Question

Let $X_1, X_2, \ldots, X_n$ be a random sample from $\text{Exp}(\theta)$ distribution, where $\theta \in (0, \infty)$. If $\bar{X} = \dfrac{1}{n}\sum_{i=1}^n X_i$, then a 95% confidence interval for $\theta$ is
 

• A. $\left(0, \dfrac{\chi^2_{2n, 0.95}}{n\bar{X}}\right)$
• B. $\left(\dfrac{\chi^2_{2n, 0.95}}{n\bar{X}}, \infty\right)$
• C. $\left(0, \dfrac{\chi^2_{2n, 0.05}}{2n\bar{X}}\right)$
• D. $\left(\dfrac{\chi^2_{2n, 0.05}}{2n\bar{X}}, \infty\right)$

Hint:

For exponential samples, $\frac{2n\bar{X}}{\theta}$ follows a $\chi^2$ distribution, making it straightforward to derive confidence intervals for $\theta$.

Answer 1

The Correct Option is C

Step 1: Distribution of $\sum X_i$.
For $X_i \sim \text{Exp}(\theta)$, we have \[ \sum_{i=1}^{n} X_i \sim \text{Gamma}(n, \theta). \] Equivalently, \[ \frac{2n\bar{X}}{\theta} \sim \chi^2_{2n}. \]

Step 2: Confidence interval for $\theta$.
For a 95% confidence level: \[ P\left(\chi^2_{2n, 0.05} \le \frac{2n\bar{X}}{\theta} \le \chi^2_{2n, 0.95}\right) = 0.95. \] Rearranging for $\theta$, \[ P\left(\frac{2n\bar{X}}{\chi^2_{2n, 0.95}} \le \theta \le \frac{2n\bar{X}}{\chi^2_{2n, 0.05}}\right) = 0.95. \]

Step 3: Upper and lower bounds.
The lower bound corresponds to $\frac{\chi^2_{2n, 0.05}}{2n\bar{X}}$ if expressed inversely in terms of rate parameter, giving \[ \boxed{\left(\frac{\chi^2_{2n, 0.05}}{2n\bar{X}}, \infty\right)}. \]

Step 4: Conclusion.
Hence, the correct 95% confidence interval is as in option (D).