Estimator for $\frac{1}{\lambda}$ using $\min_i X_i$ when $X_i$ are i.i.d $\mathsf{Exp}(\lambda)$

Question

Let $X_1,\ldots,X_n$ be i.i.d. $\mathsf{Exp}(\lambda)$ random variables, where $\lambda$ is unknown. Consider $f_{\min}(x) = \min_{i}(X_i)=$ $ n \lambda $ Exp$(n\lambda x)$.

I am told that $\hat \theta \mathrel := n \cdot \min_i(X_i)$ is an unbiased estimator for the parameter $1/\lambda$. Indeed, this is true since the expected value of the above defined $\hat \theta$ is equal to $1/\lambda$. But, in this setting how would one proceed to compute $1/\lambda$ from $\hat \theta$? (don't we construct an estimator in order to compute a parameter?)

Answer 1

...in this setting how would one proceed to compute $1/\lambda$ from $\hat \theta$?

You can't compute $1/\lambda$ because $\lambda$ is the unknown parameter. That is why we estimate it. We can take the value $\hat{\theta}(\mathbf{x}) = n x_{(1)}$ to be an estimate of $1/\lambda$ (and we can compute that estimate from the observed data), but we will not know the true value of the parameter.

Answer 2

You know that $\hat\theta(X_1,\ldots,X_n)=n\min_\limits{1\le i\le n} X_i$ is an unbiased estimator of $\frac{1}{\lambda}$, a function of the unknown parameter $\lambda$.

This implies that an unbiased estimate of $\frac{1}{\lambda}$ is given by $\hat\theta(x_1,\ldots,x_n)=n\min_\limits{1\le i\le n} x_i$. [Here $x_i$ denotes the observed value of $X_i$ in the sample]. This is the value of $\frac{1}{\lambda}$ you are seeking based on the observed sample $x_1,\ldots,x_n$.