3
$\begingroup$

Let $X_1,...,X_n$ be a sample from a distribution with pdf, $f_X(x) = e^{-x + \theta}, x \geq \theta$. Let $x_0 \geq \theta$ be given. I'm trying to find the UMVUE of $f_X(x_0) = e^{-x_0 + \theta}$. I went ahead and found that a complete sufficient statistic for $\theta$ is $X_{(1)}$. Moreover, an unbiased estimator for $f_X(x_0)$ is,

$$ T(\textbf{X}) = \begin{cases}1 & X_1 > x_0 \\ 0 & otherwise \end{cases} $$

So the UMVUE should be $\mathbb{E}[T(\textbf{X}) | X_{(1)}]$. We have,

$$ \begin{align} \mathbb{E}[T(\textbf{X}) | X_{(1)} = x] &= \mathbb{P}(X_1 > x_0 | X_{(1)} = x)\\ &= \mathbb{P}(X_1 - X_{(1)} > x_0 - X_{(1)} | X_{(1)} = x) \\ &= \mathbb{P}(X_1 - X_{(1)} > x_0 - x | X_{(1)} = x)\\ &= \mathbb{P}(X_1 - X_{(1)} > x_0 - x)\;\;\;\;\;\;\; (**)\end{align}$$

where the last equality follows from Basu's Theorem, as $X_1 - X_{(1)}$ is ancilliary for $\theta$ and $X_{(1)}$ is complete sufficient; thus they are independent. I'm stuck at finding this probability, however. How can I find the distribution of $X_1 - X_{(1)}$?

As a different approach, I went ahead found the expectation of functions of the form, $ce^{aX_{(1)}+b}$. It turns out that $g(X_{(1)}) = \frac{n-1}{n}e^{X_{(1)}-x_0}$ is unbiased for $e^{-x_0 + \theta}$ and thus must be the UMVUE. However, this was simply trial and error which fortunately led to the result. I'm not comfortable just guessing and checking. Is there a way I can proceed from $(**)$?

$\endgroup$
1
$\begingroup$

You could proceed from $(**)$ as follows: for $x \leq x_0$, \begin{align*} &\mathbb P(X_1 - X_{(1)} > x_0 - x)\\ &= 1-\mathbb P(X_1 - X_{(1)} \leq x_0 - x)\\ &= 1-\mathbb P(X_{(1)} \geq X_1 - x_0 + x)\\ &= 1-\mathbb P(X_i \geq X_1 - x_0 + x\text{ for all $i=1,\dots,n$})\\ &= 1-\mathbb P(X_i \geq X_1 - x_0 + x\text{ for all $i=2,\dots,n$})\\ &= 1-\int_{\theta}^\infty \mathbb P(X_i \geq t - x_0 + x\text{ for all $i=2,\dots,n$})f_{X_1}(t)\ dt\\ &= 1- \int_\theta^\infty f_{X_1}(t)\prod_{i=2}^n \mathbb P(X_i \geq t-x_0+x)\ dt \\ &= \dots \end{align*} However, there appears to be a problem which needs to be addressed first. When referring to a "complete sufficient statistic for $\theta$", we need to carefully consider what space of parameters $\Theta$ is being implicitly referenced. For the space $\Theta = [0,\infty)$, it is true that $X_{(1)}$ is a complete sufficient statistic, and in this case the identification of $\frac{n-1}ne^{X_{(1)}-x_0}$ as the UMVUE for $e^{-x_0+\theta}$ is correct (for $n>1$). However, this does not provide a UMVUE for $f_X(x_0)$ on this space $\Theta$, as the equation $f_X(x_0) = e^{-x_0+\theta}$ does not hold when $\theta>x_0$. The statistic $T(\mathbf X)$ also fails to be unbiased when $\theta>x_0$.

You may instead be intending to consider the space $\Theta = [0, x_0]$, but in this case $X_{(1)}$ is no longer complete, so you will need a different complete, sufficient statistic ($\min\{X_{(1)}, x_0\}$ should do the trick.) After this is sorted out, you should be able to proceed in a similar way as above.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.