For n ≥ 2, let X1, X2, ..., Xn be a random sample from a distribution with the probability density function \(f(x \mid θ)=\left\{\begin{array}{cc} θ x^{θ-1}, & 0<x<1 \\ 0, & \text { otherwise } \end{array},\right.\) where θ > 0 is an unknown parameter. Then which of the following is the uniformly minimum variance unbiased estimator for \(\frac{1}{\theta}\) ?
UMVUE for \( \frac{1}{\theta} \)
The problem asks for the Uniformly Minimum Variance Unbiased Estimator (UMVUE) for \( \frac{1}{\theta} \) based on a random sample \( X_1, X_2, \ldots, X_n \) from a distribution with probability density function (PDF):
\[ f(x \mid \theta)=\left\{\begin{array}{cc} \theta x^{\theta-1}, & 0<x<1 \\ 0, & \text { otherwise } \end{array}\right. \]
where \( \theta > 0 \).
Sufficient and Complete Statistic
First, let's find a sufficient statistic for \( \theta \). We can examine the likelihood function:
\[ L(\theta \mid x_1, \ldots, x_n) = \prod_{i=1}^n f(x_i \mid \theta) = \prod_{i=1}^n (\theta x_i^{\theta-1}) \] \[ L(\theta \mid x_1, \ldots, x_n) = \theta^n \prod_{i=1}^n x_i^{\theta-1} = \theta^n \exp\left(\sum_{i=1}^n (\theta-1) \ln x_i\right) \] \[ L(\theta \mid x_1, \ldots, x_n) = \theta^n \exp\left(\theta \sum_{i=1}^n \ln x_i - \sum_{i=1}^n \ln x_i\right) \]
This likelihood function can be written in the form \( g(T(\mathbf{x}) \mid \theta) h(\mathbf{x}) \) where \( T(\mathbf{x}) = \sum_{i=1}^n \ln x_i \) and \( h(\mathbf{x}) = 1 \). By the Factorization Theorem, \( T = \sum_{i=1}^n \ln X_i \) is a sufficient statistic for \( \theta \).
The PDF belongs to the exponential family. For an exponential family distribution, the sufficient statistic is also complete if the parameter space contains an open interval. Since \( \theta > 0 \), the parameter space is \( (0, \infty) \), which is an open interval. Thus, \( T = \sum_{i=1}^n \ln X_i \) is a complete sufficient statistic for \( \theta \).
Unbiased Estimator for \( \frac{1}{\theta} \)
According to the Lehmann-Scheffe theorem, if we can find an unbiased estimator for \( \frac{1}{\theta} \) that is a function of the complete sufficient statistic \( T \), then that estimator is the unique UMVUE.
Let's find the expected value of \( \ln X \) for a single observation \( X \):
\[ E[\ln X] = \int_0^1 \ln x \cdot f(x \mid \theta) dx = \int_0^1 \ln x \cdot \theta x^{\theta-1} dx \]
We can evaluate this integral using integration by parts with \( u = \ln x \) and \( dv = \theta x^{\theta-1} dx \). Then \( du = \frac{1}{x} dx \) and \( v = x^\theta \).
\[ E[\ln X] = \left[ (\ln x) x^\theta \right]_0^1 - \int_0^1 x^\theta \cdot \frac{1}{x} dx \] \[ E[\ln X] = \left[ x^\theta \ln x \right]_0^1 - \int_0^1 x^{\theta-1} dx \]
The first term evaluates to \( 1^\theta \ln 1 - \lim_{x \to 0^+} x^\theta \ln x = 0 - 0 = 0 \) (using L'Hopital's rule for the limit term, as shown in the thought process).
The second term is:
\[ \int_0^1 x^{\theta-1} dx = \left[ \frac{x^\theta}{\theta} \right]_0^1 = \frac{1^\theta}{\theta} - \frac{0^\theta}{\theta} = \frac{1}{\theta} \]
So, \( E[\ln X] = 0 - \frac{1}{\theta} = -\frac{1}{\theta} \).
Now consider the sample mean of \( \ln X_i \), which is \( \frac{1}{n} \sum_{i=1}^n \ln X_i \). Its expectation is:
\[ E\left[ \frac{1}{n} \sum_{i=1}^n \ln X_i \right] = \frac{1}{n} \sum_{i=1}^n E[\ln X_i] = \frac{1}{n} \sum_{i=1}^n \left(-\frac{1}{\theta}\right) = \frac{1}{n} \left(-n \frac{1}{\theta}\right) = -\frac{1}{\theta} \]
This estimator \( -\frac{1}{n} \sum_{i=1}^n \ln X_i \) has an expectation of \( -(-\frac{1}{\theta}) = \frac{1}{\theta} \).
Therefore, \( -\frac{1}{n} \sum_{i=1}^n \ln X_i \) is an unbiased estimator for \( \frac{1}{\theta} \).
UMVUE using Lehmann-Scheffe
We have established that \( T = \sum_{i=1}^n \ln X_i \) is a complete sufficient statistic for \( \theta \). The estimator \( U = -\frac{1}{n} \sum_{i=1}^n \ln X_i \) is an unbiased estimator for \( \frac{1}{\theta} \). Also, \( U \) is a function of \( T \) since \( U = -\frac{1}{n} T \).
By the Lehmann-Scheffe theorem, an unbiased estimator that is a function of a complete sufficient statistic is the unique UMVUE. Hence, \( -\frac{1}{n} \sum_{i=1}^n \ln X_i \) is the UMVUE for \( \frac{1}{\theta} \).
Comparing this with the given options:
- Option 1: \( -\frac{1}{n} \sum_{i=1}^n \ln X_i \)
- Option 2: \( -\frac{n}{\sum_{i=1}^n \ln X_i} \)
- Option 3: \( -\frac{n-1}{\sum_{i=1}^n \ln X_i} \)
- Option 4: \( -\frac{2}{n} \sum_{i=1}^n \ln X_i \)
Option 1 matches the derived UMVUE.
The final answer is \( -\frac{1}{n} \sum_{i=1}^n \ln X_i \).
Important Questions from Sampling Distributions
Let X1, X2, ..., X6 be a random sample from a gamma distribution with the probability density function
\(f(x \mid \lambda)=\left\{\begin{array}{cl} \frac{\lambda^4}{6} e^{-\lambda x} x^3, & \text { if } x>0 \\ 0, & \text { if } x \leq 0 \end{array},\right.\)
where λ > 0 is unknown. Let \(T=\sum_{i=1}^6 X_i\) and ψ be the uniformly most powerful test of size α = 0.05 for testing null hypothesis H0 : λ = 1 against alternative hypothesis H1 : λ > 1. For any positive integer v, let \(\chi_{v, α}^2\) denote the (1 - α)th quantile of \(\chi_v^2\) distribution. Then the test ψ rejects H0 if and only if
Let X1, ..., Xn be a random sample from N(μ, 1) distribution, where μ ∈ ℝ is unknown. In order to test H0 : μ = μ0 against H1 : μ > μ0, where μ0 ∈ ℝ is some specified constant, consider the following two tests:
(A) Reject H0 if and only if X̅n > c1, where c1 is such that \(P_{μ_0}\) (X̅n > c1) = α ∈ (0, 1) and X̅n = \(\frac{1}{n} \sum_{i=1}^n X_i\).
(B) Reject H0 if and only if Median {X1, ..., Xn} > c2, where c2 is such that \(P_{μ_0}\)(Median{X1, ..., Xn} > c2) = α ∈ (0, 1).
Then which of the following statements are true?
Let X1, X2, ..., Xn be a random sample from an unknown distribution with absolutely continuous cumulative distribution function (cdf) F. Let F0 be a specified absolutely continuous cdf. For testing H0 : F(x) = F0(x) for all x against H1 : F(x) ≠ F0(x) for some x, consider the following two test statistics:
\(\displaystyle T_{1, n}=\sup _{x \in \mathbb{R}}\left|\frac{1}{n} \sum_{i=1}^n I_{\left\{X_i \leq x\right\}}-F_0(x)\right| \), and \(\displaystyle T_{2, n}=\sup _{x \in \mathbb{R}} n\left|\frac{1}{n} \sum_{i=1}^n I_{\left\{X_i \leq x\right\}}-F_0(x)\right|\), where \(I_{\left\{X_i \leq x\right\}}=\left\{\begin{array}{ll}1, & \text { if } X_i \leq x \\ 0, & \text { if } X_i>x\end{array}\right.\) for i = 1, 2, ..., n.
Then which of the following statements are true?
- Which one of the following is the formula to calculate standard error of mean differences of two sample groups ?
- Standard deviation of a sampling distribution of a statistic is termed as,
