I am studying parametric statistical inference. One of the self study I have to find a sufficient, minimal and complete statistic for the $\mu$ parameter of the following p.d.f. $$ f_X(x \mid \mu) = e^{-(x - \mu)} I_{(\mu, \infty)}(x) $$ which is a exponential distribution with location parameter $\mu$.

We can write the p.d.f. of a random sample $\mathbf{x} = (x_1, \ldots, x_n)$ of $X$ as $$ f_{\mathbf{X}}(\mathbf{x} \mid \mu) = e^{-n(\bar{x} - \mu)}\,I_{(\mu, \infty)}(x_{(1)}) $$ where $x_{(1)} = \min(\mathbf{x})$.

By the Factorization Theorem I concluded that $T = X_{(1)}$ is a sufficient statistic for $\mu$. To prove that $X_{(1)}$ is also minimal I showed that the ratio $f_{\mathbf{X}}(\mathbf{x} \mid \mu)/f_{\mathbf{X}}(\mathbf{y} \mid \mu)$ does not depend on $\mu$ iff $x_{(1)} = y_{(1)}$

In regard the completeness of $X_{(1)}$ I have to prove that $E(g(T)) = 0$ for all $\mu$, i.e., there is no function of $T = X_{(1)}$ unless the $g(T) = 0$ zero function.

I have found the distribution of $T$, which is given by $$ f_T(t) = n\,e^{-n(t-\mu)} I_{(\mu, \infty)}(t). $$

However, I couldn't show that $E(g(T)) = 0$.

Is there another way to prove that $T$ is or is not a complete statistic for $\mu$?

  • $\begingroup$ Yes, I try do to this, but couldn't get the derivatives $\endgroup$
    – andre
    Sep 14 '19 at 21:51
  • $\begingroup$ $$ f_T(t) = n\,e^{-n(x-\mu)} I_{(\mu, \infty)}(t). $$ Did you mean $$ f_T(t) = n\,e^{-n(t-\mu)} I_{(\mu, \infty)}(t) \text{ ?} $$ (With $t$ rather than $x$ in the exponent?) $\qquad$ $\endgroup$ Sep 15 '19 at 17:16
  • $\begingroup$ Yes, there was a typo. $\endgroup$
    – andre
    Sep 15 '19 at 17:58

You have for any function $g$,

\begin{align} E_{\mu}(g(T))&=\int g(t)f_T(t)\,dt \\&=\int_{\mu}^\infty g(t)ne^{-n(t-\mu)}\,dt \end{align}


$$E_{\mu}(g(T))=0 \quad\,\forall\,\mu\in\mathbb R \implies \int_{\mu}^\infty g(t)e^{-nt}\,dt =0 \quad\,\forall\,\mu\in\mathbb R$$

For some $a\in(\mu,\infty)$, you can rewrite the last equation as

$$\int_{\mu}^a g(t)e^{-nt}\,dt+\int_a^\infty g(t)e^{-nt}\,dt=0\quad\,\forall\,\mu\tag{*}$$

Now differentiate both sides of $(*)$ with respect to $\mu$.

At this stage you can simply use the Fundamental theorem of calculus for the first integral.

The conclusion then follows immediately.

  • $\begingroup$ And for the second integral? The theorem can be apply? $\endgroup$
    – andre
    Sep 15 '19 at 18:00
  • $\begingroup$ @andre Does the second integral depend on $\mu$? $\endgroup$ Sep 15 '19 at 18:26
  • $\begingroup$ No, it doesn't. Thank you! $\endgroup$
    – andre
    Sep 15 '19 at 20:19

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