Consistency of $f( x| \theta ) = \exp(-(x- \theta ))$ 
Prove that the second smallest observation in a random sample of size
n from following pdf is consistent estimator of $ \theta $
$$   f( x| \theta  ) = \exp(-(x- \theta )) , \qquad x >  \theta   $$

I just need a hint on how to start this question, because I dont have any idea on how to and where to start on this.
With the hints provided, I tried it further:
$P( |X_{(2)} - \theta | > \ e  ) <  \dfrac{\operatorname{Var}(X_{(2)})}{\ e^{2}}$  is the required condition for consistency
Using order statistics, the pdf of $\ X_{(2)}  =  \dfrac{n(n-1)[1- \ exp(-(x- \theta ))]\ exp(-(x- \theta ))^{n-1}}{2}$
Now I need to find the $\operatorname{Var}(X_{(2)})$
Which I am struck at, need a little assistance!

Apply the definitions.
An estimator is a definite mathematical rule $t$ to compute a value given any finite dataset $(x_1,x_2,\ldots, x_n).$
When datasets are obtained by sampling a distribution $F,$ they produce independent and identically distributed random samples $(X_1,X_2,\ldots, X_n)$ for any $n\ge 1.$
Consider a family of distributions $\mathfrak F.$  Let $\theta$ be some property of these distributions.  All that means is for any $F\in\mathfrak F,$ $\theta(F)$ is a definite number.
An estimator $t$ is consistent at $F$ for a property $\theta$ when $t(X_1,X_2,\ldots,X_n)$ converges to $\theta$ in probability.  (This means the chance that $t(X_1,\ldots, X_n)$ differs from $\theta$ by any arbitrarily small but positive amount will eventually approach zero as $n$ grows large.)
An estimator $t$ is consistent at the family $\mathfrak F$ for a property $\theta$ when it is consistent at all $F\in\mathfrak F$ for $\theta.$
This all means you can employ a consistent estimator to obtain a reasonable guess of the value of $\theta$ even when you don't know which $F\in\mathfrak{F}$ is generating the samples.
Reformulate the question.
Suppose every $F\in\mathfrak F$ has a greatest lower bound.  A lower bound  is some finite number $\theta(F)$ (depending perhaps on $F$) for which $\Pr(X \le \theta(F)) = 0$ when $X$ has $F$ for its distribution.  $\theta$ is a greatest lower bound when, in addition to being a lower bound, it's always the case that $\Pr(X \ge \theta(F)) \gt 0.$
The exponential family in the question is a good example of this condition: it (implicitly) assigns zero probability to any values less than $\theta$ while for any $x\gt \theta$ the chance it assigns to values less than or equal to $x$ is given by $1-\exp(\theta-x),$ which is positive.
Here is the essence of the problem:

In any sample $\mathbf X = (X_1,\ldots, X_n),$ let $t_r(\mathbf X)$ be the $r^\text{th}$ smallest value among the $X_i$ (where $r$ is any fixed whole number).  Prove that $t_r$ consistently estimates the greatest lower bound $\theta.$

The solution follows immediately.
According to the definitions, we need to show that for any $F\in\mathfrak F,$ the $r^\text{th}$ smallest value in a sample of size $n$ is highly likely to be very close to the greatest lower bound $\theta(F).$
The idea should now be clear: by choosing $n$ very large, $r$ will eventually be smaller than any specified fraction of $n,$ say $n\delta.$  When $\delta$ is tiny, the $n\delta$ quantile of $F$ (no matter what $F$ might be) must be close to the greatest lower bound $\theta(F).$  Thus, the $r^\text{th}$ smallest value of a sample consistently estimates $\theta,$ QED.

If you need to see the details, here they are spelled out.
Let $\epsilon \gt 0$ indicate just how close you want $t_r$ to get to $\theta.$  Let $n$ be any sufficiently large possible sample size (so we may assume $n\ge r$).  Define $$p(\epsilon; n, r, F) = F(\theta(F)+\epsilon).$$  This is relevant because $q=1-p$ is the chance that any single value drawn from $F$ exceeds $\theta(F)+\epsilon.$  (For convenience I will drop the "$(\epsilon;n,r,F)$" after $p$ and $q.$)
Because $\theta$ is a greatest lower bound, $p \gt 0$ and therefore $q \lt 1.$
Let $\mathbf{X}_n$ be a sample of size $n$ from $F.$  Consider the chance that $t_r(\mathbf{X}_n)$ is close enough to $\theta(F).$  That is, let's contemplate the chance that the $r^\text{th}$ smallest sample value does not exceed $\theta(F)+\epsilon.$  Put another way, at most $n-r$ of the sample values may exceed $\theta(F)+\epsilon.$
We can estimate this chance.
Each of the $X_i$ individually has a chance $q$ to exceed the threshold (because they all have $F$ for their distribution). And since the $X_i$ are independent, the chance that $n-r$ or fewer of them are this large is given by the Binomial$(n,q)$ distribution evaluated at $n-r.$
We know the mean of this Binomial distribution is $nq$ and its variance is $nq(1-q).$  Thus, $n-r$ differs from the Binomial mean by
$$\frac{(n-r) - nq}{\sqrt{nq(1-q)}} = \sqrt{n}\frac{(1-q) - r/n}{\sqrt{q(1-q)}}$$
standard deviations.  Chebyshev's Inequality asserts the chance of a deviation from the mean of this magnitude (or greater) can be no larger than the reciprocal square of this value,
$$\begin{aligned}
\Pr(t_r(\mathbf{X}_n)\le\theta(F)+\epsilon) &\le \left(\sqrt{n}\frac{(1-q) - r/n}{\sqrt{q(1-q)}}\right)^{-2} \\
&= \frac{1}{n}\frac{q(1-q)}{(1-q-r/n)^2} \\
&\le \frac{1}{n}\frac{q}{1-q}\left[1 + \frac{2}{1-q}\right].
\end{aligned}$$
Clearly, no matter what $0\lt q\lt 1$ might be and no matter how large $r$ might be, the limiting value of this expression as $n$ grows large is zero.
Obviously there is no chance that $t_r(\mathbf{X}_n)$ is less than $\theta(F),$ because this estimator always equals one of the $X_i$ and since $\theta$ is a lower bound, none of the $X_i$ have any chance of being this small.  We have thereby shown that as $n$ grows large, $t_r$ is squeezed between $\theta(F)$ and $\theta(F)+\epsilon$ with arbitrarily high probability, no matter how small (but positive) $\epsilon$ might be.
This completes the proof that $t_r(\mathbf{X}_n)$ converges to $\theta(F)$ in probability for any $F\in\mathfrak F,$ and that's what it means for $t_r$ to be a consistent estimator.
