The best unbiased estimate of an exponential rate

Question

You have independent waiting times X1, \dots, Xn from an exponential with rate \lambda. The method-of-moments estimate 1/\bar X targets \lambda but is biased upward. Find the minimum-variance unbiased estimator of \lambda as a function of the sufficient total T = \sumi Xi.

Accepted Answer

The sufficient statistic and its law. The total T = \sumi Xi is sufficient (and complete) for \lambda, and it follows a Gamma distribution with shape n and rate \lambda. Any unbiased function of T will be the best unbiased estimator by Lehmann-Scheffé, so we just need to find the right function of T. Fix the bias of the reciprocal. For T \sim 	ext{Gamma}(n, \lambda), a standard expectation is \mathbb{E}\!\left[\frac{1}{T}ight] = \frac{\lambda}{n - 1} \qquad (n \ge 2). So 1/T is biased: its mean is \lambda/(n-1), not \lambda. Multiply by (n - 1) to remove the bias: \hat\lambda = \frac{n - 1…

The best unbiased estimate of an exponential rate

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