Show that an estimator $\delta = [\frac{1}{1+\lambda} + \epsilon ]X$ of $E_{\theta}(X)$ is inadmissible (with squared error loss) under each of the following conditions:

(a) $\frac{Var_{\theta}(X)}{E_\theta^2 (X)} > \lambda>0$ and $\epsilon >0$;

(b) $\frac{Var_{\theta}(X)}{E_\theta^2 (X)} < \lambda$ and $\epsilon <0$.

My idea is to use MSE. If we use the estimator $\delta = [\frac{1}{1+\lambda} + \epsilon ]X$, then the MSE of this estimator is the squared of the bias plus the variance, that is $[ E_\theta (X)-E_\theta ( \frac{1}{1+\lambda} + \epsilon )X ]^2 + (\frac{1}{1+\lambda} + \epsilon)^2 Var_\theta (X)=(\frac{\lambda}{1+\lambda}-\epsilon)^2 E_\theta (X)+(\frac{1}{1+\lambda} + \epsilon)^2 Var_\theta (X)$. Then how to do next?



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