Discussion
Let \(X_{1}, X_{2}, \ldots, X_{n}\) be independent and identically distributed random variables with probability density function given by
\[ f_{X}(x; \theta) = \begin{cases} \theta e^{-\theta(x-1)}, & x \ge 1, \\ 0 & \text{otherwise.} \end{cases} \]
Also, let \(\overline{X} = \frac{1}{n}\sum_{i=1}^{n} X_{i}\). Then the maximum likelihood estimator of \(\theta\) is
Answer: C
The maximum likelihood estimator is \(1/(\overline{X} - 1)\).
The density is an exponential shifted to start at 1. For a sample with every \(x_{i} \ge 1\) the log-likelihood is
\[ \begin{aligned} \ell(\theta) &= n\ln\theta - \theta\sum_{i=1}^{n}(x_{i} - 1) \\ \ell'(\theta) &= \frac{n}{\theta} - \sum_{i=1}^{n}(x_{i} - 1) = 0 \\ \hat{\theta} &= \frac{n}{\sum x_{i} - n} = \frac{1}{\overline{X} - 1} \end{aligned} \]
\(\ell''(\theta) = -n/\theta^{2} \lt 0\), so this is a maximum. \(1/\overline{X}\) is the estimator for the unshifted exponential, which ignores the shift by 1, and \((1/\overline{X}) - 1\) applies the shift in the wrong place. \(\overline{X}\) estimates a mean, not a rate.
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