I am having troubles with solving this task
7.4.5. Show that the first order statistic $Y_1$ of a random sample of size $n$ from the distribution having pdf $$f(x;\theta) = e^{-(x-\theta)}, \theta<x<\infty, -\infty<\theta<\infty,$$ zero elsewhere, is a complete sufficient statistic for $\theta$. Find the unique function of this statistic which is the MVUE of $\theta$.
I showed completeness and sufficiency, but I cannot get $E[Y_1]$, to find MVUE. I want to use Sheffe Lehmann theorem.
Let the random variables in question be $X_1$ to $X_n$. Note that $Y_1\gt y$ if and only if all the $X_i$ are greater than $y$. Thus for $y\gt \theta$ we have $$\Pr(Y_1\gt y)=(e^{-(y-\theta)})^n=e^{-n(y-\theta)}.$$ So for $y\gt \theta$ the cdf of $Y_1$ is $1-e^{-n(y-\theta)}$.
Now we can find the density of $Y_1$ by differentiating, and then find the expectation in the usual way.
But we really don't need to compute if we know the mean of the exponential distribution with parameter $n$. Recall this is $\frac{1}{n}$. For the mean of $Y_1$, shift right by $\theta$.