PLEASE NOTE THAT THE FUNCTION IS :

e^-(x-theta) e^-e^-(x-theta)

So if you take the log likelihood of this function it will be :

-sum(x)+n theta -e^-(sum(x)- n theta)

***PLEASE ANSWER ALL PARTS***

Let X1,..., X, be a random sample from the density f(x;0) = e-(x-6) exp(-e-(2-0), where - 2). Let X1,..., X, be a random sample from the density f(x;0) = e-(x-6) exp(-e-(2-0), where - 2)