Confidence Interval Analysis for Exponential Distribution

The question asks for the confidence coefficient of the interval $\left[ \min\{X_1, \cdots, X_n\} - \frac{\ln 4}{n}, \min\{X_1, \cdots, X_n\} + \frac{\ln 2}{n} \right]$ for the parameter $\theta$. The random sample $X_1, \dots, X_n$ is drawn from the distribution with probability density function (PDF) $f(x; \theta) = e^{-(x-\theta)}$ for $x > \theta$. The confidence coefficient is defined as the probability that the true parameter $\theta$ lies within the constructed interval, i.e., $P(\theta \in \text{CI})$.

Distribution of Minimum Order Statistic

Let $X_{(1)} = \min\{X_1, \cdots, X_n\}$. The given distribution implies $X_i > \theta$ for all $i$. Define $Y_i = X_i - \theta$. Each $Y_i$ follows a standard exponential distribution with PDF $f_Y(y) = e^{-y}$ for $y > 0$. The minimum of these transformed variables, $Y_{(1)} = \min\{Y_1, \cdots, Y_n\}$, follows an exponential distribution with rate parameter $n$. Its PDF is $f_{Y_{(1)}}(y) = n e^{-ny}$ for $y > 0$, and its cumulative distribution function (CDF) is $F_{Y_{(1)}}(y) = 1 - e^{-ny}$ for $y > 0$. Since $X_{(1)} = Y_{(1)} + \theta$, it follows that $Y_{(1)} = X_{(1)} - \theta$. As $X_{(1)} > \theta$, we know $Y_{(1)} > 0$.

Calculating Confidence Coefficient

The confidence interval is $\text{CI} = \left[ X_{(1)} - \frac{\ln 4}{n}, X_{(1)} + \frac{\ln 2}{n} \right]$. The confidence coefficient is the probability:

$ P\left( X_{(1)} - \frac{\ln 4}{n} \le \theta \le X_{(1)} + \frac{\ln 2}{n} \right) $

To evaluate this, we rearrange the inequalities to isolate $X_{(1)}$:

$ \theta - \frac{\ln 2}{n} \le X_{(1)} \le \theta + \frac{\ln 4}{n} $

Substitute $X_{(1)} = Y_{(1)} + \theta$ into these inequalities:

$ \theta - \frac{\ln 2}{n} \le Y_{(1)} + \theta \le \theta + \frac{\ln 4}{n} $

Subtracting $\theta$ from all parts simplifies the expression:

$ -\frac{\ln 2}{n} \le Y_{(1)} \le \frac{\ln 4}{n} $

Given that $Y_{(1)} > 0$, the condition becomes:

$ 0 < Y_{(1)} \le \frac{\ln 4}{n} $

Now, we compute the probability using the CDF of $Y_{(1)}$:

$ P\left( 0 < Y_{(1)} \le \frac{\ln 4}{n} \right) = F_{Y_{(1)}}\left(\frac{\ln 4}{n}\right) - F_{Y_{(1)}}(0) $

Since $Y_{(1)}$ is strictly positive, $F_{Y_{(1)}}(0) = P(Y_{(1)} \le 0) = 0$. Therefore, the probability is:

$ P\left( Y_{(1)} \le \frac{\ln 4}{n} \right) = 1 - e^{-n \left(\frac{\ln 4}{n}\right)} $

Simplifying the exponent:

$ = 1 - e^{-\ln 4} = 1 - e^{\ln(4^{-1})} = 1 - 4^{-1} = 1 - \frac{1}{4} = \frac{3}{4} $

Thus, the confidence coefficient is $\frac{3}{4}$ or 0.75.