PDF of a random variable $X$ is, $$ \begin{equation} f\left(x|\gamma\right)= \begin{cases} e^{-(x-\gamma)} & x\geq\gamma\\ 0 & \text{otherwise.} \end{cases} \end{equation} $$
Given a sample of $n$ i.i.d observations $\mathbf{x}=$ $\left(x_{1}, \ldots, x_{n}\right)$ from this distribution,
(a) Show that the estimator $\hat{\gamma} = \overline{X} - 1$ (where $\overline{X} = \frac{1}{n}\sum_{i=1}^n X_i$) is an unbiased and consistent estimator for the given distribution.
Try:
\begin{align*} E[X_1]& \stackrel{\text{using def.}}{=} \gamma+ 1 \end{align*}
\begin{align*} \mathbb E[\hat{\gamma}] &= \mathbb E[\overline{X} - 1]\\ &= \mathbb E\left[\frac{1}{n}\sum_{i=1}^n X_i\right]-1\\ &= \frac{1}{n} \sum_{i=1}^n \mathbb E[X_i] -1 \\ &= \frac{1}{n} (n\gamma+ n) -1\\ &= \gamma \end{align*}
Is it enough to say the esimator is unbiased? I couldn't understand how to show the consistence. As the definition is,
An estimator $T_n$ of parameter $\theta$ is said to be consistent, if it converges in probability to the true value of the parameter: $$\lim_{n\to \infty }T_{n}\stackrel{p}{=}\theta.$$ i.e. if, for all $\epsilon>0$ $$\lim_{n\to \infty }P(|T_n-\theta|>\epsilon)=0$$
The definition is quite complicated (to convergence in probability). I guess chebyshev's inequality can play a role here, but couldn't link that.
Any help will be appreciated. TIA.
You computations for the unbiasedness are correct. to show the consistency we use Chebickev Inequality.
First we observe that $\text{Var}(\hat{\gamma})=\text{Var}(\bar{X}_n-1)=\text{Var}(\bar{X}_n)=1/n$. Then (here $T_n = \hat{\gamma}$ and $\theta = \gamma$)
$$ \mathbb{P}(\mid \hat{\gamma} - \gamma \mid > \epsilon) = \mathbb{P}(\mid \hat{\gamma} - \mathbb{E}(\gamma) \mid > \epsilon) \le \frac{1}{\epsilon^2n^2} $$
which indeed tends to zero when $n \to \infty$.
Proof of Chebyshev's Inequality
Let $Y:=(X - \mathbb{E}(X))^2$ and $\epsilon > 0$. Since $Y$ is a nonnegative rv we can apply Markov to it, so
$$ \mathbb{P}(Y \ge \epsilon^2) \le \frac{\mathbb{E}(Y)}{\epsilon^2} = \frac{\mathbb{E}[(X - \mathbb{E}(X))^2]}{\epsilon^2} = \frac{\text{Var}(X)}{\epsilon^2} $$
Finally we obtain the desired inequality noticing that
$$ \mathbb{P}(Y \ge \epsilon^2) = \mathbb{P}((X - \mathbb{E}(X))^2 \ge \epsilon^2) = \mathbb{P}(|X - \mathbb{E}(X)| \ge \epsilon) $$