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 $\theta$?
This question requires us to find a 95% confidence interval for the parameter \(\theta\) of a given probability density function (PDF). The PDF provided is:
\(f(x|\theta) = \begin{cases} \frac{2}{\theta^2}(\theta - x), & \text{if } 0 < x < \theta \\ 0, & \text{otherwise.} \end{cases}\)
Where \(x\) is a sample point and \(\theta > 0\) is unknown.
To find the confidence interval for \(\theta\), we follow these steps:
- Given the range \(0 < x < \theta\), the maximum likelihood estimator (MLE) for \(\theta\) is based on the largest possible value that \(x\) can take, which is the observation itself \(X\).
- The distribution is a form of the uniform distribution over \((0, \theta)\), where the maximum value \(\theta\) provides a natural estimate \(X\).
- For confidence interval calculation, the pivotal quantity method can be applied. The pivotal quantity is \(T = \frac{X}{\theta}\), and for this distribution, it follows a known distribution whose behavior can be utilized for interval estimation.
- To construct the interval, note that \(T\) covers interval \((0, 1)\). For a confidence level of 95%, if the lower bound stands from sample proportion, the higher bound of \(\theta\) corresponds to \(\frac{X}{1-\sqrt{0.05}}\).
- Concretely, find the point \(\alpha\) such that \(P(T > \alpha) = 0.05\). This involves solving for \(\alpha\) in terms of probability cutoff.
Following this logic, the correct 95% confidence interval is given by:
The interval \((X, \frac{X}{1-\sqrt{0.05}})\) ensures that the coverage probability is approximately 95%.
Therefore, among the provided options, the correct interval for \(\theta\) is:
Correct Answer: \((X, \frac{X}{1-\sqrt{0.05}})\)
Important Questions from Random Variables
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 $X$ and $Y$ be two independent random variables such that the moment generating func-tions of $X$ and $Y$ are
$M_X(t) = e^{3(e^t-1)}, t \in R$,
and
$M_Y(t) = (\frac{1}{2}e^{-3t} + \frac{1}{2}e^{3t})^2, t \in R$,
respectively. Then which of the following statements are true? Let $X_1, X_2, ..., X_7$ be a random sample drawn from a continuous distribution with unknown unique median $M$. The null hypothesis $H_0: M = 2$ is tested against the alternative $H_1: M \gt 2$ at level of significance 0.05 using the right-tailed test based on the Sign test statistic $K$, which is the number of observations in the sample greater than 2. If the observed sample is $-3, -6, 1, 9, 4, 10, 12$, which of the following statements are true?
- $X_1, X_2, \cdots$ are independent identically distributed random variables having common density $f$. Assume $f(x) = f(-x)$ for all $x \in \mathbb{R}$. Which of the following statements is correct?
