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 for a set of parameters is a function of the sample data that captures all the information about the parameters contained in the sample. The Fisher-Neyman Factorization Theorem is the standard tool to find a sufficient statistic.

Step 2: Key Formula or Approach:
The Fisher-Neyman Factorization Theorem states that a statistic \(T(\mathbf{X})\) is sufficient for \(\theta\) if and only if the joint probability density function \(f(\mathbf{x}|\theta)\) can be factored into two functions: \[ f(\mathbf{x}|\theta) = g(T(\mathbf{x}), \theta) . h(\mathbf{x}) \] where \(g\) depends on the data only through the statistic \(T\), and \(h\) does not depend on the parameter \(\theta\).

Step 3: Detailed Explanation:
The probability density function (PDF) for a Beta distribution of the first kind 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 of size \(n\), \(X_1, \dots, X_n\), the joint PDF (or likelihood function) is the product of the individual PDFs: \[ L(\alpha, \beta | \mathbf{x}) = \prod_{i=1}^n \left[ \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} x_i^{\alpha-1}(1-x_i)^{\beta-1} \right] \] \[ = \left( \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \right)^n \left( \prod_{i=1}^n x_i \right)^{\alpha-1} \left( \prod_{i=1}^n (1-x_i) \right)^{\beta-1} \] Let's identify the parts for the factorization theorem. Let \( T(\mathbf{x}) = \left( \prod_{i=1}^n x_i, \prod_{i=1}^n (1-x_i) \right) \). Let's call the components \(T_1\) and \(T_2\). Then we can write the likelihood 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, - \( g(T(\mathbf{x}), (\alpha, \beta)) = \left( \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \right)^n \left(\prod x_i\right)^{\alpha-1} \left(\prod (1-x_i)\right)^{\beta-1} \). This function depends on the data only through the statistic \(T = (\prod x_i, \prod (1-x_i))\). - \( h(\mathbf{x}) = 1 \). This function does not depend on the parameters \(\alpha\) or \(\beta\). Since the joint PDF can be factored in this way, by the Fisher-Neyman Factorization Theorem, the statistic \( T = \left(\prod_{i=1}^n X_i, \prod_{i=1}^n (1-X_i)\right) \) is a sufficient statistic for \((\alpha, \beta)\).
Step 4: Final Answer:
The sufficient statistic for \((\alpha, \beta)\) is \( T = (\prod x_i, \prod (1-x_i)) \).