Showing that $X_{1:n}$ is sufficient for $\eta$, by factorization

I'm asked to show that $X_{1:n}$ (the minimum order figure) suffices for $\eta$, when it comes to an arbitrary example $(X_1, ... , X_n)$ where $X_i\sim EXP(1,\eta$) (this is both - parameter rapid circulation $EXP(\theta,\eta):$ $(1/\theta)\exp(-(x-\eta)/\theta)$, $x>\eta$ ; in this instance $\theta=1$), by utilizing the "factorization method", that is, creating $f(x_1,...,x_n;\eta)$ as $g(s,\eta)h(x_1,...,x_n)$, where $S$ is the figure ($X_{1:n}$, in this instance), $g(s;\eta)$ does not rely on $x_1,...,x_n$ other than via $s$, and also $h(x_1,...,x_n)$ does not entail $\theta$.

I have $f(x_1,...,x_n;\eta)=\exp(n\eta)\exp(-\sum_{i=1}^n x_i)$. The amount in the exponential coincides as $\sum_{i=1}^n x_{i:n}$, yet I require an expression that entails $x_{1:n}$ in one variable and also the $x_i$ remains in the various other. I do not recognize just how to do this.

I recognize there are solutions for joint pdf is of any kind of set of order statistics (fairly extensive), yet I actually do not recognize just how to continue.


2022-06-07 14:31:21
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Answers: 1

I'm mosting likely to make an enlightened hunch that $X \sim \mathrm{Exp}(1,\eta)$ suggests that $X$ has thickness $$\newcommand{\Ind}[1]{\mathbf{1}_{(#1)}} f(x) \equiv f_X(x) = e^{-(x-\eta)}\Ind{x\geq\eta} \>, $$ where $\Ind{A}$ is the indication function of the occasion $A$.

If $X_1, X_2, \ldots$ is an iid example from this circulation, after that the joint thickness is $$ \prod_{i=1}^n f(x_i) = \prod_{i=1}^n e^{-(x_i - \eta)} \Ind{x_i \geq \eta} = e^{-\sum_{i=1}^n x_i} e^{n \eta} \prod_{i=1}^n \Ind{x_i \geq \eta} = e^{-\sum_{i=1}^n x_i} e^{n \eta} \Ind{\min_i x_i \geq \eta} \> , $$ where the last equal rights adheres to given that the the item of the indications $\Ind{x_i \geq \eta}$ is one if and also just if the indication $\Ind{\min_i x_i \geq \eta}$ is one and also or else is absolutely no.

Currently, take $h(x_1,\ldots,x_n) = \exp(-\sum_i x_i)$ and also $g(s,\eta) = e^{n \eta} \Ind{s \geq \eta}$ with $s = \min_i x_i$ and also use the factorization theory.

Therefore $S = \min_{1 \leq i \leq n} X_i$ suffices for $\eta$.

2022-06-07 14:47:25