Determining an unbiased estimator

Question

Say we have a shifted exponential distribution with common density $$f(x|\theta)=\left\{\begin{matrix} e^{-(x-\theta)} & x\geq \theta\\ 0 & x<\theta \end{matrix}\right.$$

We have $\theta$ a real number as the unknown shift parameter and $\textbf{X}=(X_1,...,X_n)$ a random sample. If $X_{(1)}=min \left \{ X_1,...,X_n \right \}$, then the density $f_{(1)}(x)=ne^{-n(x-\theta)}$, where $x\geq \theta$.

I have tried to determine if this estimator $\hat{\theta}=X_{(1)}$ is unbiased or not. I got that $E[\hat{\theta}]=e^{-n\theta}(\theta+\frac{1}{n})$ which is not equal to $\theta$ so there is bias, but I am not sure if I have done this correctly.

If this estimator isn't biased, how does one then determine an unbiased estimator by making an adjustment to the estimator $\hat{\theta}=X_{(1)}$.

Answer 1

Defining $Y_{i}=X_{i}-\theta$ random variable $Y_{i}$ has standard (i.e. $\lambda=1$) exponential distribution.

Then $\min\left\{ Y_{1},\dots,Y_{n}\right\} $ has exponential distribution with parameter $\lambda=n$.

Based on: $$X_{\left(1\right)}=\min\left\{ X_{1},\dots,X_{n}\right\} =\min\left\{ \theta+Y_{1},\dots,\theta+Y_{n}\right\} =\theta+\min\left\{ Y_{1},\dots,Y_{n}\right\} $$

we find:

$$\mathbb{E}X_{\left(1\right)}=\theta+\mathbb{E}\min\left\{ Y_{1},\dots,Y_{n}\right\} =\theta+\frac{1}{n}$$

So apparantly $X_{(1)}-\frac1{n}$ will serve as unbiased estimator of $\theta$.