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Question

If the occurrence of events follows Poisson Process with mean rate λ, then inter-occurence time of events will follow:

This question was previously asked in
SSC CGL 2018 (Tier 2) Statistics Previous Year Paper (22-feb-2018)
The correct answer is

Exponential distribution

In a Poisson process with rate \(\lambda\), the probability of no event in time \(t\) is \(P(T>t)=e^{-\lambda t}\), so the inter-arrival time \(T\) has density \(f(t)=\lambda e^{-\lambda t}\) for \(t\ge 0\) — the Exponential distribution with parameter \(\lambda\).

The Poisson distribution counts the number of events in a fixed interval (not the gap); the Gamma distribution gives the time until the \(k\)-th event (general case), and Geometric is a discrete analogue. For a single inter-occurrence time the answer is Exponential distribution.

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