Point processes

evergreen#foundations#point-processes#timing

Up: 10_Foundations-MOC

Bar: state the Poisson process three ways, and explain Hawkes self-excitation + its stability condition from memory.

What a point process is

A model for random events in time (clicks, requests, purchases): the random object is the set of timestamps. Summarized by the counting process N(t)N(t) = number of events up to tt, or by the conditional intensity λ(t)\lambda(t) = instantaneous event rate given the past.

Poisson process (the baseline)

A homogeneous Poisson process with rate λ\lambda is equivalently:

  1. Counts: events in any interval of length Δ\Delta are Poisson(λΔ)\text{Poisson}(\lambda\Delta);
  2. Interarrivals: gaps are i.i.d. Exponential(λ)\text{Exponential}(\lambda), mean 1/λ1/\lambdamemoryless;
  3. Independent increments: disjoint intervals are independent.
P(N(Δ)=k)=eλΔ(λΔ)kk!P(N(\Delta)=k)=e^{-\lambda\Delta}\frac{(\lambda\Delta)^k}{k!}

Worked: λ=2\lambda=2 events/min. P(0 in 1 min)=e20.135P(\text{0 in 1 min})=e^{-2}\approx0.135; expected gap =0.5=0.5 min; expected events in 5 min =10=10.

Inhomogeneous Poisson: rate varies, λ(t)\lambda(t). Counts in [a,b][a,b] are Poisson ⁣(abλ(t)dt)\text{Poisson}\!\big(\int_a^b\lambda(t)\,dt\big). This captures time-of-day / seasonality in traffic — the deterministic daily curve.

Hawkes process (self-excitation = burstiness)

Real clickstreams are clustered: one event makes more events imminent (a session, a viral spike). Hawkes makes the past raise the intensity:

λ(t)=μ+ti<tϕ(tti),ϕ(s)=αeβs (β>α>0)\lambda(t)=\mu+\sum_{t_i<t}\phi(t-t_i),\qquad \phi(s)=\alpha\,e^{-\beta s}\ (\beta>\alpha>0)

Interpretation (branching/Galton–Watson view): each event independently triggers children; total events per "immigrant" =1/(1n)=1/(1-n). That single number controls how bursty the stream looks.

Why this matters here

By-hand exercise (meets the bar)

  1. For λ=3\lambda=3/min, compute P(N=2P(N=2 in 1 min)). (Answer: e332/2!=9e3/20.224e^{-3}3^2/2!=9e^{-3}/2\approx0.224.)
  2. A Hawkes process has α=0.6,β=1.0\alpha=0.6,\beta=1.0. Is it stable? What's the expected cluster size per immigrant? (Answer: n=0.6<1n=0.6<1, stable; cluster size 1/(10.6)=2.51/(1-0.6)=2.5.)