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 $(**)$?

  • $\begingroup$ Another option is to differentiate both sides of the equation $E_{\theta}[g(X_{(1)})]=f_X(x_0)$ wrt $\theta$ and solve for $g$. $\endgroup$ Dec 14, 2019 at 12:55

1 Answer 1


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.


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