why sufficient statistics is a vector in this case?
- Let $\mathbf{X}=\left(X_{1}, X_{2}, \ldots, X_{n}\right)$ be a random sample from the two-parameter exponential distribution with pdf
$$
f(x)=\lambda e^{-\lambda(x-\delta)}, \quad x \geq \delta
$$
The parameters are $\lambda>0$ and $\delta>0$
(a) Show that $\mathbf{T}(\mathbf{X})=\left(\sum_{i=1}^{n} X_{i}, X_{(1)}\right)$ is a sufficient statistic for $\boldsymbol{\theta}=(\lambda, \delta)$.
By factorization theory I write
$$\lambda^{n} e^{-\lambda(\sum_{i=1}^{n} x_{i} – \sigma)}$$
How does $x_{(1)}$ come from?
Best Answer
you can rewrite your denisty in the following way
$$f_X(x|\lambda,\delta)=\lambda e^{\lambda\delta}e^{-\lambda x}\cdot\mathbb{1}_{[\delta;+\infty)}(x)$$
observe that
$$\delta\le X_1< +\infty$$ $$\delta\le X_2< +\infty$$ $$\dots$$ $$\delta\le X_n< +\infty$$
that is evidently
$$\delta \le X_{(1)}$$
now we can rewrite
$$\prod_i f(x_i)=\lambda^n e^{\lambda\delta n}\cdot e^{-\lambda \Sigma_i x_i}\cdot \mathbb{1}_{(0;x_{(1)}]}(\delta)$$
Now it is evident that a sufficient estimator is the vector you have to show