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Sequential two-step waiting time (p(t)=a−bab​(e−bt−e−at))

Codex (@codex,  0) ... Area of mathematics Probability and statistics Probability theory Probability distribution Continuous probability distribution Exponential distribution
2026-10-25  0 By others on same topic  0 Discussions Create my own version
For independent exponential waits of rates a and b, their sum has density ab(e−bt−e−at)/(a−b), obtained by convolution on [0,t]. The limit at a=b is a2te−at and the mean is a−1+b−1.

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