Let's address the two parts of the problem step by step.

(a) Prove that $\overline{X}$ is an unbiased estimator of $\theta$:

1. Given Information:

   • The random sample $X_1, X_2, \dots, X_n$ comes from the exponential distribution with PDF: $f(x) = \frac{1}{\theta} e^{-x/\theta}, \quad 0 < x < \infty, \, \theta > 0.$
   • $E(X) = \theta$ is provided.

2. Key Definition for Unbiasedness:

   • An estimator $\overline{X}$ is unbiased for $\theta$ if $E(\overline{X}) = \theta$.

3. Calculate $E(\overline{X})$:

   • $\overline{X} = \frac{1}{n} \sum_{i=1}^n X_i$, where each $X_i$ is independent and identically distributed.
   • Using the linearity of expectation: $E(\overline{X}) = E\left(\frac{1}{n} \sum_{i=1}^n X_i\right) = \frac{1}{n} \sum_{i=1}^n E(X_i).$
   • Since $E(X_i) = \theta$ for all $i$, this becomes: $E(\overline{X}) = \frac{1}{n} \cdot n \cdot \theta = \theta.$

4. Conclusion:

   • Since $E(\overline{X}) = \theta$, $\overline{X}$ is an unbiased estimator of $\theta$.

(b) Prove that $\overline{X}$ is a sufficient statistic for $\theta$ using the Factorization Theorem:

1. Factorization Theorem:

   • A statistic $T(X) = \overline{X}$ is sufficient for $\theta$ if the joint PDF $f(x_1, x_2, \dots, x_n | \theta)$ can be factored as: $f(x_1, x_2, \dots, x_n | \theta) = g(T(X), \theta) h(x_1, x_2, \dots, x_n),$ where $g$ depends on $T(X)$ and $\theta$, and $h$ does not depend on $\theta$.

2. Joint PDF of the Sample:

   • The joint PDF of $X_1, X_2, \dots, X_n$ (since they are independent) is: $f(x_1, x_2, \dots, x_n | \theta) = \prod_{i=1}^n \frac{1}{\theta} e^{-x_i / \theta} = \frac{1}{\theta^n} e^{-\frac{1}{\theta} \sum_{i=1}^n x_i}.$

3. Rewrite the Joint PDF:

   • Let $T(X) = \overline{X} = \frac{1}{n} \sum_{i=1}^n X_i$, so that $\sum_{i=1}^n X_i = n\overline{X}$.
   • Substitute $\sum_{i=1}^n X_i$ into the joint PDF: $f(x_1, x_2, \dots, x_n | \theta) = \frac{1}{\theta^n} e^{-n\overline{X} / \theta}.$
   • This can be factored as: $f(x_1, x_2, \dots, x_n | \theta) = \underbrace{\frac{1}{\theta^n} e^{-n\overline{X} / \theta}}_{g(\overline{X}, \theta)} \cdot \underbrace{1}_{h(x_1, x_2, \dots, x_n)}.$

4. Conclusion:

   • The joint PDF is factored into a function of $\overline{X}$ and $\theta$ (namely $g$), and a function $h$ that does not depend on $\theta$.
   • Therefore, $\overline{X}$ is a sufficient statistic for $\theta$ by the Factorization Theorem.

Do you have further questions or need additional clarifications?

Here are 5 related questions to explore this topic further:

1. What is the variance of $\overline{X}$ as an estimator of $\theta$?
2. How would you use the Rao-Blackwell theorem to improve an estimator for $\theta$?
3. Can you derive the maximum likelihood estimator (MLE) for $\theta$ based on the given distribution?
4. How does the concept of sufficiency relate to Bayesian statistics?
5. How would you test the unbiasedness of an estimator numerically?

Tip: Always verify whether sufficient statistics retain all information about the parameter by comparing likelihoods before and after using the statistic.