Some Examples. Uniform motion. Poisson processes on the real line

Transcription

1 Some Examples Our immeiate goal is to see some examples of Lévy processes, an/or infinitely-ivisible laws on. Uniform motion Choose an fix a nonranom an efine X := for all (1) Then, {X } is a [nonranom] Lévy process with Lévy triple ( ). The process {X } enotes uniform motion in the irection of. Poisson processes on the real line If N = Poiss(λ) for some λ>, then Ee ξn = = e ξ e λ λ! = exp λ 1 e ξ (2) That is, E exp(ξn) = exp( Ψ(ξ)), where Ψ(ξ) = 1 e ξ ξ1l (1) ( ) () an () := λδ {1} (). The corresponing Lévy process is calle the Poisson process with intensity parameter λ. 9

2 1 2. Some Examples Nonstanar Brownian motion with rift The Lévy triple ( I), where I enotes the ( ) ientity matrix, belongs to a vector of i.i.. stanar-normal ranom variables, an the corresponing Lévy process is [stanar] -imensional Brownian motion. We can generalize this example easily: Choose an fix a vector, an a ( ) matrix σ, an consier the Lévy triple (σ). The corresponing Lévy exponent is Ψ(ξ) =( ξ)+ 1 2 σξ2 Therefore, Ψ is the Lévy exponent of ranom vector X in if an only if X = + σz where Z is a vector of i.i.. stanar-normal ranom variables. The corresponing Lévy process is escribe by W := + σb, where B is stanar Brownian motion [check!]. The th coorinate of W is a Brownian motion with mean an variance 2 := (σ σ), an the coorinates of W are not in general inepenent. Since lim (W /) = a.s. by the law of large numbers for Brownian motion, is calle the rift of the nonstanar Brownian motion W. Isotropic stable laws Choose an fix a number α. An isotropic stable law of inex α is the infinitely-ivisible istribution with Lévy exponent Ψ(ξ) =ξ α, where is a fixe constant. The corresponing Lévy process is calle the isotropic stable process with inex α. We consier only ranom vectors with Lévy exponent exp( ξ α ) in this iscussion. Of course, =leas to Ψ, which is the exponent of the infinitely ivisible, but trivial, ranom variable X. Therefore, we stuy only =. Also, we nee exp( Ψ(ξ)) = exp{ ξ α } 1, an this means that cannot be negative. Finally, Ψ() =, an hence we have α>. Lemma 1 (Lévy). Ψ(ξ) =ξ α is a Lévy exponent iff α ( 2]. The case α =2is the Gaussian case we just saw. An α =1is also noteworthy; the resulting istribution is the isotropic [or symmetric, in imension one] Cauchy istribution whose ensity is () = Γ +1 (+1)/ π (+1)/2 /2 2 for all (3) Proof of Lemma 1. Exercise 8 below shows that Ψ(ξ) = O(ξ 2 ) as ξ ; therefore, α 2. Since α =2is Gaussian, we limit ourselves to α<2.

3 Isotropic stable laws 11 Next let us consier α ( 2) [an of course >]. In that case, a change of variables shows that for all ξ, 1 e ξ (ξ )1l (1) () +α = (1 cos(ξ )) +α (4) ξ α [The first ientity is justifie because the left-most integral is a raial function of ξ, an hence real.] Therefore, we can choose C ( ) so that () =C (+α) is the Lévy measure with exponent Ψ(ξ) = exp( ξ α ) iff (3) on page 3 hols. Sometimes, a reasonable knowlege of the Lévy exponent of an infinitelyivisible law yiels insight into its structure. Here is a first example; Skoroho (1961) contains much more precise [an very useful] estimates of the tails of stable laws. Proposition 2. If X has an isotropic stable istribution with α ( 2), then for all β>, E(X β ) < iff β<α. In other wors, except in the case that α =2, the ecay of the tail of an isotropic stable law is slow. It is also possible to say more about the tails of the istrbution (Theorem 14, page 5). But that requires a more sophisticate analysis. Proof. Because E exp( X) = exp( α )=Ecos( X) for some >, we may apply (4) in the following form: 1 e ξ +α = (1 cos(ξ )) +α ξα (5) eplace ξ by X an take expectations to obtain E X β E (1 cos( X)) +β = Now 1 1 e α β+1 1 β+1 < 1 e α 1 e α β+1 for every β> +β Therefore, E(X β ) < iff 1 (1 e α )/ β+1 <. The result follows from this an the fact that (θ/2) 1 e θ θ for all θ ( 1).

4 12 2. Some Examples The asymmetric Cauchy istribution on the line We can specialize the preceing example to = α =1an obtain the symmetric Cauchy law µ on the line. Of course the ensity of µ is known as well. 1 But more significantly, we have learne that the Lévy triple of µ is ( ), where () 2. This suggests that perhaps we can create an asymmetric variation of the Cauchy law by consiering a Lévy triple of the form ( ) for a Lévy measure of the form () = 1 2 1l ()() l ( )() where 1 = 2 are both positive an is selecte carefully. This can in fact be one, as we will see next. Theorem 3. For every 1 2 > an θ [ 2/π 2/π] there exists an an infinitely-ivisible Borel probability measure µ on such that: (i) The Lévy triple of µ has the form ( ) for as above; an (ii) ˆµ(ξ) = exp( ξ θξ log ξ ) for ξ \{}. 2 The measure µ given above is calle the Cauchy istribution on with asymmetry parameter θ [an scale parameter ]. When θ =, µ is the symmetric Cauchy law. When θ =2/π, µ is calle the completely asymmetric Cauchy law on. The Gamma istribution on the half line It is easy to see that the Gamma (αλ) istribution on ( ) is infinitely ivisible for every α λ >. Next we ientify its Lévy triple. Proposition 4. If µ is a Gamma (αλ) istribution on + for some α λ >, then µ is infinitely ivisible with :=, σ :=, an Lévy measure an Lévy exponent Ψ respectively given by () = αe λ 1l () () Ψ(ξ) =α log 1 ξ λ where log enotes the principle branch of the logarithm. 1 It is () =1/{π(1 + 2 )} for <<. 2 I am making a little fuss about ξ \{}, since the function ξ log ξ is efine only for ξ \{}. But of course ˆµ() = 1 because µ is a probability measure. Alternatively, we can efine the function ξ log ξ cotinuously on all of [ ) by letting log be lim ξ ξ= ξ log ξ =. If so, then the state formula for ˆµ(ξ) is vali for all ξ.

5 Problems for Lecture 2 13 Aing inepenent Lévy processes Finally, let me mention the following evice which can be use to generate new Lévy processes from ol ones. Lemma 5. If {X } an { X } are inepenent Lévy processes on with respective triples ( σ ) an ( σ ), then {X + Y } is a Lévy process on with Lévy triple (AΣ M), where A := + an M := +, an Σ can be chosen in any fashion as long as Σ Σ=σ σ + σ σ. This is elementary, an can be checke by irectly verifying the efining properties of Lévy processes. However, I emphasize that: (i) There is always such a Σ; 3 an (ii) even though Σ might not be efine uniquely, Σ Σ is Problems for Lecture 2 1. Prove that if := 1 2 ( sin ) 1 2 sin, then for all ξ>, 1 e ξ + ξ1l (1) () 2 = πξ + ξ ln ξ + ξ 2 Deuce Theorem 3 from this ientity. 2. This problem outlines the proof of Proposition 4. (1) Suppose : ( ) has a continuous erivative L 1 (), an (+) := lim () an ( ) := lim () exist an are finite. Prove that for all λ ρ >, (λ) ((λ + ρ)) = (+) ( ) ln 1+ ρ λ This is the Frullani integral ientity. (2) Prove that for all λ> an ξ, 1 ξ λ = exp e λ 1 e ξ an euce Proposition 4 from this ientity. 3 (Stable scaling). Prove that if X is an isotropic stable process in with inex α ( 2], then Y := 1/α X efines a Lévy process for every >fixe. Explicitly compute the Lévy exponent of Y. Is the same result true when epens on? What happens if you consier instea an asymmetric Cauchy process on the line? 3 This follows reaily from the fact that σ σ + σ σ is a nonnegative-efinite matrix.