Poisson Representations of Measure-Valued Processes

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Poisson representations of measure-valued processes

• Poisson random measures • Branching processes • General Dawson-Watanabe superprocesses • Branching processes in random environments • Exit measures and nonlinear PDEs • Homogenization of nonlinear PDE • Conditioning on nonextinction • References • Abstract

with Eliane Rodrigues [5]

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 1 Poisson random measures S a Polish space and ν a σ-ﬁnite measure on B(S). ξ is a Poisson random measure with mean measure ν if

a) ξ is a random counting measure on S. b) For each A ∈ S with ν(A) < ∞, ξ(A) is Poisson distributed with parameter ν(A).

c) For A1,A2,... ∈ S disjoint, ξ(A1), ξ(A2),... are independent.

Lemma 1 If ν(A) < ∞, then conditioned on ξ(A), the point process Pξ(A) ξ(· ∩ A) has the same distribution as i=1 δζi , where ζ1, . . . , ζξ(A) are iid with distribution ν(A)−1ν(· ∩ A).

−1 If H : S → S0, Borel measurable, and ξb(A) = ξ(H (A)), then ξb is a Poisson random measure on S0 with mean measure νb given by −1 νb(A) = ν(H (A))

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 2 Moment identities

If ξ is a Poisson random measure with mean measure ν

R R f E[e f(z)ξ(dz)] = e (e −1)dν, P or letting ξ = i δZi ,

Y R (g−1)dν E[ g(Zi)] = e . i Similarly, Z X Y R (g−1)dν E[ h(Zj) g(Zi)] = hgdνe , j i and Z X Y 2 R (g−1)dν E[ h(Zi)h(Zj) g(Zk)] = ( hgdν) e , i6=j k

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 3 Conditionally Poisson systems

Let ξ be a random counting measure on S and Ξ be a locally ﬁnite random measure on S. ξ is conditionally Poisson with Cox measure Ξ if, conditioned on Ξ, ξ is a Poisson point process with mean measure Ξ.

− R fdξ − R (1−e−f )dΞ E[e S ] = E[e S ] for all nonnegative f ∈ M(S). If ξ is conditionally Poisson system on S × [0, ∞) with Cox measure Ξ × m where m is Lebesgue measure, then for f ∈ M(S) − R fdξ −K R (1−e−f )dΞ E[e S×[0,K] ] = E[e S ] and for f ≥ 0, 1 Z Z Ξ(f) = lim fdξ = lim e−uf(x)ξ(dx × du) a.s. K→∞ K S×[0,K] →0 S×[0,∞)

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 4 Relationship to exchangeability

Lemma 2 Suppose ξ is a conditionally Poisson random measure on S × [0, ∞) with Cox measure Ξ × m. If Ξ < ∞ a.s., then we can write P∞ ξ = i=1 δ(Xi,Ui) with U1 < U2 < ··· a.s. and {Xi} is exchangeable.

Conditioned on Ξ, {Ui} is Poisson with parameter Ξ(S) and {Xi} is iid with distribution Ξ(S)−1Ξ.

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 5 Convergence

Lemma 3 If {ξn} is a sequence of conditionally Poisson random measures on S × [0, ∞) with Cox measures {Ξn × m}. Then ξn ⇒ ξ if and only if Ξn ⇒ Ξ, Ξ the Cox measure for ξ.

If ξn → ξ in probability, then Ξn → Ξ in probability

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 6 A population model

n Consider a process with state space E = ∪n[0, r] . Qn 0 ≤ g ≤ 1, g(r) = 1, and f(u, n) = i=1 g(ui) For a > 0, and −∞ < b ≤ ra, deﬁne n Z r n 0 X X g (ui) A f(u, n) = f(u, n) 2a (g(v)−1)dv+f(u, n) (au2 −bu ) . r i i g(u ) i=1 ui i=1 i

In other words, particle levels satisfy ˙ 2 Ui(t) = aUi (t) − bUi(t), and a particle with level z gives birth at rate 2a(r − z) to a particle whose initial level is uniformly distributed between z and r.

N(t) = #{i : Ui(t) < r}

αr(n, du) the joint distribution of n iid uniform [0, r] random variables.

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 7 A calculation

R −λgn −λg 1 R r fb(n) = f(u, n)αr(n, du) = e , e = r 0 g(u)du R To calculate Cfb(n) = Arfu, n)αr(n, du), observe that Z r Z r Z r r−12a g(z) (g(v) − 1)dv = are−2λg − 2ar−1 g(z)(r − z)dz 0 z 0 and Z r Z r r−1 (az2 − bz)g0(z)dz = −r−1 (2az − b)(g(z) − 1)dz 0 0 Z r = −2ar−1 zg(z)dz + ar + b(e−λg − 1). 0 Then

−λg(n−1) −2λg −λg −λg Cfb(n) = ne are − 2are + ar + b(e − 1) = arn(fb(n + 1) − fb(n)) + (ar − b)n(fb(n − 1) − fb(n)).

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 8 Conclusion

Let Ne be a solution of the martingale problem for Cfb(n) = arn(fb(n + 1) − fb(n)) + (ar − b)n(fb(n − 1) − fb(n)), that is, Ne is a branching process with birth rate ar and death rate (ar − b).

Then there exists a solution (U1(t),...,UN(t)(t),N(t)) of the martingale problem for A such that N has the same distribution as Ne.

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 9 The limit as r → ∞

If n = O(r), then the scaling is correct for the Feller diﬀusion. Q For f(u) = i g(ui), 0 ≤ g ≤ 1, g(z) = 1, z ≥ ug, Arf converges to Z ug 0 X X g (ui) Af(u) = f(u) 2a (g(v) − 1)dv + f(u) (au2 − bu ) . i i g(u ) i ui i i

−1 If nr → y, then αr(n, du) → α(y, du) where α(y, du) is the distribution of a Poisson process on [0, ∞) with intensity y. Z −y R ∞(1−g(z))dz −yβ fb(y) = αf(y) = f(u)α(y, du) = e 0 = e g and ∞ ∞ ∞ αAf(y) = e−yβg 2ay ∫ g(z) ∫ (g(v) − 1)dvdz + y ∫ (az2 − bz)g0(z)dz 0 z 0 −yβg 2 = e (ayβg − byβg) = ayfb00(y) + byfb0(y)

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 10 Particle representation of Feller diﬀusion

Let {Ui(0)} be a conditionally Poisson process on [0, ∞) with (con- ditional) intensity Y (0). Then, {Ui(t)} is conditionally Poisson with intensity Y (t), 1 Y (t) = lim #{i : Ui(t) ≤ r}, r→∞ r and Y is a Feller diﬀusion with generator Cf(y) = ayf 00(y) + byf 0(y)

γ : N (R) → [0, ∞) 1 γ(u) = lim #{i : ui ≤ r} r→∞ r

α(y, du) Poisson process distribution on [0, ∞) with intensity y. α(y, γ−1(y)) = 1.

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 11 Extinction

Assume U1(0) < U2(0) < ···. Then for all t, all levels are above −bt U1(0)e U1(t) = a −bt 1 − b U1(0)(1 − e ) Let τ = inf{t : Y (t) = 0}

−bt −1 −yb/[(1−e−bt)a] P {τ > t} = P {U1(0) < (1 − e )a/b } = 1 − e If b ≤ 0, conditioning on nonextinction for all t is equivalent to setting U1(0) = 0. The generator becomes Z ug 0 X X g (ui) Af(u) = f(u) 2a (g(v) − 1)dv + f(u) (au2 − bu ) i i g(u ) i ui i i Z ug +f(u)2a (g(v) − 1)dv 0 and αAf(y) = ayfb00(y) + (2a + by)fb0(y)

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 12 Branching Markov processes Qn f(x, u, n) = i=1 g(xi, ui), where g : E × [0, ∞) → (0, 1] As a function of x, g is in the domain D(B) of the generator of a Markov process in E, g is continuously diﬀerentiable in u, and g(x, u) = 1 for u ≥ r. n X Bg(xi, ui) Af(x, u, n) = f(x, u, n) g(x , u ) i=1 i i n X Z r +f(x, u, n) 2a(xi) (g(xi, v) − 1)dv i=1 ui n X ∂u g(xi, ui) +f(x, u, n) (a(x )u2 − b(x )u ) i i i i i g(x , u ) i=1 i i

Each particle has a location Xi(t) in E and a level Ui(t) in [0, r].

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 13 Behavior of the process

The locations evolve independently as Markov processes with generator B, the levels satisfy ˙ 2 Ui(t) = a(Xi(t))Ui (t) − b(Xi(t))Ui(t) and particles that reach level r die.

Particles give birth at rates 2a(Xi(t))(r − Ui(t)); the initial location of a new particle is the location of the parent at the time of birth; and the initial level is uniformly distributed on [Ui(t), r].

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 14 Generator for X(t) = (X1(t),...,XN(t))

r Pn −λg(xi) −1 R − i=1 λg(xi) Setting e = r 0 g(xi, z)dz and fb(x, n) = e , and cal- culating as in the previous example, we have n n X X Cfb(x, n) = Bxi fb(x, n) + a(xi)r(fb((x, xi), n + 1) − fb(x, n)) i=1 i=1 n X + (a(xi)r − b(xi))(fb(d(x|xi), n − 1) − fb(x, n)), i=1

where Bxi is the generator B applied to fb(x, n) as a function of xi and d(x|xi) is the vector obtained from x be eliminating the ith component.

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 15 Inﬁnite population limit

Letting r → ∞, Af becomes

Z ug X Bg(xi, ui) X Af(x, u) = f(x, u) + f(x, u) 2a(x ) (g(x , v) − 1)dv g(x , u ) i i i i i i ui X ∂u g(xi, ui) +f(x, u) (a(x )u2 − b(x )u ) i i i i i g(x , u ) i i i Particle locations evolve as independent Markov processes with generator B, and levels satisfy ˙ 2 Ui(t) = a(Xi(t))Ui (t) − b(Xi(t))Ui(t)

A particle with level Ui(t) gives birth to new particles at its location Xi(t) and initial level in the interval [Ui(t) + c, Ui(t) + d] at rate 2a(Xi(t))(d − c). A particle dies when its level hits ∞.

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 16 The measure-valued limit

For µ ∈ Mf (E), let α(µ, dx × du) be the distribution of a Poisson random measure on E ×[0, ∞) with mean measure µ×m. Then setting R ∞ h(y) = 0 (1 − g(y, v))dv Z αf(µ) = f(x, u)α(µ, dx × du) = exp{− ∫ h(y)µ(dy)}, E and Z Z ∞ αAf(µ) = exp{− ∫ h(y)µ(dy)}[ Bg(y, v)dvµ(dy) E E 0 Z Z ∞ Z ∞ + 2a(y)g(y, z) (g(y, v) − 1)dvdzµ(dy) E 0 z Z Z ∞ 2 + (a(y)v − b(y)v)∂vg(y, v)dvµ(dy)] E 0 Z = exp{− ∫ h(y)µ(dy)} −Bh(y) + a(y)h(y)2 − b(y)h(y) µ(dy) E E

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 17 Representation of the Dawson-Watanabe process

It follows that the Cox measure (or more precisely, the E marginal of the Cox measure) corresponding to the particle process at time t, call it Z(t), is a solution of the martingale problem for

A = {(αf, αAf): f ∈ D}.

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 18 Branching processes in random environments[2,3,4] a and b functions of another stochastic process η, say an irreducible ﬁnite Markov chain with generator Q. Qn f(l, u, n) = f0(l)f1(u) = f0(l) i=1 g(ui), n X Z r Arf(l, u, n) = rf1(u)Qf0(l) + f(l, u, n) 2a(l) (g(v) − 1)dv i=1 ui n 0 X √ g (ui) +f(l, u, n) (a(l)u2 − rb(l)u ) , i i g(u ) i=1 i which projects to

Crfb(l, n) = rQfb(l, n) + a(l)rn(fb(l, n + 1) − fb(n)) √ +(ra(l) − rb(l))n(fb(l, n − 1) − fb(l, n)), −λ n where fb(l, n) = f0(l)e g . The process corresponding to Cr is a branching process in a random environment determined by η.

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 19 Scaling limit

Writing the process corresponding to Ar as

(η(rt),U1(t),...,UNr(t)) the process corresponding to Cr is (η(rt),Nr(t)). Note that the levels satisfy √ ˙ 2 Ui(t) = a(η(rt))Ui (t) − rb(η(rt))Ui(t). P Let π be the stationary distribution for Q. Assume that l π(l)b(l) = 0 and Qh0(l) = b(l). Then √ Z t W (r)(t) = r b(η(rs))ds 0 converges to a Brownian motion W with variance parameter X X 2 X π(k)qkl(h0(l) − h0(k)) = −2 π(l)h0(l)b(l) ≡ 2c. k l l

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 20 Limiting level processes

In the limit, the levels will satisfy √ 2 dUi(t) = (aUi(t) + cUi(t))dt + 2cUi(t)dW (t), (1) where a = P π(l)a(l). Note that the limiting levels are all driven by the same Brownian motion.

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 21 Limiting generator

Af1(u) X Z ∞ = f1(u) 2a (g(v) − 1)dv i ui 2 0 2 00 X (cui + au )g (ui) + cu g (ui) +f (u) i i 1 g(u ) i i   n 0 0 X X g (ui)g (uj) +cf1(u)  ujui  . g(ui)g(uj) j=1 i6=j

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 22 Limiting diﬀusion

Let Z ∞ Z ∞ Z ∞ 0 1 2 00 βg = (1 − g(z))dz = zg (z)dz = − z g (z)dz. 0 0 2 0 We have Z ∞ Z ∞ Z ∞ αAf(y) = e−yβg (2ay g(z) (g(v) − 1)dvdz + y (az2 + cz)g0(z)dz 0 z 0 Z ∞ Z ∞ +cy2( zg0(z)dz)2 + cy z2g00(z)dz) 0 0 −yβg 2 2 = e ((ay + cy )βg − cyβg) = Cfb(y), where Cfb(y) = (ay + cy2)fb00(y) + cyfb0(y), −1 which identiﬁes the diﬀusion limit for r Nr.

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 23 Exit measures [1,6] Let B be the generator of a diﬀusion and D a bounded domain with all points in ∂D regular.

X Bg(xi, ui) Af(x, u) = f(x, u) 1 (x ) D i g(x , u ) i i i X Z ug +f(x, u) 1D(xi)2a(xi) (g(xi, v) − 1)dv i ui X ∂u g(xi, ui) +f(x, u) 1 (x )(a(x )u2 − b(x )u ) i D i i i i i g(x , u ) i i i P Assume infx a(x) > 0, b ≤ 0, then τ = inf{t : i 1D(Xi(t) = 0} < ∞. P Deﬁne ξ = i δ(Xi(τ),Ui(τ)), and deﬁne the exit measure 1 X ZD(Γ) = lim 1Γ(Xi(τ)), r→∞ r i:Ui(τ)≤r

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 24 Nonlinear PDE

Assume {(Xi(0),Ui(0))} is a Poisson random measure with mean measure δx × m. Then for h ≥ 0, bounded and continuous, −hh,Zx i u(x) = − log E[e D ] solves the diﬀerential equation Bu = au2 − bu u = h on ∂D.

x R P −hh,Z i log(1−h/r)1[0,r]dξ U (τ)≤r log(1−h(Xi(τ))/r) e D = E[e |ZD] = E[e i |ZD]. Taking expectations,

−hh,Zx i Y h(Xi(τ)) E[e D ] = E[ (1 − )] r Ui(τ)≤r

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 25 Homogenization Bu(x) = a(x, x/)u(x)2 − b(x, x/)u(x) u = h on ∂D a(x, y), b(x, y) periodic in y, and d d 1 X ∂2g X ∂g Bg(x) = cjk(x) (x) + dj(x) (x). 2 ∂xj∂xk ∂xj j,k=1 j=1 Particle generator:

X Bg(xi, ui) Af(x, u) = f(x, u) 1 (x ) g(x , u ) D i i i i X Z ug +f(x, u) 2a(xi, xi/) (g(xi, v) − 1)dv1D(xi) i ui X ∂u g(xi, ui) +f(x, u) (a(x , x /)u2 − b(x , x /)u ) i 1 (x ) i i i i i i g(x , u ) D i i i i

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 26 Eﬀective coeﬃcients R R Let a(x) = a(x, y)πx(dy), b(x) = b(x, y)πx(dy), where πx is determined by Z d Bgb (x, y)πx(dy) = 0, g ∈ R , where d 1 X ∂2g Bgb (x, y) = cjk(x) (y). 2 ∂yj∂yk j,k=1

x −hZ,x,hi Y h(Xi(τ)) −hZ ,hi lim E[e D ] = lim E[ (1 − )] = E[e D ] →0 →0 r Ui (τ)≤r

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 27 Conditioning on non-extinction

Let a be constant and b ≡ 0. and let U∗(0) be the minimum of the initial levels. U∗(0) U∗(t) = . 1 − aU∗(0)t

1 Let τ be the time of extinction, Then {τ > T } = {U∗(0) < aT }. Con- ditioning on {τ > T } and letting T → ∞ is equivalent to conditioning on the initial Poisson process having a level at zero. The resulting generator becomes

Z rg X Bg(xi, ui) X Af(x, u) = f(x, u) + f(x, u) 2a (g(x , v) − 1)dv (2) g(x , u ) i i i i i>0 ui Z rg +f(x, u)2a (g(x0, v) − 1)dv 0 X ∂u g(xi, ui) +f(x, u) au2 i , i g(x , u ) i>0 i i

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 28 Generator for measure-valued process

The generator for the measure-valued process is given by setting Z 1 Z α0f(µ) = f(x, u)α0(µ, dx × du) = g(z, 0)µ(dz) exp{−hh, µi}, |µ| E and Z 2 1 α0Af(µ) = h−Bh(y) + ah(y) , µi g(z, 0)µ(dz) exp{−hh, µi} |µ| E 1 Z + (Bg(z, 0) − 2ag(z, 0)h(z))µ(dz) exp{−hh, µi} |µ| E

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 29 References

[1] E. B. Dynkin. Diffusions, superdiffusions and partial differential equations, volume 50 of American Mathematical Society Colloquium Publications. American Mathematical Society, Providence, RI, 2002. [2] Inge S. Helland. Minimal conditions for weak convergence to a diffusion process on the line. Ann. Probab., 9(3):429–452, 1981. [3] Niels Keiding. Extinction and exponential growth in random environments. The- oret. Population Biol., 8:49–63, 1975. [4] Thomas G. Kurtz. Diffusion approximations for branching processes. In Branching processes (Conf., Saint Hippolyte, Que., 1976), volume 5 of Adv. Probab. Related Topics, pages 269–292. Dekker, New York, 1978. [5] Thomas G. Kurtz and Eliane Rodrigues. Poisson representations of branching Markov and measure-valued branching processes. preprint. [6] Kathryn E. Temple. Particle representations for measure-valued processes with interactions and exit measures. PhD thesis, University of Wisconsin - Madison, 2005.

•First •Prev •Next •Go To •Go Back •Full Screen •Close •Quit 30 Abstract

Poisson representations of measure-valued processes Measure-valued diﬀusions and measure-valued solutions of stochastic partial diﬀer- ential equations can be represented in terms of the Cox measures of particle systems that are conditionally Poisson at each time t. The representations are useful for characterizing the processes, establishing limit theorems, and analyzing the behavior of the measure-valued processes. Examples will be given and some of the useful methodology will be described.

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