Scale functions for spectrally negative Lévy processes and their appearance in economic models

Transcription

1 Scale functions for spectrally negative Lévy processes and their appearance in economic models Andreas E. Kyprianou 1 Department of Mathematical Sciences, University of Bath 1 This is a review talk and not all my own work. The vast majority of what I say leans on collaborative with and independent work of the following individuals F. Hubalek, A. Lambert, R. Loeffen, P. Patie, M.Pistorius, Z. Palmowski, J-F. Renaud, V. Rivero, R. Song, X. Zhou. 1/ 16

2 Scale functions What is a scale function? Suppose that X = {X t : t 0} is a general spectrally negative Lévy process. 2/ 16

3 Scale functions What is a scale function? Suppose that X = {X t : t 0} is a general spectrally negative Lévy process. Use P x to denote the law of X when X 0 = x > 0 with associated expectation operator E x. 2/ 16

4 Scale functions What is a scale function? Suppose that X = {X t : t 0} is a general spectrally negative Lévy process. Use P x to denote the law of X when X 0 = x > 0 with associated expectation operator E x. Let m := E 0(X 1) and assume temporarily that m > 0. This implies that is almost surely finite in magnitude. X := inf s 0 Xs 2/ 16

5 Scale functions What is a scale function? Suppose that X = {X t : t 0} is a general spectrally negative Lévy process. Use P x to denote the law of X when X 0 = x > 0 with associated expectation operator E x. Let m := E 0(X 1) and assume temporarily that m > 0. This implies that is almost surely finite in magnitude. Let us define X := inf s 0 Xs W (x) = 1 m Px (X 0) i.e. the ruin probability with initial position x is 1 mw (x). 2/ 16

6 Scale functions What is a scale function? Suppose that X = {X t : t 0} is a general spectrally negative Lévy process. Use P x to denote the law of X when X 0 = x > 0 with associated expectation operator E x. Let m := E 0(X 1) and assume temporarily that m > 0. This implies that is almost surely finite in magnitude. Let us define X := inf s 0 Xs W (x) = 1 m Px (X 0) i.e. the ruin probability with initial position x is 1 mw (x). From now I will call W a scale function. 2/ 16

7 Scale functions Is there a better characterization of scale functions? Take its Laplace-Steiltjes transform Z 0 Z e βx W (x)dx = 1 e βx P 0( X m x)dx 0 1 = βm E0(eβX ) 3/ 16

8 Scale functions Is there a better characterization of scale functions? Take its Laplace-Steiltjes transform Z 0 Z e βx W (x)dx = 1 e βx P 0( X m x)dx 0 1 = βm E0(eβX ) The Laplace transform on the right hand side should remind you of the Wiener-Hopf factorization and in fact it has been known for a very long time that E 0(e βx ) = m β ψ(β) where E 0(e βxt ) = e ψ(β)t. 3/ 16

9 Scale functions Is there a better characterization of scale functions? Take its Laplace-Steiltjes transform Z 0 Z e βx W (x)dx = 1 e βx P 0( X m x)dx 0 1 = βm E0(eβX ) The Laplace transform on the right hand side should remind you of the Wiener-Hopf factorization and in fact it has been known for a very long time that E 0(e βx ) = m β ψ(β) where In conclusion Z 0 E 0(e βxt ) = e ψ(β)t. e βx W (x)dx = 1 ψ(β) i.e. we are just one Laplace inversion away from the ruin probability. 3/ 16

10 Scale functions What else can you do with scale functions? Define τ + a = inf{t > 0 : X t > a} and τ 0 = inf{t > 0 : X t < 0}. 4/ 16

11 Scale functions What else can you do with scale functions? Define τ + a = inf{t > 0 : X t > a} and τ 0 = inf{t > 0 : X t < 0}. Condition the event {X 0} on the event of hitting (continuously) a > 0 before entering (, 0) or its complement and we get P x (X 0) = P x (τ + a < τ 0 )P a(x 0) + 0 4/ 16

12 Scale functions What else can you do with scale functions? Define τ + a = inf{t > 0 : X t > a} and τ 0 = inf{t > 0 : X t < 0}. Condition the event {X 0} on the event of hitting (continuously) a > 0 before entering (, 0) or its complement and we get P x (X 0) = P x (τ + a < τ 0 )P a(x 0) + 0 In other words we recover a classical formula of Takács and Zoletaraev P x (τ + a < τ 0 ) = W (x) W (a) 4/ 16

13 Scale functions Push things a little further by changing measure using the transform (c 0) dp c cxt ψ(c)t = e dp Ft for c = Φ(q), the largest root of the equation ψ(θ) = q then, E x (e Φ(q)(X τ + a x) qτ +a ; τ + a < τ 0 ) = W Φ(q)(x) W Φ(q) (a) where W Φ(q) plays the role of W under P Φ(q). 5/ 16

14 Scale functions Push things a little further by changing measure using the transform (c 0) dp c cxt ψ(c)t = e dp Ft for c = Φ(q), the largest root of the equation ψ(θ) = q then, E x (e Φ(q)(X τ + a x) qτ +a ; τ + a < τ 0 ) = W Φ(q)(x) W Φ(q) (a) where W Φ(q) plays the role of W under P Φ(q). Remembering that X τ + a = a we can see that E x (e qτ + a ; τ + a < τ 0 ) = e Φ(q)(x a) W Φ(q) (x) W Φ(q) (a) = W (q) (x) W (q) (a) where W (q) (x) := e Φ(q)x W Φ(q) (x). 5/ 16

15 Scale functions Push things a little further by changing measure using the transform (c 0) dp c cxt ψ(c)t = e dp Ft for c = Φ(q), the largest root of the equation ψ(θ) = q then, E x (e Φ(q)(X τ + a x) qτ +a ; τ + a < τ 0 ) = W Φ(q)(x) W Φ(q) (a) where W Φ(q) plays the role of W under P Φ(q). Remembering that X τ + a = a we can see that E x (e qτ + a ; τ + a < τ 0 ) = e Φ(q)(x a) W Φ(q) (x) W Φ(q) (a) = W (q) (x) W (q) (a) where W (q) (x) := e Φ(q)x W Φ(q) (x). Since the Laplace exponent of (X, P Φ(q) ) is ψ Φ(q) (θ) = ψ(θ + Φ(q)) q we may compute for β > Φ(q) Z 0 e βx W (q) (x)dx = 1 ψ Φ(q) (β Φ(q)) = 1 ψ(β) q. 5/ 16

16 Scale functions Push things a little further by changing measure using the transform (c 0) dp c cxt ψ(c)t = e dp Ft for c = Φ(q), the largest root of the equation ψ(θ) = q then, E x (e Φ(q)(X τ + a x) qτ +a ; τ + a < τ 0 ) = W Φ(q)(x) W Φ(q) (a) where W Φ(q) plays the role of W under P Φ(q). Remembering that X τ + a = a we can see that E x (e qτ + a ; τ + a < τ 0 ) = e Φ(q)(x a) W Φ(q) (x) W Φ(q) (a) = W (q) (x) W (q) (a) where W (q) (x) := e Φ(q)x W Φ(q) (x). Since the Laplace exponent of (X, P Φ(q) ) is ψ Φ(q) (θ) = ψ(θ + Φ(q)) q we may compute for β > Φ(q) Z 0 e βx W (q) (x)dx = 1 ψ Φ(q) (β Φ(q)) = 1 ψ(β) q. We call W (q) the q-scale function and we are only ever one inverse Laplace transform away from it. 5/ 16

17 Scale functions Taking stock 6/ 16

18 Scale functions Taking stock So far we not really assumed much other than m > 0. In fact with a little work even this assumption can be dropped. 6/ 16

19 Scale functions Taking stock So far we not really assumed much other than m > 0. In fact with a little work even this assumption can be dropped. We have extracted a nice formula for the two sided exit problem and one might hope to see better fluctuation identities appearing. Such formulae could easily feed into a number of settings for financial and insurance models. 6/ 16

20 Scale functions Taking stock So far we not really assumed much other than m > 0. In fact with a little work even this assumption can be dropped. We have extracted a nice formula for the two sided exit problem and one might hope to see better fluctuation identities appearing. Such formulae could easily feed into a number of settings for financial and insurance models. So how far do things go? 6/ 16

21 Fast forward to some fancy formulae First passage formula: 7/ 16

22 Fast forward to some fancy formulae First passage formula: Let X τ := inf t<τ Xt where τ := inf{t > 0 : X t < 0} is the ruin time of X. 7/ 16

23 Fast forward to some fancy formulae First passage formula: Let X τ := inf t<τ Xt where τ := inf{t > 0 : X t < 0} is the ruin time of X. We are interested in the triple-law K x (q) (du, dv, dy) := E `e qτ x ; X τ du, X τ dv, X τ dy for u, y 0 and v y. 7/ 16

24 Fast forward to some fancy formulae First passage formula: Let X τ := inf t<τ Xt where τ := inf{t > 0 : X t < 0} is the ruin time of X. We are interested in the triple-law K x (q) (du, dv, dy) := E `e qτ x ; X τ du, X τ dv, X τ dy for u, y 0 and v y. Then K x (q) (du, dv, dy) = e Φ(q)(v y) {W (q) (x y) Φ(q)W (q) (x y)}π(du + v)dydv where Π is the jump measure of X and Φ(q) is the solution to ψ(θ) = q. 7/ 16

25 Fast forward to some fancy formulae First passage formula: Let X τ := inf t<τ Xt where τ := inf{t > 0 : X t < 0} is the ruin time of X. We are interested in the triple-law K x (q) (du, dv, dy) := E `e qτ x ; X τ du, X τ dv, X τ dy for u, y 0 and v y. Then K x (q) (du, dv, dy) = e Φ(q)(v y) {W (q) (x y) Φ(q)W (q) (x y)}π(du + v)dydv where Π is the jump measure of X and Φ(q) is the solution to ψ(θ) = q. If we are just interested in the undershoot-overshoots (i.e. q = 0) then we have K x (du, dv, dy) = W (x y)π(du + v)dydv 7/ 16

26 Fast forward to some fancy formulae Perpetual American put: 8/ 16

27 Fast forward to some fancy formulae Perpetual American put: The perpetual put requires one to evaluate the function v(x) = sup E x (e qτ (K e Xτ )) τ where the supremum is taken over all stopping times τ with respect to the natural filtration of X and q > 0. 8/ 16

28 Fast forward to some fancy formulae Perpetual American put: The perpetual put requires one to evaluate the function v(x) = sup E x (e qτ (K e Xτ )) τ where the supremum is taken over all stopping times τ with respect to the natural filtration of X and q > 0. Both common sense and rigour tell us that for some optimal y R. v(x) = E x (e qτ y (K e X τ y )) 8/ 16

29 Fast forward to some fancy formulae Perpetual American put: The perpetual put requires one to evaluate the function v(x) = sup E x (e qτ (K e Xτ )) τ where the supremum is taken over all stopping times τ with respect to the natural filtration of X and q > 0. Both common sense and rigour tell us that for some optimal y R. v(x) = E x (e qτ y (K e X τ y )) The evaluation of the latter can be computed from the triple law (or indeed more directly using scale functions) and the optimized over y to give Z x x Z x x v(x) = K e x x +qk W (q) (z)dz (q ψ(1)) e x x z W (q) (z)dz 0 0 where j x = log K q Φ(q) ff Φ(q) 1 q ψ(1) 8/ 16

30 Fast forward to some fancy formulae Perpetual Russian options: 9/ 16

31 Fast forward to some fancy formulae Perpetual Russian options: Perpetual russian option requires one to evaluate for q > ψ(1) u(x) = sup E(e qτ e (x X τ ) ) τ which after an exponential change of measure becomes u(x) = sup E (e pτ e (x X τ Xτ ) ) τ where p = q ψ(1) and dp /dp Ft = exp{x t ψ(1)t}. 9/ 16

32 Fast forward to some fancy formulae Perpetual Russian options: Perpetual russian option requires one to evaluate for q > ψ(1) u(x) = sup E(e qτ e (x X τ ) ) τ which after an exponential change of measure becomes u(x) = sup E (e pτ e (x X τ Xτ ) ) τ where p = q ψ(1) and dp /dp Ft = exp{x t ψ(1)t}. Again common sense and rigour tells us that the optimal strategy is first passage of x X τ X τ over a level. One computes in the end where Z x x u(x) = e x + qe x 0 W (q) (z)dz Z x x = inf{x 0 : 1 + q W (q) (z)dz qw (q) (x)}. 0 9/ 16

33 Fast forward to some fancy formulae Actuarial dividend problem: 10/ 16

34 Fast forward to some fancy formulae Actuarial dividend problem: If X it thought of as the surplus process of an insurance company (eg the classical Cramér-Lundberg process) then the process L t = (a X t) a is a dividend strategy which pays out the excess of the aggregate surplus process when it exceeds a threshold a. 10/ 16

35 Fast forward to some fancy formulae Actuarial dividend problem: If X it thought of as the surplus process of an insurance company (eg the classical Cramér-Lundberg process) then the process L t = (a X t) a is a dividend strategy which pays out the excess of the aggregate surplus process when it exceeds a threshold a. Write Y = X t L t for the aggregate process. Let be the ruin time. σ = inf{t > 0 : Y t < 0} 10/ 16

36 Fast forward to some fancy formulae Actuarial dividend problem: If X it thought of as the surplus process of an insurance company (eg the classical Cramér-Lundberg process) then the process L t = (a X t) a is a dividend strategy which pays out the excess of the aggregate surplus process when it exceeds a threshold a. Write Y = X t L t for the aggregate process. Let be the ruin time. σ = inf{t > 0 : Y t < 0} Then, for q 0, its Laplace-transform is given by Z x E x (e qσ ) = 1 + q W (q) (z)dz qw (q) (x) W (q) (a) 0 W (q) (a) 10/ 16

37 Fast forward to some fancy formulae Actuarial dividend problem: If X it thought of as the surplus process of an insurance company (eg the classical Cramér-Lundberg process) then the process L t = (a X t) a is a dividend strategy which pays out the excess of the aggregate surplus process when it exceeds a threshold a. Write Y = X t L t for the aggregate process. Let be the ruin time. σ = inf{t > 0 : Y t < 0} Then, for q 0, its Laplace-transform is given by Z x E x (e qσ ) = 1 + q W (q) (z)dz qw (q) (x) W (q) (a) 0 W (q) (a) Moreover the net present value of dividends with discount rate p > 0 has moments given by jz σ ff n «E x e pt dl t 0 = n! W (pn) (x) W (pn) (a) ny k=1 W (pk) (a) W (pk) (a) 10/ 16

38 How to find explicit scale functions Take stock again 11/ 16

39 How to find explicit scale functions Take stock again The preceding formulae are all just one Laplace-transform away from being explicit but none the less, there is clearly a Laplace inversion to take place before anything becomes explicit. 11/ 16

40 How to find explicit scale functions Take stock again The preceding formulae are all just one Laplace-transform away from being explicit but none the less, there is clearly a Laplace inversion to take place before anything becomes explicit. The only explicit scale functions in circulation (until last year) corresponded to the following Lévy processes 11/ 16

41 How to find explicit scale functions Take stock again The preceding formulae are all just one Laplace-transform away from being explicit but none the less, there is clearly a Laplace inversion to take place before anything becomes explicit. The only explicit scale functions in circulation (until last year) corresponded to the following Lévy processes Brownian motion with drift 11/ 16

42 How to find explicit scale functions Take stock again The preceding formulae are all just one Laplace-transform away from being explicit but none the less, there is clearly a Laplace inversion to take place before anything becomes explicit. The only explicit scale functions in circulation (until last year) corresponded to the following Lévy processes Brownian motion with drift α-stable processes for α (1, 2) 11/ 16

43 How to find explicit scale functions Take stock again The preceding formulae are all just one Laplace-transform away from being explicit but none the less, there is clearly a Laplace inversion to take place before anything becomes explicit. The only explicit scale functions in circulation (until last year) corresponded to the following Lévy processes Brownian motion with drift α-stable processes for α (1, 2) Cramér-Lundberg process with exponential/mixed exponential/phase-type jumps 11/ 16

44 How to find explicit scale functions Take stock again The preceding formulae are all just one Laplace-transform away from being explicit but none the less, there is clearly a Laplace inversion to take place before anything becomes explicit. The only explicit scale functions in circulation (until last year) corresponded to the following Lévy processes Brownian motion with drift α-stable processes for α (1, 2) Cramér-Lundberg process with exponential/mixed exponential/phase-type jumps How explicit can one be with scale functions? 11/ 16

45 How to find explicit scale functions Building scale functions Hidden in the Wiener-Hopf factorization is a very simple trick for constructing a plethora of examples. 12/ 16

46 How to find explicit scale functions Building scale functions Hidden in the Wiener-Hopf factorization is a very simple trick for constructing a plethora of examples. Recall that by definition Z 0 e βx W (x)dx = 1 ψ(β) 12/ 16

47 How to find explicit scale functions Building scale functions Hidden in the Wiener-Hopf factorization is a very simple trick for constructing a plethora of examples. Recall that by definition Z 0 e βx W (x)dx = 1 ψ(β) The Wiener-Hopf factorization tells us that necessarily ψ(β) = βφ(β) where e φ(β)t = E(e βht ) and H = {H t : t 0} is the descending ladder height (killed) subordinator. 12/ 16

48 How to find explicit scale functions Building scale functions Hidden in the Wiener-Hopf factorization is a very simple trick for constructing a plethora of examples. Recall that by definition Z 0 e βx W (x)dx = 1 ψ(β) The Wiener-Hopf factorization tells us that necessarily ψ(β) = βφ(β) where e φ(β)t = E(e βht ) and H = {H t : t 0} is the descending ladder height (killed) subordinator. Integration by parts thus yields Z 0 e βx W (x)dx = 1 φ(β) 12/ 16

49 How to find explicit scale functions Building scale functions Hidden in the Wiener-Hopf factorization is a very simple trick for constructing a plethora of examples. Recall that by definition Z 0 e βx W (x)dx = 1 ψ(β) The Wiener-Hopf factorization tells us that necessarily ψ(β) = βφ(β) where e φ(β)t = E(e βht ) and H = {H t : t 0} is the descending ladder height (killed) subordinator. Integration by parts thus yields Z 0 e βx W (x)dx = 1 φ(β) Classical theory now identifies W (x)dx as the renewal measure of the subordinator H about which a lot is known. 12/ 16

50 How to find explicit scale functions The basic idea Instead of choosing the Lévy process (ie. choosing ψ) choose a subordinator (ie. choose φ) whose renewal measure is known and build an associated Lévy process by defining its Laplace exponent ψ(β) = βφ(β). 13/ 16

51 How to find explicit scale functions The basic idea Instead of choosing the Lévy process (ie. choosing ψ) choose a subordinator (ie. choose φ) whose renewal measure is known and build an associated Lévy process by defining its Laplace exponent ψ(β) = βφ(β). Q: How do we know which subordinators are allowed to participate in such a relation? 13/ 16

52 How to find explicit scale functions The basic idea Instead of choosing the Lévy process (ie. choosing ψ) choose a subordinator (ie. choose φ) whose renewal measure is known and build an associated Lévy process by defining its Laplace exponent ψ(β) = βφ(β). Q: How do we know which subordinators are allowed to participate in such a relation? A: Again the Wiener-Hopf factorization helps us out and tell us that any subordinator who s jump measure is absolutely continuous with non-increasing density. 13/ 16

53 How to find explicit scale functions The basic idea Instead of choosing the Lévy process (ie. choosing ψ) choose a subordinator (ie. choose φ) whose renewal measure is known and build an associated Lévy process by defining its Laplace exponent ψ(β) = βφ(β). Q: How do we know which subordinators are allowed to participate in such a relation? A: Again the Wiener-Hopf factorization helps us out and tell us that any subordinator who s jump measure is absolutely continuous with non-increasing density. The latter condition of a decreasing Lévy density is natural because, also hidden in the Wiener-Hopf factorization, is the relation Π(x, ) = υ φ (x) where υ φ is the Lévy density associated with φ. 13/ 16

54 How to find explicit scale functions The basic idea Instead of choosing the Lévy process (ie. choosing ψ) choose a subordinator (ie. choose φ) whose renewal measure is known and build an associated Lévy process by defining its Laplace exponent ψ(β) = βφ(β). Q: How do we know which subordinators are allowed to participate in such a relation? A: Again the Wiener-Hopf factorization helps us out and tell us that any subordinator who s jump measure is absolutely continuous with non-increasing density. The latter condition of a decreasing Lévy density is natural because, also hidden in the Wiener-Hopf factorization, is the relation Π(x, ) = υ φ (x) where υ φ is the Lévy density associated with φ. This method gives us a rich supply of scale functions W but not too many concrete examples of q-scale functions W (q). The fundamental difficulty being that the latter two are related via an exponential change of measure involving Φ(q), the inverse of ψ(β) which is not always explicitly available. 13/ 16

55 How to find explicit scale functions Concrete example of the triple law: I In this example we build a spectrally negative Lévy process with Laplace exponent ψ(θ) = θγ(ν + θ + λ)/cγ(ν + θ), where θ 0, c > 0, ν 0 and λ (0, 1). 14/ 16

56 How to find explicit scale functions Concrete example of the triple law: I In this example we build a spectrally negative Lévy process with Laplace exponent ψ(θ) = θγ(ν + θ + λ)/cγ(ν + θ), where θ 0, c > 0, ν 0 and λ (0, 1). It has no Gaussian component and its Lévy measure satisfies λ Π(x, ) = c 2 Γ(1 λ) (e x 1). λ+1 e x(1 ν) 14/ 16

57 How to find explicit scale functions Concrete example of the triple law: I In this example we build a spectrally negative Lévy process with Laplace exponent ψ(θ) = θγ(ν + θ + λ)/cγ(ν + θ), where θ 0, c > 0, ν 0 and λ (0, 1). It has no Gaussian component and its Lévy measure satisfies λ Π(x, ) = c 2 Γ(1 λ) (e x 1). λ+1 A straightforward calculation shows that e x(1 ν) ψ (0+) = E(X 1) = Γ(ν + λ)/cγ(ν) 0 14/ 16

58 How to find explicit scale functions Concrete example of the triple law: I In this example we build a spectrally negative Lévy process with Laplace exponent ψ(θ) = θγ(ν + θ + λ)/cγ(ν + θ), where θ 0, c > 0, ν 0 and λ (0, 1). It has no Gaussian component and its Lévy measure satisfies λ Π(x, ) = c 2 Γ(1 λ) (e x 1). λ+1 A straightforward calculation shows that Then e x(1 ν) ψ (0+) = E(X 1) = Γ(ν + λ)/cγ(ν) 0 Z x z (ν+λ 1) e W (x) = c (e z 1) λ 1 dz. 0 Γ(λ) 14/ 16

59 How to find explicit scale functions Concrete example of the triple law: I In this example we build a spectrally negative Lévy process with Laplace exponent ψ(θ) = θγ(ν + θ + λ)/cγ(ν + θ), where θ 0, c > 0, ν 0 and λ (0, 1). It has no Gaussian component and its Lévy measure satisfies λ Π(x, ) = c 2 Γ(1 λ) (e x 1). λ+1 A straightforward calculation shows that Then And e x(1 ν) ψ (0+) = E(X 1) = Γ(ν + λ)/cγ(ν) 0 Z x z (ν+λ 1) e W (x) = c (e z 1) λ 1 dz. 0 Γ(λ) P x ( X τ > u, X τ dv, X τ dy) = λ e (x y)ν (1 e (x y) ) λ 1 e (u+v)(1 ν) dydv. Γ(1 λ)γ(λ) (e (u+v) 1) λ+1 14/ 16

60 How to find explicit scale functions Concrete example of the triple law: II In this example we are able to handle the classical tempered stable process (one sided CGMY) ψ(θ) = (θ + c) α c α, where θ 0, c 0 and α (1, 2). 15/ 16

61 How to find explicit scale functions Concrete example of the triple law: II In this example we are able to handle the classical tempered stable process (one sided CGMY) ψ(θ) = (θ + c) α c α, where θ 0, c 0 and α (1, 2). The tempered stable process has no Gaussian component, is a process of unbounded variation and its Lévy measure is given by 1 Π(dx) = dx, x > 0. Γ( α) x 1+α e cx 15/ 16

62 How to find explicit scale functions Concrete example of the triple law: II In this example we are able to handle the classical tempered stable process (one sided CGMY) ψ(θ) = (θ + c) α c α, where θ 0, c 0 and α (1, 2). The tempered stable process has no Gaussian component, is a process of unbounded variation and its Lévy measure is given by 1 Π(dx) = dx, x > 0. Γ( α) x 1+α e cx Note that E(X 1) = ψ (0+) = αc α / 16

63 How to find explicit scale functions Concrete example of the triple law: II In this example we are able to handle the classical tempered stable process (one sided CGMY) ψ(θ) = (θ + c) α c α, where θ 0, c 0 and α (1, 2). The tempered stable process has no Gaussian component, is a process of unbounded variation and its Lévy measure is given by 1 Π(dx) = dx, x > 0. Γ( α) x 1+α e cx Note that E(X 1) = ψ (0+) = αc α 1 0 This is a fortunate example where it is an easy exercise to show that. Φ(q) = (q + c α ) 1/α c 15/ 16

64 How to find explicit scale functions Concrete example of the triple law: II In this example we are able to handle the classical tempered stable process (one sided CGMY) ψ(θ) = (θ + c) α c α, where θ 0, c 0 and α (1, 2). The tempered stable process has no Gaussian component, is a process of unbounded variation and its Lévy measure is given by 1 Π(dx) = dx, x > 0. Γ( α) x 1+α e cx Note that E(X 1) = ψ (0+) = αc α 1 0 This is a fortunate example where it is an easy exercise to show that. Then where Φ(q) = (q + c α ) 1/α c W (q) (x) = e cx x α 1 E α,α((q + c α )x α ), E α,β (x) = X n=0 x n Γ(αn + β) is the two-parameter Mittag-Leffler function. 15/ 16

65 How to find explicit scale functions ctd... Finally we have the full triple law E x (e qτ ; X τ du, X τ dv, X τ y) = 1 α ) 1/α c)(x v) Γ( α) e((q+c e c(u+v) (u + v) 1+α n e (q+cα ) 1/αx x α 1 E α,α((q + c α )x α ) o e (q+cα ) 1/α (x y) (x y) α 1 E α,α((q + c α )(x y) α ) dudv 16/ 16