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Interarrival time

An interarrival time is the duration between consecutive events. See how completed gaps, a censored final wait and a fixed observation window affect a rate estimate.

An interarrival time is the elapsed time between two consecutive events. If TiT_i is the time of event ii, its interarrival time is Xi=Ti−Ti−1X_i=T_i-T_{i-1}. Arrival times are clock readings; interarrival times are durations.

For a quant studying order messages, trades or credit events, this distinction turns timestamps into a sequence of waits. The sequence can reveal changing activity or clustering that a single daily count conceals. An event must first be defined consistently: a quote update and a completed trade are different things to count.

Read the gaps, then read the stopping rule

Consider a synthetic record of qualifying events. Start the clock at an observed event, T0=0T_0=0, and record subsequent events at 0.4, 1.1 and 1.6 hours. Observation ends at the predeclared time 2 hours. The origin event anchors the clock; it is not counted as a new event in (0,2](0,2].
Consecutive eventsCompleted interarrival time
0 to 0.4 hours0.4 hours = 24 minutes
0.4 to 1.1 hours0.7 hours = 42 minutes
1.1 to 1.6 hours0.5 hours = 30 minutes

The sample mean of the three completed gaps is

xˉ=0.4+0.7+0.53=815 hours=32 minutes.\bar x=\frac{0.4+0.7+0.5}{3} =\frac{8}{15}\text{ hours}=32\text{ minutes}.
From the last event to the end of observation, another 0.4 hours pass without an event. This is a right-censored wait: we know X4>0.4X_4>0.4 hours, but do not know its eventual length. Calling it a fourth completed gap would replace an unknown duration with its observed lower bound. Omitting it from a rate calculation would discard exposure during which no event occurred.
Synthetic events at 0, 0.4, 1.1 and 1.6 hours create completed gaps of 0.4, 0.7 and 0.5 hours. Observation ends at 2 hours, after a censored 0.4-hour wait. The completed-gap mean is 32 minutes; its reciprocal is 1.875 per hour. Three subsequent events over the full two-hour window give a homogeneous Poisson rate MLE of 1.5 per hour.
Synthetic events at 0, 0.4, 1.1 and 1.6 hours create completed gaps of 0.4, 0.7 and 0.5 hours. Observation ends at 2 hours, after a censored 0.4-hour wait. The completed-gap mean is 32 minutes; its reciprocal is 1.875 per hour. Three subsequent events over the full two-hour window give a homogeneous Poisson rate MLE of 1.5 per hour.
Original synthetic timeline. Filled dots are events; the end bar marks the observation boundary. The final 24 minutes contribute exposure, although they do not complete another gap.

A probability law needs an assumption

The definition of an interarrival time does not imply a particular distribution or independence. Under a homogeneous Poisson process with constant rate λ>0\lambda>0, the successive gaps are independent and identically distributed with an exponential distribution:
Xi∼Exp⁡(λ),E[Xi]=1λ,P(Xi>x)=e−λx,x≥0.X_i\sim\operatorname{Exp}(\lambda),\qquad \mathbb{E}[X_i]=\frac{1}{\lambda},\qquad \mathbb{P}(X_i>x)=e^{-\lambda x},\quad x\ge0.
As a separate stipulated model, take λ=2\lambda=2 events per hour. The theoretical mean gap is 0.5 hours, or 30 minutes, and the probability that a gap exceeds half an hour is e−1≈0.367879e^{-1}\approx0.367879, about 36.79%. These are model quantities, not estimates obtained from the three synthetic gaps. Rates have inverse-time units; the product λx\lambda x is dimensionless.
For the construction linking independent exponential gaps to event counts, Defining the Poisson Process: Three Equivalent Constructions has a public preview; the full lesson requires membership.

Why the two rate calculations differ

The reciprocal of the completed-gap average is

1xˉ=31.6=1.875 events per hour.\frac{1}{\bar x}=\frac{3}{1.6}=1.875\text{ events per hour}.
That is the exponential rate maximum likelihood estimate for a design that records three complete waits and stops at the third arrival. Our stated design instead continues to a fixed two-hour endpoint.
Under the homogeneous Poisson model, each completed wait contributes a density factor λe−λxi\lambda e^{-\lambda x_i}. The unfinished wait contributes a survival factor e−λ(0.4)e^{-\lambda(0.4)}. Multiplying gives the likelihood, conditional on the origin event,
L(λ)=(∏i=13λe−λxi)e−λ(0.4)=λ3e−2λ.L(\lambda) =\left(\prod_{i=1}^3\lambda e^{-\lambda x_i}\right)e^{-\lambda(0.4)} =\lambda^3e^{-2\lambda}.
All durations here are expressed in hours. The log-likelihood, up to a parameter-independent constant, is 3log⁡λ−2λ3\log\lambda-2\lambda. Its derivative vanishes at
λ^=32=1.5 events per hour.\hat\lambda=\frac{3}{2}=1.5\text{ events per hour}.
The same maximiser follows from the Poisson distribution for the three events counted in the full window. The lower rate reflects the additional event-free exposure. It does not contradict the 32-minute completed-gap average: the two calculations use different observation endpoints. The survival contribution follows the usual likelihood treatment of censored observations.
In real records, also check the beginning of the window. If observation starts between events and the preceding event time is unknown, the first observed delay is a forward waiting time, not a measured full interarrival gap. These have the same exponential law in the homogeneous Poisson case, but need not have the same law more generally. Time-of-day effects, dependence between arrivals and timestamp resolution all matter before treating observed gaps as independent exponential data.

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  • arrival times
  • waiting times
  • point processes
  • censoring
  • poisson arrivals
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