Fitting an exponential, the MLE for a failure rate

Question

You log the gaps between incoming orders on a quiet venue and model them as $n$ independent draws from an exponential distribution with rate $\lambda$ , whose density is $f(x) = \lambda e^{-\lambda x}$ for $x > 0$ .

Find the maximum likelihood estimate of $\lambda$ , and say whether it is unbiased.

Accepted Answer

Build and maximize the likelihood. The joint density of the sample is L(\lambda) = \prod{i=1}^n \lambda e^{-\lambda xi} = \lambda^n e^{-\lambda \sumi xi}. Taking logs gives \ell(\lambda) = n\ln\lambda - \lambda\sumi xi. Differentiate and set to zero: \ell'(\lambda) = \frac{n}{\lambda} - \sumi xi = 0 \implies \hat\lambda = \frac{n}{\sumi xi} = \frac{1}{\bar x}. The second derivative -n/\lambda^2 < 0 confirms a maximum. So the best-fit rate is just one over the average gap, completely intuitive. Is it unbiased? No. The mean 	heta = 1/\lambda is estimated cleanly by \bar x (which is unbiased), …

Fitting an exponential, the MLE for a failure rate

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