Let $\mathbf{X}=(X_1,X_2,…,X_n)^T$ be a simply sample of random variable $X$ whose distribution belongs to family $\mathcal{P}=\{ f(x; \lambda, \eta, \mu ), 0<\lambda, \eta <\infty, -\infty <\mu <\infty \};$ where density function is $$ f(x; \lambda, \eta, \mu )= \frac{\lambda^{\eta}}{\Gamma(\eta)}x^{\eta -1} e^{-\lambda(x-\mu)}, x\geq \mu.$$ I need to find a sufficient statistic of parameters $\eta$ and $\lambda$ when $\mu$ is known.
So O think that density function I can write like this $$ f(x)=\prod _{i=1}^n\big( \frac{\lambda^{\eta}}{\Gamma(\eta)} \big) x_i^{\eta-1}e^{-\lambda(x_i-\mu)}=\big( \frac{\lambda^{\eta}}{\Gamma(\eta)}\big)^n \big(\prod _{i=1}^n x_i \big)^{\eta -1} e^{-\lambda\sum (x_i-\mu)}.$$
The joint density of the sample takes the form required by the Fisher–Neyman factorization theorem, by letting
$$ W(X)=1$$ and $q(T; \theta)=\big( \frac{\lambda^{\eta}}{\Gamma(\eta)}\big)^n \big(\prod _{i=1}^n x_i \big)^{\eta -1} e^{-\lambda \sum ( x_i-\mu)} $ where $L_X(\theta)=q(T; \theta) * W(X)$ is likelihood function. Then $T=(T_1;T_2)=\big( \prod_{i=1}^n x_i; \sum_{i=1}^n x_i \big)$ is a sufficient statistic.
Is this right or I made a mistake?
Best Answer
Density function I can write like this $$ f(x)=\prod _{i=1}^n\big( \frac{\lambda^{\eta}}{\Gamma(\eta)} \big) x_i^{\eta-1}e^{-\lambda(x_i-\mu)}=\big( \frac{\lambda^{\eta}}{\Gamma(\eta)}\big)^n \big(\prod _{i=1}^n x_i \big)^{\eta -1} e^{-\lambda\sum (x_i-\mu)}.$$
The joint density of the sample takes the form required by the Fisher–Neyman factorization theorem, by letting $$ W(X)=1$$ and $q(T; \theta)=\big( \frac{\lambda^{\eta}}{\Gamma(\eta)}\big)^n \big(\prod _{i=1}^n x_i \big)^{\eta -1} e^{-\lambda \sum ( x_i-\mu)} $ where $L_X(\theta)=q(T; \theta) * W(X)$ is likelihood function. Then $T=(T_1;T_2)=\big( \prod_{i=1}^n x_i; \sum_{i=1}^n x_i \big)$ is a sufficient statistic.