Verify that certain statistic is complete (namely, $X_{(1)}$ for $f_\theta(x)=e^{-(x-3\theta)}I_{(3\theta,\infty)}(x)$)

Question

Given the density function of a certain population

$$f_\theta(x)=e^{-(x-3\theta)}I_{(3\theta,\infty)}(x)$$

I am asked to find a complete statistic for a random simple sample $(X_1,X_2,\ldots,X_n)$ of it.

From the nature of the given density function, I would like to prove that $X_{(1)}$ is complete.

To do so, the density of $X_{(1)}$ is given by

$$g(x)=ne^{-n(x-3\theta)}$$

And if $\operatorname E[f(S)]=0$, where $S=X_{(1)}$, then

$$\operatorname E[f(S)]=\int_{3\theta}^\infty f(x)ne^{-n(x-3\theta)}\, dx=0$$

How to proceed from this point, and prove that $f=0$ almost everywhere? Thanks in advance.

Answer 1

Firstly, the CDF of shifted exponential distribution is $$ F_{X_1}(x)=I_{(3\theta,\infty)}(x)\int_{3\theta}^x e^{-(t-3\theta)}\, dt = - e^{-(t-3\theta)} \big|_{3\theta}^x\cdot I_{(3\theta,\infty)}(x)=(1-e^{-(x-3\theta)})I_{(3\theta,\infty)}(x). $$ It cannot be $1+e^{-(x-3\theta)}$ as assumed in your calculations since this value is greater than $1$, and $F_{X_1}(x)$ is a probability.

Next, the CDF of $X_{(1)}$ is $$ F_{X_{(1)}}(x)=1-(1-F_{X_1}(x))^n = (1-e^{-n(x-3\theta)})I_{(3\theta,\infty)}(x). $$ Finally, the PDF of $X_{(1)}$ is $$ g(x)=ne^{-n(x-3\theta)}I_{(3\theta,\infty)}(x) $$

To prove completeness we need to suggest that for any $\theta\in\mathbb R$ $$ \int_{3\theta}^\infty f(x)ne^{-nx}e^{3n\theta}\, dx =0 $$ and derive that $f(x)=0$ a.e.

Note that multiplier $e^{3n\theta}$ cannot be zero. Therefore $$ \int_{3\theta}^\infty f(x)ne^{-nx}e^{3n\theta}\, dx \equiv 0 \iff \int_{3\theta}^\infty f(x)ne^{-nx}\, dx \equiv 0 $$ Since the last integral equals zero for any $\theta$, we can subtract two integrals for different $\theta$ and get: for any $a<b$ $$\tag{1}\label{1} \int_{a}^b f(x)ne^{-nx}\, dx = 0 $$ If you can apply Radon-Nikodym theorem here, do it and complete. Other case we need to follow the rest way.

Prove that for any Borel set $B$ $$\tag{2}\label{2} \int\limits_B f(x)ne^{-nx}\, dx = 0. $$ To do it, let $$\mathcal A=\left\{B\in \mathfrak B(\mathbb R): \int\limits_B f(x)ne^{-nx}\, dx = 0\right\}.$$ We collect into $\mathcal A$ all Borel sets with the property: integrals over these sets are zero.

Prove that $\mathfrak B(\mathbb R)\subseteq\mathcal A$. Indeed, any interval $(a,b)$ belongs to $\mathcal A$ due to (\ref{1}). Next, you can check that $\mathcal A$ is a sigma-algebra. Therefore it contains Borel sigma-algebra since Borel sigma-algebra is the smallest sigma-algebra that contains any interval on $\mathbb R$.

So, the proved (\ref{2}).

Finally, consider Borel sets $B_1=\{x: f(x)>0\}$ and $B_2=\{x: f(x)<0\}$. From (\ref{2}) we have $$ \int\limits_{B_1} f(x)ne^{-nx}\, dx = 0 \text{ and } \int\limits_{B_2} f(x)ne^{-nx}\, dx = 0. $$ If Lebesgue measure $\lambda(B_i)$ is not equal to zero than the first integral is strictly positive, and the second is strictly negative. So, both measures are zero and $f(x)=0$ almost everywhere.