1. **Problem statement:** Given a random sample $X_1, X_2, \ldots, X_n$ from the probability density function (pdf) $$f(x;\theta) = \theta e^{-\theta x}, \quad x > 0, \theta > 0,$$ find: (a) the maximum likelihood estimator (MLE) of $\theta$, (b) the expectation $\mu = E[X]$, (c) whether the MLE $\hat{\theta}$ is biased or unbiased, (d) the MLE of $\mu$, (e) compute $\hat{\mu}$ given the sample $2.4, 5.3, 4.8, 3.6, 7.4$. 2. **(a) Find the MLE of $\theta$: ** The likelihood function for the sample is $$L(\theta) = \prod_{i=1}^n \theta e^{-\theta x_i} = \theta^n e^{-\theta \sum_{i=1}^n x_i}.$$ The log-likelihood is $$\ell(\theta) = n \ln \theta - \theta \sum_{i=1}^n x_i.$$ Taking derivative and setting to zero: $$\frac{d\ell}{d\theta} = \frac{n}{\theta} - \sum_{i=1}^n x_i = 0 \implies \hat{\theta} = \frac{n}{\sum_{i=1}^n x_i}.$$ 3. **(b) Find the expectation $\mu = E[X]$: ** For the exponential distribution, $$E[X] = \frac{1}{\theta}.$$ 4. **(c) Is $\hat{\theta}$ biased or unbiased?** Since $\hat{\theta} = \frac{n}{\sum X_i}$ and $\sum X_i$ is Gamma distributed, $$E[\hat{\theta}] = E\left[\frac{n}{\sum X_i}\right] \neq \theta,$$ so $\hat{\theta}$ is a biased estimator. 5. **(d) Find the MLE of $\mu = E[X]$: ** Since $\mu = \frac{1}{\theta}$, the MLE of $\mu$ is $$\hat{\mu} = \frac{1}{\hat{\theta}} = \frac{\sum_{i=1}^n x_i}{n} = \bar{X}.$$ 6. **(e) Compute $\hat{\mu}$ for the sample $2.4, 5.3, 4.8, 3.6, 7.4$: ** Calculate the sample mean: $$\bar{X} = \frac{2.4 + 5.3 + 4.8 + 3.6 + 7.4}{5} = \frac{23.5}{5} = 4.7.$$ **Final answers:** - MLE of $\theta$: $\hat{\theta} = \frac{n}{\sum X_i}$ - Expectation: $\mu = \frac{1}{\theta}$ - $\hat{\theta}$ is biased - MLE of $\mu$: $\hat{\mu} = \bar{X}$ - Computed $\hat{\mu}$ for sample: $4.7$