# Find Uniform Minimum Variance Unbiased estimator (UMVU) using Lehmann Scheffé - showing statistic is complete

Let $$X_1,...,X_n$$ be independent copies of a real-valued random variable $$X$$ where $$X$$ has Lebesque density

\begin{align*} p_\theta(x) = \begin{cases} \exp(\theta-x),\quad x>\theta \\ 0, \quad\quad\quad\quad\;\ x\leq \theta, \end{cases} \end{align*} where $$\theta\in \mathbb{R}$$ is an unknown parameter. Let $$S:=\min(X_1,...,X_n)$$.

Find the Uniform Minimum Variance Unbiased (UMVU) estimator of $$\theta$$.

I already know that $$S$$ is sufficient for $$\theta$$ and that $$T:=S-1/n$$ is an unbiased estimator of $$\theta.$$ My idea is to apply the Lehmann-Scheffé thm. since then the UMVU is given by

\begin{align*} \mathbb{E}[T|S]=\mathbb{E}[S-1/n|S]=S-1/n. \end{align*}

Is this the correct approach? If yes, for applying Lehmann-Scheffé, I would also need that S is a complete statistic. How do I show this properly?

Edit: I tried to show completeness by definition, i.e. I setup the equation $$\mathbb{E}_\theta[g(S)]=0 \;\forall \theta$$ for some function $$g$$ and now want to show that $$g(S)=0 \; \mathbb{P}_\theta$$-a.s. for all $$\theta$$. Since the $$X_i$$ are iid it is easy to see that the cdf is $$F_S(x)=1-(1-P_\theta(x))^n$$, where $$P_\theta(x)$$ is the cdf of $$X_i$$. By taking the derivative we get the pdf for $$S$$: $$f_S(x)=n\cdot p_\theta(x)(1-P_\theta (x))^{n-1}$$. $$P_\theta (x)$$ can be easily calculated and we get \begin{align*} f_S(x)=n\cdot e^{n(\theta-x)}. \end{align*}

Hence, $$\mathbb{E}_\theta[g(S)]=\int_\theta^\infty g(x)ne^{n(\theta-x)}dx$$ has to be $$0$$.

Is it now enough to say that $$g(S)=0 \; \mathbb{P}_\theta$$-a.s. for all $$\theta$$, since the exponential function is always positive? Or is there a more rigorous way to show it?

• This is the correct approach. To show completeness, did you setup the equation $E_{\theta}(g(S))=0$ for all $\theta$ and for some function $g$? Can you show, as per definition, that $g(S)=0$ with probability 1? – StubbornAtom Aug 6 '19 at 17:26
• Ok, so $\mathbb{E}_\theta (g(S))=\int_\theta^{\infty} g(x)e^{\theta-x}dx = e^{\theta}\int_\theta^{\infty}g(x)e^{-x}dx$ which is $0$ iff $\int_\theta^{\infty}g(x)e^{-x}dx=0$ for all $\theta$. Since the exponential function is always positive, this is only possible if $g(S)=0$ $\mathbb{P}_\theta$-a.s. for all $\theta$ and therefore, by definition, $S$ is complete. Is this the correct justification or can one show it more rigorous? – CauchySchwarz Aug 6 '19 at 18:29
• You have to use pdf of $S$ in the integral. – StubbornAtom Aug 6 '19 at 18:53
• Oh, of course! So the pdf of $S$ is $f_{\min(X_1,...,X_n)}(x)=p_{\theta}(x)(1-P_\theta(x))^{n-1}$ and therefore $\mathbb{E}_\theta(g(S))=\int_\theta^{\infty} e^{\theta-x}g(x)(1-(1-e^{\theta-x}))^{n-1}dx=\int_{\theta}^{\infty}g(x)e^{\theta-x}(e^{(\theta-x)(n-1)})dx=\int_{\theta}^{\infty}g(x)e^{n(\theta-x)}dx$ and the argumentation from above would then still hold? – CauchySchwarz Aug 6 '19 at 19:06
• Please add your work in the post, not in comments. How did you derive the pdf of $S$? – StubbornAtom Aug 6 '19 at 19:52

For some measurable function $$g$$, suppose

$$\mathbb E_{\theta}\left[g(S)\right]=\int_{\theta}^\infty g(x)ne^{-n(x-\theta)}\,dx=0\quad\,\forall\,\theta\in\mathbb R$$

That is, $$\int_{\theta}^\infty g(x)e^{-nx}\,dx=0\quad\forall\,\theta$$

Now for some $$a\in(\theta,\infty)$$, we can rewrite the last equation as

$$\int_{\theta}^a g(x)e^{-nx}\,dx+\int_a^\infty g(x)e^{-nx}\,dx=0\quad\forall\,\theta$$

Differentiating both sides of the last equation with respect to $$\theta$$, we get

$$g(\theta)e^{-n\theta}=0\quad\forall\,\theta$$

Now that $$e^{-n\theta}>0$$ for each $$\theta$$, you can conclude that $$g$$ is exactly zero almost everywhere.

This perhaps is a more convincing argument.