lets say $\ Y $ distributed exponential with the following pdf:
$$\ f_{\theta, \tau} = \theta \cdot e^{-\theta(y-\tau)}\mathbb I \{y \ge \tau\} , \theta > 0 $$
and I'm trying to find MLE when both $\ \theta, \tau $ are unknowns.
If I did everything correctly then the log likelihood function is $$\ l(\theta, \tau ; y) = n\cdot ln(\theta) + \theta \cdot \tau \cdot n - \theta \sum y_i $$ derivative w.r.t $\ \theta $ is $$\ \frac{\partial l}{\partial \theta} = \frac{n}{\theta} + \tau \cdot n - \sum y_i \Rightarrow \theta = \frac{n}{\sum y_i - \tau \cdot n}$$
and derivative w.r.t to $\ \tau $ is just $\ \frac{\partial l}{\partial \tau} = n \Rightarrow n = 0 $ so I did something wrong here but can't figure out what ?
Thanks.
$\mathbb{I}(\cdot)$ is an indicator function I presume. In that case, your distribution does not have two parameters! It is only $\theta>0.$ Your distribution can actually be written in the following way to make thing clear :
$f_\theta(y)=\theta\cdot e^{-\theta(y-\tau)},$ $y\geq\tau,$ $\theta>0.$