# FInding a complete and sufficient statistic

I am attempting to learn how to find a complete and sufficient statistic. So, I am working on this problem for class:

Let $$X_1, \cdot\cdot\cdot,X_n$$ be a random sample from the pdf $$f(x_i|u)=e^{-(x-\mu)}$$, where $$- \infty < \mu < x_i <\infty$$. Show that $$X_{1} = min_i\{X_i\}$$ is a complete sufficient statistic.

Here is what I have done so far. First I tried to show it is sufficient by factorization theorem:

• $$\prod_{i=1}^{n} f(x_1,...,x_n|\theta) = exp^{-\sum_{i=1}^{n}x_i+n\mu} I\{\mu.

• $$g(T(X|\theta)=exp^{-\sum_{i=1}^{n}x_i+n\mu}$$

• $$h(X) = 1$$.

First why is $$min_i\{X_i\}$$ a sufficient in this case? I know that for a statistic to be complete it must also satisfy the condition that for all $$g(.)$$ the expectation of $$E[g(T)]=0$$. and this should happen with probability 1. So then I must take the expectation of a function of T to show the completeness, but I am not sure if how I proceeded is correct. Thanks in advance!

Sufficiency: The minimum is a sufficient statistic by the Neyman Factorization Theorem. $$g_\mu$$ has $$\mu$$ interacting only with the sufficient statistic being the minimum of the order statistics.

$$\begin{split} f_\mu(x) &= e^{n\mu-\sum_ix_i}\prod_iI[\mu

Completeness: As you've written, you need to show that for $$\mathbb{E}[g(T)]=0$$, then $$\forall\mu:\mathbb{P}_\mu(g(T)=0)=1$$. We will use the pdf for the minimum of a set of $$n$$ iid random variables.

$$p(t)=\frac{d}{dt}(1-(1-F(t))^n) = ne^{-n(t-\mu)}$$

$$\begin{split} \mathbb{E}_\mu[g(T)] & = \int_\mu^\infty g(t)p(t)dt \\ & = \int_\mu^\infty g(t)ne^{-n(t-\mu)}dt = 0 \end{split}$$

It remains to show

$$n\int_\mu^\infty g(t)e^{-n(t-\mu)}dt = 0\equiv \int_\mu^\infty g(t)e^{-nt}dt = 0$$

$$\frac{d}{d\mu}\int_\mu^\infty g(t)e^{-nt}dt =\frac{d}{d\mu} 0\rightarrow g(\mu)e^{-n\mu}=0$$

This needs to hold for any value of $$\mu$$, and hence, we conclude that $$g=0$$ in all cases.

• Thanks, I was able to get this from my class book, and I missing understanding how to obtain the pdf of ordered statistics. Thanks for your comment. Commented Mar 1 at 15:23