Test your intuition! Complete graphs Infinite trees Seen on Gil Kalai’s blog You have a box with n red balls and n blue balls. You take out each time a ball at random. If the ball was red, you put it back in the box and take out a blue ball. If the ball was blue, you put it back in the box and take out a red ball. Igor Kortchemski Preys & Predators 5 / 465
Test your intuition! Complete graphs Infinite trees Seen on Gil Kalai’s blog You have a box with n red balls and n blue balls. You take out each time a ball at random. If the ball was red, you put it back in the box and take out a blue ball. If the ball was blue, you put it back in the box and take out a red ball. You keep doing it until left only with balls of the same color. How many balls will be left (as a function of n )? Igor Kortchemski Preys & Predators 5 / 465
Test your intuition! Complete graphs Infinite trees Seen on Gil Kalai’s blog You have a box with n red balls and n blue balls. You take out each time a ball at random. If the ball was red, you put it back in the box and take out a blue ball. If the ball was blue, you put it back in the box and take out a red ball. You keep doing it until left only with balls of the same color. How many balls will be left (as a function of n )? 1) Roughly ✏ n for some ✏ > 0 . 2) Roughly √ n . 3) Roughly log n . 4) Roughly a constant. 5) Some other behavior. Igor Kortchemski Preys & Predators 5 / 465
Test your intuition! Complete graphs Infinite trees Seen on Gil Kalai’s blog You have a box with n red balls and n blue balls. You take out each time a ball at random. If the ball was red, you put it back in the box and take out a blue ball. If the ball was blue, you put it back in the box and take out a red ball. You keep doing it until left only with balls of the same color. How many balls will be left (as a function of n )? 1) Roughly ✏ n for some ✏ > 0 . 2) Roughly √ n . 3) Roughly log n . 4) Roughly a constant. 5) Some other behavior. Other formulation (O.K. Corral problem, Williams & McIlroy, 1998) . There are two groups of n gunmen that shoot at each other. Once a gunman is hit he stops shooting, and leaves the place happily and peacefully. How many gunmen will be left after all gunmen in one team have left? Igor Kortchemski Preys & Predators 5 / 465
Test your intuition! Complete graphs Infinite trees Figure: Excerpt of the film “Gunfight at the O.K. Corral” (1957) Igor Kortchemski Preys & Predators 6 / 465
Test your intuition! Complete graphs Infinite trees Vu sur le blog de Gil Kalai You have a box with n red balls and n blue balls. You take out each time a ball at random. If the ball was red, you put it back in the box and take out a blue ball. If the ball was blue, you put it back in the box and take out a red ball. You keep as before until left only with balls of the same color. How many balls will be left (as a function of n )? 1) Roughly ✏ n for some ✏ > 0 . 2) Roughly √ n . 3) Roughly log n . 4) Roughly a constant. 5) Some other behavior. Other formulation (O.K. Corral problem, Williams & McIlroy, 1998) . There are two groups of n gunmen that shoot at each other. Once a gunman is hit he stops shooting, and leaves the place happily and peacefully. How many gunmen will be left after all gunmen in one team have left? Igor Kortchemski Preys & Predators 7 / 465
Test your intuition! Complete graphs Infinite trees Vu sur le blog de Gil Kalai You have a box with n red balls and n blue balls. You take out each time a ball at random. If the ball was red, you put it back in the box and take out a blue ball. If the ball was blue, you put it back in the box and take out a red ball. You keep as before until left only with balls of the same color. How many balls will be left (as a function of n )? 1) Roughly ✏ n for some ✏ > 0 . 2) Roughly √ n . 3) Roughly log n . 4) Roughly a constant. 5) Some other behavior. Other formulation (O.K. Corral problem, Williams & McIlroy, 1998) . There are two groups of n gunmen that shoot at each other. Once a gunman is hit he stops shooting, and leaves the place happily and peacefully. How many gunmen will be left after all gunmen in one team have left? Igor Kortchemski Preys & Predators 7 / 465
Test your intuition! Complete graphs Infinite trees Kingman & Volkov’s solution (1/3) If urn A has m balls and urn B has n balls, the probability that a ball is removed from A is m + n . n Igor Kortchemski Preys & Predators 8 / 465
Test your intuition! Complete graphs Infinite trees Kingman & Volkov’s solution (1/3) If urn A has m balls and urn B has n balls, the probability that a ball is removed from A is m + n . But n n 1 /m 1 /m + 1 /n = P ( Exp ( 1 /m ) < Exp ( 1 /n )) . m + n = Igor Kortchemski Preys & Predators 8 / 465
Test your intuition! Complete graphs Infinite trees Kingman & Volkov’s solution (2/3) Let ( X i , Y i ) i > 1 be independent random variables such that X i are Y i exponential random variables with mean i . Igor Kortchemski Preys & Predators 9 / 465
Test your intuition! Complete graphs Infinite trees Kingman & Volkov’s solution (2/3) Let ( X i , Y i ) i > 1 be independent random variables such that X i are Y i exponential random variables with mean i . Consider a piece of wood represented by the interval [ − n , n ] and made of 2 n pieces such that length ([ i − 1, i ]) = X i , length ([ − i , − i + 1 ]) = Y i ( 1 6 i 6 n ) . X 1 X 2 X 3 X 4 Y 4 Y 3 Y 2 Y 1 Igor Kortchemski Preys & Predators 9 / 465
Test your intuition! Complete graphs Infinite trees Kingman & Volkov’s solution (2/3) Let ( X i , Y i ) i > 1 be independent random variables such that X i are Y i exponential random variables with mean i . Consider a piece of wood represented by the interval [ − n , n ] and made of 2 n pieces such that length ([ i − 1, i ]) = X i , length ([ − i , − i + 1 ]) = Y i ( 1 6 i 6 n ) . X 1 X 2 X 3 X 4 Y 4 Y 3 Y 2 Y 1 Light both ends, and stop the fire when the origin is reached. Igor Kortchemski Preys & Predators 9 / 465
Test your intuition! Complete graphs Infinite trees Kingman & Volkov’s solution (2/3) Let ( X i , Y i ) i > 1 be independent random variables such that X i are Y i exponential random variables with mean i . Consider a piece of wood represented by the interval [ − n , n ] and made of 2 n pieces such that length ([ i − 1, i ]) = X i , length ([ − i , − i + 1 ]) = Y i ( 1 6 i 6 n ) . Light both ends, and stop the fire when the origin is reached. Let R ( n ) be the number of remaining pieces. Igor Kortchemski Preys & Predators 9 / 465
Test your intuition! Complete graphs Infinite trees Kingman & Volkov’s solution (2/3) Let ( X i , Y i ) i > 1 be independent random variables such that X i are Y i exponential random variables with mean i . Consider a piece of wood represented by the interval [ − n , n ] and made of 2 n pieces such that length ([ i − 1, i ]) = X i , length ([ − i , − i + 1 ]) = Y i ( 1 6 i 6 n ) . Light both ends, and stop the fire when the origin is reached. Let R ( n ) be the number of remaining pieces. Then R ( n ) has the same law as the number of remaining balls in the urn/gunman problem. Igor Kortchemski Preys & Predators 9 / 465
Test your intuition! Complete graphs Infinite trees Kingman & Volkov’s solution (3/3) In order to estimate the number R ( n ) of remaining pieces, first estimate the remaining length L ( n ) : � � n n � � X X L ( n ) = X i − Y i � . � � � � � i = 1 i = 1 Igor Kortchemski Preys & Predators 10 / 465
Test your intuition! Complete graphs Infinite trees Kingman & Volkov’s solution (3/3) In order to estimate the number R ( n ) of remaining pieces, first estimate the remaining length L ( n ) : � � n n � � X X L ( n ) = X i − Y i � . � � � � � i = 1 i = 1 Then n n ! n X X X 2 j 2 ' n 3 . Var = X i − Y i i = 1 i = 1 j = 1 Igor Kortchemski Preys & Predators 10 / 465
Test your intuition! Complete graphs Infinite trees Kingman & Volkov’s solution (3/3) In order to estimate the number R ( n ) of remaining pieces, first estimate the remaining length L ( n ) : � � n n � � X X L ( n ) = X i − Y i � . � � � � � i = 1 i = 1 Then n n ! n X X X 2 j 2 ' n 3 . Var = X i − Y i i = 1 i = 1 j = 1 Hence L ( n ) ' n 3 / 2 . Igor Kortchemski Preys & Predators 10 / 465
Test your intuition! Complete graphs Infinite trees Kingman & Volkov’s solution (3/3) In order to estimate the number R ( n ) of remaining pieces, first estimate the remaining length L ( n ) : � � n n � � X X L ( n ) = X i − Y i � . � � � � � i = 1 i = 1 Then n n ! n X X X 2 j 2 ' n 3 . Var = X i − Y i i = 1 i = 1 j = 1 Hence L ( n ) ' n 3 / 2 . Set S k = X 1 + · · · + X k . Igor Kortchemski Preys & Predators 10 / 465
Test your intuition! Complete graphs Infinite trees Kingman & Volkov’s solution (3/3) In order to estimate the number R ( n ) of remaining pieces, first estimate the remaining length L ( n ) : � � n n � � X X L ( n ) = X i − Y i � . � � � � � i = 1 i = 1 Then n n ! n X X X 2 j 2 ' n 3 . Var = X i − Y i i = 1 i = 1 j = 1 Hence L ( n ) ' n 3 / 2 . Set S k = X 1 + · · · + X k . We have E [ S k ] ' k 2 Igor Kortchemski Preys & Predators 10 / 465
Test your intuition! Complete graphs Infinite trees Kingman & Volkov’s solution (3/3) In order to estimate the number R ( n ) of remaining pieces, first estimate the remaining length L ( n ) : � � n n � � X X L ( n ) = X i − Y i � . � � � � � i = 1 i = 1 Then n n ! n X X X 2 j 2 ' n 3 . Var = X i − Y i i = 1 i = 1 j = 1 Hence L ( n ) ' n 3 / 2 . Set S k = X 1 + · · · + X k . We have E [ S k ] ' k 2 , so S k ' k 2 . Igor Kortchemski Preys & Predators 10 / 465
Test your intuition! Complete graphs Infinite trees Kingman & Volkov’s solution (3/3) In order to estimate the number R ( n ) of remaining pieces, first estimate the remaining length L ( n ) : � � n n � � X X L ( n ) = X i − Y i � . � � � � � i = 1 i = 1 Then n n ! n X X X 2 j 2 ' n 3 . Var = X i − Y i i = 1 i = 1 j = 1 Hence L ( n ) ' n 3 / 2 . Set S k = X 1 + · · · + X k . We have E [ S k ] ' k 2 , so S k ' k 2 . But, if the left part burns first, S R ( n ) ' L ( n ) . Igor Kortchemski Preys & Predators 10 / 465
Test your intuition! Complete graphs Infinite trees Kingman & Volkov’s solution (3/3) In order to estimate the number R ( n ) of remaining pieces, first estimate the remaining length L ( n ) : � � n n � � X X L ( n ) = X i − Y i � . � � � � � i = 1 i = 1 Then n n ! n X X X 2 j 2 ' n 3 . Var = X i − Y i i = 1 i = 1 j = 1 Hence L ( n ) ' n 3 / 2 . Set S k = X 1 + · · · + X k . We have E [ S k ] ' k 2 , so S k ' k 2 . But, if the left part burns first, S R ( n ) ' L ( n ) . Hence R ( n ) 2 ' n 3 / 2 Igor Kortchemski Preys & Predators 10 / 465
Test your intuition! Complete graphs Infinite trees Kingman & Volkov’s solution (3/3) In order to estimate the number R ( n ) of remaining pieces, first estimate the remaining length L ( n ) : � � n n � � X X L ( n ) = X i − Y i � . � � � � � i = 1 i = 1 Then n n ! n X X X 2 j 2 ' n 3 . Var = X i − Y i i = 1 i = 1 j = 1 Hence L ( n ) ' n 3 / 2 . Set S k = X 1 + · · · + X k . We have E [ S k ] ' k 2 , so S k ' k 2 . But, if the left part burns first, S R ( n ) ' L ( n ) . Hence R ( n ) 2 ' n 3 / 2 so that R ( n ) ' n 3 / 4 . Igor Kortchemski Preys & Predators 10 / 465
Test your intuition! Complete graphs Infinite trees This “decoupling” idea is called the Athreya–Karlin embedding, and is useful to study more general Pólya urn schemes. Igor Kortchemski Preys & Predators 11 / 465
Test your intuition! Complete graphs Infinite trees I. Test your intuition! II. Prey & predators on a complete graph III. Preys & predators on an infinite tree √ Igor Kortchemski Preys & Predators 12 / 17
Test your intuition! Complete graphs Infinite trees We consider K N + 2 , a complete graph on N + 2 vertices, and start the dynamics with one I vertex, one R vertex and N S vertices. √ Igor Kortchemski Preys & Predators 13 / 17
Test your intuition! Complete graphs Infinite trees We consider K N + 2 , a complete graph on N + 2 vertices, and start the dynamics with one I vertex, one R vertex and N S vertices. √ Igor Kortchemski Preys & Predators 13 / 17
Test your intuition! Complete graphs Infinite trees We consider K N + 2 , a complete graph on N + 2 vertices, and start the dynamics with one I vertex, one R vertex and N S vertices. Set E N ext = { at a certain moment, there are no more S vertices } . √ Igor Kortchemski Preys & Predators 13 / 17
Test your intuition! Complete graphs Infinite trees We consider K N + 2 , a complete graph on N + 2 vertices, and start the dynamics with one I vertex, one R vertex and N S vertices. Set E N ext = { at a certain moment, there are no more S vertices } . � E N � Question. How does P behave as N → ∞ ? ext √ Igor Kortchemski Preys & Predators 13 / 17
Test your intuition! Complete graphs Infinite trees We consider K N + 2 , a complete graph on N + 2 vertices, and start the dynamics with one I vertex, one R vertex and N S vertices. Set E N ext = { at a certain moment, there are no more S vertices } . � E N � Question. How does P behave as N → ∞ ? ext Theorem (K. ’13) . We have if 0 λ ∈ P ( E N ext ) λ = if − → N → ∞ if 1 λ > . √ Igor Kortchemski Preys & Predators 13 / 17
Test your intuition! Complete graphs Infinite trees We consider K N + 2 , a complete graph on N + 2 vertices, and start the dynamics with one I vertex, one R vertex and N S vertices. Set E N ext = { at a certain moment, there are no more S vertices } . � E N � Question. How does P behave as N → ∞ ? ext Theorem (K. ’13) . We have if λ ∈ ( 0, 1 ) 0 P ( E N ext ) λ = 1 if − → N → ∞ if 1 λ > 1. √ Igor Kortchemski Preys & Predators 13 / 17
Test your intuition! Complete graphs Infinite trees We consider K N + 2 , a complete graph on N + 2 vertices, and start the dynamics with one I vertex, one R vertex and N S vertices. Set E N ext = { at a certain moment, there are no more S vertices } . � E N � Question. How does P behave as N → ∞ ? ext Theorem (K. ’13) . We have if λ ∈ ( 0, 1 ) 0 P ( E N 1 ext ) λ = 1 if − → 2 N → ∞ if 1 λ > 1. √ Igor Kortchemski Preys & Predators 13 / 17
Test your intuition! Complete graphs Infinite trees Decoupling using Yule processes √ Igor Kortchemski Preys & Predators 14 / 17
Test your intuition! Complete graphs Infinite trees Transition rates Let S t , I t , R t be the population sizes at time t . Total rate of { S , I } → { I , I } : √ Igor Kortchemski Preys & Predators 15 / 17
Test your intuition! Complete graphs Infinite trees Transition rates Let S t , I t , R t be the population sizes at time t . Total rate of { S , I } → { I , I } : λ · S t · I t . √ Igor Kortchemski Preys & Predators 15 / 17
Test your intuition! Complete graphs Infinite trees Transition rates Let S t , I t , R t be the population sizes at time t . Total rate of { S , I } → { I , I } : λ · S t · I t . Total rate { R , I } → { R , R } : √ Igor Kortchemski Preys & Predators 15 / 17
Test your intuition! Complete graphs Infinite trees Transition rates Let S t , I t , R t be the population sizes at time t . Total rate of { S , I } → { I , I } : λ · S t · I t . Total rate { R , I } → { R , R } : I t · R t . √ Igor Kortchemski Preys & Predators 15 / 17
Test your intuition! Complete graphs Infinite trees Transition rates Let S t , I t , R t be the population sizes at time t . Total rate of { S , I } → { I , I } : λ · S t · I t . Total rate { R , I } → { R , R } : I t · R t . Hence, at time t , the probability that { S , I } → { I , I } happens before { R , I } → { R , R } is λ S t I t λ S t = . λ S t I t + I t R t λ S t + R t √ Igor Kortchemski Preys & Predators 15 / 17
Test your intuition! Complete graphs Infinite trees Transition rates Let S t , I t , R t be the population sizes at time t . Total rate of { S , I } → { I , I } : λ · S t · I t . Total rate { R , I } → { R , R } : I t · R t . Hence, at time t , the probability that { S , I } → { I , I } happens before { R , I } → { R , R } is λ S t I t λ S t = . λ S t I t + I t R t λ S t + R t y We are going to be able to decouple the evolutions of S and R. √ Igor Kortchemski Preys & Predators 15 / 17
Test your intuition! Complete graphs Infinite trees Coupling and decoupling via two Yule processes √ Igor Kortchemski Preys & Predators 16 / 17
Test your intuition! Complete graphs Infinite trees Yule processes Definition (Yule process) In a Yule process ( Y ( t )) t > 0 of parameter λ , starting with one individual, each individual lives a random time distributed according to a Exp ( λ ) random variable, and at its death gives birth to two individuals √ Igor Kortchemski Preys & Predators 17 / 17
Test your intuition! Complete graphs Infinite trees Yule processes Definition (Yule process) In a Yule process ( Y ( t )) t > 0 of parameter λ , starting with one individual, each individual lives a random time distributed according to a Exp ( λ ) random variable, and at its death gives birth to two individuals, and Y ( t ) denotes the total number of individuals at time t . √ Igor Kortchemski Preys & Predators 17 / 17
Test your intuition! Complete graphs Infinite trees Yule processes Definition (Yule process) In a Yule process ( Y ( t )) t > 0 of parameter λ , starting with one individual, each individual lives a random time distributed according to a Exp ( λ ) random variable, and at its death gives birth to two individuals, and Y ( t ) denotes the total number of individuals at time t . y In particular, the intervals between each discontinuity are distributed according to independent Exp ( λ ) , Exp ( 2 λ ) , Exp ( 3 λ ) , . . . random variables. √ Igor Kortchemski Preys & Predators 17 / 17
Test your intuition! Complete graphs Infinite trees Coupling with two Yule processes Let ( R ( t )) t > 0 be a Yule process of parameter 1 , and ( S N ( t )) t > 0 a Yule process of parameter λ , time-reversed at its N -th jump. √ Igor Kortchemski Preys & Predators 18 / 17
Test your intuition! Complete graphs Infinite trees Coupling with two Yule processes Let ( R ( t )) t > 0 be a Yule process of parameter 1 , and ( S N ( t )) t > 0 a Yule process of parameter λ , time-reversed at its N -th jump. E N c E N ext ext S N S N R R T T Figure: Ex. N = 7 , where red crosses represent infections and purple ones recoveries. √ Igor Kortchemski Preys & Predators 18 / 17
Test your intuition! Complete graphs Infinite trees Coupling with two Yule processes Let ( R ( t )) t > 0 be a Yule process of parameter 1 , and ( S N ( t )) t > 0 a Yule process of parameter λ , time-reversed at its N -th jump. The prey-predator dynamics can be described by using R and S N , which describe in what order the infections and recoveries happen! E N c E N ext ext S N S N R R T T Figure: Ex. N = 7 , where red crosses represent infections and purple ones recoveries. √ Igor Kortchemski Preys & Predators 18 / 17
Test your intuition! Complete graphs Infinite trees Coupling with two Yule processes Let ( R ( t )) t > 0 be a Yule process of parameter 1 , and ( S N ( t )) t > 0 a Yule process of parameter λ , time-reversed at its N -th jump. The prey-predator dynamics can be described by using R and S N , which describe in what order the infections and recoveries happen! E N c E N ext ext S N S N R R T T Figure: Ex. N = 7 , where red crosses represent infections and purple ones recoveries. T is the time when a type of vertices (S or I) disappears. √ Igor Kortchemski Preys & Predators 18 / 17
Test your intuition! Complete graphs Infinite trees Coupling with two Yule processes Let ( R ( t )) t > 0 be a Yule process of parameter 1 , and ( S N ( t )) t > 0 a Yule process of parameter λ , time-reversed at its N -th jump. The prey-predator dynamics can be described by using R and S N , which describe in what order the infections and recoveries happen! E N c E N ext ext S N S N R R T T Figure: Ex. N = 7 , where red crosses represent infections and purple ones recoveries. T is the time when a type of vertices (S or I) disappears. T is the smallest between: the first moment when there are more discontinuities of R than discontinuities of S N (I disappears first, c E N ext ) the N -th discontinuity of S N (S disappears first, E N ext ) √ Igor Kortchemski Preys & Predators 18 / 17
Test your intuition! Complete graphs Infinite trees Identification of the critical parameter λ = 1 √ Igor Kortchemski Preys & Predators 19 / 17
Test your intuition! Complete graphs Infinite trees Notation. Denote by S N ( 1 ) , S N ( 2 ) , . . . , S N ( N ) the discontinuities S N and by R ( 1 ) , . . . , R ( N ) the discontinuities of R ( t ) . √ Igor Kortchemski Preys & Predators 20 / 17
Test your intuition! Complete graphs Infinite trees Notation. Denote by S N ( 1 ) , S N ( 2 ) , . . . , S N ( N ) the discontinuities S N and by R ( 1 ) , . . . , R ( N ) the discontinuities of R ( t ) . c E N E N ext ext S (1) S (2) S (3) S (5) (1) S (2) S (5) S (4) S (6) S (7) S (3) (4) S (6) (7) S S S N N N N N N N N N N N N N N S N S N R R T T R (1) R (2) R (3) R (4) R (5) R (1) R (7) R (2) R (3) R (4) R (5) R (6) R (7) Figure: Example for N = 7 , where the crosses represent discontinuities. √ Igor Kortchemski Preys & Predators 20 / 17
Test your intuition! Complete graphs Infinite trees Notation. Denote by S N ( 1 ) , S N ( 2 ) , . . . , S N ( N ) the discontinuities S N and by R ( 1 ) , . . . , R ( N ) the discontinuities of R ( t ) . Proposition S N ( N ) has the same distribution as R ( N ) has the same distribution c E N E N ext ext S (1) S (2) S (3) S (5) (1) S (2) S (5) S (4) S (6) S (7) S (3) (4) S (6) (7) S S S N N N N N N N N N N N N N N S N S N R R T T R (1) R (2) R (3) R (4) R (5) R (1) R (7) R (2) R (3) R (4) R (5) R (6) R (7) Figure: Example for N = 7 , where the crosses represent discontinuities. √ Igor Kortchemski Preys & Predators 20 / 17
Test your intuition! Complete graphs Infinite trees Notation. Denote by S N ( 1 ) , S N ( 2 ) , . . . , S N ( N ) the discontinuities S N and by R ( 1 ) , . . . , R ( N ) the discontinuities of R ( t ) . Proposition S N ( N ) has the same distribution as Exp ( λ N ) + Exp ( λ ( N − 1 )) + · · · + Exp ( λ ) . R ( N ) has the same distribution c E N E N ext ext S (1) S (2) S (3) S (5) (1) S (2) S (5) S (4) S (6) S (7) S (3) (4) S (6) (7) S S S N N N N N N N N N N N N N N S N S N R R T T R (1) R (2) R (3) R (4) R (5) R (1) R (7) R (2) R (3) R (4) R (5) R (6) R (7) Figure: Example for N = 7 , where the crosses represent discontinuities. √ Igor Kortchemski Preys & Predators 20 / 17
Test your intuition! Complete graphs Infinite trees Notation. Denote by S N ( 1 ) , S N ( 2 ) , . . . , S N ( N ) the discontinuities S N and by R ( 1 ) , . . . , R ( N ) the discontinuities of R ( t ) . Proposition S N ( N ) has the same distribution as Exp ( λ N ) + Exp ( λ ( N − 1 )) + · · · + Exp ( λ ) . R ( N ) has the same distribution Exp ( 1 ) + Exp ( 2 ) + · · · + Exp ( N ) . c E N E N ext ext S (1) S (2) S (3) S (5) (1) S (2) S (5) S (4) S (6) S (7) S (3) (4) S (6) (7) S S S N N N N N N N N N N N N N N S N S N R R T T R (1) R (2) R (3) R (4) R (5) R (1) R (7) R (2) R (3) R (4) R (5) R (6) R (7) Figure: Example for N = 7 , where the crosses represent discontinuities. √ Igor Kortchemski Preys & Predators 20 / 17
Test your intuition! Complete graphs Infinite trees Notation. Denote by S N ( 1 ) , S N ( 2 ) , . . . , S N ( N ) the discontinuities S N and by R ( 1 ) , . . . , R ( N ) the discontinuities of R ( t ) . Proposition S N ( N ) has the same distribution as Exp ( λ N ) + Exp ( λ ( N − 1 )) + · · · + Exp ( λ ) . R ( N ) has the same distribution Exp ( 1 ) + Exp ( 2 ) + · · · + Exp ( N ) . A typical situation for λ > 1 : A typical situation for λ < 1 : S N ( N ) S N ( N ) S N S N R R R ( N ) T T R ( N ) √ Igor Kortchemski Preys & Predators 20 / 17
Test your intuition! Complete graphs Infinite trees Notation. Denote by S N ( 1 ) , S N ( 2 ) , . . . , S N ( N ) the discontinuities S N and by R ( 1 ) , . . . , R ( N ) the discontinuities of R ( t ) . Proposition S N ( N ) has the same distribution as Exp ( λ N ) + Exp ( λ ( N − 1 )) + · · · + Exp ( λ ) . R ( N ) has the same distribution Exp ( 1 ) + Exp ( 2 ) + · · · + Exp ( N ) . A typical situation for λ > 1 : A typical situation for λ < 1 : S N ( N ) S N ( N ) S N S N R R R ( N ) T T R ( N ) Hence λ ∈ ( 0, 1 ) 0 if P ( E N 1 ext ) λ = 1 if − → 2 N → ∞ 1 if λ > 1. √ Igor Kortchemski Preys & Predators 20 / 17
Test your intuition! Complete graphs Infinite trees Study of the final state of the system √ Igor Kortchemski Preys & Predators 21 / 17
Test your intuition! Complete graphs Infinite trees Definition Denote by S ( N ) , I ( N ) , R ( N ) the number of S , I , R vertices at the first time T when a type ( S or I ) of vertices disappears. √ Igor Kortchemski Preys & Predators 22 / 17
Test your intuition! Complete graphs Infinite trees Definition Denote by S ( N ) , I ( N ) , R ( N ) the number of S , I , R vertices at the first time T when a type ( S or I ) of vertices disappears. c E N E N ext ext S N ( N ) S N ( N ) S N S N R R R ( N ) R ( N ) T T S ( N ) = 2 , I ( N ) = 0 , R ( N ) = 7 S ( N ) = 0 , I ( N ) = 3 , R ( N ) = 6 Figure: Ex. N = 7 , where red crosses represent infections and purple ones recoveries. √ Igor Kortchemski Preys & Predators 22 / 17
Test your intuition! Complete graphs Infinite trees Definition Denote by S ( N ) , I ( N ) , R ( N ) the number of S , I , R vertices at the first time T when a type ( S or I ) of vertices disappears. c E N E N ext ext S N ( N ) S N ( N ) S N S N R R R ( N ) R ( N ) T T S ( N ) = 2 , I ( N ) = 0 , R ( N ) = 7 S ( N ) = 0 , I ( N ) = 3 , R ( N ) = 6 Figure: Ex. N = 7 , where red crosses represent infections and purple ones recoveries. Question. What can be said of the asymptotic behavior of S ( N ) , I ( N ) , R ( N ) as N → ∞ ? √ Igor Kortchemski Preys & Predators 22 / 17
Test your intuition! Complete graphs Infinite trees Definition Denote by S ( N ) , I ( N ) , R ( N ) the number of S , I , R vertices at the first time T when a type ( S or I ) of vertices disappears. c E N E N ext ext S N ( N ) S N ( N ) S N S N R R R ( N ) R ( N ) T T S ( N ) = 2 , I ( N ) = 0 , R ( N ) = 7 S ( N ) = 0 , I ( N ) = 3 , R ( N ) = 6 Figure: Ex. N = 7 , where red crosses represent infections and purple ones recoveries. Question. What can be said of the asymptotic behavior of S ( N ) , I ( N ) , R ( N ) as N → ∞ ? This should be related to the asymptotic behavior of Yule processes. √ Igor Kortchemski Preys & Predators 22 / 17
Test your intuition! Complete graphs Infinite trees Yule processes and terminal value Proposition Let ( Y ( t )) t > 0 be a Yule process of parameter λ . 1) We have the convergence a . s . e − λ t Y t E , − → t → ∞ where E is a Exp ( 1 ) random variable, called terminal value of Y . √ Igor Kortchemski Preys & Predators 23 / 17
Test your intuition! Complete graphs Infinite trees Yule processes and terminal value Proposition Let ( Y ( t )) t > 0 be a Yule process of parameter λ . 1) We have the convergence a . s . e − λ t Y t E , − → t → ∞ where E is a Exp ( 1 ) random variable, called terminal value of Y . 2) For t > 0 and k > 1 , we have P ( Y t = k ) = e − λ t ( 1 − e − λ t ) k − 1 . √ Igor Kortchemski Preys & Predators 23 / 17
Test your intuition! Complete graphs Infinite trees Yule processes and terminal value Proposition Let ( Y ( t )) t > 0 be a Yule process of parameter λ . 1) We have the convergence a . s . e − λ t Y t E , − → t → ∞ where E is a Exp ( 1 ) random variable, called terminal value of Y . 2) For t > 0 and k > 1 , we have P ( Y t = k ) = e − λ t ( 1 − e − λ t ) k − 1 . Corollary if τ N denotes the N -th jump time of Y , then a . s . λτ N − ln ( N ) − ln ( E ) − → N → ∞ √ Igor Kortchemski Preys & Predators 23 / 17
Test your intuition! Complete graphs Infinite trees Number of susceptible individuals remaining Theorem (K. ’13) . (i) Fix λ ∈ ( 0, 1 ) . (ii) Fix λ = 1 . (iii) Fix λ > 1 . √ Igor Kortchemski Preys & Predators 24 / 17
Test your intuition! Complete graphs Infinite trees Number of susceptible individuals remaining Theorem (K. ’13) . (i) Fix λ ∈ ( 0, 1 ) . (ii) Fix λ = 1 . (iii) Fix λ > 1 . Then S ( N ) converges in probability towards √ Igor Kortchemski Preys & Predators 24 / 17
Test your intuition! Complete graphs Infinite trees Number of susceptible individuals remaining Theorem (K. ’13) . (i) Fix λ ∈ ( 0, 1 ) . (ii) Fix λ = 1 . (iii) Fix λ > 1 . Then S ( N ) converges in probability towards 0 as N → ∞ . √ Igor Kortchemski Preys & Predators 24 / 17
Test your intuition! Complete graphs Infinite trees Number of susceptible individuals remaining Theorem (K. ’13) . (i) Fix λ ∈ ( 0, 1 ) . Then S ( N ) ( d ) Exp ( 1 ) λ . − → N 1 − λ N → ∞ (ii) Fix λ = 1 . (iii) Fix λ > 1 . Then S ( N ) converges in probability towards 0 as N → ∞ . √ Igor Kortchemski Preys & Predators 24 / 17
Test your intuition! Complete graphs Infinite trees Number of susceptible individuals remaining Theorem (K. ’13) . (i) Fix λ ∈ ( 0, 1 ) . Then S ( N ) ( d ) Exp ( 1 ) λ . − → N 1 − λ N → ∞ (ii) Fix λ = 1 . Then for every i > 0 , ⇣ S ( N ) = i ⌘ 1 / 2 i + 1 . P − → N → ∞ (iii) Fix λ > 1 . Then S ( N ) converges in probability towards 0 as N → ∞ . √ Igor Kortchemski Preys & Predators 24 / 17
Test your intuition! Complete graphs Infinite trees Idea of the proof: case λ = 1 On the event c E N ext , S N ( N ) S N ln ( E / E ) R R ( N ) T √ Igor Kortchemski Preys & Predators 25 / 17
Test your intuition! Complete graphs Infinite trees Idea of the proof: case λ = 1 On the event c E N ext , S N ( N ) S N ln ( E / E ) R R ( N ) T Let E be the terminal value of the Yule process associated with S N , and E is the terminal value of R . √ Igor Kortchemski Preys & Predators 25 / 17
Test your intuition! Complete graphs Infinite trees Idea of the proof: case λ = 1 On the event c E N ext , S N ( N ) S N ln ( E / E ) R R ( N ) T Let E be the terminal value of the Yule process associated with S N , and E is the terminal value of R . We have S N ( N ) ' ln ( N ) − ln ( E ) , R ( N ) ' ln ( N ) − ln ( E ) √ Igor Kortchemski Preys & Predators 25 / 17
Test your intuition! Complete graphs Infinite trees Idea of the proof: case λ = 1 On the event c E N ext , S N ( N ) S N ln ( E / E ) R R ( N ) T Let E be the terminal value of the Yule process associated with S N , and E is the terminal value of R . We have S N ( N ) ' ln ( N ) − ln ( E ) , R ( N ) ' ln ( N ) − ln ( E ) , with E / E > 1 . √ Igor Kortchemski Preys & Predators 25 / 17
Test your intuition! Complete graphs Infinite trees Idea of the proof: case λ = 1 On the event c E N ext , S N ( N ) S N ln ( E / E ) R R ( N ) T Let E be the terminal value of the Yule process associated with S N , and E is the terminal value of R . We have S N ( N ) ' ln ( N ) − ln ( E ) , R ( N ) ' ln ( N ) − ln ( E ) , with E / E > 1 . Thus, S ( N ) ' value of a Yule process of parameter λ at time ln ( E / E ) , conditionnally on E / E > 1 . √ Igor Kortchemski Preys & Predators 25 / 17
Test your intuition! Complete graphs Infinite trees Idea of proof: case λ 2 ( 0, 1 ) S N ( N ) S N 1 ln ( N ) 1 − λ R R ( N ) T √ Igor Kortchemski Preys & Predators 26 / 17
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