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Question: Let $T_1$ and $T_2$ be independent unbiased estimators of a parameter $\Theta$.

Assume that $\operatorname{Var}(T_2) = \operatorname{Var}(T_1)$.

Using the MSE critertion, define which is a better estimator for $\Theta^2$:

$T_1^2\text{ or }T_1 * T_2$


So I found out that: $\begin{align} & E(T_{1}^2) = \operatorname{Cov}(T_1,T_1)+E(T_1)*E(T_1) = Var(T_1)+\Theta^2\\ \implies & \operatorname{Bias}(T_1^2) = E(T_1^2)-\Theta^2=Var(T_1) \end{align}$ $\begin{align} & E(T_1 * T_2) = \operatorname{Cov}(T_1,T_2) +E(T_1)*E(T_2) = \Theta^2 \\ \implies & \operatorname{Bias}(T_1*T_2) = 0 \end{align}$ $\begin{align} \operatorname{MSE}(T_{1}^2) &= \operatorname{Var}(T_{1}^2) + \operatorname{Bias}^2(T_1^2) = \operatorname{Var}(T_{1}^2)+ \operatorname{Var}^{2}(T_{1}) \\ \operatorname{MSE}(T_{1}*T_2) &= \operatorname{Var}(T_{1}*T_{2}) + \operatorname{Bias}(T_1*T_2) = ... = 2\Theta^2Var(T_1) +\operatorname{Var}^2(T_1) \end{align} $

But still I don't know how to compare between $\operatorname{Var}(T_{1}^2)$ and $2\Theta^2\operatorname{Var}(T_1)$.

I'm not a LaTeX expert, so I hope that it is somewhat readable...Thanks alot!

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    MSE stands for Minimum Square Error2013-06-07

1 Answers 1

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I just give an answer as the persons visiting the question shouldnt go through all the conversations.

Without imposing another conditions on the parameter $\Theta$ or the types of these two estimators, namely $T_1T_2$ and $T_1^2$ one can not make an objective comparison between their $MSE$.

My suggestion would be to put some constain such as Cramer-Rao lower bound $(CRLB)$ or the relation between $\Theta$ and $Var(T_1^2)$ to make it a valid and a nice question.

Ok as there might be some people who might be interested, the question is a valid one if you assume that $T_1$ and $T_2$ are minimum variance unbiased estimator(s), $MVUE$. In this case $T_1T_2$ is a better estimator than $T_1^2$.

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    @ByronSchmuland OK I edit.2012-09-05