Let $X_1,\cdots,X_n$ be independent random variables with density$$f_{X_i}(x;\theta)=\begin{cases}e^{i\theta-x},&i\theta\leqslant x\\
0,&\text{otherwise}\end{cases}$$
where $-\infty<\theta<\infty$, $i=1,2,\cdots,n$.
Find the complete and sufficient statistic for $\theta$ and compute the unique minimum variance unbiased estimator of $\theta$.
My approach:
$$f_{X_i}(x;\theta)= e^{i\theta-x}I_{(i\theta\leqslant x)}\\
L(x;\theta)=\prod_{i=1}^{n} e^{i\theta-x}I_{(i\theta\leqslant x)}$$
which on solving tells me that $Y= \min\dfrac{X_i}{i}$ is the sufficient statistic of $\theta$. Also, how to prove completeness?
Now, my question is in order to find UMVUE, we need to have pdf of $Y$. How to go about that?
Best Answer
The function $L$ should be viewed as a function of $\theta$ with $x_1,\ldots,x_n$ fixed and your notation doesn't make that clear. Also, $x$ should not remain the same as $i$ changes.
For $y\ge\theta$ we have: \begin{align} \Pr(Y>y) & = \Pr( X_1> y\ \&\ X_2 > 2y\ \&\ \cdots \ \&\ X_n > ny) \\[8pt] & = e^{\theta-y} \cdot e^{2(\theta-y)} \cdot e^{3(\theta-y)} \cdots e^{n(\theta-y)} \\[8pt] & = e^{(1+2+3+\cdots+n)(\theta-y)} \\[8pt] & = \int_y^\infty (1+2+3+\cdots+n)e^{(1+2+3+\cdots+n)(\theta-x)} \, dx. \end{align} Thus the function of $x$ that is integrated here is the probability density function of the random variable $Y$ (for $x>\theta$).
Consequently \begin{align} \operatorname EY & = \int_\theta^\infty x\cdot (1+2+3+\cdots+n) e^{(1+2+3+\cdots+n)(\theta-x)} \, dx \\[8pt] & = \frac 1 {1+2+3+\cdots+n} + \theta. \end{align} Since the fraction in the last line above does not depend on $\theta,$ you get an unbiased estimator of $\theta.$
Completeness would mean there is no nonzero function $g$ such that \begin{align} \operatorname Eg(Y) & = \int_\theta^\infty g(x) (1+2+3+\cdots+n) e^{(1+2+3+\cdots+n)(\theta-x)} \, dx \\[8pt] & = \int_0^\infty g(x+\theta) (1+2+3+\cdots+n) e^{-(1+2+3+\cdots+n) x} \, dx \end{align} remains equal to $0$ as $\theta$ changes.