1 / 24

Segue from time series to point processes. Y = 0,1 E(Y) = Prob{Y = 1}

Segue from time series to point processes. Y = 0,1 E(Y) = Prob{Y = 1} (Y 1 , Y 2 ) E(Y 1 Y 2 } = Prob{( Y 1 ,Y 2 ) = (1,1)} {Y(t)} case. mean level: c Y (t) = Prob{Y(t) = 1}

gaerwn
Download Presentation

Segue from time series to point processes. Y = 0,1 E(Y) = Prob{Y = 1}

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Segue from time series to point processes. Y = 0,1 E(Y) = Prob{Y = 1} (Y1, Y2 ) E(Y1 Y2} = Prob{( Y1 ,Y2) = (1,1)} {Y(t)} case. mean level: c Y(t) = Prob{Y(t) = 1} product moment: Prob{Y(t1)=1, Y(t2) = 1} = E{Y(t 1)Y(t2)} Naïve interpretations Stationary case : cYY(t1 – t 2) Can use acf, ccf, … i,e, stationary series R functions

  2. Can approximate point process data {τj , j=1,…,J } by a 0-1 tt.s. data points isolated, pick small Δt time series Y(t/ Δt) T = J/Δt can be large

  3. Point processes on the line. Nerve firing.

  4. Stochastic point process. Building blocks Process on R {N(t)}, t in R, with consistent set of distributions Pr{N(I1)=k1 ,..., N(In)=kn } k1 ,...,kn integers  0 I's Borel sets of R. Consistentency example. If I1 , I2 disjoint Pr{N(I1)= k1 , N(I2)=k2 , N(I1 or I2)=k3 } =1 if k1 + k2 =k3 = 0 otherwise Guttorp book, Chapter 5

  5. Points: ... -1  0  1  ... discontinuities of {N} N(t) = #{0 < j  t} Simple: j k if j  k points are isolated dN(t) = 0 or 1 Surprise. A simple point process is determined by its void probabilities Pr{N(I) = 0} I compact

  6. Conditional intensity. Simple case History Ht = {j  t} Pr{dN(t)=1 | Ht } = (t:)dt  r.v. Has all the information Probability points in [0,T) are t1 ,...,tN Pr{dN(t1)=1,..., dN(tN)=1} = (t1)...(tN)exp{- (t)dt}dt1 ... dtN [1-(h)h][1-(2h)h] ... (t1)(t2) ...

  7. Dirac delta. Picture a r.v. , U, = 0 with probability 1 then E{g(U)} = g(0) Picture a r.v. , V, with distribution N(0, σ2), σ small then E{g(V)}approaches g(0) as σ decreases, g cts at 0 Picture that U has a density δ(u), a generalized function then E{g(U)} = ∫ g(u) δ(u) du Properties: ∫ δ(u) du = 1, δ(u) = 0 for u ≠ 0 dH(u) = δ(u) du for H the Heavyside function

  8. Parameters. Suppose points are isolated dN(t) = 1 if point in (t,t+dt] = 0 otherwise 1. (Mean) rate/intensity. E{dN(t)} = pN(t)dt = Pr{dN(t) = 1} j g(j) =  g(s)dN(s) E{j g(j)} =  g(s)pN(s)ds Trend: pN(t) = exp{+t} Cycle: exp{cos(t+)}

  9. Product density of order 2. Pr{dN(s)=1 and dN(t)=1} = E{dN(s)dN(t)} = [(s-t)pN(t) + pNN (s,t)]dsdt Factorial moment

  10. Autointensity. Pr{dN(t)=1|dN(s)=1} = (pNN (s,t)/pN (s))dt s  t = hNN(s,t)dt = pN (t)dt if increments uncorrelated

  11. Covariance density/cumulant density of order 2. cov{dN(s),dN(t)} = qNN(s,t)dsdt st = [(s-t)pN(s)+qNN(s,t)]dsdt generally qNN(s,t) = pNN(s,t) - pN(s) pN(t) st

  12. Identities. 1. j,k  g(j ,k ) =  g(s,t)dN(s)dN(t) Expected value. E{ g(s,t)dN(s)dN(t)} =  g(s,t)[(s-t)pN(t)+pNN (s,t)]dsdt =  g(t,t)pN(t)dt +  g(s,t)pNN(s,t)dsdt

  13. 2. cov{ g(j ),  g(k )} = cov{ g(s)dN(s),  h(t)dN(t)} =  g(s) h(t)[(s-t)pN(s)+qNN(s,t)]dsdt =  g(t)h(t)pN(t)dt +  g(s)h(t)qNN(s,t)dsdt

  14. Product density of order k. t1,...,tkall distinct Prob{dN(t1)=1,...,dN(tk)=1} =E{dN(t1)...dN(tk)} = pN...N (t1,...,tk)dt1 ...dtk = Prob{dN(t1)=1,...,dN(tk)=1} E{N(t) (k)} = ∫0t… ∫0t pN...N (t1,...,tk)dt1 ...dtk

  15. Cumulant density of order k. t1,...,tkdistinct cum{dN(t1),...,dN(tk)} = qN...N (t1 ,...,tk)dt1 ...dtk

  16. Stationarity. Joint distributions, Pr{N(I1+t)=k1 ,..., N(In+t)=kn} k1 ,...,kn integers  0 do not depend on t for n=1,2,... Rate. E{dN(t)=pNdt Product density of order 2. Pr{dN(t+u)=1 and dN(t)=1} = [(u)pN + pNN (u)]dtdu

  17. Autointensity. Pr{dN(t+u)=1|dN(t)=1} = (pNN (u)/pN)du u  0 = hN(u)du Covariance density. cov{dN(t+u),dN(t)} = [(u)pN + qNN (u)]dtdu

  18. Mixing. cov{dN(t+u),dN(t)} small for large |u| |pNN(u) - pNpN| small for large |u| hNN(u) = pNN(u)/pN ~ pN for large |u|  |qNN(u)|du < 

  19. Algebra/calculus of point processes. Consider process {j, j+u}. Stationary case dN(t) = dM(t) + dM(t+u) Taking "E", pNdt = pMdt+ pMdt pN = 2 pM

  20. Taking "E" again,

  21. Association. Measuring? Due to chance? Are two processes associated? Eg. t.s. and p.p. How strongly? Can one predict one from the other? Some characteristics of dependence: E(XY)  E(X) E(Y) E(Y|X) = g(X) X = g (), Y = h(),  r.v. f (x,y)  f (x) f(y) corr(X,Y)  0

  22. Bivariate point process case. Two types of points (j ,k) Crossintensity. Prob{dN(t)=1|dM(s)=1} =(pMN(t,s)/pM(s))dt Cross-covariance density. cov{dM(s),dN(t)} = qMN(s,t)dsdt no ()

More Related