Solutions to group problems

1.gi,i-1=i gii=-i() gi,i+1=iHence -gi,i+1/gii==1-(-gi,i-1/gii).Thus the jump chain goes up one step with probability and down one step with probability 0 is an absorbing state.2. Consider a 0-1 process. It has jump chain with transition matrixand stationary distribution (1/2,1/2), but the stationary distribution of the process itself is

0 1

1 0

⎛⎝⎜

⎞⎠⎟

βα + β

αα + β

⎛⎝⎜

⎞⎠⎟

3. This is a birth and death process with n=l and n=m. We know the stationary distribution is

i−1

μii=1

n

∏λ i−1

μii=1

n

∏n=0

∞

∑=

lm

⎛⎝⎜

⎞⎠⎟

n

1−l

m

∝ exp(−nlogl

m⎛⎝⎜

⎞⎠⎟)

Poisson process

Birth process with rate independent of the stateInfinitesimal generator

Time between events?

G =

− L L L

0 − L L

0 0 − L

0 0 0 − L

L L L L L

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

Poisson process, cont.

dp0k (t)

dt=−p0k (t) + p0,k−1(t)

G(s; t) =EsX(t) = poi (t)si

i∑

∂G(s; t)

∂dt= −λG(s; t) + λsG(s; t)

∂G(s; t)

∂t= −λ(1− s)G(s; t)

G(s; t) =A(s)e−(1−s )t

G(s;0) =1⇒ A(s) =1

pij (s,s + t) =(t)j−i

(j−i)!e−t; j≥i

Siméon Denis Poisson (1781-1840)

Rudolf Julius Emanuel Clausius (1822-1888)

Ladislaus Josephowitsch Bortkiewicz (1868-1931)

Independent increments

X~Po(), Y~Po() independent, what is the distribution of X+Y?

Write X(t,t+s]=X(t+s)-X(t)

independent of j.

So # events in (0,t] is independent of # events in (t,t+s], Xt has independent increments

EsX+Y =EsXEsY =e(s−1)e(s−1) =e(+)(s−1)

P(X(t, t + s] =l X(t) =j) =P(X(t+ s) =j+ l X(t) =j)

=pj,j+l (s) =p0,l (s)

Counting process

N(A) = # points in A

If A = (s,t] then N(A)=X(t)-X(s)

Renyi’s theorem(s):

N is the counting process corresponding to a Poisson process of rate iff

(i) P(N(A)=0)=e-|A| for all A

or

(ii) N(A) and N(B) are independent for all A disjoint from B

Subsampling

Suppose we delete points in a Poisson process independently with probability 1-p. How does that affect the infinitesimal generator?

Poisson process of rate p.

P1,n+1(t+ Δt)P1n(t)

=pΔt+ o(Δt)

G =

−p p 0 L

0 −p p L

0 0 −p L

L L L L

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

Volcanic eruptions

Recording of volcanic erupotions has gotten more complete over the last decades

pt

If Xt~Po(pt), Yt=Xt/pt is a reconstruction

Lots of variability in early centuries

Nonhomogeneous Poisson process

Xt~Po((t)) where

Time change theorem:

Let Yt be a unit rate Poisson process. Then

Proof: Note that (t) is monotone. Let s= (t). Then P(Ys=k)=ske-s/k!

=((t))ke-(t)/k!=P(Xt=k).

(t) = λ(s)ds0

t

∫

Y(t) =d

Xt

General definition

Consider points in some space S, subset of Rd. They constitute a Poisson point pattern if

(i) N(A)~Po((A))

(ii) N(A) is independent of N(B) for disjoint A and B

(•) is called the mean measure.

If we call(s) the intensity function.

(A) = λ(s)dsA∫

Spatial case

Complete spatial randomness

Clustering and regularityTo get a clustered process, start with a Poisson spatial process, then add new points iid around the original points

To get a regular process, delete points from a Poisson process that are closer than d together

Real point patterns

Linhares experimental forest, Brazil

Control plot

Clear-cut plot

A conditional property

Let N be a Poisson counting process with intensity (x). Suppose A is a set with , N(A)=n, and let Q(B)=(B)/(A) be a cumulative distribution. It has density (x)/(A) Then the points in A have the same distribution as n points drawn independently from the distribution Q.

0 < (A) < ∞

Proof

Let A1,...,Ak be a partition of A. Then if n1+...+nk=n we have

i.e., a multinomial distribution.

P(N(A1) =n1, ...,N(Ak ) =nk N(A) =n)

=P(N(Ai ) =ni )

i=1

k

∏P(N(A) =n)

=(Ai )

nie−(Ai ) / ni !i=1

k

∏(A)ne−(A ) / n!

=n!

n1!L nk !Q(A1)

n1L Q(Ak )nk

Some facts about Poisson point patterns

Superposition: The overlay of two independent Poisson patterns is a Poisson pattern with mean function the sum of the mean functions

Coloring: Consider a Poisson pattern with intensity (x) in which point independently is colored either green (with probability (x)) or purple (with probability 1-(x)). Then the green points form a Poisson process with intensity (x)(x), and the purple points an independent one with intensity (x)(1-(x))