Try
Consider two ways to produce two hourly counts. In the first, generate a new Poisson(2) count for each hour independently. In the second, generate one Poisson(2) count and copy it into the second hour. Both columns separately have exactly the same theoretical histogram. Pairing the columns reveals a different story.
| Question about a pair | Independent windows | Copied count |
|---|
| Mean of either count | 2 | 2 |
| Chance both are zero | exp(−4) | exp(−2) |
| Covariance of the pair | 0 | 2 |
| Variance of their sum | 4 | 8 |
Cover the last three rows and try to reconstruct them. For copied counts X = Y = Z, use X + Y = 2Z. Multiplying a variable by two multiplies its variance by four, because variance measures squared deviations.
Predict
Now return to independent exponential waits. Predict whether counts in disjoint windows are independent, and what evidence would prove it. The answer needs a joint probability for every collection of counts, not two similar histograms or one zero covariance.
Return to joint independence and covariance or the waiting-time construction as needed. No prerequisite completion is required.
Experiment
At rate λ = 3 per hour, split two hours into a half-hour window and a one-and-a-half-hour window. Their mean counts are 3/2 and 9/2. Under the construction, the probability of one arrival in the first and two in the second is the product of their two Poisson masses.
If someone tells you that the total was three, the question changes. Divide that joint probability by the Poisson(6) mass at three. The result is
The first interval occupies one quarter of the combined duration. More generally, given total n, its count is Binomial(n,1/4). Before conditioning the counts are independent; after conditioning they must sum to n. Knowing one then determines the other. State the conditioning event beside every claim about independence.
Understand
Partition (0,H] into successive intervals with positive durations h₁,…,hₘ. Let kⱼ be the proposed count in interval j and K = k₁ + ⋯ + kₘ. The arrival construction gives density λᴷ exp(−λH) for K ordered arrival epochs together with no next arrival before H.
Within interval j, the region for kⱼ ordered epochs has volume hⱼᵏʲ/kⱼ!. The interval order is already fixed, so the volumes multiply. This gives
Each factor is a normalized Poisson mass. The factorization for every possible count vector proves independent increments; dependence on hⱼ alone proves that the law is unchanged when a window is shifted in time. Disjoint windows with gaps are handled by putting the gaps into the partition and summing their counts out. This proof uses the independent, constant-rate waiting-time construction. It does not infer the construction from observed marginal counts.
Attach an independent label to each arrival
Suppose each arrival is labeled D with probability p and C otherwise. The choices must be independent across arrivals and independent of the arrival process. This is called independent thinning: retaining only D arrivals produces a lower-rate stream.
Let μ = λh be the total mean count in one window. To obtain d D labels and c C labels, first obtain d + c arrivals and then choose which d receive D. The joint probability is
Thus the two label counts are independent Poisson variables with means μp and μ(1 − p). Applying the same argument to multiple windows establishes independent increments for the thinned stream. At p = 0 or 1 one stream is empty with certainty; handle those cases directly instead of relying on ambiguous power notation.
For a nearby check, use a three-hour window, λ = 2 per hour and p = 1/3. The mean D count is two. The probability of no D arrival is exp(−2), not the probability exp(−6) of no arrival of any kind.
Same individual marking probability, different process
Draw one Bernoulli(p) variable B for the whole window and give every arrival its value. Then D = BN and C = (1 − B)N. The D mean remains pμ, but its variance becomes pμ + p(1 − p)μ². The extra term comes from the shared decision. Moreover DC is always zero, so their covariance is −p(1 − p)μ². This violates the independent-mark assumption even though each arrival individually has D probability p.
This is why copying one window, retaining a hidden mode or sharing a mark can change a joint law without changing selected marginal summaries. The archived MIT Poisson-process lecture supplies the source context for distinguishing count laws from process assumptions. The constructions and calculations above are original cases.
Apply
Imagine each arrival enters an independent logical route that eventually assigns D with probability 2/5. If arrivals occur at rate 3 per hour, arrivals during two hours eventually labeled D have a Poisson(12/5) count under the stated independence assumptions.
This does not tell you how many cases finish by the two-hour deadline. The route has logical transitions, with no assigned service durations, waiting-line policy or shared capacity. Write an additional assumption you would need for a completion-by-deadline calculation. Do not turn route steps into hours just because both appear in the same investigation.
Continue
In finite absorption and tail certificates, derive an eventual D probability from a route instead of taking it as given. For a different dependence example, return to observation protocols and random reveal.
