Let $X_1, X_2, \ldots, X_n$ be independent and identically distributed random variables with probability density function $$f(x) = \begin{cases} e^{-(x-\theta)}, & \text{if } x > \theta, \\ 0, & \text{otherwise.} \end{cases}$$ If $(X_{(1)} - \frac{1}{n} \log_e 10, X_{(1)})$ is a $100\beta\%$ confidence interval of $\theta$ where $X_{(1)} = \min\{X_i : 1 \le i \le n\}$, then the value of $\beta$ is
Understanding the Distribution and Minimum
The random variables $X_i$ follow a shifted exponential distribution with parameter $\theta$. The probability density function (PDF) is given by $f(x) = e^{-(x-\theta)}$ for $x > \theta$.
Let $Y_i = X_i - \theta$. Then $Y_i$ follows a standard exponential distribution with rate $\lambda=1$, i.e., $Y_i \sim \text{Exp}(1)$. Its PDF is $f_Y(y) = e^{-y}$ for $y > 0$.
The minimum of the sample, $X_{(1)} = \min\{X_i\}$, relates to the minimum of $Y_i$, denoted $Y_{(1)} = \min\{Y_i\}$, as $X_{(1)} = Y_{(1)} + \theta$. Therefore, $X_{(1)} - \theta = Y_{(1)}$.
For $n$ independent exponential random variables $Y_i \sim \text{Exp}(1)$, their minimum $Y_{(1)}$ follows an exponential distribution with rate $n$, i.e., $Y_{(1)} \sim \text{Exp}(n)$.
Thus, the transformation $X_{(1)} - \theta$ follows an exponential distribution with rate $n$. The cumulative distribution function (CDF) is:
$ P(X_{(1)} - \theta \le x) = 1 - e^{-nx}, \quad \text{for } x > 0 $
Confidence Interval Derivation
The given confidence interval for $\theta$ is $(L, U) = (X_{(1)} - \frac{1}{n} \log_e 10, X_{(1)})$.
The confidence level is $100\beta\%$, which means we need the probability that this interval contains $\theta$ to be $\beta$:
$ P(L < \theta < U) = \beta $
Substituting the interval limits:
$ P(X_{(1)} - \frac{1}{n} \log_e 10 < \theta < X_{(1)}) = \beta $
The inequality $\theta < X_{(1)}$ holds true because $X_{(1)}$ is the minimum of $X_i$, and all $X_i > \theta$.
The inequality $X_{(1)} - \frac{1}{n} \log_e 10 < \theta$ can be rewritten as:
$ X_{(1)} - \theta < \frac{1}{n} \log_e 10 $
So, the probability we need to calculate is:
$ P(X_{(1)} - \theta < \frac{1}{n} \log_e 10) = \beta $
Calculating the Value of $\beta$
Using the CDF of $X_{(1)} - \theta \sim \text{Exp}(n)$ derived earlier:
$ \beta = P(X_{(1)} - \theta \le \frac{1}{n} \log_e 10) $
$ \beta = 1 - e^{-n \times (\frac{1}{n} \log_e 10)} $
$ \beta = 1 - e^{-\log_e 10} $
Using the property $e^{-\log_e a} = e^{\log_e (1/a)} = 1/a$:
$ \beta = 1 - \frac{1}{10} $
$ \beta = 1 - 0.1 $
$ \beta = 0.9 $
Therefore, the value of $\beta$ is $0.9$. This corresponds to a $90\%$ confidence interval.
Important Questions from Random Variables
Let $X$, $Y$, and $Z$ be independent Normal random variables with means $-1$, $0$, and $1$, respectively, and variances $1$, $1$, and $3$, respectively. Which of the following random variables has a Cauchy distribution with location parameter $0$ and scale parameter $1$?
- Consider a finite population of size $N = 100$. Let $T_1$ be the sample mean of a study variable based on a sample of size $n$ ($1 < n < N$) under simple random sampling with replacement scheme. Let $T_2$ be the sample mean of the same study variable based on a sample of size $n$ under simple random sampling without replacement scheme. If $Var(T_1) = 9Var(T_2)$, then the sample size $n$ equals
Let $X_1$ and $X_2$ be a random sample from Uniform$[0, \theta]$ distribution, where $\theta > 0$. For testing the hypothesis
$H_0: \theta = 1$ against $H_1: \theta = 2$,
consider a test which rejects $H_0$ if $X_1 + X_2 > \frac{4}{5}$. Then, the probability of type-I error is- Let $\{Y_n: n \ge 1\}$ be a sequence of independent and identically distributed random variables, where $Y_1 \sim \text{Bernoulli}(\frac{1}{2})$. Define $Z = \sum_{n=1}^\infty \frac{4Y_n}{5^n}$. Then, which of the following statements is true?
Let $X$ be a single sample from an absolutely continuous distribution with probability density function
$f(x|\theta) = \begin{cases} \frac{2}{\theta^2}(\theta - x), & \text{if } 0 < x < \theta \\ 0, & \text{otherwise,} \end{cases}$
where $\theta > 0$ is unknown. Which of the following intervals is a $95\%$ confidence interval for $\theta$?
