Show that admissibility of a decision rule under weighted square error loss implies it's admissibility under square error loss.

Weighted Square error loss = $\frac {(d(x) - \theta)^2}{\theta*(1-\theta)}$
Square error loss = $(d(x) - \theta)^2$

I am not sure how to start.

| cite | improve this question | | | | |
  • 1
    $\begingroup$ Consider how the weighting changes the risk functions. (You will need to assume $0\lt\theta\lt 1$.) $\endgroup$ – whuber Nov 18 '14 at 14:20

The general definition of admissibility is as follows:

An estimator $\delta_0$ is inadmissible for the risk function $R(\cdot,\cdot)$ if there exists at least one other estimator $\delta_1$ such that$$\forall\theta\quad R(\theta,\delta_1)\le R(\theta,\delta_0)$$with strict inequality for at least one $\theta$. It is admissible otherwise.

If an estimator $\delta_0$ is not admissible under squared error loss, it thus means there exists at least another estimator $\delta_1$ such that $$\forall\theta\quad\mathbb{E}_\theta[(\delta_1(X)-\theta)^2]{\le}\mathbb{E}_\theta[(\delta_0(X)-\theta)^2]$$ with strict inequality for at least one $\theta$. Dividing both sides of the inequality by $\theta(1-\theta)$implies$$\forall\theta\quad\frac{\mathbb{E}_\theta[(\delta_1(X)-\theta)^2]}{\theta(1-\theta)}{\le}\frac{\mathbb{E}_\theta[(\delta_0(X)-\theta)^2]}{\theta(1-\theta)}$$ with strict inequality for at least one $\theta$. Hence inadmissibility for the weighted loss.

| cite | improve this answer | | | | |

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.