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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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  • $\begingroup$ Thanks for editing, axblount! The question looks much better now. $\endgroup$
    – Shahar
    Sep 4, 2012 at 18:37
  • $\begingroup$ I liked the question but I also ended up with something dependent on $\theta$. Are you sure that there isnt extra information? $\endgroup$ Sep 4, 2012 at 20:26
  • $\begingroup$ Meanwhile, I found the MSE of $T_1T_2$ as $2E[T_1^2]-\theta^4$ $\endgroup$ Sep 4, 2012 at 20:29
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    $\begingroup$ ahh OK I found! look! I will just give u the hint and thats it!! I believe you will understand and solve it.MVUE $\endgroup$ Sep 5, 2012 at 17:44
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    $\begingroup$ It is okay. Dont worry about it. Yes If you impose MVUE criterion, you end up with $Var(T_1^2)>4\Theta^2Var(T_1)$ which is similar to your Prof.'s criterion with $\Theta=1$. I think your conclusion seems right; for $-\sqrt(2)<\Theta<\sqrt{2}$ choose $T_1T_2$ else $T_1^2$ $\endgroup$ Sep 6, 2012 at 15:35

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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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  • $\begingroup$ What is CRLB?${}$ $\endgroup$
    – user940
    Sep 5, 2012 at 16:54
  • $\begingroup$ @ByronSchmuland it is currently an unvalid question and CRLB is a nice constraint. If you reduce the space of constaints then you will be able to make the problem as a valid one. Every person can choose another criteria but CRLB will be one of the most reasonable as it lower bounds the variance. $\endgroup$ Sep 5, 2012 at 17:49
  • $\begingroup$ What does the acronym "CRLB" mean? I have never seen it, and maybe other readers are wondering as well. $\endgroup$
    – user940
    Sep 5, 2012 at 17:58
  • $\begingroup$ @ByronSchmuland OK I edit. $\endgroup$ Sep 5, 2012 at 17:59

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