Problem
Let $Y_1, Y_2, . . . , Y_n$ be i.i.d. with p.d.f.
$f(y; θ) = \frac1θ e^\frac{-y}θ , y>0 $
Consider two estimators for the parameter θ,
$\hat θ_1 = n \min_{i=1,2,..,n} Y_i $
$\hat θ_2 = \frac{Y_1 + Y_2 + . . . + Y_n}n$
Show that $\hat θ_1$ is indeed an unbiased estimator, and find its efficiency relative to $\hat θ_2$
My progress
The expected value of $f$ is $θ$ because it is an exponetial distribution with the rate parameter of $\frac1θ$
Now I need to find the the variance and mean of $\hat θ_1$ but I have no idea how to do that because we have a min function if someone could explain that to me.
For $\hat θ_2$ I can write it a $Y_1$ which is just $f$ because we are adding $n$ $Y$ components and dividing by $n$. So the mean is $θ$ and variance is ${θ^2}$
The CDF of the min of $n$ iid rv is
$$F_{min}=1-(1-F)^n$$
Thus in your case, setting $U=min(\mathbf{y})$
$$F_U(u)=1-e^{-\frac{yn}{\theta}}$$
and consequently, derivating,
$$f_U(u)=\frac{n}{\theta}e^{-\frac{yn}{\theta}}$$
Now you have all the tools to proceed by yourself
$\hat{\theta}_2$ is the sample mean...thus its mean is $\theta$ the same of the population's one but its variance is $\frac{\theta^2}{n}$
In fact:
$$\mathbb{V}\Bigg[\frac{1}{n}\Sigma_i Y_i\Bigg]=\frac{1}{n^2}n\theta^2$$
(The expectation is a linear operator but the variance is not. It's a quadratic operator)