Thread: Constructing one-sided CI from MLE

1. Constructing one-sided CI from MLE

$X_1,X_2,...,X_n$ is a sample from the distribution whose density is:

$f_{X}(x) = e^{-x-\theta} \mbox{ if }x \geq \theta$

Based on the MLE estimator of $\theta$ construct a one-sided confidence interval for the unknown parameter at confidence level $1-\alpha$

Here's what I have so far.

$f(x_1,...,x_n|\theta) = e^{-x_{i} + \theta}\cdot\cdot\cdot e^{-x_{n} + \theta}$

$=e^{\sum_{1}^{n} -x_i + n\theta}
$

$\log{f(x_1...x_n|\theta)}=\sum_{i=1}^{n}-x_i + n\theta$
$\frac{d}{d\theta}f(x_1...x_n|\theta) = n$

$n=0 ?$

I'm not too sure where to go from here. Any help? Thanks.

2. YOU are ignoring the indicator function
BESIDES TWO TYPOs (the i and the negative sign in front of theta)
THE MLE is the smallestorder stat.
YOU need to use common sense, not calculus (incorrectly)
to maximize the likelihood function

$f(x_1,...,x_n|\theta) = e^{-x_{1} + \theta}\cdot\cdot\cdot e^{-x_{n} + \theta}I(X_{(1)}>\theta)$

$= e^{-\sum x_{i} + n\theta}I(X_{(1)}>\theta)$

This is smallest, ZERO, when $X_{(1)}<\theta$

while you want $e^{-\sum x_{i} + n\theta}I(X_{(1)}>\theta)$ as large as possible WITH RESPECT TO THETA

So, you will need $X_{(1)}>\theta$

so to make $e^{-\sum x_{i} + n\theta}I(X_{(1)}>\theta)$ as large as possible
you want $n\theta$ or $\theta$ as big as possible
and it CANNOT exceed the data, hence the MLE (is our sufficient statistic by the way) $X_{(1)}$

3. Thank you! I was wondering what X_{(1)} means? Is that just X_1?

Also, what is the sufficient statistic? I looked it up on wikipedia but I don't really understand it.

4. Originally Posted by BERRY
Thank you! I was wondering what X_{(1)} means? Is that just X_1?

Also, what is the sufficient statistic? I looked it up on wikipedia but I don't really understand it.

$X_{(1)}$ is the minimum of $X_1, X_2,...,X_n$