On a prey-predator model

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On a prey-predator model

Igor Kortchemski

CNRS & CMAP, École polytechnique

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Test your intuition! Complete graphs Infinite trees

What is this about?

On a graph, we are interested in the following prey-predator model (introduced by Bordenave ’12) :

- each vertex is either occupied by a prey, or a predator, or is vacant, - at fixed rate > 0, each prey propagates to every vacant neighbour, - at fixed rate 1, each predator propagates to every neighbouring prey.

Motivations :

Model of two competing species, or model of first-passage percolation with destruction.

Other possible analogies: vacant vertex ()

prey ()

predator ()

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What is this about?

- each vertex is either occupied by a prey, or a predator, or is vacant,

- at fixed rate > 0, each prey propagates to every vacant neighbour, - at fixed rate 1, each predator propagates to every neighbouring prey.

Motivations :

prey ()

predator ()

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What is this about?

- each vertex is either occupied by a prey, or a predator, or is vacant, - at fixed rate > 0, each prey propagates to every vacant neighbour,

- at fixed rate 1, each predator propagates to every neighbouring prey.

Motivations :

prey ()

predator ()

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What is this about?

Motivations :

prey ()

predator ()

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What is this about?

Motivations :

prey ()

predator ()

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What is this about?

Motivations :

Other possible analogies:

vacant vertex

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What is this about?

Motivations :

vacant vertex () vacant vertex

prey () healthy cell

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What is this about?

Motivations :

vacant vertex normal individual

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What is this about?

Motivations :

vacant vertex () Susceptible (S) individual

prey () Infected (I) individual

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Here, {I, S} ! {I, I}, {R, I} !¹ {R, R}.

Other type of models studied in the literature:

– SIR model (Kermack—McKendrick ’27), where {I, S} ! {I, I}, I !¹ R – Daley–Kendall (’65) rumour propagation model, where

{I, S} !¹ {I, I}, {R, I} !¹ {R, R}, {I, I} !¹ {R, R}.

– Maki–Thompson (’73) directed rumour propagation model, where (I, S) !¹ (I, I), (R, I) !¹ (R, R), (I, I) !¹ (I, R).

– Williams Bjerknes (’71) tumor growth model (or biased voter model), where (I, S) ! (I, I), (S, I) !¹ (S, S).

– Kordzakhia (’05), where {I, S} ! {I, I}, {R, I} !¹ {R, R}, {R, S} !¹ {R, R}.

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Here, {I, S} ! {I, I}, {R, I} !¹ {R, R}.

– SIR model (Kermack—McKendrick ’27), where {I, S} ! {I, I}, I !¹ R

– Daley–Kendall (’65) rumour propagation model, where {I, S} !¹ {I, I}, {R, I} !¹ {R, R}, {I, I} !¹ {R, R}.

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Here, {I, S} ! {I, I}, {R, I} !¹ {R, R}.

{I, S} !¹ {I, I}, {R, I} !¹ {R, R}, {I, I} !¹ {R, R}.

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Here, {I, S} ! {I, I}, {R, I} !¹ {R, R}.

{I, S} !¹ {I, I}, {R, I} !¹ {R, R}, {I, I} !¹ {R, R}.

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Here, {I, S} ! {I, I}, {R, I} !¹ {R, R}.

{I, S} !¹ {I, I}, {R, I} !¹ {R, R}, {I, I} !¹ {R, R}.

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Here, {I, S} ! {I, I}, {R, I} !¹ {R, R}.

{I, S} !¹ {I, I}, {R, I} !¹ {R, R}, {I, I} !¹ {R, R}.

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Outline

I. Test your intuition!

II. Preys & predators on a complete graph III. Preys & predators on an infinite tree

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Outline

II. Preys & predators on a complete graph

III. Preys & predators on an infinite tree

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Outline

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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 p

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?

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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)?

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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)?

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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)?

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Seen on Gil Kalai’s blog

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Seen on Gil Kalai’s blog

4) Roughly a constant.

5) Some other behavior.

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Seen on Gil Kalai’s blog

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Vu sur le blog de Gil Kalai

You keep as before until left only with balls of the same color. How many balls will be left (as a function of n)?

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Vu sur le blog de Gil Kalai

You keep as before until left only with balls of the same color. How many balls will be left (as a function of n)?

Other formulation (O.K. Corral problem, Williams & McIlroy, 1998) . There are

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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+nⁿ .

But

n

m + n = 1/m

1/m + 1/n = P (Exp(1/m) < Exp(1/n)) .

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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+nⁿ . But

n

m + n = 1/m

1/m + 1/n = P (Exp(1/m) < Exp(1/n)) .

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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 2n 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.

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Kingman & Volkov’s solution (2/3)

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.

X

₁

X

₂

X

₃

X

₄

Y

₁

Y

₂

Y

₃

Y

₄

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Kingman & Volkov’s solution (2/3)

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.

X

₁

X

₂

X

₃

X

₄

Y

₁

Y

₂

Y

₃

Y

₄

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Kingman & Volkov’s solution (2/3)

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.

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Kingman & Volkov’s solution (2/3)

Light both ends, and stop the fire when the origin is reached. Let be the

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Kingman & Volkov’s solution (3/3)

In order to estimate the number R(n) of remaining pieces, first estimate the remaining length L(n):

L(n) =

Xn i=1

X_i -

Xn i=1

Y_i .

Then

Var

Xn i=1

X_i -

Xn i=1

Y_i

!

=

Xn j=1

2j² ' n³.

Hence

L(n) ' n^3/2.

Set S_k = X₁ + · · · + X_k. We have E [S_k] ' k², so S_k ' k². 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.

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Kingman & Volkov’s solution (3/3)

L(n) =

Xn i=1

X_i -

Xn i=1

Y_i .

Then

Var

Xn i=1

X_i -

Xn i=1

Y_i

!

=

Xn j=1

2j² ' n³.

Hence

L(n) ' n^3/2.

R(n)² ' n^3/2 so that R(n) ' n^3/4.

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Kingman & Volkov’s solution (3/3)

L(n) =

Xn i=1

X_i -

Xn i=1

Y_i .

Then

Var

Xn i=1

X_i -

Xn i=1

Y_i

!

=

Xn j=1

2j² ' n³.

Hence

L(n) ' n^3/2.

R(n)² ' n^3/2 so that R(n) ' n^3/4.

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Kingman & Volkov’s solution (3/3)

L(n) =

Xn i=1

X_i -

Xn i=1

Y_i .

Then

Var

Xn i=1

X_i -

Xn i=1

Y_i

!

=

Xn j=1

2j² ' n³.

Hence

L(n) ' n^3/2.

Set S = X + · · · + X .

We have E [S_k] ' k², so S_k ' k². 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.

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Kingman & Volkov’s solution (3/3)

L(n) =

Xn i=1

X_i -

Xn i=1

Y_i .

Then

Var

Xn i=1

X_i -

Xn i=1

Y_i

!

=

Xn j=1

2j² ' n³.

Hence

L(n) ' n^3/2.

Set S_k = X₁ + · · · + X_k. We have E [S_k] ' k²

, so S_k ' k². 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.

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Kingman & Volkov’s solution (3/3)

L(n) =

Xn i=1

X_i -

Xn i=1

Y_i .

Then

Var

Xn i=1

X_i -

Xn i=1

Y_i

!

=

Xn j=1

2j² ' n³.

Hence

L(n) ' n^3/2.

Set S = X + · · · + X . We have E [S ] ' k², so S ' k².

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.

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Kingman & Volkov’s solution (3/3)

L(n) =

Xn i=1

X_i -

Xn i=1

Y_i .

Then

Var

Xn i=1

X_i -

Xn i=1

Y_i

!

=

Xn j=1

2j² ' n³.

Hence

L(n) ' n^3/2.

part burns first, S ' L(n).

Hence

R(n)² ' n^3/2 so that R(n) ' n^3/4.

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Kingman & Volkov’s solution (3/3)

L(n) =

Xn i=1

X_i -

Xn i=1

Y_i .

Then

Var

Xn i=1

X_i -

Xn i=1

Y_i

!

=

Xn j=1

2j² ' n³.

Hence

L(n) ' n^3/2.

Set S = X + · · · + X . We have E [S ] ' k², so S ' k². But, if the left

so that R(n) ' n^3/4.

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Kingman & Volkov’s solution (3/3)

L(n) =

Xn i=1

X_i -

Xn i=1

Y_i .

Then

Var

Xn i=1

X_i -

Xn i=1

Y_i

!

=

Xn j=1

2j² ' n³.

Hence

L(n) ' n^3/2.

part burns first, S ' L(n). Hence

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This “decoupling” idea is called the Athreya–Karlin embedding, and is useful to study more general Pólya urn schemes.

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II. Prey & predators on a complete graph

III. Preys & predators on an infinite tree

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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}.

Question. How does P E^N_ext behave as N ! 1 ?

We have

P(E^N_ext) !

N!1

8>

<

>:

0 if 2

(0, 1)

1 2

if =

1

1 if >

1

. Theorem (K. ’13).

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Set

We have

P(E^N_ext) !

N!1

8>

<

>:

0 if 2

(0, 1)

1 2

if =

1

1 if >

1

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Set

We have

P(E^N_ext) !

N!1

8>

<

>:

0 if 2

(0, 1)

1 2

if =

1

1 if >

1

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Set

We have

P(E^N_ext) !

N!1

8>

<

>:

0 if 2

(0, 1)

1 2

if =

1

1 if >

1

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Set

We have

P(E^N_ext) !

N!1

8>

<

>:

0 if 2

(0, 1)

1 2

if =

1

1 if >

1

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Set

We have

P(E^N_ext) !

N!1

8>

<

>:

0 if 2 (0, 1)

1 2

if = 1 1 if > 1.

Theorem (K. ’13).

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Set

We have

P(E^N_ext) !

N!1

8>

<

>:

0 if 2 (0, 1)

1

2 if = 1

1 if > 1.

Theorem (K. ’13).

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Decoupling using Yule processes

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Transition rates

Let S_t, I_t, R_t be the population sizes at time t. Total rate of {S, I} ! {I, I} :

· 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

S_tI_t

S_tI_t + I_tR_t = S_t

S_t + R_t .

y We are going to be able to decouple the evolutions of S and R.

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Transition rates

Let S_t, I_t, R_t be the population sizes at time t.

Total rate of {S, I} ! {I, I} : · S_t · I_t.

Total rate {R, I} ! {R, R} : I_t · R_t.

S_tI_t

S_t + R_t .

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Transition rates

Total rate of {S, I} ! {I, I} : · S_t · I_t. Total rate {R, I} ! {R, R} :

I_t · R_t.

S_tI_t

S_t + R_t .

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Transition rates

Total rate of {S, I} ! {I, I} : · S_t · I_t. Total rate {R, I} ! {R, R} : I_t · R_t.

S_tI_t

S_t + R_t .

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Transition rates

S_tI_t

S_t + R_t .

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Transition rates

S_tI_t

S_t + R_t .

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Coupling and decoupling via two Yule processes

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Yule processes

Definition (Yule process)

In a Yule process (Y(t))_t_>₀ of parameter , starting with one individual, each individual lives a random time distributed according to a Exp( ) 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( ), Exp(2 ), Exp(3 ), . . . random variables.

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Yule processes

Definition (Yule process)

variable, and at its death gives birth to two individuals, and Y(t) denotes the total number of individuals at time t.

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Yule processes

Definition (Yule process)

variable, and at its death gives birth to two individuals, and Y(t) denotes the total number of individuals at time t.

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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 , 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!

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, ^cE^N_ext)

the N-th discontinuity of S_N (S disappears first, E^N_ext)

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Coupling with two Yule processes

T T

E^N_ext

cE^N_ext

S_N S_N

R R

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Coupling with two Yule processes

T T

E^N_ext

cE^N_ext

S_N S_N

R R

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Coupling with two Yule processes

T T

E^N_ext

cE^N_ext

S_N S_N

R R

T is the time when a type of vertices (S or I) disappears.

T is the smallest between:

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Coupling with two Yule processes

T T

E^N_ext

cE^N_ext

S_N S_N

R R

T is the time when a type of vertices (S or I) disappears.

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Identification of the critical parameter = 1

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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( N) + Exp( (N - 1)) + · · · + Exp( ).

R(N) has the same distribution

Exp(1) + Exp(2) + · · · + Exp(N).

Hence

P(E^N_ext) !

N!1

8>

<

>:

0 if 2 (0, 1)

1

2 if = 1

1 if > 1.

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Notation.

Proposition

Exp( N) + Exp( (N - 1)) + · · · + Exp( ).

Exp(1) + Exp(2) + · · · + Exp(N).

T T

E^N_ext

S_N S_N

R R

SN(1) S

N(2) S

N(3) S

N(4) S

N(5) S

N(6) S

N(7)

R(1) R(2) R(3) R(4) R(5) R(7)

cE^N_ext

S_N(1) S_N(2) S

N(3) S

N(4) S_N(5) S

N(6) S

N(7)

R(1) R(2) R(3) R(4) R(5) R(6) R(7)

Figure: Example for N = 7, where the crosses represent discontinuities.

Hence

P(E^N_ext) !

N!1

8>

<

>:

0 if 2 (0, 1)

1

2 if = 1

1 if > 1.

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Notation.

Proposition

Exp( N) + Exp( (N - 1)) + · · · + Exp( ).

Exp(1) + Exp(2) + · · · + Exp(N).

T T

E^N_ext

S_N S_N

R R

SN(1) S

N(2) S

N(3) S

N(4) S

N(5) S

N(6) S

N(7)

R(1) R(2) R(3) R(4) R(5) R(7)

cE^N_ext

S_N(1) S_N(2) S

N(3) S

N(4) S_N(5) S

N(6) S

N(7)

R(1) R(2) R(3) R(4) R(5) R(6) R(7)

Hence

P(E^N_ext) !

N!1

8>

<

>:

0 if 2 (0, 1)

1

2 if = 1

1 if > 1.

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Notation.

Proposition

S_N(N) has the same distribution as Exp( N) + Exp( (N - 1)) + · · · + Exp( ). R(N) has the same distribution

Exp(1) + Exp(2) + · · · + Exp(N).

T T

E^N_ext

S_N S_N

R R

SN(1) S

N(2) S

N(3) S

N(4) S

N(5) S

N(6) S

N(7)

R(1) R(2) R(3) R(4) R(5) R(7)

cE^N_ext

S_N(1) S_N(2) S

N(3) S

N(4) S_N(5) S

N(6) S

N(7)

R(1) R(2) R(3) R(4) R(5) R(6) R(7)

Hence

P(E^N_ext) !

N!1

8>

<

>:

0 if 2 (0, 1)

1

2 if = 1

1 if > 1.

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Notation.

Proposition

S_N(N) has the same distribution as Exp( N) + Exp( (N - 1)) + · · · + Exp( ). R(N) has the same distribution Exp(1) + Exp(2) + · · · + Exp(N).

T T

E^N_ext

S_N S_N

R R

SN(1) S

N(2) S

N(3) S

N(4) S

N(5) S

N(6) S

N(7)

R(1) R(2) R(3) R(4) R(5) R(7)

cE^N_ext

S_N(1) S_N(2) S

N(3) S

N(4) S_N(5) S

N(6) S

N(7)

R(1) R(2) R(3) R(4) R(5) R(6) R(7)

Hence

P(E^N_ext) !

N!1

8>

<

>:

0 if 2 (0, 1)

1

2 if = 1

1 if > 1.