Given the total, exponential proportions forget the rate

Question

You have independent waiting times X1, \dots, Xn from an exponential with rate \lambda, and you condition on their total T = \sumi Xi. Explain why, once the total is fixed, the proportions Xi / T carry no information about \lambda, and connect this to the total being a sufficient statistic.

Accepted Answer

The idea in words. A sufficient statistic is one that, once known, leaves behind randomness with no trace of the parameter. For the exponential, the claim is that the total waiting time is sufficient: fix it, and the way it splits among the individual gaps is a rate-free pattern. Why the proportions forget \lambda. Rescaling helps. If every waiting time is measured by a faster clock (a larger \lambda), all the gaps shrink by the same factor, and so does their total. But the fractions X1/T, X2/T, \dots are unchanged by a common rescaling. Concretely, the vector of proportions (X1/T, \dots, Xn/…

Given the total, exponential proportions forget the rate

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