Let $X_1, X_2, \cdots, X_n$ for $n \geq 5$ be a random sample from the distribution with probability density function $f(x; \theta) = \begin{cases} e^{-(x-\theta)} & \text{if } x > \theta \\ 0 & \text{otherwise} \end{cases}$ for $\theta > 0$. The confidence coefficient of the confidence interval $\left[ \min\{X_1, \cdots, X_n\} - \frac{\ln 4}{n}, \min\{X_1, \cdots, X_n\} + \frac{\ln 2}{n} \right]$ for $\theta$, is
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.
Important Questions from Random Variables
- A mobile manufacturing company uses two brands of batteries for its mobiles. The life (in years) of batteries of Brand I follows an exponential distribution with the probability density function
$ f(x) = \begin{cases} e^{-x}, & \text{if } x>0, \\ 0, & \text{otherwise,} \end{cases} $
and that of Brand II follows a gamma distribution with the probability density function
$ g(x) = \begin{cases} \frac{x}{4} e^{-x/2}, & \text{if } x>0, \\ 0, & \text{otherwise.} \end{cases} $
The company uses the batteries of Brands I and II in proportion of $20\%$ and $80\%$ respectively, in its mobiles. The probability that a randomly selected mobile has the battery life more that $2$ years is - Consider a discrete random variable $X$ with the probability mass function
$ P(X = 0) = \frac{\theta}{3}, \ P(X = 1) = 1 - \frac{\theta}{2}, \ P(X = 2) = \frac{\theta}{6}, $
where $\theta \in (0,1)$ is an unknown parameter. In a random sample of size $90$ from this distribution, the observed counts for $X = 0, 1$ and $2$ are $20, 60$ and $10$, respectively. Then, the maximum likelihood estimate of $\theta$ is - Let $X$ be a random sample of size $1$ from the probability density function
$ f(x|\theta) = \begin{cases} \frac{3}{\theta^3} (\theta - x)^2, & \text{if } 0<x<\theta, \\ 0, & \text{otherwise.} \end{cases} $
If $ \left(\frac{X}{1-\lambda_1}, \frac{X}{1-\lambda_2}\right) $ is a confidence interval for $\theta$ with confidence coefficient $1 - \alpha$, where $\lambda_i \in (0,1), \ i = 1,2, \ \lambda_1<\lambda_2$, and $\alpha \in (0,1)$, then which of the following statements is true? - Let $X_1, X_2, . . ., X_n$ be a random sample from a continuous distribution with the common probability density function
$ f(x|\theta) = \begin{cases} \frac{2\theta^2}{x^{\theta+1}}, & \text{if } x>2, \\ 0, & \text{otherwise,} \end{cases} $
where $\theta (> 0)$ is an unknown parameter. Suppose $P(Y>\chi^2_{m,\beta}) = \beta$, where $Y \sim \chi^2_m$. For testing $H_0: \theta = 1$ against $H_1 : \theta>1$, a uniformly most powerful test of size $\alpha, \ 0<\alpha<1$, will reject $H_0$ if - Suppose we want to estimate the population mean $\bar{Y}$ of a variable for a finite population of size $85$, with $34$ Statisticians and $51$ Biologists. We consider the following sampling scheme:
A stratified random sample with $2$ strata of Statisticians (Stratum-1) and Biologists (Stratum-2), where $12$ Statisticians and $15$ Biologists are drawn from Stratum-1 and Stratum-2, respectively, using SRSWOR scheme.
Denote $\bar{y}_S, \bar{y}_B$, and $\bar{y}$ as the mean of the variable among the Statistician sample, Biologist sample, and the combined sample, respectively. Which of the following is an unbiased estimator of $\bar{Y}$?
