Question

If \(X \sim \beta_1(\alpha, \beta)\) such that parameters \(\alpha, \beta\) are unknown, then the sufficient statistic for \((\alpha, \beta)\) is

• A. \( T = (\sum x_i, \sum (1-x_i)) \)
• B. \( T = (\prod x_i, \sum (1-x_i)) \)
• C. \( T = (\sum x_i, \prod (1-x_i)) \)
• D. \( T = (\prod x_i, \prod (1-x_i)) \)

Hint:

For distributions in the exponential family, the sufficient statistic can be read directly from the form of the PDF. The Beta distribution is in the two-parameter exponential family, and its sufficient statistics are \( \sum \ln(X_i) \) and \( \sum \ln(1-X_i) \), which are one-to-one functions of \( \prod X_i \) and \( \prod (1-X_i) \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A sufficient statistic encapsulates all relevant information from a sample needed to estimate parameters. The Fisher-Neyman Factorization Theorem is used to identify sufficient statistics.

Step 2: Key Formula or Approach:
The Fisher-Neyman Factorization Theorem states that \(T(\mathbf{X})\) is sufficient for \(\theta\) if and only if the joint probability density function \(f(\mathbf{x}|\theta)\) can be expressed as: \[ f(\mathbf{x}|\theta) = g(T(\mathbf{x}), \theta) . h(\mathbf{x}) \] where \(g\) depends on the data only through \(T\), and \(h\) is independent of \(\theta\).

Step 3: Detailed Explanation:
The PDF for a Beta distribution is: \[ f(x; \alpha, \beta) = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} x^{\alpha-1}(1-x)^{\beta-1}, \quad 0<x<1 \] For a random sample \(X_1, \dots, X_n\), the joint PDF (likelihood function) is: \[ L(\alpha, \beta | \mathbf{x}) = \prod_{i=1}^n \left[ \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} x_i^{\alpha-1}(1-x_i)^{\beta-1} \right] \] \[ = \left( \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \right)^n \left( \prod_{i=1}^n x_i \right)^{\alpha-1} \left( \prod_{i=1}^n (1-x_i) \right)^{\beta-1} \] To apply the factorization theorem, let \( T(\mathbf{x}) = \left( \prod_{i=1}^n x_i, \prod_{i=1}^n (1-x_i) \right) \), with components \(T_1\) and \(T_2\). Then, the likelihood can be written as: \[ L(\alpha, \beta | \mathbf{x}) = \left[ \left( \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \right)^n T_1^{\alpha-1} T_2^{\beta-1} \right] . 1 \] Here, - \( g(T(\mathbf{x}), (\alpha, \beta)) = \left( \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \right)^n \left(\prod x_i\right)^{\alpha-1} \left(\prod (1-x_i)\right)^{\beta-1} \), depending on the data only through \(T = (\prod x_i, \prod (1-x_i))\). - \( h(\mathbf{x}) = 1 \), independent of \(\alpha\) and \(\beta\). Therefore, by the Fisher-Neyman Factorization Theorem, \( T = \left(\prod_{i=1}^n X_i, \prod_{i=1}^n (1-X_i)\right) \) is a sufficient statistic for \((\alpha, \beta)\).
Step 4: Final Answer:
The sufficient statistic for \((\alpha, \beta)\) is \( T = (\prod x_i, \prod (1-x_i)) \).