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

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

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

Here's what I have so far.

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

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

$

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

$\displaystyle \frac{d}{d\theta}f(x_1...x_n|\theta) = n$

$\displaystyle n=0 ?$

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