Chapter 20 Bessel Processes 1 n We follow the discussion of section one of chapter XI of [17]. Let Bt = (B ; :::; B ) be an uncorrelated d-dimensional Brownian motion. Then the process q 1 2 n 2 ρt = jBtj = (Bt ) + ··· + (Bt ) (20.1) 2 is called an n-dimensional Bessel process and Zt = ρt is called a squared Bessel process. Using Ito's formula, n X i i dZt = 2 Bt dBt + n dt (20.2) i=1 Consider the process n X Z t Bi W := s dBi (20.3) t ρ s i=1 0 s For n > 1, ρt > 0 almost surely and for d = 1 the set fs j ρs = 0g has Lebesgue measure zero almost surely such that Wt is well defined. Because of the quadratic variation n X Z t (Bi)2 hW ;W i = s ds = t (20.4) t t ρ2 i=1 0 s Wt coincides with a one dimensional Brownian motion. Thus Zt is the solution of the SDE p dZt = 2 Zt dWt + n dt (20.5) Although the square root is only H¨oldercontinuous, but not Lipshitz, equation (20.5) has a unique solution for all n ≥ 0 (not necessarily an integer) and Z0 = x ≥ 0, see for example [17], section 3 of chapter IX. Thus Zt is a time homogenous Markov process which n n is determined by its transition function qt(x; y) = qt (x; y) = q (0; x; t; y) given by n s;x;n q (s; x ; t; y) dy = P Zt 2 [y; y + dy) (20.6) 163 164 Chapter 20 s;x;n s;x;n where Zt denotes the solution of (20.5) with initial condition Zs = x. Finally we introduce the probability density n;x n qt (y) := q (0; x ; t; y) (20.7) such that Z 0;x;n n;x E F (Zt ) = F (y) qt (y) dy (20.8) R There is the following Theorem 20.1: n;x a) Let qt be the probability density for an n-dimensional squared Bessel process start- + + ing at x 2 R at time zero given by (20.7), n 2 R . Then for all n1; n2; x1; x2 ≥ 0 n1;x1 n2;x2 n1+n2;x1+x2 qt ∗ qt = qt (20.9) That is, R qn1;x1 (z) qn2;x2 (y − z) dz = qn1+n2;x1+x2 (y) (20.10) R t t t 0;x;n b) Let µ be some measure on R and let Zt denotes the solution of (20.5) with initial 0;x;n condition Z0 = x. Then there exist some numbers Aµ and Bµ such that h R 1 0;x;n i − 0 Zt dµ(t) x n E e = Aµ Bµ (20.11) c) The transition function (20.6) for the n-dimensional squared Bessel process is given n n by q (s; x ; t; y) = qt−s(x; y) where ν p n 1 y x+y xy 2 − 2t qt (x; y) = 2t x e Iν t ; x; y ≥ 0 (20.12) n P1 1 z 2k+ν Here ν = 2 − 1 and Iν(z) = k=0 k! Γ(k+ν+1) 2 is the Bessel function with n;0 imaginary argument. In particular, the probability density qt is given by n y n;0 1 y 2 −1 − qt (y) = n e 2t ; y ≥ 0 (20.13) 2t Γ( 2 ) 2t Proof: a) We have qn1;x1 ∗ qn2;x2 (y) dy = R qn1;x1 (z) qn2;x2 (y − z) dz dy t t Rz t t = R PZ0;x1;n1 2 [z; z + dz) PZ0;x2;n2 2 [y − z; y − z + dy) Rz t t = R PZ0;x1;n1 2 [z; z + dz) Pz + Z0;x2;n2 2 [y; y + dy) Rz t t 0;x1;n1 0;x2;n2 = P Zt + Zt 2 [y; y + dy) 0;x1+x2;n1+n2 = P Zt 2 [y; y + dy) (20.14) n1+n2;x1+x2 = qt (y) dy (20.15) Chapter 20 165 0;x1;n1 0;x2;n2 The equality in (20.14) follows from the fact that Yt := Zt + Zt has the same 0;x1+x2;n1+n2 distribution as Zt . Namely, Yt is a solution of the SDE q q 0;x1;n1 1 0;x2;n2 2 dYt = 2 Zt dWt + Zt dWt + (n1 + n2) dt p =: 2 Yt dWt + (n1 + n2) dt (20.16) where we defined t p p Z 0;x1;n1 0;x2;n2 Zt 1 Zt 2 Wt := p dW + p dW (20.17) Yt t Yt t 0 t 0;x1;n1 0;x2;n2 Since the quadratic variation hW ;W i = R Zs +Zs ds = t, the process W is a t t 0 Ys t Brownian motion and part (a) follows. b) Suppose that x = x1 + x2 and n = n1 + n2. Let h − R 1 Z0;x;ndµ(t)i φ(x; n) = E e 0 t (20.18) Then, with part (a) h − R 1(Z0;x1;n1 +Z0;x2;n2 )dµ(t)i φ(x1 + x2; n1 + n2) = E e 0 t t = φ(x1; n1) φ(x2; n2) (20.19) which gives φ(x; n) = φ(x; 0) φ(0; n) = φ(1; 0)x φ(0; 1)n (20.20) This proves part (b). n c) We compute the Laplace transform of qt (x; y), 0;x;n Z −λZt −λy 0;x;n E e = e P Zt 2 [y; y + dy) R Z 1 −λy n = e qt (x; y) dy (20.21) 0 To obtain the left hand side ofp (20.21), we put dµ(s) = λδ(s−t)ds in part (b) and consider φ(x; 1) = Ax B . Then, if B0; x denotes a one dimensional Brownian motion starting at p µ µ t x at time zero, p h −λ R 1 Z0;x;1 δ(s−t)dsi h −λZ0;x;1 i h −λ(B0; x)2 i φ(x; 1) = E e 0 s = E e t = E e t p 2 x p R −λy2 1 − ( x−y) e− 2t R −(λ+ 1 )y2+ x y = e p e 2t dy = p e 2t t R 2πt 2πt R x x − 2t 2 1 x x 1 e q π 4t (λ+ ) 1 − 2t + 2t 1+2λt = p 1 e 2t = p e 2πt λ+ 2t 1+2λt 1 − λ x = p e 1+2λt (20.22) 1+2λt 166 Chapter 20 which gives 1 − λ B = p ;A = e 1+2λt (20.23) µ 1+2λt µ and therefore 1 Z 1 λ −λy n − 1+2λt x e qt (x; y) dy = n e (20.24) 0 (1 + 2λt) 2 However, this equation is solved by (20.12). Next next theorem computes the numbers Aµ and Bµ in (20.11) of the last theorem. The measure is supposed to be a sum of a Dirac measure plus an absolute continuous part. 0;x;n Theorem 20.2: Let Xs = Xs be an n-dimensional squared Bessel process starting at x ≥ 0 at time zero. Let ρ = ρ(s) be some real density and a 2 R. Then h a X − 1 R t X ρ(s)dsi n 1 f 0(0)x E e 2 t 2 0 s = [f(t)] 2 e 2 (20.25) where the function f(s) is the unique solution of the differential equation f 00(s) − ρ(s) f(s) = 0 ; s 2 (0; t) (20.26) with boundary conditions f(0) = 1 (20.27) f 0(t) − a f(t) = 0 (20.28) 1 Xt− hXit Proof: First we recall that, if Xt is some martingale, then also Yt = e 2 is a 1 1 martingale since, by Ito, dYt = Yt(dXt − 2 dhXit) + 2 YtdhXit = YtdXt. Now let Xt be the n-dimensional squared Bessel process starting at x ≥ 0 at time zero. Then Mt := Xt − nt (20.29) is a martingale and, for some function F = F (s), also Z t Nt := F (s) dMs (20.30) 0 Z t Z t = F (s) dXs − n F (s) ds 0 0 Z t Z t = F (t)Xt − F (0)X0 − XsdF (s) − n F (s) ds (20.31) 0 0 Chapter 20 167 is a martingale, and, by the above remark, also 1 Nt− hNit Yt := e 2 R t F (s) dM − 1 R t F 2(s)dhMi = e 0 s 2 0 s (20:31) F (t)X −F (0)x−R t X dF (s)−n R t F (s) ds− 1 R t F 2(s)dhXi = e t 0 s 0 2 0 s (20:5) F (t)X −F (0)x−R t X dF (s)−n R t F (s) ds−2 R t F 2(s)X ds = e t 0 s 0 0 s F (t)X −F (0)x−R t(F 0+2F 2)(s)X ds−n R t F (s) ds = e t 0 s 0 (20.32) is a martingale. Thus we have E[Yt] = Y0 = 1 or, h F (t)X −R t(F 0+2F 2)(s)X dsi F (0)x+n R t F (s) ds E e t 0 s = e 0 (20.33) Now let F = λf 0=f = λ(log f)0 for some function f. Then 0 2 f 00f−(f 0)2 2 (f 0)2 F + 2F = λ f 2 + 2λ f 2 f 00 2 (f 0)2 = λ f + (2λ − λ) f 2 λ= 1 2 1 f 00 = 2 f (20.34) and (20.33) becomes 0 00 0 h f (t) X −R t f (s)X dsi f (0) x+ n log( f(t) ) E e 2f(t) t 0 2f s = e 2f(0) 2 f(0) (20.35) Because of (20.26-20.28), the theorem follows.