Due to 58) and Lemma again, we have Y t; t n ) Y t; t n ) n 1 r m=0 C m t m+1 t m ), t n t t n + θ n. In order to approximate Y t; t n ) for t n t t n + θ n, we consider the equation dy n dt = y k n yn, y n t n ) = Y t n ; t n ) = V t n ). 60) By repeating the procedure 49)-56) we arrive at V t) := y n t) + σ wt) wt n)), t n t t n+1, 61) satisfying V t) V t) = Rn t) r with n m=0 D 1 + D ) t m+1 t m ), t n t t n+1, 6) V t m ) R n t) := Y t; t n ) y n t), t n t t n+1, and in particular 63) n V tn+1 ) V t n+1 ) r D 1 + D ) t m+1 t m ). V t m ) m=0 13

14 Remark 8 In principle it is possible to use the distribution function Q see 43)) for constructing V t) for t n < t < t n+1. However, we rather consider for t n t t n+1 the approximation Ṽ t) := y n t) + σ w nt), t n t t n+1, where a) for t n+1 < t 0 + T, w is an arbitrary continuous function satisfying wt n ) = 0, wt n+1 ) = wt n+1 ) wt n ) = rξ n, max t n t t n+1 w n t) r, and b) for t n+1 = t 0 + T, one may take wt) 0. As a result we get similar to 44) an insignificant increase of the error, n V t) Ṽ t) r D 1 + D ) t m+1 t m ) + σr, t n < t < t n+1. V t m ) m=0 Let us consolidate the above procedure in a concise way. 5.1 Simulation algorithm Set V t 0 ) = V t 0 ). Let the point t n, V t n )) be known for an n 0. Simulate independent random variables ξ n with P ξ n = ±1) = 1/, and θ n as described in subsection 4.1. If t n + θ n < t 0 + T, set t n+1 = t n + θ n, else set t n+1 = t 0 + T. Solve equation 60) on the interval [t n, t n+1 ] with solution y n and set V t n+1 ) = y n t n+1 ) + σ { rξn if t n+1 < t 0 + T, 0 if t n+1 = t 0 + T. So, under the assumption 57) we obtain the estimate 6) possibly enlarged with a term σr). The next theorem shows that if a trajectory of V t) under consideration is positive on [t 0, t 0 + T ], then the algorithm is convergent on this trajectory. We recall that in the case kλ σ almost all trajectories are positive, hence in this case the proposed method is almost surely convergent. Theorem 9 Let 4kλ σ i.e., 0). Then for any positive trajectory V t) > 0 on [t 0, t 0 + T ] the proposed method is convergent on this trajectory. In particular, there exist η > 0 depending on the trajectory V ) only, and r 0 > 0 depending on η such that V t m ) η rσ, for m = 0, 1,,... for any r < r 0. So in particular 57) is fulfilled for all m = 0, 1,..., and the estimate 6) implies that for any r < r 0, V tn+1 ) V t n+1 ) r D 1 + D ) T, n = 0, 1,,..., ν. η 14

15 Proof. Let us define η := 1 min t 0 t t 0 +T r 0 := min η σ, V t) and η ) D 1 + D T η, 64) and let r < r 0. We then claim that for all m, V t m ) η rσ. 65) For m = 0 we trivially have V t 0 ) = V t 0 ) η η r 0 σ rσ. Now suppose by induction that V tm+1 ) V t m+1 ) r D 1 + D V t j ) η for j = 0,..., m.then due to 63) we have η ) T r 0 D 1 + D because of 64). Thus, since V t m+1 ) η, it follows that proves 65) and the convergence for r 0. η ) T η V t m+1 ) η rσ. This Remark 10. In the case where 4kλ σ > kλ trajectories will reach zero with positive probability, that is convergence on such trajectories is not guaranteed by Theorem 9. So it is important to develop some method for continuing the simulations in cases of very small V t m ). One can propose different procedures, for instance, one can proceed with standard SDE approximation methods relying on some known scheme suitable for small V e.g. see [3]). However, the uniformity of the simulation would be destroyed in this way. We therefore propose in the next section a uniform simulation method that may be started in a value V t m ) close to zero. 6 Simulation of trajectories near to zero Henceforth we assume that > 0. Let us suppose that V t n ) = y n σr and consider conditions that guarantee that V t n+1 ) σr under t n+1 t 0 + T. Of course in the case ξ n = 1 this is trivially fulfilled, and we thus consider the case ξ n = 1, yielding V t n+1 ) = y n t n+1 ) σr = [y ne kt n+1 t n) + k 1 e kt n+1 t n) )] 1/ σr. We so need yne kt n+1 t n) + k 1 e kt n+1 t n) ) 9σ r, i.e. 4 yn ) e kt n+1 t n) 9σ r k 4 k. 66) 15

16 Since we are interested in properties of algorithms when r 0, we may further assume w.l.o.g. that 9σ r /4 /k < 0, i.e. r < 3 kσ. 67) Under assumption 67), 66) is obviously fulfilled when y n /k. If y n < /k we need 9σ r e kt n+1 t n) 4 k yn, hence t n+1 t n 1 k ln k y n, k which is fulfilled if 9σ r k 4 y n 3 σr. 68) Note that 67) is equivalent with 3σr/ < /k, and so 68) is the condition we were looking for. Conversely, if σr y n < 3σr/, then V t n+1 ) < σr with positive probability. In view of the above considerations, one may carry out the algorithm of Subsection 5.1 as long as 68) is fulfilled. Let us say that n was the last step where 68) was true. Then the aggregated error of V t n+1 ) due to the algorithm up to step n may be estimated by cf. 57) and 58)), V tn+1 ) V t n+1 ) n r D 1 + D ) n t ym m+1 t m ) = r D 1 + D ) t m+1 t m ). 69) V t m ) m=0 Let us recall that our primal goal is a scheme where V t) V t) 0, almost surely and uniformly in t 0 t t 0 + T. In this respect, and in particular in the case σ > kλ where trajectories may attain zero with positive probability, it is not recommended to carry out scheme 5.1 all the way through until 68) is not satisfied anymore. Indeed, if the trajectory attains zero, the worst case almost sure error bound would then be when all y m would be close to 3σr/, hence of order O1/r). That is, no convergence on such trajectories. We therefore propose to perform scheme 5.1 up to a stopping) index m, defined by V t k ) 1 Ara > 3 σr, k = 0, 1,..., m, and V t m+1 ) < 1 Ara, 70) where A is a positive constant and 0 < a < 1/ is to be determined suitably. A pragmatic choice would be a = 1/3 see Remark 11). Due to 70) and 69) with n replaced by m we then have, m V tm+1 ) V t m+1 ) r for some constant D 3 > 0. m=0 m=0 D 1 + 4D ) t A r a m+1 t m ) : D 3 r 1 a. 71) 16

17 Let us now fix a realization t m+1 := t m+1, and consider two solutions of equation 1) starting at the moment t m+1 from v := V t m+1 ) known value) and v = V t m+1 ) true but unknown value), denoted by V tm+1,v and V tm+1,v, respectively, Let ϑ x, 0 x < A r a, be the first time at which the solution V tm+1,xt + t m+1 ) of 1) attains the level A r a, hence 0 V tm+1,xt m+1 + t) < A r a, 0 t < ϑ x, V tm+1,xt m+1 + ϑ x ) = A r a. A construction of the distribution function of ϑ x is worked out in Section 6.1. Let us now denote t m+ := t m+1 + ϑ with ϑ = ϑ v. For simplicity and w.l.og. we assume that t m+ < t 0 + T ). We then naturally set V t m+ ) = V tm+1,vt m+1 + ϑ) = A r a. The solutions V tm+1,v and V tm+1,v correspond to two solutions Y t, t m+1 ) and Y t, t m+1 ) of 0) with ϕt) = wt) wt m+1 ), t t m+1, starting in Y t m+1, t m+1 ) = v and Y t m+1, t m+1 ) = v, respectively. Due to Lemma, see Remark 3, and 71) it thus follows that V tm+1,vt) V tm+1,vt) = Y t, t m+1 ) Y t, t m+1 ) V tm+1 ) V t m+1 ) and in particular D 3 r 1 a, t m+1 t t m+, 7) V tm+ ) V t m+ ) D 3 r 1 a. In contrast to the previous steps we now specify the behavior of V t) on [t m+1, t m+ ] by V t) = V tm+1,vt), t m+1 t t m+, 73) which we actually do not know. However, we do know that V t m+ ) = V tm+1,vt m+ ) = A r a, and that V is bounded on [t m+1, t m+ ] by A r a. Therefore, if we just take a straight line Lt) that connects the points t m+1, v) and t m+, Ar a ) as an approximation for V t), then V t) Lt) Ar a, t m+1 t t m+. By 7) and 73) we then also have V t) Lt) Ar a, t m+1 t t m+. Thus, the accuracy of the approximation to V for 0 t t m+1 outside the band 0, 1 ) Ara is of order Or 1 a ), and for t m+1 < t < t m+ inside the band 0, Ar a ) of order Or a ). But, at the boundary point V t m+ ) = A r a the accuracy is of order Or 1 a ) again. Finally, the scheme may be continued from the state V t m+ ) = Ar a with the algorithm of Subsection 5.1. Remark 11 From the above construction it is clear that for a = 1/3 in 70) the accuracy for 0 t t m+1 outside the band 0, 1 ) Ara, and for t m+1 < t < t m+ inside the band 0, Ar a ) are of the same order. However, an exponent 0 < a < 1/3 would give a higher accuracy outside the band 0, 1 ) Ara and at the exit points of the band 0, Ar a ), while inside the band the accuracy is worse but uniformly bounded by Ar a. 17

18 6.1 Simulation of ϑ x In order to carry out the above simulation method for trajectories near zero we have to find the distribution function of ϑ x = ϑ x,l, where ϑ x,l is the first-passage time of the trajectory X 0,x s), to the level l. For this it is more convenient to change notation and to write 1) in the form dxs) = kλ Xs))ds + σ Xdws), X0) = x, 74) where without loss of generality we take the initial time to be s = 0. The function ut, x) := P ϑ x,l < t), is the solution of the first boundary value problem of parabolic type [16], Ch. 5, Sect. 3) with initial data and boundary conditions u t = 1 σ x u + kλ x) u, t > 0, 0 < x < l, 75) x x u0, x) = 0, 76) ut, 0) is bounded, ut, l) = 1. 77) To get homogeneous boundary conditions we introduce v = u 1. The function v then satisfies: v t = 1 σ x v + kλ x) v, t > 0, 0 < x < l, x x 78) v0, x) = 1; vt, 0) is bounded, vt, l) = 1. 79) The problem 78)-79) can be solved by the method of separation of variables. In this way the Sturm-Liouville problem for the confluent hypergeometric equation the Kummer equation) arises. This problem is rather complicated however. Below we are going to solve an easier problem as a good approximation to 78)-79). Along with 74), let us consider the equations dx + s) = kλds + σ X + dws), X + 0) = x, 80) dx s) = kλ l)ds + σ X dws), X 0) = x, 81) with 0 l < λ. It is not difficult to prove the following inequalities X s) Xs) X + s). 8) According to 8), we consider three boundary value problems: first 75)-77) and next similar ones for the equations u + t u t From 8) it follows that = 1 σ x u + x = 1 σ x u x + kλ u+, t > 0, 0 < x < l, x + kλ l) u, t > 0, 0 < x < l. 83) x u t, x) ut, x) u + t, x), 18

19 hence v t, x) vt, x) v + t, x), where v = u 1, v + = u + 1. As the band 0 < x < l = A r a, for a certain a > 0, is narrow due to small enough r, the difference v + v will be small and so we can consider the following problem v + t = 1 σ x v + x + kλ v+, t > 0, 0 < x < l, 84) x v + 0, x) = 1; v + t, 0) is bounded, v + t, l) = 0, 85) as a good approximation of 78)-79). Henceforth we write v := v +. By separation of variables we get as elementary independent solutions to 84), T t)x x), where T t) + µt t) = 0, i.e. T t) = T 0 e µt, µ > 0, and 86) 1 σ xx + kλx + µx = 0, X 0+) is bounded, X l) = 0. 87) It can be verified straightforwardly that the solution of 87) can be obtained in terms of Bessel functions of the first kind e.g. see [4]), X x) = X γ ± x) := x γ J ±γ σ 1 ) 8µx = x γ Ox ±γ ) if x 0, with γ := 1 kλ σ. 88) Since X x) has to be bounded for x 0 we may take regardless the sign of γ!)) X x) = Xγ x) =: X γ x) = x γ J γ σ 1 ) 8µx. 89) In our setting we have > 0, i.e. γ < 1/4. The following derivation of a Fourier-Bessel series for v is standard but included for convenience of the reader. Denote the positive zeros of J ν by π ν,m, for example, J 1/ x) = πx sin x and π 1/,m = mπ, m = 1,,... 90) Then the homogeneous) boundary condition X γ l) = 0 yields and we have σ 1 8µl = π γ,m, i.e., µ m := σ π γ,m 8l X γ,m x) := x γ J γ σ 1 ) ) x 8µ m x = x γ J γ π γ,m. l 91) By the well-known orthogonality relation 1 0 zj γ π γ,k z)j γ π γ,k z)dz = δ k,k J γ+1π γ,k ), 19

20 we get by setting z = x/l Now set l 0 J γ π γ,m x l )J γπ γ,m l 0 x l )dx = lδ m,m J γ+1π γ,m ), X γ,m x)x γ,m x)x γ dx = lδ m,m J γ+1π γ,m ). vt, x) = hence β m e µmt X γ,m x), 0 x l. 9) m=1 For t = 0 we have due to the initial condition v0, x) = 1, So for any p = 1,,..., Further it holds that l 0 1 = β m X γ,m x). m=1 l X γ,p x)x γ dx = β p lj γ+1π γκ,p), i.e. 0 l 0 β p = X γ,px)x γ dx lj γ+1π γ,p ). 93) X γ,p x)x γ dx = l 0 ) x x γ J γ π γ,p dx l 1 = l γ+1 z γ+1 J γ π γ,p z) dz 0 = l γ+1 J γ+1 π γ,p ) π γ,p by well-known identities for Bessel functions e.g. see [4]), and 93) thus becomes β p =, p = 1,,... 94) l γ π γ,p J γ+1 π γ,p ) So, from v = u 1, 86) 89), 91), 94), and 9) we finally obtain J x ) [ ut, x) = 1 x γ l γ γ π γ,m l π γ,m J γ+1 π γ,m ) exp σ π γ,m ] t, 8l 0 x l. 95) m=1 Example 1 For γ = 1/4 we get from 95) by 90) straightforwardly, ut, x) = 1 + π l x 1) m m=1 m sin πm x l ) ] exp [ σ π m t. 8l 0

21 For solving 83) we set λ := λ l, and then apply the Fourier-Bessel series 95) with γ replaced by γ := 1 kλ σ = γ + kl σ. 96) Example 13 We now consider some numerical examples concerning u + = u in 95) and u given by 95) due to 96). Note that actually in 95) the function u only depends on σ, l, and γ. That is, u depends on σ, l, and the product kλ. Let us consider a CIR process with σ = 1, λ = 1, k = 0.75, and let us take l = 0.1. We then compare u +, which is given by 95) for γ = 0.5 due to 88) see Example 1), with u given by 95) for γ = due to 96). The results are depicted in Figure 1. The sums corresponding to 95) are computed with five terms more terms did not give any improvement). Normalization of ut, x) For practical applications it is useful to normalize 95) in the following way. Let us treat γ as essential but fixed parameter, introduce as new parameters and consider the function ũ t, x) := 1 x γ m=1 that is connected to 95) via x l = x, 0 < x 1, σ t 8l J γ π γ,m x ) = t, t 0, π γ,m J γ+1 π γ,m ) exp [ π γ,m t ], 0 < x 1, t 0, ũ t, x) = ũ σ t 8l, x l ) = u8l t σ, l x). For simulation of ϑ x we need to solve the equation uϑ x, x) = U, where U Uniform[0, 1]. For this we set x = x/l and solve the normalized equation ũ ϑ x, x) = U, and then take Note that ϑ x = 8l σ ϑ x. P ϑ x < t) = P ϑ x < σ t 8l ) = ũσ t 8l, x l ). We have plotted in Figure the normalized function ũ t, x) for γ = 1/4. 1

22 Figure 1: Upper panel u + 0.1, x), lower panel u + 0.1, x) u 0.1, x), for 0 x Figure : Normalized distribution function ũ t, x) for γ = 1/4

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