1
$\begingroup$

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

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

$\endgroup$
1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.