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Limit theorems for chain-binomial population models Phil Pollett Department of Mathematics The University of Queensland http://www.maths.uq.edu.au/˜pkp AUSTRALIAN RESEARCH COUNCIL Centre of Excellence for Mathematics and Statistics of Complex Systems

Limit theorems for chain-binomial population models...Collaborators Fionnuala Buckley Department of Mathematics University of Queensland ∗Buckley, F.M. and Pollett, P.K. (2010) Limit

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  • Limit theorems for chain-binomialpopulation models

    Phil Pollett

    Department of MathematicsThe University of Queensland

    http://www.maths.uq.edu.au/˜pkp

    AUSTRALIAN RESEARCH COUNCILCentre of Excellence for Mathematicsand Statistics of Complex Systems

  • Collaborators

    Fionnuala BuckleyDepartment of MathematicsUniversity of Queensland

    ∗Buckley, F.M. and Pollett, P.K. (2010) Limit theorems for discrete-timemetapopulation models. Probability Surveys 7, 53-83.

  • Collaborators

    Ross McVinishDepartment of MathematicsUniversity of Queensland

    ∗McVinish, R. and Pollett, P.K. (2010) Limits of large metapopulationswith patch dependent extinction probabilities. Advances in Applied Prob-ability 42, 1172-1186.

  • Metapopulations

  • Metapopulations

    Colonization

  • Metapopulations

  • Metapopulations

    Local Extinction

  • Metapopulations

  • Metapopulations

  • Metapopulations

  • Metapopulations

  • Metapopulations

    Total Extinction

  • Metapopulations

  • Metapopulations

  • Metapopulations

    Colonization

    from the mainland

  • Metapopulations

  • SPOM

    A Stochastic Patch Occupancy Model (SPOM)

  • SPOM

    A Stochastic Patch Occupancy Model (SPOM)

    Suppose that there are n patches.

  • SPOM

    A Stochastic Patch Occupancy Model (SPOM)

    Suppose that there are n patches.

    Let X(n)t = (X(n)1,t , . . . , X

    (n)n,t ), where X

    (n)i,t is a binary variable

    indicating whether or not patch i is occupied.

  • SPOM

    A Stochastic Patch Occupancy Model (SPOM)

    Suppose that there are n patches.

    Let X(n)t = (X(n)1,t , . . . , X

    (n)n,t ), where X

    (n)i,t is a binary variable

    indicating whether or not patch i is occupied.

    For each n, (X(n)t , t = 0, 1, . . . ) is assumed to be a Markovchain.

  • SPOM

    A Stochastic Patch Occupancy Model (SPOM)

    Suppose that there are n patches.

    Let X(n)t = (X(n)1,t , . . . , X

    (n)n,t ), where X

    (n)i,t is a binary variable

    indicating whether or not patch i is occupied.

    For each n, (X(n)t , t = 0, 1, . . . ) is assumed to be a Markovchain.

    Colonization and extinction happen in distinct, successivephases.

  • SPOM - Phase structure

    For many species the propensity for colonization and localextinction is markedly different in different phases of theirlife cycle.

  • SPOM - Phase structure

    For many species the propensity for colonization and localextinction is markedly different in different phases of theirlife cycle. Examples:

    The Vernal pool fairy shrimp (Branchinecta lynchi) and

    the California linderiella (Linderiella occidentalis), both

    listed under the Endangered Species Act (USA)

    The Jasper Ridge population of Bay checkerspot

    butterfly (Euphydryas editha bayensis), now extinct

  • SPOM - Phase structure

    Colonization and extinction happen in distinct, successivephases.

    t− 1 t t+ 1 t+ 2

    t− 1 t t+ 1 t+ 2

  • SPOM - Phase structure

    Colonization and extinction happen in distinct, successivephases.

    t− 1 t t+ 1 t+ 2

    We will we assume that the population is observed aftersuccessive extinction phases (CE Model).

  • SPOM - Phase structure

    Colonization and extinction happen in distinct, successivephases.

    Colonization: unoccupied patches become occupiedindependently with probability c(n−1

    ∑n

    i=1X(n)i,t ), where

    c : [0, 1] → [0, 1] is continuous, increasing and concave.

  • Examples of c(x)

    c(x) = cx, where c ∈ (0, 1] is the maximum colonizationpotential.

  • Examples of c(x)

    c(x) = cx, where c ∈ (0, 1] is the maximum colonizationpotential.

    c(x) = c, where c ∈ (0, 1] is a fixed colonizationpotential—mainland colonization dominant.

  • Examples of c(x)

    c(x) = cx, where c ∈ (0, 1] is the maximum colonizationpotential.

    c(x) = c, where c ∈ (0, 1] is a fixed colonizationpotential—mainland colonization dominant.

    c(x) = c0 + cx, where c0 + c ∈ (0, 1] (mainland and islandcolonization).

  • Examples of c(x)

    c(x) = cx, where c ∈ (0, 1] is the maximum colonizationpotential.

    c(x) = c, where c ∈ (0, 1] is a fixed colonizationpotential—mainland colonization dominant.

    c(x) = c0 + cx, where c0 + c ∈ (0, 1] (mainland and islandcolonization).

    c(x) = 1− exp(−xβ) (β > 0).

  • SPOM - Phase structure

    Colonization and extinction happen in distinct, successivephases.

    Colonization: unoccupied patches become occupiedindependently with probability c(n−1

    ∑n

    i=1X(n)i,t ), where

    c : [0, 1] → [0, 1] is continuous, increasing and concave.

  • SPOM - Phase structure

    Colonization and extinction happen in distinct, successivephases.

    Colonization: unoccupied patches become occupiedindependently with probability c(n−1

    ∑n

    i=1X(n)i,t ), where

    c : [0, 1] → [0, 1] is continuous, increasing and concave.

    Extinction: occupied patch i remains occupiedindependently with probability Si (random).

  • SPOM

    Thus, we have a Chain Bernoulli structure:

    X(n)i,t+1

    d= Bin

    (

    X(n)i,t + Bin

    (

    1−X(n)i,t , c(

    1n

    ∑n

    j=1X(n)j,t

    )

    )

    , Si

    )

  • SPOM

    Thus, we have a Chain Bernoulli structure:

    X(n)i,t+1

    d= Bin

    (

    X(n)i,t +Bin

    (

    1−X(n)i,t , c(

    1n

    ∑n

    j=1X(n)j,t

    )

    )

    , Si

    )

  • SPOM

    Thus, we have a Chain Bernoulli structure:

    X(n)i,t+1

    d= Bin

    (

    X(n)i,t + Bin

    (

    1−X(n)i,t , c(

    1n

    ∑n

    j=1X(n)j,t

    )

    )

    , Si

    )

  • SPOM

    Thus, we have a Chain Bernoulli structure:

    X(n)i,t+1

    d= Bin

    (

    X(n)i,t + Bin

    (

    1−X(n)i,t , c(

    1n

    ∑n

    j=1X(n)j,t

    )

    )

    , Si

    )

  • SPOM

    Thus, we have a Chain Bernoulli structure:

    X(n)i,t+1

    d= Bin

    (

    X(n)i,t + Bin

    (

    1−X(n)i,t , c(

    1n

    ∑n

    j=1X(n)j,t

    )

    )

    , Si

    )

  • SPOM

    n = 30, Si ∼Beta(25.2, 19.8) (ESi = 0.56) and c(x) = 0.7x

    0 0 0 0 1 0 1 1 0 1 0 1 0 0 0 0 1 1 1 0 1 0 1 0 0 0 1 0 0 0

    c(x) = c(1130) = 0.7× 0.36̇ = 0.256̇

  • SPOM

    n = 30, Si ∼Beta(25.2, 19.8) (ESi = 0.56) and c(x) = 0.7x

    0 0 0 0 1 0 1 1 0 1 0 1 0 0 0 0 1 1 1 0 1 0 1 0 0 0 1 0 0 0C 1 0 0 0 1 1 1 1 0 1 0 1 0 0 0 0 1 1 1 1 1 1 1 0 0 0 1 0 1 0

  • SPOM

    n = 30, Si ∼Beta(25.2, 19.8) (ESi = 0.56) and c(x) = 0.7x

    0 0 0 0 1 0 1 1 0 0 0 1 0 0 0 0 1 1 1 0 1 0 1 0 0 0 1 0 0 0C 1 0 0 0 1 1 1 1 0 1 0 1 0 0 0 0 1 1 1 1 1 1 1 0 0 0 1 0 1 0

  • SPOM

    n = 30, Si ∼Beta(25.2, 19.8) (ESi = 0.56) and c(x) = 0.7x

    0 0 0 0 1 0 1 1 0 0 0 1 0 0 0 0 1 1 1 0 1 0 1 0 0 0 1 0 0 0C 1 0 0 0 1 1 1 1 0 1 0 1 0 0 0 0 1 1 1 1 1 1 1 0 0 0 1 0 1 0

    0.60 0.56 0.63 0.62 0.52 0.61 0.68 0.49 0.49 0.49 0.500.41 0.59 0.63 0.60 0.61

  • SPOM

    n = 30, Si ∼Beta(25.2, 19.8) (ESi = 0.56) and c(x) = 0.7x

    0 0 0 0 1 0 1 1 0 1 0 1 0 0 0 0 1 1 1 0 1 0 1 0 0 0 1 0 0 0C 1 0 0 0 1 1 1 1 0 1 0 1 0 0 0 0 1 1 1 1 1 1 1 0 0 0 1 0 1 0E 0 0 0 0 1 0 0 1 0 1 0 1 0 0 0 0 1 0 1 1 1 1 0 0 0 0 0 0 1 0

  • SPOM

    n = 30, Si ∼Beta(25.2, 19.8) (ESi = 0.56) and c(x) = 0.7x

    0 0 0 0 1 0 1 1 0 1 0 1 0 0 0 0 1 1 1 0 1 0 1 0 0 0 1 0 0 0C 1 0 0 0 1 1 1 1 0 1 0 1 0 0 0 0 1 1 1 1 1 1 1 0 0 0 1 0 1 0E 0 0 0 0 1 0 0 1 0 1 0 1 0 0 0 0 1 0 1 1 1 1 0 0 0 0 0 0 1 0

    c(x) = c(1030) = 0.7× 0.3̇ = 0.23̇

  • SPOM

    n = 30, Si ∼Beta(25.2, 19.8) (ESi = 0.56) and c(x) = 0.7x

    0 0 0 0 1 0 1 1 0 1 0 1 0 0 0 0 1 1 1 0 1 0 1 0 0 0 1 0 0 0C 1 0 0 0 1 1 1 1 0 1 0 1 0 0 0 0 1 1 1 1 1 1 1 0 0 0 1 0 1 0E 0 0 0 0 1 0 0 1 0 1 0 1 0 0 0 0 1 0 1 1 1 1 0 0 0 0 0 0 1 0C 0 0 1 0 1 0 0 1 1 1 0 1 0 0 1 0 1 1 1 1 1 1 0 0 0 0 0 0 1 0

  • SPOM

    n = 30, Si ∼Beta(25.2, 19.8) (ESi = 0.56) and c(x) = 0.7x

    0 0 0 0 1 0 1 1 0 1 0 1 0 0 0 0 1 1 1 0 1 0 1 0 0 0 1 0 0 0C 1 0 0 0 1 1 1 1 0 1 0 1 0 0 0 0 1 1 1 1 1 1 1 0 0 0 1 0 1 0E 0 0 0 0 1 0 0 1 0 1 0 1 0 0 0 0 1 0 1 1 1 1 0 0 0 0 0 0 1 0C 0 0 1 0 1 0 0 1 1 1 0 1 0 0 1 0 1 1 1 1 1 1 0 0 0 0 0 0 1 0

  • SPOM

    n = 30, Si ∼Beta(25.2, 19.8) (ESi = 0.56) and c(x) = 0.7x

    0 0 0 0 1 0 1 1 0 1 0 1 0 0 0 0 1 1 1 0 1 0 1 0 0 0 1 0 0 0C 1 0 0 0 1 1 1 1 0 1 0 1 0 0 0 0 1 1 1 1 1 1 1 0 0 0 1 0 1 0E 0 0 0 0 1 0 0 1 0 1 0 1 0 0 0 0 1 0 1 1 1 1 0 0 0 0 0 0 1 0C 0 0 1 0 1 0 0 1 1 1 0 1 0 0 1 0 1 1 1 1 1 1 0 0 0 0 0 0 1 0E 0 0 0 0 1 0 0 1 0 1 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 0

  • SPOM

    n = 30, Si ∼Beta(25.2, 19.8) (ESi = 0.56) and c(x) = 0.7x

    0 0 0 0 1 0 1 1 0 1 0 1 0 0 0 0 1 1 1 0 1 0 1 0 0 0 1 0 0 0C 1 0 0 0 1 1 1 1 0 1 0 1 0 0 0 0 1 1 1 1 1 1 1 0 0 0 1 0 1 0E 0 0 0 0 1 0 0 1 0 1 0 1 0 0 0 0 1 0 1 1 1 1 0 0 0 0 0 0 1 0C 0 0 1 0 1 0 0 1 1 1 0 1 0 0 1 0 1 1 1 1 1 1 0 0 0 0 0 0 1 0E 0 0 0 0 1 0 0 1 0 1 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 0

  • SPOM

    n = 30, Si ∼Beta(25.2, 19.8) (ESi = 0.56) and c(x) = 0.7x

    0 0 0 0 1 0 1 1 0 1 0 1 0 0 0 0 1 1 1 0 1 0 1 0 0 0 1 0 0 0C 1 0 0 0 1 1 1 1 0 1 0 1 0 0 0 0 1 1 1 1 1 1 1 0 0 0 1 0 1 0E 0 0 0 0 1 0 0 1 0 1 0 1 0 0 0 0 1 0 1 1 1 1 0 0 0 0 0 0 1 0C 0 0 1 0 1 0 0 1 1 1 0 1 0 0 1 0 1 1 1 1 1 1 0 0 0 0 0 0 1 0E 0 0 0 0 1 0 0 1 0 1 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 0...C 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0E 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

  • SPOM - Homogeneous case

    Compare this with the homogeneous case, where Si = s(non-random) is the same for each i, and we merely countthe number N (n)t of occupied patches at time t.

    We have the following Chain Binomial structure:

    N(n)t+1

    d= Bin

    (

    N(n)t + Bin

    (

    n−N (n)t , c(

    1nN

    (n)t

    )

    )

    , s)

  • SPOM - Homogeneous case

    Compare this with the homogeneous case, where Si = s(non-random) is the same for each i, and we merely countthe number N (n)t of occupied patches at time t.

    We have the following Chain Binomial structure:

    N(n)t+1

    d= Bin

    (

    N(n)t + Bin

    (

    n−N (n)t , c(

    1nN

    (n)t

    )

    )

    , s)

  • SPOM - Homogeneous case

    Compare this with the homogeneous case, where Si = s(non-random) is the same for each i, and we merely countthe number N (n)t of occupied patches at time t.

    We have the following Chain Binomial structure:

    N(n)t+1

    d= Bin

    (

    N(n)t + Bin

    (

    n−N (n)t , c(

    1nN

    (n)t

    )

    )

    , s)

  • SPOM - Homogeneous case

    Compare this with the homogeneous case, where Si = s(non-random) is the same for each i, and we merely countthe number N (n)t of occupied patches at time t.

    We have the following Chain Binomial structure:

    N(n)t+1

    d= Bin

    (

    N(n)t +Bin

    (

    n−N (n)t , c(

    1nN

    (n)t

    )

    )

    , s)

  • SPOM - Homogeneous case

    Compare this with the homogeneous case, where Si = s(non-random) is the same for each i, and we merely countthe number N (n)t of occupied patches at time t.

    We have the following Chain Binomial structure:

    N(n)t+1

    d= Bin

    (

    N(n)t + Bin

    (

    n−N (n)t , c(

    1nN

    (n)t

    )

    )

    , s)

  • SPOM - Homogeneous case

    Compare this with the homogeneous case, where Si = s(non-random) is the same for each i, and we merely countthe number N (n)t of occupied patches at time t.

    We have the following Chain Binomial structure:

    N(n)t+1

    d=Bin

    (

    N(n)t + Bin

    (

    n−N (n)t , c(

    1nN

    (n)t

    )

    )

    , s)

  • SPOM - Homogeneous case

    Compare this with the homogeneous case, where Si = s(non-random) is the same for each i, and we merely countthe number N (n)t of occupied patches at time t.

    We have the following Chain Binomial structure:

    N(n)t+1

    d= Bin

    (

    N(n)t + Bin

    (

    n−N (n)t , c(

    1nN

    (n)t

    )

    )

    , s)

  • A deterministic limit

    Theorem [BP] If N (n)0 /np→ x0 (a constant), then

    N(n)t /n

    p→ xt, for all t ≥ 1,

    with (xt) determined by xt+1 = f(xt), where

    f(x) = s(x+ (1− x)c(x)).

    [BP] Buckley, F.M. and Pollett, P.K. (2010) Limit theorems for discrete-timemetapopulation models. Probability Surveys 7, 53-83.

  • Stability

    xt+1 = f(xt), where f(x) = s(x+ (1− x)c(x)).

    Stationarity : c(0) > 0. There is a unique fixed pointx∗ ∈ [0, 1]. It satisfies x∗ ∈ (0, 1) and is stable.Evanescence: c(0) = 0 and 1 + c ′(0) ≤ 1/s. Now 0 is theunique fixed point in [0, 1]. It is stable.

    Quasi stationarity : c(0) = 0 and 1 + c ′(0) > 1/s. Thereare two fixed points in [0, 1]: 0 (unstable) and x∗ ∈ (0, 1)(stable).

    [Notice that if c(0) = 0, we require c ′(0) > 0 for quasistationarity.]

  • CE Model - Evanescence

    0 5 10 15 20 25 30 35 40 45 500

    10

    20

    30

    40

    50

    60

    70

    80

    90

    100CE Model simulation (n = 100, s = 0.56, c(x) = cx with c =0.7)

    t

    Num

    ber

    ofocc

    upie

    dpatc

    hes

  • CE Model - Quasi stationarity

    0 10 20 30 40 50 60 70 80 90 1000

    10

    20

    30

    40

    50

    60

    70

    80

    90

    100CE Model simulation (n = 100, s = 0.8, c(x) = cx with c =0.7)

    t

    Num

    ber

    ofocc

    upie

    dpatc

    hes

  • A Gaussian limit

    Theorem [BP] Further suppose that c(x) is twicecontinuously differentiable, and let

    Z(n)t =

    √n(N

    (n)t /n− xt).

    If Z(n)0d→ z0, then Z(n)• converges weakly to the Gaussian

    Markov chain Z• defined by

    Zt+1 = f′(xt)Zt + Et (Z0 = z0),

    with (Et) independent and Et ∼ N(0, v(xt)), where

    v(x) = s[

    (1− s)x+ (1− x)c(x)(

    1− sc(x))]

    .

  • CE Model - Quasi stationarity

    0 10 20 30 40 50 60 70 80 90 1000

    10

    20

    30

    40

    50

    60

    70

    80

    90

    100CE Model simulation (n = 100, s = 0.8, c(x) = cx with c =0.7)

    t

    Num

    ber

    ofocc

    upie

    dpatc

    hes

  • CE Model - Quasi stationarity

    0 500 1000 1500 2000 2500 3000 3500 40000

    10

    20

    30

    40

    50

    60

    70

    80

    90

    100CE Model simulation (n =100, s =0.8, c(x) = cx with c =0.7)

    t

    Num

    ber

    ofocc

    upie

    dpatc

    hes

  • CE Model - Quasi-stationary distribution

    0 500 1000 1500 2000 2500 3000 3500 40000

    10

    20

    30

    40

    50

    60

    70

    80

    90

    100CE Model simulation (n =100, s =0.8, c(x) = cx with c =0.7)

    t

    Num

    ber

    ofocc

    upie

    dpatc

    hes

  • CE Model - Gaussian approximation

    0 500 1000 1500 2000 2500 3000 3500 40000

    10

    20

    30

    40

    50

    60

    70

    80

    90

    100CE Model simulation (n =100, s =0.8, c(x) = cx with c =0.7)

    t

    Num

    ber

    ofocc

    upie

    dpatc

    hes

    Nx∗ = 64.2857

  • A deterministic limit

    Returning to the general case, where patch survivalprobabilities are random and patch dependent , and wekeep track of which patches are occupied . . .

    X(n)i,t+1

    d= Bin

    (

    X(n)i,t + Bin

    (

    1−X(n)i,t , c(

    1n

    ∑n

    j=1X(n)j,t

    )

    )

    , Si

    )

  • A deterministic limit

    Returning to the general case, where patch survivalprobabilities are random and patch dependent , and wekeep track of which patches are occupied . . .

    X(n)i,t+1

    d= Bin

    (

    X(n)i,t + Bin

    (

    1−X(n)i,t , c(

    1n

    ∑n

    j=1X(n)j,t

    )

    )

    , Si

    )

    First, . . .

    Notation: If σ is a probability measure on [0, 1) and let s̄kdenote its k-th moment, that is,

    s̄k =∫ 1

    0xkσ(dx).

  • A deterministic limit

    Theorem Suppose there is a probability measure σ anddeterministic sequence {d(0, k)} such that

    1n

    ∑n

    i=1 Ski

    p→ s̄k and 1n∑n

    i=1 Ski X

    (n)i,0

    p→ d(0, k)

    for all k = 0, 1, . . . , T . Then, there is a (deterministic)triangular array {d(t, k)} such that, for all t = 0, 1, . . . , T andk = 0, 1, . . . , T − t,

    1n

    ∑n

    i=1 Ski X

    (n)i,t

    p→ d(t, k),

    where

    d(t+ 1, k) = d(t, k + 1) + c (d(t, 0)) (s̄k+1 − d(t, k + 1)) .

  • A deterministic limit d(0,k)

    0 1 2 3 T

    t0

    1

    2

    3

    T

    k

  • A deterministic limit d(1,k)

    0 1 2 3 T

    t0

    1

    2

    3

    T

    k

  • A deterministic limit d(2,k)

    0 1 2 3 T

    t0

    1

    2

    3

    T

    k

  • A deterministic limit d(3,k)

    0 1 2 3 T

    t0

    1

    2

    3

    T

    k

  • A deterministic limit d(t,k)

    0 1 2 3 T

    t0

    1

    2

    3

    T

    k

  • A deterministic limit d(t,0)

    0 1 2 3 T

    t0

    1

    2

    3

    T

    k

  • Remarks

    Typically, we are only interested in d(t, 0), being theasymptotic proportion of occupied patches at time t:

    1n

    ∑n

    i=1X(n)i,t

    p→ d(t, 0).

  • Remarks: d(t,0)

    0 1 2 3 T

    t0

    1

    2

    3

    T

    k

  • Remarks: d(t,k)

    0 1 2 3 T

    t0

    1

    2

    3

    T

    k

  • Remarks: d(t,0)

    0 1 2 3 T

    t0

    1

    2

    3

    T

    k

  • Remarks

    Typically, we are only interested in d(t, 0), being theasymptotic proportion of occupied patches at time t:

    1n

    ∑n

    i=1X(n)i,t

    p→ d(t, 0).

  • Remarks

    Typically, we are only interested in d(t, 0), being theasymptotic proportion of occupied patches at time t:

    1n

    ∑n

    i=1X(n)i,t

    p→ d(t, 0).

    However, we may still interpret the ratio d(t, k)/d(t, 0)(k ≥ 1) as the k-th moment of the conditional distributionof the patch survival probability given that the patch isoccupied. (From these moments, the conditionaldistribution could then be reconstructed.)

  • A deterministic limit

    Theorem Suppose there is a probability measure σ anddeterministic sequence {d(0, k)} such that

    1n

    ∑n

    i=1 Ski

    p→ s̄k and 1n∑n

    i=1 Ski X

    (n)i,0

    p→ d(0, k)

    for all k = 0, 1, . . . , T . Then, there is a (deterministic)triangular array {d(t, k)} such that, for all t = 0, 1, . . . , T andk = 0, 1, . . . , T − t,

    1n

    ∑n

    i=1 Ski X

    (n)i,t

    p→ d(t, k),

    where

    d(t+ 1, k) = d(t, k + 1) + c (d(t, 0)) (s̄k+1 − d(t, k + 1)) .

  • Homogeneous case

    When s̄k = s̄k1 for all k, that is the patch survivalprobabilities are the same, then it is possible to simplify

    d(t+ 1, k) = d(t, k + 1) + c (d(t, 0)) (s̄k+1 − d(t, k + 1)) .

    We can show by induction that d(t, k) = s̄k1xt, where

    xt+1 = s̄1 (xt + (1− xt) c(xt)) .

    Compare this with the earlier [BP] result....

  • A deterministic limit

    Theorem [BP] If N (n)0 /np→ x0 (a constant), then

    N(n)t /n

    p→ xt, for all t ≥ 1,

    with (xt) determined by xt+1 = f(xt), where

    f(x) = s(x+ (1− x)c(x)).

    [BP] Buckley, F.M. and Pollett, P.K. (2010) Limit theorems for discrete-timemetapopulation models. Probability Surveys 7, 53-83.

  • Stability

    Theorem Any fixed point d = (d(0), d(1), . . . ) is given by

    d(k) =∫ 1

    0c(ψ)xk+1

    1−x+c(ψ)xσ(dx),

    where ψ (= d(0)) solves

    R(ψ) =∫ 1

    0c(ψ)x

    1−x+c(ψ)xσ(dx) = ψ. (1)

    If c(0) > 0, there is a unique ψ > 0. If c(0) = 0 and

    c ′(0)∫ 1

    0x

    1−xσ(dx) ≤ 1,

    then ψ = 0 is the unique solution to (1). Otherwise, (1) hastwo solutions, one of which is ψ = 0.

  • Stability

    Theorem If c(0) = 0 and

    c ′(0)∫ 1

    0x

    1−xσ(dx) ≤ 1,

    then d(k) ≡ 0 is a stable fixed point. Otherwise, thenon-zero solution to

    R(ψ) =∫ 1

    0c(ψ)x

    1−x+c(ψ)xσ(dx) = ψ

    provides the stable fixed point through

    d(k) =∫ 1

    0c(ψ)xk+1

    1−x+c(ψ)xσ(dx).

  • CE Model (homogeneous) - Evanescence

    0 5 10 15 20 25 30 35 40 45 500

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    100CE Model simulation (n = 100, s = 0.56, c(x) = cx with c =0.7)

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  • CE Model - Evanescence

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    c′(0)

    ∫1

    0

    x

    1−xσ(dx) =0.9383

    0 0.5 10

    2

    4

    6Beta(25.2,19.8)

  • CE Model - Quasi stationarity

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    c′(0)

    ∫1

    0

    x

    1−xσ(dx) =1.2572

    0 0.5 10

    1

    2

    3Beta(4.368,3.432)

  • CE Model - Quasi stationarity

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    100CE Model simulation (n = 100, s1 = 0.56, c(x) = cx with c =0.7)

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    c′(0)

    ∫1

    0

    x

    1−xσ(dx) = ∞

    0 0.5 10

    0.5

    1

    1.5

    2Beta(1.176,0.924)

  • CE Model - Quasi stationarity

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    c′(0)

    ∫1

    0

    x

    1−xσ(dx) = ∞

    0 0.5 10

    2

    4

    6

    8Beta(0.784,0.616)

  • CE Model - Quasi stationarity

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    c′(0)

    ∫1

    0

    x

    1−xσ(dx) = ∞

    0 0.5 10

    10

    20

    30Beta(0.3304,0.2596)

  • CE Model - Quasi stationarity

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    c′(0)

    ∫1

    0

    x

    1−xσ(dx) = ∞

    0 0.5 10

    2

    4

    6

    8Beta(0.784,0.616)

  • CE Model - Quasi stationarity

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    100CE Model simulation (n = 100, s1 = 0.6, c(x) = cx with c =0.7)

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    c′(0)

    ∫1

    0

    x

    1−xσ(dx) =2.1

    0 0.5 10

    0.5

    1

    1.5

    2Beta(3,2)

    CollaboratorsCollaboratorsMetapopulationsMetapopulationsMetapopulationsMetapopulationsMetapopulationsMetapopulationsMetapopulationsMetapopulationsMetapopulationsMetapopulationsMetapopulationsMetapopulationsMetapopulationsSPOMSPOMSPOMSPOMSPOMSPOM - Phase structureSPOM - Phase structure

    SPOM - Phase structureSPOM - Phase structureSPOM - Phase structureExamples of $�oldsymbol {c} �oldsymbol {(} �oldsymbol {x} �oldsymbol {)}$Examples of $�oldsymbol {c} �oldsymbol {(} �oldsymbol {x} �oldsymbol {)}$Examples of $�oldsymbol {c} �oldsymbol {(} �oldsymbol {x} �oldsymbol {)}$Examples of $�oldsymbol {c} �oldsymbol {(} �oldsymbol {x} �oldsymbol {)}$

    SPOM - Phase structureSPOM - Phase structureSPOMSPOMSPOMSPOMSPOMSPOMSPOMSPOMSPOMSPOMSPOMSPOMSPOMSPOMSPOMSPOMSPOM - Homogeneous caseSPOM - Homogeneous caseSPOM - Homogeneous caseSPOM - Homogeneous caseSPOM - Homogeneous caseSPOM - Homogeneous caseSPOM - Homogeneous caseA deterministic limitStabilityCE Model - EvanescenceCE Model - Quasi stationarityA Gaussian limitCE Model - Quasi stationarityCE Model - Quasi stationarityCE Model - Quasi-stationary distributionCE Model - Gaussian approximationA deterministic limitA deterministic limit

    A deterministic limitA deterministic limit $�oldsymbol {d}�oldsymbol {(}�oldsymbol {0}�oldsymbol {,}�oldsymbol {k}�oldsymbol {)}$A deterministic limit $�oldsymbol {d}�oldsymbol {(}�oldsymbol {1}�oldsymbol {,}�oldsymbol {k}�oldsymbol {)}$A deterministic limit $�oldsymbol {d}�oldsymbol {(}�oldsymbol {2}�oldsymbol {,}�oldsymbol {k}�oldsymbol {)}$A deterministic limit $�oldsymbol {d}�oldsymbol {(}�oldsymbol {3}�oldsymbol {,}�oldsymbol {k}�oldsymbol {)}$A deterministic limit $�oldsymbol {d}�oldsymbol {(}�oldsymbol {t}�oldsymbol {,}�oldsymbol {k}�oldsymbol {)}$A deterministic limit $�oldsymbol {d}�oldsymbol {(}�oldsymbol {t}�oldsymbol {,}�oldsymbol {0}�oldsymbol {)}$RemarksRemarks: $�oldsymbol {d}�oldsymbol {(}�oldsymbol {t}�oldsymbol {,}�oldsymbol {0}�oldsymbol {)}$Remarks: $�oldsymbol {d}�oldsymbol {(}�oldsymbol {t}�oldsymbol {,}�oldsymbol {k}�oldsymbol {)}$Remarks: $�oldsymbol {d}�oldsymbol {(}�oldsymbol {t}�oldsymbol {,}�oldsymbol {0}�oldsymbol {)}$RemarksRemarksA deterministic limitHomogeneous caseA deterministic limitStabilityStabilityCE Model (homogeneous)- EvanescenceCE Model - EvanescenceCE Model - Quasi stationarityCE Model - Quasi stationarityCE Model - Quasi stationarityCE Model - Quasi stationarityCE Model - Quasi stationarityCE Model - Quasi stationarity