Connect a waiting-time construction, arrival epochs and interval counts before simulating.
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A sensor records the gaps between successive events: 0.2, 0.4, 0.1 and 0.6 minutes. Before choosing a distribution, add those gaps. The arrival times are 0.2, 0.6, 0.7 and 1.3 minutes. Three events occur in the first minute. The fourth gap is not the fourth arrival time.
This is a worked sequence, not measured data or the result of the live simulator. Change only the final gap to 0.2 minute on paper. The last arrival moves to 0.9 minute, so the first-minute count becomes four. Keep the same time scale when comparing the two sequences.
Predict
Suppose gaps are independent and follow one exponential law with rate λ = 2 events per minute. Predict the chance of no arrival in one minute and the mean time of the second arrival. A rate is not the number guaranteed to arrive; some intervals contain no event and others contain many.
Let W₁, W₂, … be the waiting times, each with density λ exp(−λw) for w greater than zero. Density has units of inverse time. Integrating it gives the survival probability P(W₁ > t) = exp(−λt), where λt is dimensionless. Thus the no-arrival prediction is exp(−2), approximately 0.1353.
An arrival epoch is Sₙ = W₁ + ⋯ + Wₙ; the subscript n counts events. For two independent waits at rate 2, convolving their densities gives
fS2(s)=∫0s2e−2u2e−2(s−u)du=4se−2s,s>0.
The integration variable u is the first wait; s − u is the second. Every split between zero and s contributes the same exponential factor. The extra factor s counts the length of that integration interval. The mean second-arrival time is 1 minute, but the chance that the second arrival is still in the future at one minute is 3 exp(−2), approximately 0.4060. A mean is not a deadline.
Try the nearby case λ = 4 per minute. Both waits are halved in scale: the mean second-arrival time is 1/2 minute and its density is 16s exp(−4s). Keep minutes and inverse minutes attached to the appropriate quantities.
Understand
For an integer n at least one, convolution gives the general arrival-epoch density
fSn(s)=(n−1)!λnsn−1e−λs,s>0.
Here (n − 1)! is the product of the positive integers up to n − 1, with 0! = 1. The n = 1 case is the exponential density. To prove the next case, multiply this density at u by the next independent wait density λ exp(−λ(s − u)), then integrate u from zero to s. The integral of uⁿ⁻¹ is sⁿ/n. This produces the stated formula with n replaced by n + 1. It is a derivation, not a fit to a graph.
Does this construction allow infinitely many arrivals before a finite horizon H? The density provides a bound:
P(Sn≤H)≤n!(λH)n⟶0.
We bounded the exponential factor by one and integrated the remaining power. The events “the nth arrival occurs by H” decrease as n grows, so the probability that all of them occur is zero. Almost surely, every fixed finite horizon contains finitely many arrivals. This still does not put a deterministic upper limit on the count: a simulator needs its own resource cap and an honest overflow status.
To derive a count law, let N(h) count epochs in (0,h]. Exactly k arrivals means ordered times s₁ through sₖ inside the window and no next arrival before h. For k greater than zero, their joint density including that final survival factor is λᵏ exp(−λh). The ordered region has volume hᵏ/k!: its k! possible orderings partition the h-by-⋯-by-h cube apart from ties of zero volume. Therefore
P(N(h)=k)=e−λhk!(λh)k,k=0,1,2,….
For k = 0, the same answer follows directly from the first-wait survival probability. This is a Poisson law with mean λh. The construction assumes independent waits with one fixed rate. Exponential marginal waits alone are insufficient: if W₁ = W₂ = Z for an Exp(1) variable Z, then S₂ = 2Z, with CDF 1 − exp(−t/2), which differs from the independent-wait CDF 1 − exp(−t)(1 + t).
Deterministic boundary times have zero arrival probability in this ideal continuous model. Finite software still needs a declared boundary rule; the investigation uses windows (0,h] and (h,2h].
Apply
Write a two-column audit: which conclusions came from a supplied sampling assumption, and which were proved from it? Include independent waits, rate units, the second-epoch density, nonexplosion, and the count mass. Then change the rate after the first minute. Explain why that new process needs a piecewise calculation instead of silently retaining the original exponential survival law.
Continue to independent increments and thinning to prove the joint law of counts in several windows. A one-window Poisson formula alone does not establish those windows are independent.
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Practice and reasoning
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An original arrival model has iid waiting times W with rate λ=3 per minute. Derive the density, survival function and mean of W. Find P(W>1/2 minute) and the same probability when time is measured in seconds. Explain why the rate and the mean count over half a minute have different units.
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Let W₁,W₂ be independent Exp(2 per minute) waits and S₂=W₁+W₂. Derive the density of S₂ by convolution, then its survival function, mean and variance. Compute P(S₂>1 minute).
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For iid Exp(λ) waits with λ>0, prove by induction that Sₙ has density λⁿsⁿ⁻¹e^(−λs)/(n−1)! for s>0 and integer n≥1. Use this density to prove that infinitely many arrivals cannot occur before a fixed finite H. You may use that aⁿ/n!→0 for fixed a.
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Define N(h) as the number of arrival epochs in (0,h], h>0, from iid Exp(λ) waits. Derive P(N(h)=k) for all integers k≥0 using the first k arrival times and the absence of a further arrival before h. Explain the ordered-region volume hᵏ/k!.
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Compare independent Exp(1) waits W₁,W₂ with a changed model W₁=W₂=Z, where Z is Exp(1). Both waits have the same individual law. Derive P(S₂≤t) for both models, t≥0. Explain which construction proof fails after the change.
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A different stipulated process has independent Poisson counts with rate 1 per minute in (0,1] and rate 3 per minute after time 1. Derive the first-arrival survival probability at time t≥0. Compare P(T>3/2|T>1) with P(T>1/2). Explain why a single constant-rate exponential law is inappropriate.
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