Shifted exponential mean by invariance

Question

Latencies follow a shifted exponential with density $f(x \mid a, \lambda) = \lambda e^{-\lambda(x - a)}$ for $x \ge a$ , both parameters unknown. The mean of this model is $E[X] = a + 1/\lambda$ . Five latencies are 3, 6, 4, 10, 7.

What is the maximum likelihood estimate of the mean latency $E[X]$ ?

Accepted Answer

The threshold MLE is \hat a = \mini xi = 3. The rate MLE is \hat\lambda = \dfrac{n}{\sumi (xi - \hat a)}, so its reciprocal is the mean excess: \frac{1}{\hat\lambda} = \frac{1}{n}\sumi (xi - \hat a) = \bar x - \hat a. By invariance, the MLE of the mean is \widehat{E[X]} = \hat a + \frac{1}{\hat\lambda} = \hat a + (\bar x - \hat a) = \bar x. The threshold cancels, leaving just the sample mean. With data 3, 6, 4, 10, 7 summing to 30 over n = 5, \widehat{E[X]} = \bar x = \frac{30}{5} = 6 	ext{ ms}. Takeaway. Even in a shifted exponential the MLE of the overall mean collapses to the plain sample…

Shifted exponential mean by invariance

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