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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?

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  • $\begingroup$ en.wikipedia.org/wiki/Sufficient_statistic#Gamma_distribution $\endgroup$ – d.k.o. Feb 12 at 11:40
  • $\begingroup$ Of course, you need to write down the joint density of $X_1,\ldots,X_n$ (likelihood function) and use the Factorization theorem. $\endgroup$ – StubbornAtom Feb 12 at 15:19
  • $\begingroup$ @StubbornAtom I just updated my question with my answer. Can you check if I did it right? $\endgroup$ – Atstovas Feb 12 at 15:38
  • $\begingroup$ Exponent of $e$ should be $-\lambda \sum (x_i-\mu)$, so your answer changes a little. And $W(X)$ is strictly speaking an indicator variable $\mathbf1_{x_1,\ldots,x_n\ge \mu}$, inherently present in the joint density. $\endgroup$ – StubbornAtom Feb 12 at 15:43
  • $\begingroup$ @StubbornAtom I do not understand what you mean.. $\endgroup$ – Atstovas Feb 12 at 16:06
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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.

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