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
- 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 $\bar{Y}$?
