Random walks are completely determined by their trace on the positive half-line

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Random walks

are completely determined by their trace

on the positive half-line

Mateusz Kwa±nicki

Wrocªaw University of Science and Technology mateusz.kwasnicki@pwr.edu.pl

Guanajuato, Nov 29, 2017

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Introduction Detailed statement Idea of the proof Algebra Complex analysis

Main Theorem

Random walks

are completely determined by their trace

on the positive half-line

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A few remarks

• Acknowledgement: Loïc Chaumont from Angers, France.

• SPA 2017 conference

(The 39th Conference on Stochastic Processes and their Applications, Moscow, Jul 2428, 2017)

Loïc Chaumont, Ron Doney

On distributions determined by their upward, space-time WienerHopf factor

arXiv:1702.00067 V. Vigon

Simpliez vos Lévy en titillant la factorisation de WienerHopf

PhD thesis, INSA de Rouen, 2001

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A few remarks

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A few remarks

arXiv:1702.00067

V. Vigon

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A few remarks

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Random walks

• Arandom walk X_n is a sequence of partial sums of i.i.d.

random variables:

X_n = ∆X₁+ ∆X₂ + . . . + ∆X_n,

where ∆X₁, ∆X₂, . . . are independent and identically distributed on R.

• We say that a random walk Xn is non-trivial if P(X1 >0) 6= 0.

• We write A= B^d if P(A > t) = P(B > t) for all t ∈ R.

• Of course if X1 d

= Y₁, then Xn d

= Y_n for all n = 1, 2, . . .; in this case we say that Xn and Yn are identical.

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Random walks

random variables:

X_n = ∆X₁+ ∆X₂ + . . . + ∆X_n,

= Y₁, then Xn d

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Random walks

random variables:

X_n = ∆X₁+ ∆X₂ + . . . + ∆X_n,

= Y₁, then Xn d

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Random walks

random variables:

X_n = ∆X₁+ ∆X₂ + . . . + ∆X_n,

= Y₁, then Xn d

= Y_n for all n = 1, 2, . . .;

in this case we say that Xn and Yn are identical.

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Main theorem

Theorem

If Xn and Yn are non-trivial random walks, and P(Xⁿ> t) = P(Yⁿ > t)

for all n = 1, 2, . . . and all t ∈(0, ∞), then the same is true for all t ∈R(that is, Xn and Yn are identical).

• This was proved by Chaumont and Doney under additional conditions on Xn and Yn:

I if X1 has exponential moments; or

I if P(X1 > t) is completely monotone on (0, ∞); or

I if X₁ has analytic density function on (0, ∞).

• This covers a majority of interesting examples.

• It is often enough to take n = 1, 2 in the assumption.

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Main theorem

Theorem

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Main theorem

Theorem

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Main theorem

Theorem

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Simple reformulation

Theorem (equivalent version)

If Xn and Yn are non-trivial random walks, and max{0, Xn}=^d max{0, Yn} for all n = 1, 2, . . ., then

X_n= Y^d _n for all n = 1, 2, . . .

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Some uctuation theory

• Dene Xn =max{0, X1, X₂, . . . , X_n}.

• Spitzer's formula: if |w| < 1 and Im z > 0, then

∞

X

n=0

E exp(izXn)wⁿ=exp

^∞ X

n=0

E exp(iz max{0, Xn})

n wⁿ

.

• Knowing the distributions of Xn is thus equivalent to knowing the distributions of max{0, Xn}.

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Some uctuation theory

• Dene Xn =max{0, X1, X₂, . . . , X_n}.

∞

X

n=0

E exp(izXn)wⁿ=exp

^∞ X

n=0

n wⁿ

.

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Some uctuation theory

• Dene Xn =max{0, X1, X₂, . . . , X_n}.

∞

X

n=0

E exp(izXn)wⁿ=exp

^∞ X

n=0

n wⁿ

.

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Another reformulation

If Xn and Yn are non-trivial random walks, and X_n = Y^d _n

for all n = 1, 2, . . ., then

X_n= Y^d _n for all n = 1, 2, . . .

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Some more uctuation theory

• Let N be the smallest number n such that Xn>0

(the rst ladder time).

• Let H be the the value of Xn for n = N

(the rst ladder height).

• Knowing the joint distribution of N and H, one can reconstruct the distributions of Xn for n = 1, 2, . . .

• The characteristic function of (N, H) is essentially the upward space-time WienerHopf factor.

If Xn and Yn are non-trivial random walks with equal upward space-time WienerHopf factors, then Xnand Yn are identical.

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Some more uctuation theory

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Some more uctuation theory

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Some more uctuation theory

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Some more uctuation theory

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Lévy processes

• A Lévy process is, in some sense, a random walk in continuous time.

• Formally, Xt is a Lévy process if it has independent and stationary increments, and càdlàg paths.

Corollary

If Xt and Yt are non-trivial Lévy processes, and max{0, Xt}=^d max{0, Yt}

(or Xt d

= Y_t)

for all t > 0, then

X_t = Y^d _t for all t > 0.

• Conjectured by Vigon, proved under extra assumptions by Chaumont and Doney

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Lévy processes

Corollary

(or Xt d

= Y_t)

for all t > 0, then

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Lévy processes

Corollary

(or Xt d

= Y_t)

for all t > 0, then

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Lévy processes

Corollary

If Xt and Yt are non-trivial Lévy processes, and max{0, Xt}=^d max{0, Yt} (or Xt d

= Y_t) for all t > 0, then

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Lévy processes

Corollary

If Xt and Yt are non-trivial Lévy processes, and max{0, Xt}=^d max{0, Yt} (or Xt d

= Y_t) for all t > 0, then

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A variant for measures

• All measures below are nite signed Borel measures on R.

• The convolution of measures µ and ν is given by (µ ∗ ν)(A) =

Z

R

µ(A − x )ν(dx ). Convolutive powers of µ are denoted by µⁿ.

• We say that a measure µ isnon-trivial if the restriction of µto (0, ∞) is a non-zero measure.

Theorem (extended version)

If µ and ν are non-trivial measures and µⁿ(A) = νⁿ(A)

for all Borel A ⊆ (0, ∞) and n = 1, 2, . . ., then µ = ν.

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A variant for measures

Z

R

µ(A − x )ν(dx ).

Convolutive powers of µ are denoted by µⁿ.

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A variant for measures

Z

R

µ(A − x )ν(dx ).

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A variant for measures

Z

R

µ(A − x )ν(dx ).

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Change of notation

• We assume that

µⁿ(A) = νⁿ(A) for all Borel A ⊆ (0, ∞) and n = 1, 2, . . .

• Considering n = 1, we see that the restrictions of µ and ν to (0, ∞) agree.

• Denote:

α = 1(0,∞)µ = 1(0,∞)µ, β = 1(−∞,0]µ,

γ = 1(−∞,0]ν.

• Now µ = α + β and ν = α + γ.

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Change of notation

• We assume that

• Denote:

α = 1(0,∞)µ = 1(0,∞)µ, β = 1(−∞,0]µ,

γ = 1(−∞,0]ν.

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Change of notation

• We assume that

• Denote:

α = 1(0,∞)µ = 1(0,∞)µ, β = 1(−∞,0]µ,

γ = 1(−∞,0]ν.

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Change of notation

• We assume that

• Denote:

α = 1(0,∞)µ = 1(0,∞)µ, β = 1(−∞,0]µ,

γ = 1(−∞,0]ν.

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Idea of the proof

• α is a non-zero measure concentrated on (0, ∞), β and γ are concentrated on (−∞, 0].

• The proof consists of two steps:

1(0,∞)(α + β)ⁿ = 1(0,∞)(α + γ)ⁿ for all n = 1, 2, . . . w

w

(simple algebra)

1⁽0,∞)(αⁿ∗ β) = 1⁽0,∞)(αⁿ∗ γ)for all n = 0, 1, . . .

w w

(complex analysis) β = γ

.

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Idea of the proof

1(0,∞)(α + β)ⁿ= 1(0,∞)(α + γ)ⁿ for all n = 1, 2, . . . w

w

(simple algebra)

1(0,∞)(αⁿ∗ β) = 1(0,∞)(αⁿ∗ γ)for all n = 0, 1, . . .

w w

(complex analysis) β = γ

.

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Idea of the proof

1(0,∞)(α + β)ⁿ= 1(0,∞)(α + γ)ⁿ for all n = 1, 2, . . . w

w

(simple algebra)

1(0,∞)(αⁿ∗ β) = 1(0,∞)(αⁿ∗ γ)for all n = 0, 1, . . . w

w

(complex analysis) β = γ.

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• We prove that

1(0,∞)(α + β)ⁿ= 1(0,∞)(α + γ)ⁿ for all n = 1, 2, . . . w

w

1(0,∞)(αⁿ∗ β^k) = 1(0,∞)(αⁿ∗ γ^k)for all n = 0, 1, . . . and k = 1, 2, . . .

• Induction with respect to n.

• For n = 0:

1(0,∞)(β^k) = 0 = 1(0,∞)(γ^k).

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• We prove that

1(0,∞)(α + β)ⁿ= 1(0,∞)(α + γ)ⁿ for all n = 1, 2, . . . w

w

• For n = 0:

1(0,∞)(β^k) = 0 = 1(0,∞)(γ^k).

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• We prove that

1(0,∞)(α + β)ⁿ= 1(0,∞)(α + γ)ⁿ for all n = 1, 2, . . . w

w

• For n = 0:

1(0,∞)(β^k) = 0 = 1(0,∞)(γ^k).

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• Suppose that

1(0,∞)(αⁿ∗ β^k) = 1(0,∞)(αⁿ∗ γ^k)for n = 0, 1, . . . , N − 1 and k = 1, 2, . . .

• By the binomial formula,

zero on (0, ∞) by the assumption

z }| {

(α + β)^N+¹− (α + γ)^N+¹

= α^N+¹− α^N+¹

| {z }

zero on R

(j =0)

+ (N +1) α^N∗ β − α^N ∗ γ

(j =1) +

N

X

j =2

N +1 j

α^N+^1−j∗ β^j − α^N+^1−j∗ γ^j

| {z }

zero on (0, ∞) by the induction hypothesis

+ β^N+¹− γ^N+¹

| {z }

zero on (0, ∞)

. (j = N +2)

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• Suppose that

z }| {

(α + β)^N+1− (α + γ)^N+1

= α^N+¹− α^N+¹

| {z }

zero on R

(j =0)

+ (N +1) α^N ∗ β − α^N ∗ γ

(j =1) +

N

X

j =2

N +1 j

α^N+^1−j∗ β^j − α^N+^1−j ∗ γ^j

| {z }

+ β^N+¹− γ^N+¹

| {z }

zero on (0, ∞)

. (j = N +2)

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• Suppose that

z }| {

(α + β)^N+1− (α + γ)^N+1

= α^N+¹− α^N+¹

| {z }

zero on R

(j =0)

+ (N +1) α^N ∗ β − α^N ∗ γ

(j =1) +

N

X

j =2

N +1 j

α^N+^1−j∗ β^j − α^N+^1−j ∗ γ^j

| {z }

+ β^N+¹− γ^N+¹

| {z }

zero on (0, ∞)

. (j = N +2)

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• Suppose that

z }| {

(α + β)^N+1− (α + γ)^N+1

= α^N+¹− α^N+¹

| {z }

zero on R

(j =0)

+ (N +1) α^N ∗ β − α^N ∗ γ

(j =1) +

N

X

j =2

N +1 j

α^N+^1−j∗ β^j − α^N+^1−j ∗ γ^j

| {z }

+ β^N+¹− γ^N+¹

| {z }

zero on (0, ∞)

. (j = N +2)

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• Suppose that

z }| {

(α + β)^N+1− (α + γ)^N+1

= α^N+¹− α^N+¹

| {z }

zero on R

(j =0)

+ (N +1) α^N ∗ β − α^N ∗ γ

(j =1) +

N

X

j =2

N +1 j

α^N+^1−j∗ β^j − α^N+^1−j ∗ γ^j

| {z }

+ β^N+¹− γ^N+¹

| {z }

zero on (0, ∞)

. (j = N +2)

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• Suppose that

z }| {

(α + β)^N+1− (α + γ)^N+1

= α^N+¹− α^N+¹

| {z }

zero on R

(j =0)

+ (N +1) α^N ∗ β − α^N ∗ γ

(j =1) +

N

X

j =2

N +1 j

α^N+^1−j∗ β^j − α^N+^1−j ∗ γ^j

| {z }

+ β^N+¹− γ^N+¹

| {z }

zero on (0, ∞)

. (j = N +2)

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• Thus, 0 = (N + 1) α^N ∗ β − α^N∗ γ

on (0, ∞).

• This is the desired result for n = N, k = 1.

• Larger values of k: induction within induction.

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• Thus, 0 = (N + 1) α^N ∗ β − α^N∗ γ

on (0, ∞).

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• Thus, 0 = (N + 1) α^N ∗ β − α^N∗ γ

on (0, ∞).

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• Suppose that

1(0,∞)(α^N ∗ β^k) = 1(0,∞)(α^N ∗ γ^k) for k = 1, 2, . . . , K − 1.

• Then,

σ ∗ β = (1(−∞,0]σ) ∗ β + (1(0,∞)σ) ∗ β

1(0,∞)(α^N ∗ β^K)

=

1(0,∞) ( α^N ∗ β^{K −}¹

| {z }

σ

) ∗ β

= 1(0,∞) 1(0,∞)(α^N ∗ β^{K −}¹) ∗ β

=

1(0,∞) (α^N∗ γ^{K −}¹) ∗ β

= 1(0,∞) 1(0,∞)(α^N ∗ γ^{K −}¹) ∗ β

=

1(0,∞) (α^N∗ β) ∗ γ^{K −1}

= 1(0,∞) 1(0,∞)(α^N ∗ β) ∗ γ^{K −1}

=

1(0,∞) (α^N ∗ γ) ∗ γ^{K −}¹

= 1(0,∞) 1(0,∞)(α^N ∗ γ) ∗ γ^{K −}¹

=

1(0,∞)(α^N ∗ γ^K).

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• Suppose that

1(0,∞)(α^N ∗ β^k) = 1(0,∞)(α^N ∗ γ^k) for k = 1, 2, . . . , K − 1.

• Then,

σ ∗ β = (1(−∞,0]σ) ∗ β + (1(0,∞)σ) ∗ β

1(0,∞)(α^N ∗ β^K)

=

1(0,∞) ( α^N ∗ β^{K −}¹

| {z }

σ

) ∗ β

= 1(0,∞) 1(0,∞)(α^N ∗ β^{K −}¹) ∗ β

=

1(0,∞) (α^N∗ γ^{K −}¹) ∗ β

= 1(0,∞) 1(0,∞)(α^N ∗ γ^{K −}¹) ∗ β

=

1(0,∞) (α^N∗ β) ∗ γ^{K −}¹

= 1(0,∞) 1(0,∞)(α^N ∗ β) ∗ γ^{K −}¹

=

1(0,∞) (α^N ∗ γ) ∗ γ^{K −}¹

= 1(0,∞) 1(0,∞)(α^N∗ γ) ∗ γ^{K −}¹

=

1(0,∞)(α^N ∗ γ^K).

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• Suppose that

1(0,∞)(α^N ∗ β^k) = 1(0,∞)(α^N ∗ γ^k) for k = 1, 2, . . . , K − 1.

• Then, σ ∗ β = (1(−∞,0]σ) ∗ β + (1(0,∞)σ) ∗ β 1(0,∞)(α^N ∗ β^K)

=

1(0,∞) ( α^N ∗ β^{K −}¹

| {z }

σ

) ∗ β

= 1(0,∞) 1(0,∞)(α^N ∗ β^{K −}¹) ∗ β

=

1(0,∞) (α^N∗ γ^{K −}¹) ∗ β

= 1(0,∞) 1(0,∞)(α^N ∗ γ^{K −}¹) ∗ β

=

1(0,∞) (α^N∗ β) ∗ γ^{K −}¹

= 1(0,∞) 1(0,∞)(α^N ∗ β) ∗ γ^{K −}¹

=

1(0,∞) (α^N ∗ γ) ∗ γ^{K −}¹

= 1(0,∞) 1(0,∞)(α^N∗ γ) ∗ γ^{K −}¹

=

1(0,∞)(α^N ∗ γ^K).

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• Suppose that

1(0,∞)(α^N ∗ β^k) = 1(0,∞)(α^N ∗ γ^k) for k = 1, 2, . . . , K − 1.

• Then,

σ ∗ β = (1(−∞,0]σ) ∗ β + (1(0,∞)σ) ∗ β

1(0,∞)(α^N ∗ β^K)

=

1(0,∞) ( α^N ∗ β^{K −}¹

| {z }

σ

) ∗ β

= 1(0,∞) 1(0,∞)(α^N ∗ β^{K −}¹) ∗ β

=

1(0,∞) (α^N∗ γ^{K −}¹) ∗ β

=

1⁽0,∞) 1⁽0,∞)(α^N ∗ γ^{K −}¹) ∗ β

=

1(0,∞) (α^N∗ β) ∗ γ^{K −}¹

= 1(0,∞) 1(0,∞)(α^N ∗ β) ∗ γ^{K −}¹

=

1(0,∞) (α^N ∗ γ) ∗ γ^{K −}¹

= 1(0,∞) 1(0,∞)(α^N∗ γ) ∗ γ^{K −}¹

=

1(0,∞)(α^N ∗ γ^K).

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• Suppose that

1(0,∞)(α^N ∗ β^k) = 1(0,∞)(α^N ∗ γ^k) for k = 1, 2, . . . , K − 1.

• Then,

σ ∗ β = (1(−∞,0]σ) ∗ β + (1(0,∞)σ) ∗ β

1(0,∞)(α^N ∗ β^K)

=

1(0,∞) ( α^N ∗ β^{K −}¹

| {z }

σ

) ∗ β

= 1(0,∞) 1(0,∞)(α^N ∗ β^{K −}¹) ∗ β

=

1⁽0,∞) (α^N∗ γ^{K −}¹) ∗ β

= 1⁽0,∞) 1⁽0,∞)(α^N ∗ γ^{K −}¹) ∗ β

=

1(0,∞) (α^N∗ β) ∗ γ^{K −}¹

= 1(0,∞) 1(0,∞)(α^N ∗ β) ∗ γ^{K −}¹

=

1(0,∞) (α^N ∗ γ) ∗ γ^{K −}¹

= 1(0,∞) 1(0,∞)(α^N∗ γ) ∗ γ^{K −}¹

=

1(0,∞)(α^N ∗ γ^K).

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• Suppose that

1(0,∞)(α^N ∗ β^k) = 1(0,∞)(α^N ∗ γ^k) for k = 1, 2, . . . , K − 1.

• Then,

σ ∗ β = (1(−∞,0]σ) ∗ β + (1(0,∞)σ) ∗ β

1(0,∞)(α^N ∗ β^K)

=

1(0,∞) ( α^N ∗ β^{K −}¹

| {z }

σ

) ∗ β

= 1(0,∞) 1(0,∞)(α^N ∗ β^{K −}¹) ∗ β

=

1⁽0,∞) (α^N∗ γ^{K −}¹) ∗ β

= 1⁽0,∞) 1⁽0,∞)(α^N ∗ γ^{K −}¹) ∗ β

=

1(0,∞) (α^N∗ β) ∗ γ^{K −}¹

= 1(0,∞) 1(0,∞)(α^N ∗ β) ∗ γ^{K −}¹

=

1(0,∞) (α^N ∗ γ) ∗ γ^{K −}¹

= 1(0,∞) 1(0,∞)(α^N∗ γ) ∗ γ^{K −}¹

=

1(0,∞)(α^N ∗ γ^K).

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• Suppose that

1(0,∞)(α^N ∗ β^k) = 1(0,∞)(α^N ∗ γ^k) for k = 1, 2, . . . , K − 1.

• Then,

σ ∗ β = (1(−∞,0]σ) ∗ β + (1(0,∞)σ) ∗ β

1(0,∞)(α^N ∗ β^K)

=

1(0,∞) ( α^N ∗ β^{K −}¹

| {z }

σ

) ∗ β

= 1(0,∞) 1(0,∞)(α^N ∗ β^{K −}¹) ∗ β

=

1⁽0,∞) (α^N∗ γ^{K −}¹) ∗ β

= 1⁽0,∞) 1⁽0,∞)(α^N ∗ γ^{K −}¹) ∗ β

=

1(0,∞) (α^N∗ β) ∗ γ^{K −}¹

= 1(0,∞) 1(0,∞)(α^N ∗ β) ∗ γ^{K −}¹

=

1(0,∞) (α^N ∗ γ) ∗ γ^{K −}¹

= 1(0,∞) 1(0,∞)(α^N∗ γ) ∗ γ^{K −}¹

=

1(0,∞)(α^N ∗ γ^K).

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• Suppose that

1(0,∞)(α^N ∗ β^k) = 1(0,∞)(α^N ∗ γ^k) for k = 1, 2, . . . , K − 1.

• Then,

σ ∗ β = (1(−∞,0]σ) ∗ β + (1(0,∞)σ) ∗ β

1(0,∞)(α^N ∗ β^K)

=

1(0,∞) ( α^N ∗ β^{K −}¹

| {z }

σ

) ∗ β

= 1(0,∞) 1(0,∞)(α^N ∗ β^{K −}¹) ∗ β

=

1⁽0,∞) (α^N∗ γ^{K −}¹) ∗ β

= 1⁽0,∞) 1⁽0,∞)(α^N ∗ γ^{K −}¹) ∗ β

=

1(0,∞) (α^N∗ β) ∗ γ^{K −}¹

= 1(0,∞) 1(0,∞)(α^N ∗ β) ∗ γ^{K −}¹

=

1(0,∞) (α^N ∗ γ) ∗ γ^{K −}¹

=

1(0,∞) 1(0,∞)(α^N∗ γ) ∗ γ^{K −}¹

=

1(0,∞)(α^N ∗ γ^K).

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• Suppose that

1(0,∞)(α^N ∗ β^k) = 1(0,∞)(α^N ∗ γ^k) for k = 1, 2, . . . , K − 1.

• Then,

σ ∗ β = (1(−∞,0]σ) ∗ β + (1(0,∞)σ) ∗ β

1(0,∞)(α^N ∗ β^K)

=

1(0,∞) ( α^N ∗ β^{K −}¹

| {z }

σ

) ∗ β

= 1(0,∞) 1(0,∞)(α^N ∗ β^{K −}¹) ∗ β

=

1⁽0,∞) (α^N∗ γ^{K −}¹) ∗ β

= 1⁽0,∞) 1⁽0,∞)(α^N ∗ γ^{K −}¹) ∗ β

=

1(0,∞) (α^N∗ β) ∗ γ^{K −}¹

= 1(0,∞) 1(0,∞)(α^N ∗ β) ∗ γ^{K −}¹

=

1(0,∞) (α^N ∗ γ) ∗ γ^{K −}¹

= 1(0,∞) 1(0,∞)(α^N∗ γ) ∗ γ^{K −}¹

=

1(0,∞)(α^N ∗ γ^K).

(62)

• Suppose that

1(0,∞)(α^N ∗ β^k) = 1(0,∞)(α^N ∗ γ^k) for k = 1, 2, . . . , K − 1.

• Then,

σ ∗ β = (1(−∞,0]σ) ∗ β + (1(0,∞)σ) ∗ β

1(0,∞)(α^N ∗ β^K)

=

1(0,∞) ( α^N ∗ β^{K −}¹

| {z }

σ

) ∗ β

= 1(0,∞) 1(0,∞)(α^N ∗ β^{K −}¹) ∗ β

=

1⁽0,∞) (α^N∗ γ^{K −}¹) ∗ β

= 1⁽0,∞) 1⁽0,∞)(α^N ∗ γ^{K −}¹) ∗ β

=

1(0,∞) (α^N∗ β) ∗ γ^{K −}¹

= 1(0,∞) 1(0,∞)(α^N ∗ β) ∗ γ^{K −}¹

=

1(0,∞) (α^N ∗ γ) ∗ γ^{K −}¹

= 1(0,∞) 1(0,∞)(α^N∗ γ) ∗ γ^{K −}¹

=

1(0,∞)(α^N ∗ γ^K).

(63)

• We prove that

1(0,∞)(αⁿ∗ β) = 1(0,∞)(αⁿ∗ γ)for all n = 0, 1, . . . w

w

β = γ.

• Equivalently: it is not possible to have

1(0,∞)(αⁿ∗ (β − γ)) =0 for all n = 0, 1, . . . and α 6= 0, β − γ 6= 0.

• We proceed by contradiction.

(64)

• We prove that

1(0,∞)(αⁿ∗ β) = 1(0,∞)(αⁿ∗ γ)for all n = 0, 1, . . . w

w

β = γ.

(65)

• We prove that

1(0,∞)(αⁿ∗ β) = 1(0,∞)(αⁿ∗ γ)for all n = 0, 1, . . . w

w

β = γ.

(66)

• We know that αⁿ∗ (β − γ)is concentrated on (−∞, 0]

for all n = 1, 2, . . .

• Dene analytic extensions of characteristic functions: f (z) =

Z

(0,∞)

e^iztα(dt) (Im z > 0) g (z) =

Z

(−∞,0]

e^izt(β − γ)(dt) (Im z 6 0) h_n(z) =

Z

(−∞,0]

e^izt(αⁿ∗ (β − γ))(dt) (Im z 6 0).

• We know that

(f (z))ⁿg (z) = h_n(z) for z ∈ R.

(67)

for all n = 1, 2, . . .

• Dene analytic extensions of characteristic functions:

f (z) = Z

(0,∞)

Z

(−∞,0]

e^izt(β − γ)(dt) (Im z 6 0) h_n(z) =

Z

(−∞,0]

e^izt(αⁿ∗ (β − γ))(dt) (Im z 6 0).

• We know that

(f (z))ⁿg (z) = h_n(z) for z ∈ R.

(68)

for all n = 1, 2, . . .

• Dene analytic extensions of characteristic functions:

f (z) = Z

(0,∞)

Z

(−∞,0]

e^izt(β − γ)(dt) (Im z 6 0) h_n(z) =

Z

(−∞,0]

e^izt(αⁿ∗ (β − γ))(dt) (Im z 6 0).

• We know that

(f (z))ⁿg (z) = hn(z) for z ∈ R.

Random walks are completely determined by their trace on the positive half-line

Random walks

are completely determined by their trace

on the positive half-line

Mateusz Kwa±nicki

Random walks

are completely determined by their trace

on the positive half-line

A few remarks

A few remarks

A few remarks

A few remarks

Random walks

Random walks

Random walks

Random walks

Main theorem

Main theorem

Main theorem

Main theorem

Simple reformulation

Some uctuation theory

Some uctuation theory

Some uctuation theory

Another reformulation

Some more uctuation theory

Some more uctuation theory

Some more uctuation theory

Some more uctuation theory

Some more uctuation theory

Lévy processes

Lévy processes

Lévy processes

Lévy processes

Lévy processes

A variant for measures

A variant for measures

A variant for measures

A variant for measures

Change of notation

Change of notation

Change of notation

Change of notation

Idea of the proof

Idea of the proof

Idea of the proof

Some uctuation theory

Some uctuation theory

Some uctuation theory

Some more uctuation theory

Some more uctuation theory

Some more uctuation theory

Some more uctuation theory

Some more uctuation theory