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: Identify the sampling distribution

For exponential distribution with mean $\theta$, the sum $T = \sum_{i=1}^n X_i$ follows a Gamma distribution with shape parameter $n$ and scale parameter $\theta$.

A key property is that: $$\frac{2T}{\theta} = \frac{2n\bar{X}}{\theta} \sim \chi^2_{2n}$$

Step 2: Set up the confidence interval

For a 95% confidence level, we use the chi-square distribution to establish bounds. Since we're dealing with a scale parameter $\theta$ that must be positive, we construct a one-sided upper confidence interval.

The appropriate form uses: $$P\left(\frac{2n\bar{X}}{\theta} \leq \chi^2_{2n,0.05}\right) = 0.95$$

Step 3: Solve for $\theta$

Rearranging the inequality: $$\frac{2n\bar{X}}{\theta} \leq \chi^2_{2n,0.05}$$

$$\theta \geq \frac{2n\bar{X}}{\chi^2_{2n,0.05}}$$

Since $\theta > 0$, the 95% confidence interval is: $$\left(0, \frac{\chi^2_{2n,0.05}}{2n\bar{X}}\right]$$

This provides an upper confidence bound for the exponential parameter $\theta$ based on the observed sample mean.

Answer: (C) $\left(0, \frac{\chi^2_{2n,0.05}}{2n\bar{X}}\right]$