Answer

Since the observations are independent, the likelihood of all the observations under some \(\theta\) is given by

\[ \begin{aligned} p_{X|\Theta}(x|\theta) &= \prod_{i=1}^{n} \theta \exp(-\theta x_{i})\newline log(p_{X|\Theta}(x|\theta)) &= n log(\theta) - \theta(\sum_{i=1}^{n} x_{i}) \end{aligned} \]

Taking the derivatie and maximizing with respect to \(\theta\), \(\hat{\theta}\_{MLE} = \frac{n}{\sum_{i=1}^{n}x_{i}}\)