Question

If $X_{1}, X_{2}, \dots, X_{n}$ be independent random variable each from Gamma($\alpha, \beta$) then the jointly sufficient statistics for vector ($\alpha, \beta$) is

• A. $(\sum_{i=1}^{n} X_{i}, \prod_{i=1}^{n} X_{i})$
• B. $(\prod_{i=1}^{n} X_{i}, \sum_{i=1}^{n} X_{i})$
• C. $(\prod_{i=1}^{n} X_{i}, \prod_{i=1}^{n} X_{i})$
• D. $(\sum_{i=1}^{n} X_{i}, \sum_{i=1}^{n} X_{i})$

Hint:

For members of the exponential family like the Gamma distribution, the sufficient statistics are simply the sums of the terms that appear in the exponent of the canonical form.

Answer 1

The Correct Option is A

We use the Neyman-Fisher Factorization Theorem to identify the sufficient statistics from the joint density function.

Step 1: \color{redWrite the PDF of a Gamma Distribution
The PDF for $X \sim Gamma(\alpha, \beta)$ is:
$f(x; \alpha, \beta) = \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x}$.

Step 2: \color{redForm the Joint Likelihood Function
For $n$ i.i.d. observations:
$L(\alpha, \beta) = \prod_{i=1}^n \left[ \frac{\beta^\alpha}{\Gamma(\alpha)} x_i^{\alpha-1} e^{-\beta x_i} \right]$
$L(\alpha, \beta) = \frac{\beta^{n\alpha}}{[\Gamma(\alpha)]^n} \left( \prod_{i=1}^n x_i \right)^{\alpha-1} e^{-\beta \sum x_i}$.

Step 3: \color{redApply the Factorization Theorem
The factorization theorem states that $T$ is sufficient for $\theta$ if $L(\theta) = g(T(x), \theta) h(x)$.
In our case, the likelihood depends on the sample only through two terms:
1. $\sum_{i=1}^n x_i$
2. $\prod_{i=1}^n x_i$

Step 4: \color{redIdentify the Sufficient Statistic Vector
The vector $T = (\sum X_i, \prod X_i)$ contains all the information needed to estimate the parameters $\alpha$ and $\beta$.
Note: Technically, $(\sum X_i, \sum \ln X_i)$ is also a common representation, but since $\sum \ln X_i = \ln(\prod X_i)$, the product is equivalent.
The correct statistic is $(\sum X_i, \prod X_i)$.