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

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


1 Answer 1


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.

  • $\begingroup$ 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 $X_{(1)}$ $\endgroup$
    – Pedros
    Commented May 2, 2019 at 17:58
  • $\begingroup$ MLE of $\beta$ is $X_{(1)}$, so yes that is correct by invariance of MLE. $\endgroup$ Commented May 2, 2019 at 18:01

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