Branching random walk with stretched exponential tails Piotr - PowerPoint PPT Presentation

Branching random walk with stretched exponential tails Piotr Dyszewski (TUM & UWr) June 5, 2020

M n X ′ X 0 ◮ the particles reproduce according to a Galton-Watson process with reproduction mean m > 1 ◮ the displacements are iid copies of X such that P [ X > t ] ∼ e − t r , r ∈ ( 0, 1 ) ◮ M n = the position of the rightmost particle

Theorem (P .D, N. Gantert, T. Höfelsauer) Take X such that E X = 0 and E X 2 = 1 . Suppose that (. . . ) and that X has a stretched exponential tail, i.e. P [ X > t ] ∼ e − t r , r ∈ ( 0, 1 ) . 0, 2 Then there are constants γ , σ , α such that: for r ∈ � � , 3 1 M n − α n r → d F ( x ) = E � − γ We − x �� � exp 1 r − 1 σ n � 2 � and for r ∈ 3 , 1 1 M n − α n → 1 r 2 σ , n 2 − 1 r where W is a martingale limit associated with the underlying Galton-Watson process.

Let ( Z n ) be a Galton-Watson process with reproduction mean m > 1. Assume for simplicity P [ Z 1 = 0 ] = 0. n ≥ 0 N n be the corresponding Galton-Watson tree. Let T ⊆ � ∅ | x | = 1 0 1 | x | = 2 00 10 11 12 | x | = 3 000 001 002 003

∅ X 0 X 1 | x | = 1 0 1 X 10 X 12 X 00 X 11 | x | = 2 00 10 11 12 X 000 X 003 X 001 X 002 | x | = 3 000 001 002 003 | x | = n X y , y ∈ T iid For x ∈ T , V ( x ) = ∑ X y y ≤ x M n = max V ( x ) | x | = n

P [ X > t ] ∼ e − t r , r ∈ ( 0, 1 ) P [ X 1 + X 2 > t ] ∼ P [ max { X 1 , X 2 } > t ] S n = ∑ 1 ≤ k ≤ n X k , X ∗ n = max 1 ≤ k ≤ n X k P [ S n > t n ] ∼ ? � � � � 1 1 X ∗ r − s ≈ sup P [ S n − 1 > s ] P S n > cn s P n > cn r − s 2 � � � r � 1 r − s ≈ sup − n exp cn 2 n s � 1 r − rc r − 1 n 2 − 1 � � S n − 1 > rc r − 1 n 2 − 1 � X ∗ ≈ P n > cn P r r � 1 � � � n 3 − 2 �� − c r n + O ∼ n exp P S n > cn r r

If | x | = n , then n d ∑ V ( x ) = S n = X k . k = 1 � � � � 1 1 = P ∃ | x | = n , V ( x ) > cn P M n > cn r r    ∑ � ≥ 1 = P ✶ � 1  V ( x ) > cn r | x | = n    ∑ � 1 �  = E [ Z n ] P ≤ E ✶ � S n > cn r 1 � V ( x ) > cn r | x | = n � 1 � = m n P S n > cn r ∼ m n exp {− c r n ( 1 + o ( 1 )) } M n 1 → α = log ( m ) r 1 n r

The first term in the asymptotic expansion of M n is related to the biggest displacement, i.e. N n = max X y . | y |≤ n Proposition 1 N n − α n r → d H d = E [ exp {− γ ′ We − x } ] 1 r − 1 σ n Proof. x n → ∞ � � P [ X ≤ x n ] Y n P [ N n ≤ x n ] = E [ P [ N n ≤ x n | ( Z n ) ] ] = E ∼ E [ exp {− Y n P [ X > x n ] } ] Y n = # {| x | ≤ n } = ∑ n k = 1 Z k .

Proof continued . m − n Z n is a positive martingale Z n m n = W > 0 lim n → ∞ m m − 1 W m n Y n ∼ 1 1 r − 1 x r + σ n x n = α n m � r � m − 1 W m n exp � � 1 1 r − 1 x r + σ n Y n P [ X > x n ] ∼ − α n m m − 1 W m n exp � � − α r n − α r − 1 r σ x ∼ ∼ γ ′ W exp {− x } σ = α 1 − r m γ ′ = m − 1 r 1 � � N n − α n r = P [ N n ≤ x n ] → E [ exp {− γ ′ We − x } ] ≤ x P 1 r − 1 σ n

M n N n → α → α 1 1 n n r r � � 1 V ( x ) = ∑ X v : X v ≪ α n = o ( . . . ) max r | x | = n v ≤ x # {| v | ≤ n } = Y n ≈ m n � 1 � ≈ m o ( n ) | v | ≤ n : X v ≈ α n # r � 1 � | v | ≤ n : X v ≈ α n is a small subset of {| v | ≤ n } r

1 ≈ α n r V ( x ) = R 1 ( x ) + V 0 ( x ) + N ( x ) + R 2 ( x ) = V 0 ( x ) + N ( x )

M n ≈ max { V 0 ( x ) + N ( x ) } 1 N ( x ) ≈ α n r d � 1 V 0 ( x ) = S n − o ( n ) � X k ≪ α n � r The contribution of V 0 ( x ) is at most � � n 2 − 1 V 0 ( x ) = O r on the other hand d 1 1 r − 1 H ( 1 + o ( 1 )) r + σ n N n = max X v = α n | v |≤ n

� r − 1 � 3 , n 2 − 1 1 If r < 2 r = o n and so d 1 1 r − 1 H ( 1 + o ( 1 )) r + σ n M n ≈ N n = α n 1 1 M n − α n ∼ N n − α n r r → d H d = E [ exp {− γ ′ We − x } ] 1 1 r − 1 r − 1 σ n σ n M n − N n → 0 1 r − 1 n

r − 1 = o � � 1 n 2 − 1 If r > 2 3 , n r 1 ≈ α n r M n ≈ max { V 0 ( x ) + N ( x ) } 1 M n − α n → 1 r n 2 − 1 2 σ r

r − 1 = √ 3 , n 2 − 1 1 If r = 2 r = n n M n ≈ max { N ( x ) + V 0 ( x ) } d d N x = Φ ( · ) = N ( 0, 1 ) . 1 � 1 � M n − α n N ( x ) − α n + 1 r r σ √ σ √ ≈ max σ N x n n  � Y n  � 1 � � 1 M n − α n N ( x ) − α n + 1 r r σ √ ≤ t ≈ E σ √ σ N > t P  P  n n � � �� � γ ′ W e − y ( 1 − Φ ( σ ( t − y )) dy ≈ E exp � − γ We − t �� � = E exp

Theorem (P .D, N. Gantert, T. Höfelsauer) Take X such that E X = 0 and E X 2 = 1 . Suppose that (. . . ) and that X has a stretched exponential tail, i.e. P [ X > t ] ∼ e − t r , r ∈ ( 0, 1 ) . 0, 2 Then there are constants γ , σ , α such that: for r ∈ � � , 3 1 M n − α n r → d F ( x ) = E � − γ We − x �� � exp , 1 r − 1 σ n � 2 � and for r ∈ 3 , 1 1 M n − α n → 1 r n 2 − 1 2 σ r where W is a martingale limit associated with the underlying Galton-Watson process.

References I Piotr Dyszewski, Nina Gantert, and Thomas Höfelsauer, The maximum of a branching random walk with stretched exponantial tails , https://arxiv.org/abs/2004.03871 Nina Gantert, The maximum of a branching random walk with semiexponential increments , Annals of Probability 28 (2000), no. 3, 1219–1229. A. V. Nagaev, Integral limit theorems taking large deviations into account when Cramér’s condition does not hold. I , Theory of Probability & Its Applications 14 (1969), no. 1, 51–64.