# admissibility under weighted error loss implies admissibility under square error loss

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.

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

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.