Discussion
Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample of size \(n\ (n \geq 1)\) from a distribution with the probability density function
\[ \begin{aligned} f(x) &= \frac{1}{2\beta^{3}}\, e^{3x - \frac{e^{x}}{\beta}}, \\ x &\in \mathbb{R}, \end{aligned} \]
where \(\beta \gt 0\) is an unknown parameter. Which one of the following is a maximum likelihood estimator of \(\beta\)?
Answer: C
The maximum likelihood estimator is \(\frac{1}{3n}\sum_{i=1}^{n} e^{X_{i}}\).
The log-likelihood is
\[ \begin{aligned} \ell(\beta) &= -n\log 2 - 3n\log\beta \\ &\quad + 3\sum x_{i} - \frac{1}{\beta}\sum e^{x_{i}} \end{aligned} \]
\[ \begin{aligned} \ell'(\beta) &= -\frac{3n}{\beta} + \frac{\sum e^{x_{i}}}{\beta^{2}} = 0 \\ \Rightarrow \hat{\beta} &= \frac{\sum e^{x_{i}}}{3n} \end{aligned} \]
\(\ell'\) is positive below this value and negative above it, so it is the maximum. (Substituting \(U = e^{X}\) shows \(U\) is gamma with shape 3 and scale \(\beta\), mean \(3\beta\), which explains the 3.)
\(\frac{3}{n}\sum e^{X_{i}}\) multiplies by 3 instead of dividing. \(\frac{1}{n}\sum X_{i}\) and \(\frac{1}{n}\sum X_{i}^{3}\) ignore the fact that the data enter the likelihood only through \(\sum e^{x_{i}}\).
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