MLE for an exponential rate, and its hidden bias

Question

You observe X1, \dots, Xn i.i.d. from an exponential distribution with rate \lambda, so the density is f(x) = \lambda e^{-\lambda x} for x \ge 0 (mean 1/\lambda). Derive the MLE of \lambda. Is it unbiased?

Accepted Answer

Log-likelihood. \ell(\lambda) = \sum{i=1}^n \big(\log \lambda - \lambda Xi\big) = n\log\lambda - \lambda \sum{i=1}^n Xi. Maximize. Differentiate and set to zero: \frac{d\ell}{d\lambda} = \frac{n}{\lambda} - \sum{i=1}^n Xi = 0 \implies \hat\lambda = \frac{n}{\sumi Xi} = \frac{1}{\bar X}. The second derivative -n/\lambda^2 < 0 confirms a maximum. The MLE of the rate is one over the sample mean, which is intuitive since the mean is 1/\lambda. Bias check. The sum S = \sumi Xi follows a Gamma distribution with shape n and rate \lambda, and a standard result gives \mathbb{E}[1/S] = \frac{\lambda}{n…

MLE for an exponential rate, and its hidden bias

Your answer

More Statistics questions