# Complete statistics for $f_X(x) = e^{-(x - \mu)} I_{\mu, \infty}(x)$

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 $$\mu$$?

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

Therefore,

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

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.

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