I have the i.i.d. exponential random variables $X_1, \dots, X_n$ with the density functions
$$f(x; \sigma, \tau)= \begin{cases} \dfrac{1}{\sigma} e^{-(x - \tau)/\sigma} &\text{if}\, x\geq \tau\\ 0 &\text{otherwise} \end{cases}$$
I want to (1) calculate the MLEs for $\sigma$ and $\tau$, (2) show that they are biased, and (3) correct the bias and in doing so find unbiased estimators.
I have that the MLEs for $\tau$ and $\sigma$ are $$\tau^{\text{MLE}} = \min\left\{x_{i}\right\}\qquad \sigma^{\text{MLE}} = \frac{\sum_{i=1}^{n}\left(x_{i}-\min\left\{x_{i}\right\}\right)}{n}$$
From this Youtube video, I found that $$E[\min{\{x_i\}}] = \dfrac{\sigma}{n}$$
I also calculated that $$E\left[ \dfrac{\sum\limits_{i = 1}^n (x_i - \min{\{x_i\} })}{n} \right] = E\left[ \dfrac{1}{n} \sum_{i = 1}^n x_i \right] - \dfrac{1}{n} E[\min{\{x_i\} }] = \bar{x} - \dfrac{\sigma}{n^2}$$
So since we have that
$$E[\min{\{x_i\}}] = \dfrac{\sigma}{n} \not= \tau$$
and
$$E\left[ \dfrac{\sum\limits_{i = 1}^n (x_i - \min{\{x_i\} })}{n} \right] = E\left[ \dfrac{1}{n} \sum_{i = 1}^n x_i \right] - \dfrac{1}{n} E[\min{\{x_i\} }] = \bar{x} - \dfrac{\sigma}{n^2} \not= \sigma,$$
we can conclude that the MLEs for the parameters $\tau$ and $\sigma$ are biased, right?
Is my work here correct? If so, then how do I now 'unbias' them?
I did not look at the Youtube video but I do not agree...indeed the expectation is independent of $\tau$ that is not nice...
$$F_X(x)=1-e^{-(x-\tau)/\sigma}$$
and thus
$$F_{\min}(z)=1-e^{-n(z-\tau)/\sigma}$$
and
$$f_{\min}(z)=\frac{n}{\sigma}e^{-n(z-\tau)/\sigma}$$
which means that the expectation is
$$E[X_{(1)}]=\int_{\tau}^{\infty}\frac{n}{\sigma}ze^{-n(z-\tau)/\sigma}dz=\tau+\frac{\sigma}{n}$$
concluding:
$X_{(1)}$ is biased for $\tau$ but $T=X_{(1)}-\frac{\sigma}{n}$ is unbiased