TITLE: Is $T = \dfrac{n(\bar X - X_{(1)})} {n-1}$ an unbiased and consistent estimator of $b$. Problem in mathematical statistics.
QUESTION [0 upvotes]: Good morning everybody! I am in an urgent need of help/leads (absolute 0 in math statistics).
1) A random variable has a Cumulative distribution function $F(x)=1-e^{-(x-a)/b}$, where $x > a$.
Is $T = \dfrac{n(\bar X - X_{(1)})} {n-1}$ an unbiased and consistent estimator of $b$.
2) Find the estimator $b$ by a method of Maximum Likelihood from problem 1. Compare the efficiency of estimators from both problems.
Thanks in advance.

REPLY [1 votes]: Disclaimer. Here is a partial quick-and-dirty answer with minimal effort. The OP is invited to fill-in the blanks, and correct typos and thinkos...

The PDF of $X$ is $f(x) = F'(x)= (1/b)\exp(-(x-a)/b)1[\![x > a]\!]$ and it's mean is
$$
\mathbb E X = \int_{\mathbb R} xf(x)dx = \int_{a}^\infty (x/b)\exp(-(x-a)/b)dx = a + b
$$
$(A)$ Given an iid sample $X_1,X_2,\ldots,X_n$ from $F$, sort it so that $a < X_{(1)} \le X_{(2)} \le \ldots \le X_{(n)}$, and set $T_n := \dfrac{n(\overline{X}_n-X_{(1)})}{n-1}$. Note that $\overline{X}_n,X_{(1)},\ldots,X_{(n)},T_n$ are all random variables. As $n$ grows, $X_{(1)}$ will accumulate around $a$ on average (this can be made more formal if needed...), i.e $\mathbb E X_{(1)} \longrightarrow a$ in the limit as $n \longrightarrow \infty$. Let $\overline{X}_n = (1/n)\sum_{i=1}^n X_i$ be the sample mean. Then $\mathbb E \overline{X}_n = \mathbb E X = a + b$, and  so 
$\mathbb E T_n = \dfrac{n(\mathbb E\overline{X}_n-\mathbb EX_{(1)})}{n-1} = \dfrac{n(a + b - \mathbb E X_{(1)})}{n-1} \longrightarrow \dfrac{nb}{n-1} \longrightarrow b$ in the limit as $n \longrightarrow \infty$.

Thus $T_n$ is a consistent estimator for the parameter $b$.

(B) MLE estimation of $b$. The negative log-likelihood of the sample $X_1,\ldots,X_n$ is
$$
\begin{split}
\ell_n(a,b) &:= -\sum_{i=1}^n \log f(X_i)=-\sum_{i=1}^n \log((1/b)\exp(-(X_i-a)/b)) \\
&= n\log b +\frac{1}{b}\sum_{i=1}^n(X_i-a) = n\log b + \frac{n}{b}(\overline{X}_n-a).
\end{split}
$$
Differentiating w.r.t to $b$ and setting to zero gives $0 = \dfrac{\partial \ell_n(a,b)}{\partial b} = \dfrac{n}{b}-\dfrac{n(\bar{X}_n-a)}{b^2}$, from which $b=\hat{b} = \overline{X}_n-a$.
Moreover,

$\hat{b}$ consistently estimates the parameter $b$ since $\mathbb E \hat{b} = \mathbb E \overline{X}_n - a = a+b - a = b$.
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