So this is a question and solution to an old exam I was practicing for my statistics course.In the solution for 1a), they say that $L(\mathbf X;a,k)=k^ne^{\sum_{i=1}^nX_i}e^{nka} \mathbf 1(X_{(1)} \geq a)$. Why do they use the $\mathbf 1$ there, and what does that $X_{(1)}$ mean?
I calculated $L(\mathbf X;a,k)=k^ne^{\sum_{i=1}^nX_i}e^{nka}(X_i \geq a)$, but than you get in trouble for the sufficient statistic for $a$.
Thanks in advance!
$X_{(1)}$ means the first order statistic of the $X_i$, in other words the minimum of the observations; similarly $X_{(n)}$ would the $n$th order statistic i.e. the $n$ lowest value
$\mathbf 1(X_{(1)} \geq a)$ is an indicator function taking the value $1$ when the event in the bracket occurs and the value $0$ when it does not, so here is $1$ when the minimum (and so all) $X_i$ are greater than equal to $a$ but $0$ when any of the $X_i$ are less than $a$