I just want to check something.
Let $X_1,\ldots,X_n$ be i.i.d with pdf
$$f(x;\theta,\delta) = \frac{1}{\theta}e^{-\frac{x-\delta}{\theta}}\,\mathbb{I}\{x\ge\delta \},\space\theta>0$$
I am able to see that the MLE of $\theta$ and $\delta$ can be found by the following process.
$$L[\theta,\delta]=\theta^{-n}\exp\left[{-\frac{1}{\theta}\sum_{i=1}^{n}(X_i}-n\delta)\right]\mathbb{I}\{X_{(1)}\ge\delta \}$$ Thus $\hat{\delta}=X_{(1)}$ is quite clear.
To find $\hat{\theta}$ is what I want to make sure I am doing right.
$$l[\theta,\delta]=-n\ln \theta-\frac{1}{\theta}\sum_{i=1}^{n}X_i+\frac{n\delta}{\theta}+\ln [\mathbb{I}\{X_{(1)}\ge\delta \}]$$
Taking the partial derivatives with respect to $\theta$ I get $$\frac{\partial}{\partial\theta} l[\theta,\delta] = \frac{-n}{\theta}+\frac{\sum_{i=1}^{n} X_i}{\theta^2}-\frac{n\delta}{\theta^2}$$
and you can solve for
$$\frac{\partial}{\partial\theta} l[\hat{\theta},X_{(1)}]=0$$
So when I do that I get
$$\hat{\theta}=\bar{X}-X_{(1)}$$
I am a little worried because I have notes saying that the MLE is $\bar{X}$ instead of what I got.
I would appreciate your input.
A simple way to see which estimator for $\theta$ is correct is to choose, for example, $\delta = 100$ and $\theta = 1$. Then $$f_X(x) = e^{-(x-100)} \mathbb 1 (x > 100).$$ Then $X = Y + 100$ where $Y \sim \operatorname{Exponential}(1)$ and $\operatorname{E}[X] = 100 + \operatorname{E}[Y] = 101$. So the sample mean $\bar X$, which is guaranteed to exceed $100$ as all observations must be greater than $100$, is not a good estimator of $\theta = 1$, nor does it get better with increasing sample size. So you can easily see that $\hat \theta = \bar X$ cannot be the MLE.