2
$\begingroup$

for two real random $n$-vector $y$ and $x$ and a random $n$-vector $e$ with distribution $F$ independent of $x$ we know (1) that the estimator

$$\text{med}\left( \frac{y_i}{x_i}\right)$$

is minimax estimator for $\alpha$ in the model:

$$y_i=\alpha x_i+e_i$$

My question is the following: does this result implies that the estimator

$$\text{med}\left( y_i x_i\right)$$

is minimax estimator for $\alpha$ in the model

$$y_i=\alpha/x_i+e_i$$

(1) Martin,R.D. Yohai, V.J. and Zamar,R.H., Mini-max bias, robust regression. Ann. Stat. (1989)

$\endgroup$
2
$\begingroup$

Yes. Let $z_i = 1/x_i$. Use the initial result on the model $y_i=\alpha z_i + e_i$, for which all assumptions seem to be satisfied, and you get the result you want.

$\endgroup$

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.