Problem: Let $X_1, \ldots, X_n$ be a random sample from an exponential distribution with parameter $\theta$. Find a MLE for $P[X > x]$ with $x>0$ given (fixed).
My attempt: I first calculated: $$ P[X > x] = \int_x^{\infty} \theta e^{-\theta y} dy = e^{-\theta x}. $$ I also know that $$ L(\theta; \vec{x}) = \prod_{i=1}^n f(x_i; \theta) = \theta^n \exp \left(- \theta \sum_{i=1}^n x_i \right).$$ Then $$ \ln L(\theta; \vec{x}) = n \ln(\theta) - \theta \sum_{i=1}^n x_i. $$ How to proceed now? Normally when we are looking for the MLE for a parameter $\theta$, we differentiate with respect to that parameter. But now I'm asked to find the MLE for a probability. Do I just differentiate with respect to $e^{-\theta x}$?
Help is appreciated.
For any (measurable) function $g(\theta)$, $\hat{g}_n = g(\hat{\theta})$, this called the invariance property of the MLE estimator. Hence, in your case $$ \hat{p}_n = e^{-\hat{\theta}_n x}. $$
Where $\hat{\theta}_n = \frac{1}{\bar{X}_n}$ is the MLE of $\theta$.