How to find unbiased estimator which is based on maximum likelihood for this problem? 
Let $X_1, \ldots, X_n$ ($n \ge 2$) be a random sample from a distribution
having p.d.f.
$$f(x\mid \underline{\theta}) = \frac{1}{\sigma}e^{-(x-\mu)/\sigma}\;\;\; \text{if }
 x>\mu \text{ and } 0 \text{ otherwise} \tag{1}$$
where $\underline{\theta} = (\mu, \sigma) \in \mathbb{R} \times
 \mathbb{R}^+$.
Let $g(\underline{\theta}) = \mu$. Find an unbiased estimator of $g(\underline{\theta})$ which is based on the Maximum
Likelihood Estimator (MLE)

The Maximum Likelihood function is
$$L(\theta) = \frac{1}{\sigma^n}e^{-(1/\sigma)\sum_1^n(x_i-\mu)}\;\;\; \text{if } x_{(1)} > \mu \text{ and } 0 \text{ otherwise}$$
where $x_{(1)} = \min\{x_1, x_2, \ldots x_n\}$
Clearly, $L(\theta) $ is maximum when $\mu = x_{(1)}$
$ l(\theta) = \ln(L(\theta)) = -n\ln(\sigma) - \frac1\sigma\sum_1^n(x_i-\mu)$
$\Rightarrow l_\theta = -\frac{n}\sigma + \frac{1}{\sigma^2}\sum_1^n(x_i-\mu)$
$l_\theta = 0 \Rightarrow \sigma = \frac1n\sum_1^n(x_i-x_{(1)} )$
Hence, the MLE is:  $$\delta_M = \left(X_{(1)}, \frac1n\sum_1^n(X_i-X_{(1)} )\right)$$
Let $F(x)$ be distribution obtained from $f(x)$
Let $Y = X_{(1)}$ and $T = \sum_1^n(X_i-X_{(1)} )$
I want to calculate $f_Y$ and $f_T$
I know that, $f_Y(y) = n [1-F(y)]^{n-1}f(y)$ where $F$ is distribution function obtained from $f$ given in $(1)$
$$ \implies f_Y(y) = n[1-(1-e^{-(y-\mu)/\sigma})]^{(n-1)}\frac{1}{\sigma}e^{-(x-\mu)/\sigma}  $$
$$\implies f_Y(y) = \frac{n}\sigma e^{-n(y-\mu)/\sigma}$$
$$\implies E(Y) = \mu + \frac{\sigma}n$$
Now, to proceed further, I need to find $f_T$.
This is where I am getting stuck. I don't know how to find  $f_T $ where $T = \sum_1^n(X_i-X_{(1)} )$.
Please help me... Related information/links will be much appreciated
 A: Since $X_i-\mu$ are i.i.d Exponential with mean $\sigma$ for all $i$,  $X_{(1)}-\mu$ is Exponential with mean $\frac{\sigma}{n}$.
Therefore, $$E\left[\sum_{i=1}^n(X_i-X_{(1)})\right]=\sum_{i=1}^n E\left[X_i\right]-nE\left[X_{(1)}\right]=(n-1)\sigma$$
Now from $E\left[X_{(1)}-\frac{\sigma}{n}\right]=\mu$ you get an unbiased estimator of $\mu$ based on the MLE by replacing $\sigma$ with its unbiased estimator:
$$E_{\mu,\sigma}\left[X_{(1)}-\frac1{n(n-1)}\sum_{i=1}^n (X_i-X_{(1)})\right]=\mu\quad,\,\forall\,\mu,\sigma$$
A: In order to derive the distribution of
$$T=\sum_{i=1}^n X_i-X_{(1)}$$
One can observe that:
Reminding that the negative exponential distribution owns the memoryless property, we can think of $X_1,...,X_n$ as the lifetime of n devices which are turned on at the time $\mu$.
At the time when the first device is off, the remaining devices are "good as new", that is, the remaining lifetime $Y_i-Y_{(1)}$ is always Exponentially distributed with the same mean and variance. Thus in the summation $T=\sum_{i=1}^n X_i-X_{(1)}$, one of the $n$ terms is zero (corresponding to the first device off), and the other $(n-1)$ terms are iid exp distributed. Thus $T$ has the same distribution as the sum of $(n-1)$ exponential rv's. As this sum is a Gamma distribution, this can be standarized to a $\chi_{2(n-1)}^2$.
