I'm looking to find the estimate of $\mu$ for $n$ data using the method of moments and the maximum likelihood for the pdf given by $f(x) = \begin{cases} e^{-(x-\mu)}, & \text{if} \, x \geq \mu \\ 0, & \text{if} \, x \lt \mu\\ \end{cases}$
For the method of moments, I estimate
$\mu = \int_{\mu}^{\infty} x e^{\mu-x}dx, \text{for} \ x \geq \mu \ $ and
$\mu = \int_{-\infty}^{\mu} x 0 dx = 0, \text{for} \ x < \mu$
For the first cast, I simplify:
$\mu = e^{\mu}\int_{\mu}^{\infty} x e^{-x}dx = e^{\mu}(-e^{-x}x-\int-e^{-x}dx)$
$ \mu = 0 - e^{\mu}(-e^{-\mu}\mu-e^{-\mu}) = \mu e^0+ e^0$
$\mu = \mu + 1 \ $ Which is obviously erroneous.
For the maximum likelihood method, I first find log likelihood $l(\mu)$:
$\mathcal{l}(\mu) = \sum_i^n (\mu - x_i)ln(e) = n \mu - x_n $ with a derivative $n$. But since this is for $x \geq \mu$, these are not positive likelihoods. Since is it 0 elsewhere, and $ln(0) = -\infty$, I'm not sure how to approach this. EDIT: yes, these are positive likelihoods.
Am I integrating incorrectly in the method of moments case? Am I misunderstanding the max likelihood method? Thank you for any tips.
For the maximum likelihood estimate observe that the joint density or likelihood function is given by $$ \begin{align} f(x_1\dotsc, x_n, \mu)&=I(x_1\geq \mu\dotsc, x_n\ge \mu) \times\prod_{i=1}^ne^{-{(x_i-\mu)}}\\ &=I(\min x_i\geq \mu )\times\prod_{i=1}^ne^{-{(x_i-\mu)}}\\ &\propto I(\min x_i\geq \mu )\times e^{n\mu} \end{align} $$ where the last line is proportional up to a positive constant and $I$ denotes indicator function. From the last expression it is clear that the MLE estimate is $\hat{\mu}=\min x_i.$