When the clock and the arrivals sample different hours
A reader-first guide to ordinary time averages, intensity-weighted arrival averages and the limit of a familiar Poisson shortcut.
Choose the question you need to answer
Define the two averages
Work the four-block example
Test the assumption
Use evidence and limits
Question, check and continue
1. Two averages can answer two different questions
Imagine a service desk observed for one hour. A clock asks, “What fraction of the hour was the system crowded?” A customer asks, “What fraction of arrivals reached it while it was crowded?” Those sound like the same question, but a time-varying arrival rate gives the two samples different weights.
A time average gives every minute equal weight. An arrival average gives a minute more weight when more arrivals are expected in that minute. If demand is concentrated in the crowded part of the hour, the customer-facing fraction can be much larger than the clock-facing fraction even when both calculations are correct.
This distinction fills the gap between the stationary PASTA lesson and the time-varying backlog lesson. Here the central question is the sampling measure itself.
2. Define the observation before taking an average
Let g(t) describe the quantity visible at clock time t. It might be a crowded/not-crowded indicator, an externally scheduled service condition, or a modelled transient mean queue length. Over a window from 0 to T, the time average is (1/T) times the integral of g(t).
Let arrivals follow a nonhomogeneous Poisson process with deterministic intensity λ(t). The expected number of arrivals in the window is Λ(T)=∫λ(t)dt. Conditional on an arrival occurring in the window, its time has density λ(t)/Λ(T). The corresponding arrival-time average of g is therefore ∫g(t)λ(t)dt divided by Λ(T).
The formulas differ only in their weights: dt/T for the clock and λ(t)dt/Λ(T) for the arrival. They coincide when λ(t) is constant, or in special cases where g happens to be uncorrelated with the changing weights. Neither equality should be assumed merely because the process is Poisson.
3. Build a four-block model that can be checked by hand
Divide one hour into four equal 15-minute blocks. Suppose the desk is classed as crowded during blocks 2 and 3, and not crowded during blocks 1 and 4. The clock therefore sees crowding for two of four equal blocks: the time-average crowded fraction is 2/4=0.50.
Now use hourly arrival intensities 2, 6, 10 and 2 across the four blocks. Multiplying each rate by 0.25 hour gives expected arrivals 0.5, 1.5, 2.5 and 0.5. The expected total is 5 arrivals. Of those, 1.5+2.5=4 are expected in crowded blocks.
The arrival-weighted crowded fraction is therefore 4/5=0.80. The desk is crowded for half the time, yet an arrival is expected to land in a crowded block four times out of five. There is no contradiction: the two questions sample the hour differently.
4. Put the complete ledger in one table
Writing every weight beside the state prevents a common mistake: averaging the four intensities first and then forgetting when each intensity applied. The order of the blocks matters because g(t) and λ(t) are multiplied before they are added.
| Block | Duration (h) | λ per hour | Expected arrivals | Crowded indicator | Expected crowded arrivals |
|---|---|---|---|---|---|
| 1 | 0.25 | 2 | 0.5 | 0 | 0 |
| 2 | 0.25 | 6 | 1.5 | 1 | 1.5 |
| 3 | 0.25 | 10 | 2.5 | 1 | 2.5 |
| 4 | 0.25 | 2 | 0.5 | 0 | 0 |
| Total | 1.00 | — | 5.0 | Time average 0.50 | 4.0 |
The correct sequence is: define the window, divide it where either the state or intensity changes, compute each block’s exposure, and only then form the required denominator. A single daily average rate cannot reconstruct this ledger.
5. Derive the arrival-time density rather than memorising it
For a nonhomogeneous Poisson process, a small interval [t,t+dt) has approximately λ(t)dt arrivals in expectation. Across the whole window, the expected count is Λ(T). If one arrival from the window is selected, the chance that it came from that small interval is therefore proportional to λ(t)dt.
Normalising those proportional weights gives f(t)=λ(t)/Λ(T). Integrating f across the window gives one, so it is a genuine probability density. This is the continuous version of dividing each block’s expected arrivals by the expected total in the worked ledger.
This derivation is also a useful check on units. λ has units of arrivals per unit time; Λ is an expected number of arrivals; λ/Λ has units of inverse time, as a time density should.
6. Do not turn this into a blanket nonstationary PASTA claim
The stationary PASTA result has a precise setting. In a time-varying system, the phrase “arrivals see time averages” becomes ambiguous because there is more than one sensible time average: an ordinary clock average, a phase-specific average, a cycle average and an intensity-weighted average.
The calculation above is exact for the distribution of nonhomogeneous Poisson arrival times and for a deterministic or suitably predictable function g(t). If g(t) is the queue length generated by those same arrivals, arrival epochs and system state are linked through the queue’s dynamics. A transient queue generally requires its own forward equations, simulation or replicated observations.
Use careful language: an arrival sample weights periods by intensity. Do not claim that every nonstationary Poisson queue automatically inherits the stationary distributional equality from PASTA.
7. Reverse the schedule and watch the customer view flip
Keep the same crowded blocks, 2 and 3, and keep the same total intensity across the hour. Change only the rates to 10, 2, 2 and 6 per hour. Expected arrivals are now 2.5, 0.5, 0.5 and 1.5, still totalling 5.
Only 0.5+0.5=1 expected arrival falls in a crowded block, so the arrival-weighted crowded fraction becomes 1/5=0.20. The clock average remains 0.50 because the desk is crowded for the same two blocks.
This changed-context check is the core transfer test. Equal total arrivals and equal crowded minutes do not determine the arrival experience. Their alignment in time does.
8. Compare the two schedules explicitly
| Measure | Schedule A: 2,6,10,2 | Schedule B: 10,2,2,6 |
|---|---|---|
| Expected arrivals | 5 | 5 |
| Crowded time fraction | 0.50 | 0.50 |
| Expected arrivals in crowded blocks | 4 | 1 |
| Arrival-weighted crowded fraction | 0.80 | 0.20 |
| Difference from time fraction | +0.30 | −0.30 |
A report that shows only five expected arrivals per hour and fifty per cent crowded time would hide the entire mechanism. The paired schedule is the evidence.
In a real study, replace the binary flag with a measured quantity such as modelled wait, queue length or probability of delay. The same weighting principle applies, but the evidence must still show how that quantity was obtained.
9. Estimate both averages from data without mixing denominators
For a clock estimate, sample the system at times chosen independently of arrivals, or integrate a high-frequency state record over the observation window. Divide accumulated state exposure by observed time.
For an arrival estimate, record the state immediately before each arrival when that is the defined observation. Sum the observed values and divide by the number of eligible arrivals. If the state is queue length, “immediately before” matters because counting the new customer would shift every observation by one.
Use several comparable days or simulation replications. Report the intensity pattern, the observation window, missing intervals and whether arrivals were filtered. Confidence intervals must respect time dependence; treating every minute or arrival as independent can understate uncertainty.
10. Limits, edge cases and failure modes
If Λ(T)=0, no arrival average exists for that window. If intensity is nearly zero in a period, arrival data reveal little about the state there even though a clock average can describe it well.
A very high short spike can dominate the arrival average while occupying little time. Conversely, a long quiet failure can dominate the time average while affecting few arrivals. Neither measure is universally more truthful; each belongs to a different decision.
Scheduled appointments, balking, retrials, batch arrivals and state-dependent arrivals break parts of the simple model. For batches, decide whether the sampling unit is an event or a job. For retries, decide whether repeat attempts or unique customers define the population.
11. Questions that expose mistaken shortcuts
Does a Poisson label guarantee equality with the ordinary time average? No. A nonhomogeneous intensity selects clock times unevenly.
Can I average the block rates first? Only if g is constant or the required weighted product is preserved. Averaging λ before multiplying by g generally destroys the alignment.
Should staffing use the time average or arrival average? Usually neither alone. Staffing needs the time profile of arrivals and service, while customer-experience reporting may use an arrival-based measure. Show both when they answer materially different questions.
Is an observed difference proof that arrivals caused crowding? No. It may reflect shared time-of-day structure. Causal claims need a design that separates the mechanisms.
12. Build a defensible reading and evidence route
Start with the Queueing Theory root, then compare stationary Poisson arrival sampling, time-varying capacity and backlog, and event rates versus job counts. Return to the Gold Standard of Education and the How X Works hub for the wider transfer method.
Authoritative technical reading: the Caltech counting-process notes state that, conditional on the number of arrivals, nonhomogeneous Poisson arrival times have density λ(t)/∫λ; the University of Minnesota Poisson-process notes define the intensity and mean-value function.
Practical evidence check: show the full time grid, rate, duration, state definition and both denominators. Recalculate after shifting the same intensity totals to different blocks. If the conclusion changes, timing—not the daily total—is doing the explanatory work.