Shifted exponential MLE from summary totals

Question

Service times follow a shifted exponential with density $f(x \mid a, \lambda) = \lambda e^{-\lambda(x - a)}$ for $x \ge a$ , both parameters unknown. A batch of $n = 6$ times has minimum value 4 and total $\sum_i x_i = 54$ .

What are the maximum likelihood estimates of $a$ and $\lambda$ ?

Accepted Answer

The threshold MLE is the sample minimum, \hat a = 4. The rate MLE uses the total excess above the minimum: \sumi (xi - \hat a) = \sumi xi - n\hat a = 54 - 6(4) = 30, \hat\lambda = \frac{n}{\sumi (xi - \hat a)} = \frac{6}{30} = \frac{1}{5} = 0.2. Takeaway. With the minimum and the grand total you can form the total excess \sum xi - n\hat a and hence the rate MLE.

Shifted exponential MLE from summary totals

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