Solutions to group problems

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Solutions to group problems

• gi,i-1i? gii-i(???) gi,i1i?
• Hence -gi,i1/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 matrix
• and stationary distribution
• (1/2,1/2), but the stationary distribution of the
  process itself is

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• 3. This is a birth and death process with ?nl
  and ?nm. We know the stationary distribution is

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Poisson process

• Birth process with rate independent of the state
• Infinitesimal generator
• Time between events?

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Poisson process, cont.
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• Siméon Denis Poisson (1781-1840)
• Rudolf Julius Emanuel Clausius (1822-1888)
• Ladislaus Josephowitsch Bortkiewicz (1868-1931)

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Independent increments

• XPo(?), YPo(?) independent, what is the
  distribution of XY?
• Write X(t,tsX(ts)-X(t)
• independent of j.
• So events in (0,t is independent of
  events in (t,ts, Xt has independent increments

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Counting process

• N(A) points in A
• If A (s,t then N(A)X(t)-X(s)
• Renyis theorem(s)
• N is the counting process corresponding to a
  Poisson process of rate ? iff
• P(N(A)0)e-?A for all A
• or
• (ii) N(A) and N(B) are independent for all A
  disjoint from B

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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.

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Volcanic eruptions

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

pt
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• If XtPo(?pt), YtXt/pt is a reconstruction
• Lots of variability in early centuries

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Nonhomogeneous Poisson process

• XtPo(?(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(Ysk)ske-s/k!
• (?(t))ke-?(t)/k!P(Xtk).

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General definition

• Consider points in some space S, subset of Rd.
  They constitute a Poisson point pattern if
• N(A)Po(?(A))
• N(A) is independent of N(B) for disjoint A and B
• ?() is called the mean measure.
• If we call??(s) the intensity function.

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Spatial case

• Complete spatial randomness

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Clustering and regularity

• To 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

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Real point patterns

• Linhares experimental forest, Brazil
• Control plot
• Clear-cut plot

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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.

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Proof

• Let A1,...,Ak be a partition of A. Then if
  n1...nkn we have
• i.e., a multinomial distribution.

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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))