# Finding a sufficient statistic when density function is given

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?

• en.wikipedia.org/wiki/Sufficient_statistic#Gamma_distribution – d.k.o. Feb 12 at 11:40
• Of course, you need to write down the joint density of $X_1,\ldots,X_n$ (likelihood function) and use the Factorization theorem. – StubbornAtom Feb 12 at 15:19
• @StubbornAtom I just updated my question with my answer. Can you check if I did it right? – Atstovas Feb 12 at 15:38
• 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. – StubbornAtom Feb 12 at 15:43
• @StubbornAtom I do not understand what you mean.. – Atstovas Feb 12 at 16:06

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.