# Unbiased Estimator based on Sufficient Statistic

suppose $$X_1, ... , X_n$$ are iid with pdf $$f(x|\beta) = e^{-(x-\beta))}I_{(\beta, \infty)}(x)$$

and the pdf of ( the smallest order statistic) $$X_{(1)}$$ is given by

$$f_{X_1}(x)$$ = n $$*$$ $$e^{n(\beta-x)}$$ , $$\beta \leq x$$

Question is below:

If our goal is to find a function of $$X_{(1)}$$ , for example $$g(X_{(1)})$$ so that , that function is unbiased estimator of $$\beta$$.

which means we want $$E_{\beta}[g(X_{(1)})]$$ = $$\beta$$

where $$E_{\beta}[g(X_{(1)})]$$ $$=$$ $$\int_{\theta}^{\infty}g(x)f_{X_1}(x)dx$$ $$=\int_{\beta}^{\infty}(x)*n*e^{n(\beta-x)}dx$$ = $$\beta$$ $$+$$ $$\frac{1}{n}$$ is the last calculation of the integral right?

also does that mean that an unbiased estimator of $$\beta$$ which is a function of $$X_{(1)}$$ equals to

$$g(X_{(1)})$$ = $$X_{(1)}$$ $$-$$ $$\frac{1}{n}$$ ? ?

Since $$X_i-\beta$$ are i.i.d $$\mathsf{Exp}(1)$$, we have $$\min_i(X_i-\beta)=X_{(1)}-\beta\sim \mathsf{Exp}$$ with mean $$1/n$$.
Your conclusion is correct, but there is no need for guesswork. Just work out $$E\left[X_{(1)}\right]$$ from the pdf of $$X_{(1)}$$, i.e. find $$\int xf_{X_{(1)}}(x)\,dx$$ directly. If you are starting with $$E_{\beta}\left[g(X_{(1)})\right]=\beta$$, then you have to differentiate this equation with respect to $$\beta$$ to solve for $$g(\cdot)$$, which of course gives the same answer.
• One more question please. Is t true that the MLE of $e^{\beta}$ is $e^{X_{(1)}}$ since $f(x|\beta) = e^{-(x-\beta))}I_{(\beta, \infty)}(x)$. ( also MLE of of ${\beta}$ is $X_{(1)}$ Commented May 2, 2019 at 17:58
• MLE of $\beta$ is $X_{(1)}$, so yes that is correct by invariance of MLE. Commented May 2, 2019 at 18:01