Probability distribution of $Y_1=min(X_i)$ Let $Y_1<Y_2<\ldots<Y_n$ denote the order statistics  of a random sample of size $n$ from the distribution with pdf :
$$f(x;\theta)=e^{-(x-\theta)}I_{(\theta,\infty)}(x)$$
Here we use the indicator function of a set $A$ defined by 
$I_A$ =
\begin{cases}
1,  & x\in A \\
0, & x\notin A   \\
\end{cases}


*

*Why the range of $I$ is from $\theta$ to $\infty$, i.e., $I_{(\theta,\infty)}$ ? not $I_{(0,\infty)}$ or $I_{(0,\theta)}$ or elsewhere ?

*Also i have not understood how the pdf of $Y_1=min(X_i)$ is
$$f_{Y_1}(y_1;\theta)=ne^{-n(y_1-\theta)}I_{(\theta,\infty)}(y_1)$$ ?
 A: The definition of the indicator function $I_A(x)$ is such that the subscript $A$ represents a set.  In particular, it is the set of $x$-values for which the indicator function equals $1$.  So if we let $A = (\theta,\infty)$, the open interval of reals greater than $\theta$, then we simply have $$I_{(\theta,\infty)}(x) = \begin{cases} 1, & x > \theta \\ 0, & x \le \theta. \end{cases}$$  That's what we want.  If we wrote $I_{(0,\infty)}(x)$, then this is $1$ if $x > 0$ and $0$ otherwise, and that doesn't make sense in the context of the given probability density.  The set $A$ does not represent a "range" of $I$.  It does, however, indicate the support of the probability density.
Once you understand $f(x;\theta)$, we note that $$\Pr[Y_1 > y_1] = \Pr[\min(X_1, \ldots, X_n) > y_1] = \Pr[(X_1 > y_1) \cap (X_2 > y_1) \cap \cdots \cap (X_n > y_1)].$$  The leftmost equality is true by definition of $Y_1$ as the minimum of the sample $X_1, \ldots, X_n$.  The rightmost equality is implied by the independence of the observations, and the fact that if the smallest observation is larger than $y_1$, then this is the same as saying that all of the observations are larger than $y_1$, and vice versa.  Now, since each observation is drawn from the same distribution, it follows that $$\Pr\left[\bigcap_{i=1}^n  \; (X_i > y_1)\right] = \prod_{i=1}^n \Pr[X_i > y_1] = \left(\Pr[X_1 > y_1]\right)^n.$$  Consequently, the density of $Y_1$ is $$\begin{align*} f_{Y_1}(y_1) &= \frac{dF_{Y_1}}{dy_1} \\ &= \frac{d}{dy_1}\left[ 1 - \Pr[Y_1 > y_1] \right] \\ &= -\frac{d}{dy_1}\left[(1 - F(y_1;\theta))^n\right] \\ &= -n(1 - F(y_1;\theta))^{n-1} f(y_1;\theta), \end{align*}$$ where $F(x;\theta)$ is the CDF of the density $f(x;\theta)$.
