From mean gap to MLE of the arrival rate

Question

Order arrivals at a venue are modeled so that the gap between consecutive orders is exponential with unknown mean $\theta$ , density $f(x \mid \theta) = \frac{1}{\theta} e^{-x/\theta}$ . From a long log the sample mean gap is $\bar x = 8$ seconds. The arrival rate is $\lambda = 1/\theta$ orders per second.

Using maximum likelihood, what is the estimated arrival rate $\lambda$ ?

Accepted Answer

The MLE of the exponential mean is the sample mean, \hat	heta = \bar x = 8 seconds. The rate is a one-to-one function of the mean, \lambda = 1/	heta, so by the invariance property of maximum likelihood the MLE of the rate is that same function of \hat	heta: \hat\lambda = \frac{1}{\hat	heta} = \frac{1}{\bar x} = \frac{1}{8} = 0.125 	ext{ orders per second}. You would reach the identical answer by writing the likelihood directly in terms of \lambda and maximizing; invariance saves that work. Takeaway. Maximum likelihood is invariant under reparameterization: estimate the mean, then just in…

From mean gap to MLE of the arrival rate

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