Laplace transform. Lecture notes. Ante Mimica. Department of Mathematics. Univeristy of Zagreb. Croatia

Transcription

1 Laplace transform Lecture notes Ante Mimica Department of Mathematics Univeristy of Zagreb Croatia These lecture notes follow the course given in period April 27 - May at Technische Universität Dresden.

2

3 Contents Contents 3 1 Definition and basic properties 5 2 Continuity theorem and applications Completely monotone functions Tauberian theorems Further developments Bibliography 23 3

4

5 1 Definition and basic properties Definition 1.1 Let µ be a measure on. The Laplace transform Lµ of µ is defined by Lµ(λ) = e λx µ(dx) for λ > σ, where σ = inf{λ R : e λx µ(dx) < }. Remark 1.2 (i) If µ is a finite measure, then σ. (ii) Assume that Lµ(a) < for some a R. Then ν(dx) = e ax Lµ(a) µ(dx) defines a probability measure on [, ) with the Laplace transform e λx ν(dx) = 1 e (λ+a)x µ(dx) = Lµ(λ+a). Lµ(a) Lµ(a) (iii) Some special cases: (iiia) µ(dx) = f(x)dx Lµ(λ) = Lf(λ) = e λx f(x)dx. (iiib) let X be a non-negative random variable on a probability space (Ω,F,P) and denote by µ the law of X; that is µ(b) = P(X B). Then Lµ(λ) = E[e λx ] for λ > σ. 5

6 1 Definition and basic properties Example 1.3 (a) f(x) = x α, α > Lf(λ) = e x( x ) α dx λ λ = Γ(α+1), σ λ 1+α =. (b) f(x) = e ax, a R Lf(λ) = (c) f(x) = e x2, σ =, since (d) f(x) = e x2, σ = +, since e (a+λ)x dx = 1 a+λ, σ = a. e λx x2 dx converges for any λ R. e λx+x2 dx does not converge for any λ R. Proposition 1.4 (Properties of LT) Letµbeameasureonansassume that Lµ is finite (, ) (i.e. σ ). (i) Then Lµ C ((, )) and for all n N (Lµ) (n) (λ) = ( 1) n x n e λx µ(dx). (ii) µ has finite n-th moment if and only if (Lµ) (n) (+) exists and it is finite. In particular, µ is finite if and only if Lµ(+) exists and it is finite. (iii) If γ : is defined by γ(x) = ax, a >, then L(µ γ 1 )(λ) = Lµ(aλ), λ >. (iv) If ν is a measure such that Lν(λ) exists for λ >, then L(µ ν) = LµLν, where the convolution of measures µ and ν is defined by (µ ν)(b) := 1 B (x+y)µ(dx)ν(dy) for a Borel measurable set B. 6

7 Proof. (i) Lµ(λ+h) Lµ(λ) h = e hx 1 h e λx µ(dx). Since e hx 1 h hx e hx xe λ 4 x ce λ 2 x for h λ h 4 for some constant c > depending only on λ >, we can use dominated convergence theorem to conclude that (Lµ) e hx 1 (λ) = e λx µ(dx) = xe λx µ(dx). h For higher derivatives we proceed in a similar manner. (ii) An application of monotone convergence theorem yields ( 1) n (Lµ) (n) (+) = lim x n e λx µ(dx) = λ + and the claim follows. (iii) L(µ γ 1 )(λ) = e λx (µ γ 1 )(dx) = e λ =ax {}}{ x n µ(dx) γ(x) µ(dx) = Lµ(ax). (iv) L(µ ν)(λ) = = Lµ(λ)Lν(λ). e λx (µ ν)(dx) = e λ(x+y) µ(dx)ν(dy) Definition 1.5 For a measure µ on R its distribution function F : R is defined by F(x) = µ((,x]), x R. We call x R a continuity point of µ (or F) if F is continuous at x. 7

8 1 Definition and basic properties Remark 1.6 Note that F is always right continuous with left limits and nondecreasing function and it has at most countable number of discontinuities. If µ is a measure on, then F(x) = for x < and F(x) = µ([,x]) for x. Proposition 1.7 Let µ be a measure on [, ) with the distribution function F. Then for any λ > σ and x > following integration by parts formula holds e λy µ(dy) = e λx F(x)+ λe λy F(y)dy. In particular, [,x] If µ is a probability measure, then [,x] e λy F(y)dy = Lµ(λ) λ. e λx (1 F(x))dx = 1 Lµ(λ) λ for all λ >. Proof. By Fubini theorem [,x] e λy µ(dy) = = 1 [,x z](y) {}}{ 1 [,x] (y)1 [y, ) (z)λe λz dzµ(dy) λe λz µ([,x z])dz = e λx µ([,x])+ = e λx F(x)+ x x λe λz µ([,z])dz λe λz F(z)dz. Second claim follows by letting x and using monotone convergence theorem, since we can find ε > so that λ ε > σ and hence > Lµ(λ ε) e (λ ε)y µ(dy) e εx e λx F(x) [,x] yielding e λx F(x) x. The last claim follows directly from the previous. 8

9 Proposition 1.8 (Uniqueness and inversion of LT) A finite measure µ on [, ) is uniquely determined by its Laplace transform. More precisely, for any x we have µ([,x]) 1 µ({x}) = lim 2 ( 1) knk k! (Lµ)(k) (n). (1.1) k nx In particular, at every continuity point x of µ we have µ([,x]) = lim k nx ( 1) knk k! (Lµ)(k) (n). (1.2) Proof. By Proposition 1.4 (a), for any x > and n N, ( 1) knk (nt) k k! (Lµ)(k) (λ) = e nt µ(dt). k! Note, k nx k nx k nx (nt) k e nt = P(X nx), k! where X has Poisson distribution with mean nt. First we consider case x = t. Let {Y i : i 1} be a sequence of independent Poisson random variables with mean t. By the central limit theorem, P(X nt) = P For x < t, ( Y Y n nt nt ) 1 2π e y2 2 dy = 1 2. P(X nx) = P(n(t x) nt X) P(n(t x) X nt ) VarX n 2 (t x) = t 2 n(t x). 2 and, similarly, for x > t, P(X nx) = 1 P(X nt > n(x t)) 1 P( X nt n(x t)) t 1 n(t x)

10 1 Definition and basic properties The last three displays show that k nx (nt) k k! e nt 1 [,x) (t) {x}(t) and, thus by the dominated convergence theorem ( lim ( 1)knk k! (Lµ)(k) (n) = 1 [,x) (t)+ 1 ) 2 1 {x}(t) = µ([,x]) 1 2 µ({x}). µ(dt) Remark 1.9 If f : R is a continuous function such that for some b R supe bx f(x) <, x then the Post-Widder inversion formula holds ( 1) f(x) = lim n n! ( n x ) n+1(lf) (n) ( n x ), x >. 1

11 2 Continuity theorem and applications Theorem 2.1 (Continuity theorem) Let (µ n ) be a sequence of measures on and denote by F n their distribution functions. (i) If µ is a measure on with the distribution function F such that F n F at continuity points of F and if there is a such that sup n 1 Lµ n (a) <, then Lµ n (λ) Lµ(λ) for all λ > a. (ii) If there is a R such that φ(λ) = lim Lµ n (λ) exists for all λ > a, then φ is the Laplace transform of a measure µ and if F is its distribution function, then F n F at all continuity points of F. Proof. (i) Let A := suplµ n (a) <. Using Proposition 1.7 and dominated n 1 convergence theorem, for any λ > a and point of continuity x >, x e λy µ n (dy) = λe λy F n (y)ds+e λx F n (x) e [,x] [,x] λy µ(dy), since F n (y) e ax Lµ n (a) Ae ax for all y x. Let λ > a and ε >. For any point of continuity x > of F satisfying Ae (λ a)x ε we have e λy µ n (dy) Lµ n (λ) [,x] e λy µ n (dy)+e (λ a)x e ay µ n (dy) e λy µ n (dy)+ε. [,x] (x, ) This implies e λy µ(dy) liminf Lµ n(λ) limsuplµ n (λ) [,x] [,x] [,x] e λy µ(dy)+ε 11

12 2 Continuity theorem and applications and, letting x (along points of continuity of F) we obtain Lµ(λ) liminf Lµ n(λ) limsuplµ n (λ) Lµ(λ)+ε. Since ε > was arbitrarywe conclude that lim Lµ n (λ) exists and equals Lµ(λ) for any λ > a. (ii) Take λ > a. By Remark 1.2 (ii), ν n (dx) = e λ x Lµ n (λ ) µ n(dx) are probability measures with the Laplace transforms Lµ n (λ +λ) Lµ n (λ ) for n N. Denoting by G n the distribution function of ν n for n N, we can apply Helly s selection theorem to obtain a right-continuous non-decreasing function G and a subsequence such that G nk G at all continuity points of G. Let ν be the k measure on corresponding to G (such that G is the distribution function of ν). Since Lν n () = 1 for all n N, we can apply (i) to conclude that φ(λ +λ) φ(λ ) = lim k Lµ nk (λ +λ) Lµ nk (λ ) = Lν(λ +λ). In the same way, we obtain that any subsequence (G nk ) of (G n ) has a subsequence (G nkl ) converging to G at all continuity points of G. This implies that (G n ) converges to G at all continuity points of G(see 1 ). Define µ by µ(dx) = φ(λ )e λ x ν(dx), Note that it has the same continuity points as ν since, by Proposition 1.7, [ µ([,x]) = φ(λ ) e λx G(x)+ x ] λ e λt G(t)dt. 1 Here we use the following reasoning from analysis. If (a n ) is a sequence of real numbers such that every sequence has a subsequence converging to some fixed number L R, then (a n ) converges to L as well. If this were not true, we could find ε > and a subsequence (a nk ) such that a nk L ε for all k N which would lead to a contradiction, since we could not find any further subsequence of (a nk ) converging to L. 12

13 2.1 Completely monotone functions Also, for any λ > λ Lµ(λ) = φ(λ) e (λ λ )x ν(dx) = φ(λ ) φ(λ +(λ λ )) φ(λ ) = φ(λ). Then for any continuity point x > of the distribution function F of µ, by the dominated convergence theorem and Proposition 1.7, it follows that F n (x) = Lµ n (λ )e λy ν n (dy) = Lµ n (λ )e λx G n (x) Lµ n (λ )λ e λy G n (y)dy [,x] [,x] φ(λ )e λx G(x) φ(λ )λ e λy G(y)dy = φ(λ )e λy ν(dy) = µ([,x]) = F(x). [,x] [,x] Corollary 2.2 Let (µ n ) n be a sequence of probability measures such that Lµ n (λ) φ(λ) for all λ >. If φ(+) = 1, then there exists a probability measure µ on such that Lµ = φ and µ n ([,x]) µ([,x]) at continuity points x > of µ. Proof. Since sup n 1 Lµ n () = supµ n ( = 1, Theorem 2.1 (ii) yiedls a measure n 1 µ on such that µ n ([,x]) µ([,x]) at continuity points of µ satisfying Lµ(λ) = φ(λ) for all λ >. Using Proposition 1.4 (i) we get that 1 = φ(+) = µ(). 2.1 Completely monotone functions In this section we explore range of the Laplace transform. 13

14 2 Continuity theorem and applications Definition 2.3 Afunctionf : (, ) (, )iscompletelymonotonefunction if it has derivatives of all orders and satisfies ( 1) n f (n) (λ) for all λ > and n N {}. Theorem 2.4 (Bernstein-Hausdorff-Widder) A function f: (, ) (, ) is completely monotone if and only if there exists a measure µ on such that f(λ) = Lµ(λ) = e λx µ(dx) for all λ >. Proof. By Proposition 1.4 (i), for any n N and λ > we have ( 1) n f (n) (λ) = x n e λx µ(dx). Let f be a completely monotone function such that f(+) <. Since ( 1) n f (n) is nonincreasing for any n N, we have ( 1) n f (n) 2 λ λ 2/λ Iterating this inequality we obtain ( 1) n f (n) (y)dy 2 λ ( 1)n 1 f (n 1) ( λ 2 ). 2n(n+1) ( 1) n f (n) 2 (y) f( y y n 2 ) 2 n(n+1) 2 f(+), y >. n y n Having this inequality we can use integration by parts and change of variable y = n t to get f(λ) f( ) = λ f (y)dy = =... = ( 1)n (n 1)! = = n/λ λ λ (y λ)f (y)dy (y λ) n 1 f (n) (y)dy ( n f (n) ( n t λ)n 1( 1)n t ) ndt (n 1)! t 2 ( 1 λt ) n 1 1 [,n](λt) ( 1)n f (n) ( n t )(n t )n+1 n n! dt. 14

15 2.1 Completely monotone functions Define µ n (dt) = ( 1)n f (n) ( n t )(n t )n+1 dt. n! By monotone convergence theorem ( µ n () = lim 1 λt ) n 1 1 λ + [,n](λt)µ n (dt) = f(+) f( ) n for all n N and so we may use Helly s selection principle to obtain a nondecreasing right continuous function F and a subsequence (µ nk ) such that µ nk ([,x]) k F(x) at continuity points x of F. Let µ be the measure corresponding to µ. By the continuity theorem (Theorem 2.1), Lµ nk (λ) k Lµ(λ) It can be shown that ( 1 x n) n 11[,n] (x) e λx uniformly in x >, hence, implying that Lµ nk (λ) ( 1 λt ) nk 1 1 [,nk ](λt)µ nk (dt) n k k f(λ) f( ) = Lµ (λ). Therefore, in this case it is enough to put µ = µ +f( )δ, where δ is point mass at. In general case we set f a (λ) := f(a+λ) for a > and note that f a is also a completely monotone function such that f a (+) < and so there is a measure µ a such that f a = Lµ a. Since it is possible do this for any a >, uniqueness theorem for the Laplace transform (Proposition 1.8) implies that for < a < b from Lµ b (λ) = f(a+(λ+b a)) = e λx e λ(b a) µ a (dx), 15

16 2 Continuity theorem and applications we deduce that e ax µ a (dx) = e bx µ b (dx). Hence, for any a > we may consistently define measure µ(dx) = e ax µ a (dx). Then for λ > and a = λ 2 we get f(λ) = f( λ 2 + λ 2 ) = e λ 2 x µ λ/2 (dx) = e λx µ(dx). Proposition 2.5 Let f be a completely monotone function. (a) If g is a completely monotone function, then f g is also completely monotone. (b) If h : (, ) (, ) is such that h is completely monotone, then f h is also completely monotone. Proof. (a) By Theorem 2.4, there exist measures µ and ν such that f = Lµ and g = Lν. Then Proposition 1.4 (iv) implies that fg = LµLν = L(µ ν). Now it is enough to observe that L(µ ν) is completely monotone by Theorem 2.4. (b) We prove this by mathematical induction. First, Assume that for some n N (h f) = (f h) }{{} h }{{}. ( 1) k ( f h) (k) for all k {1,...,n} (2.1) and all functions f, h : (, ) (, ) such that h and f are completely monotone. Since f and h are completely monotone, by Leibniz formula for higher derivatives and (2.1), ( 1) (n+1) (h f) (n+1) = ( 1) n [(( f ) h) h ] (n) n ( n = ( 1) n k = n k= ( n k k= ) (( f) h) (k) (h ) (n k) ) ( 1) k (( f ) h) (k) } {{ } ( 1) n k (h ) }{{ (n k). } 16

17 2.2 Tauberian theorems 2.2 Tauberian theorems We have seen that continuity theorem enables us to deduce convergence of distribution functions from the convergence of the Laplace transforms of the corresponding measures. In this section we will see that under certain conditions behavior of the Laplace transform around the origin determines the behavior of the distribution function near the infinity. Such type of relation, describing behavior of measure µ in terms of transform Lµ is often called Tauberian. Reverse relation is known as Abelian. Theorem 2.6 Let ρ and let µ be a measure on such that its Laplace transform Lµ is defined on (, ). The following claims are equivalent (i) (ii) Lµ( λ t ) Lµ( 1 t ) λ ρ t for all λ >, µ([,tx]) µ([,t]) t xρ for all x >. Moreover, if the relations above hold, then Lµ( 1 t ) µ([,t]) Γ(1+ρ). (2.2) t Proof. (i) = (ii) (Tauberian theorem) By Proposition 1.4 (iii) with γ(y) = y t, for any λ > and t > we obtain Lµ( λ t ) ( ) µ γ 1 Lµ( 1 t ) = L Lµ( 1 t ) (λ). Hence, ( ) { µ γ 1 L Lµ( 1 t ) (λ) λ ρ e λy y ρ 1 Γ(ρ) = dy, ρ > t Lδ (λ), ρ = and so, by the continuity theorem (Theorem 2.1), we conclude that µ([,tx]) Lµ( 1 t ) t { x y ρ 1 Γ(ρ) dy, ρ > [,x] δ (dy), ρ = = xρ Γ(1+ρ) for any x >. (2.3) 17

18 2 Continuity theorem and applications In particular, taking x = 1 in (2.3) we get (2.2) and then µ([,tx]) µ([,t]) = µ([,tx]) Lµ( 1 t ) x ρ for any x >. µ([,t]) t Lµ( 1 t ) (ii) = (i) (Abelian theorem) Similarly as before, by Proposition 1.4 (iii) with γ(y) = y t ( ) µ γ 1 L (λ) = Lµ(λ t ) for λ,t >. µ([,t]) µ([,t]) In order to apply continuity theorem first we are going to prove that for some t > Lµ( 1 t sup ) t>t µ([,t]) <. From (ii) it follows that there is t > such that Then for any a > Lµ( a t ) µ([,t])+ µ([,2t]) 2 2 ρ µ([,t]) for every t t. (2.4) n=1 [2 n 1 t,2 n t] e ay t µ(dy) µ([,t])+ e a2n 1 µ([,2 n t]) yielding together with multiple application of (2.4) Lµ( a t ) µ([,t]) 1+ e a2n 1µ([,2n t]) 1+ e a2n 1 2 n(1+ρ) <. µ([,t]) n=1 Therefore, we may use continuity theorem (Theorem 2.1) to conclude that Lµ( λ t ) ( ) { µ γ 1 µ([,t]) = L (λ) e λy ρy ρ 1 dy, ρ > µ([,t]) t Lδ (λ), ρ = = Γ(1+ρ) λ ρ for all λ > a, implying that the convergence actually holds for all λ >, since a > was arbitrary. As in the first part of the proof, by taking λ = 1 we obtain (2.2) and then Lµ( λ t ) Lµ( 1 t ) = Lµ( λ t ) µ([,t]) Lµ( 1 t ) µ([,t]) n=1 n=1 t λ ρ for all λ >. 18

19 2.2 Tauberian theorems Example 2.7 By (2.2), µ([,t]) lnt as t Lµ(λ) ln 1 λ as λ +. Definition 2.8 A function g: (, ) (, ) varies regularly at infinity with index ρ R if g(tx) lim t g(t) = tρ for all x >. If l: (, ) (, ) varies regluarly with index ρ =, then it is said to vary slowly at infinity, i.e. l(tx) lim t l(t) = 1 for all x >. Using this terminology we can restate Theorem 2.6 as follows. Theorem 2.9 Let µ be a measure on, l: (, ) (, ) a slowly varying function at infinity and ρ. The following relations are equivalent: (i) Lµ(λ) λ ρ l( 1 λ ), λ + 1 (ii) µ([,t]) Γ(1+ρ) tρ l(t), t. Proof. If (i) holds, then Lµ( λ t ) Lµ( 1 t ) λ ρl(t λ ) l(t) λ ρ as t, which is (i) in Theorem 2.6. Hence, from (2.2) we deduce µ([,t]) 1 Γ(1+ρ) Lµ(1 t ) 1 Γ(1+ρ) tρ l(t) as t. In the other case we proceed similarly. 19

20 2 Continuity theorem and applications Proposition 2.1 (Monotone density theorem) Let µ be a measure on such that µ([,t]) = t m(s)ds, t >, where m : (, ) (, ) is ultimately monotone, i.e. there exists t > such that m is monotone on (t, ). If there exist ρ R and a slowly varying function l : (, ) (, ) such that µ([,t]) t ρ l(t), as t, (2.5) then m(t) ρt ρ 1 l(t), as t. (2.6) Proof. Let us assume that m is eventualy nondecreasing and let < a < b. Then for t > large enough, (b a)tm(at) t ρ l(t) µ((at,bt]) t ρ l(t) Since l varies slowly at infinity we have µ((at,bt]) t ρ l(t) Therefore, it follows from (2.7) that (b a)tm(bt) t ρ l(t). (2.7) = µ([,bt]) (bt) ρ l(bt) bρl(bt) l(t) µ([,at]) (at) ρ l(at) aρl(at) l(t) t bρ a ρ. lim sup t m(at) t ρ 1 l(t) bρ a ρ b a and by taking a = 1 and letting b 1+ we get Similarly, showing that lim sup t lim inf t m(t) t ρ 1 l(t) lim b ρ 1 b 1+ b 1 = ρ. m(t) t ρ 1 l(t) lim 1 a ρ a 1 1 a = ρ lim t m(t) t ρ 1 l(t) = ρ. 2

21 2.2 Tauberian theorems Remark 2.11 It can be proved that (2.5) and (2.6) in Proposition 2.1 are equivalent when ρ >. The direction that we have not proved is known as Karamata theorem (see [1, Proposition 1.5.8]). Example 2.12 Let µ be a probability measure on [, ) with the distribution function. Then, by Proposition 1.7, e λx [1 F(x)]dx = 1 Lµ(λ), λ >. λ Since 1 F is monotone, Theorem 2.9 and Proposition 2.1 and previous remark imply that for a slowly varying function l and ρ > the following relations are equivalent 1 Lµ(λ) λ 1 ρ l( 1 ρ λ ), λ + and 1 F(t) Γ(1+ρ) }{{} = 1 Γ(ρ) t ρ 1 l(t), t. (2.8) Example 2.13 (Stable distributions) Letα (,1)andφ(λ) = e λα. Then φ is a completely monotone function by Proposition 2.5 (b) as a composition of a completely monotone function and a function with a completely monotone derivative. Moreover, since φ() = 1, by Theorem 2.4 φ is the Laplace transform of a probability measure µ. Let (X n ) be a sequence of independent random variables on a probability space (Ω,F,P) with law µ. Then the Laplace transform of the random variable X X n is n 1/α ] E [e λx Xn n 1/α = φ(λn 1/α ) n = φ(λ), showing by the uniqueness of the Laplace transform (Proposition 1.8) that the law of X X n is again µ. Note that n 1/α 1 Lµ(λ) = 1 e λα λ α, λ +. If X has the law µ, then (2.8) with ρ = 1 α yields P(X > t) 1 Γ(1 α) t1 α 1 = t α Γ(1 α), t. 21

22 2 Continuity theorem and applications 2.3 Further developments If we consider example of the gamma distribution then γ(dx) = xα 1 Γ(α) e x dx (α > ), Lµ(λ) = 1 e (1+λ)x x α 1 dx = Γ(α) 1 (1+λ) α implying that σ = 1. It can be proved that the tail of the measure µ(t, ) = x α 1 t Γ(α) e x dx satisfies lnµ(t, ) lim t t More generally, we have the following result. = 1 = σ. Theorem 2.14 Let f : (, ) (, ) be a completely monotone function and let µ be the measure such that f = Lµ (see Theorem 2.4). Assume that f(σ +2λ) lim supλlnf(σ +λ) = and limsup λ + λ + f(σ +λ) < 1. (2.9) Then Proof. See [3, Theorem 1.2]. 1 lim t t lnµ(t, ) = σ. Remark 2.15 Condition (2.9) is satisfied if function λ f(σ + λ) varies regularly at with index ρ <. 22

23 Bibliography [1] N. H. Bingham, C. M. Goldie, J. L. Teugels. Regular variation, Cambridge University Press, Cambridge, [2] W. Feller. An Introduction to Probability Theory and Its Applications, Volume II, John Wiley & Sons, New York, [3] A. Mimica. Exponential decay of measures and Tauberian theorems, preprint, 214., [4] R. L. Schilling, R. Song, Z. Vondraček. Bernstein functions: Theory and applications, 2nd edition, De Gruyter, Berlin,